\(\int \frac {x^{7/2} (A+B x^2)}{(a+b x^2)^2} \, dx\) [375]

Optimal result

Integrand size = 22, antiderivative size = 310 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}+\frac {\sqrt [4]{a} (5 A b-9 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}} \]

[Out]

-1/10*(5*A*b-9*B*a)*x^(5/2)/a/b^2+1/2*(A*b-B*a)*x^(9/2)/a/b/(b*x^2+a)+1/8*a^(1/4)*(5*A*b-9*B*a)*arctan(1-b^(1/
4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(13/4)*2^(1/2)-1/8*a^(1/4)*(5*A*b-9*B*a)*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4
))/b^(13/4)*2^(1/2)+1/16*a^(1/4)*(5*A*b-9*B*a)*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(13/4)*
2^(1/2)-1/16*a^(1/4)*(5*A*b-9*B*a)*ln(a^(1/2)+x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(13/4)*2^(1/2)+1/2*
(5*A*b-9*B*a)*x^(1/2)/b^3

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {468, 327, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt [4]{a} (5 A b-9 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} b^{13/4}}+\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}}+\frac {\sqrt {x} (5 A b-9 a B)}{2 b^3}-\frac {x^{5/2} (5 A b-9 a B)}{10 a b^2}+\frac {x^{9/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

[In]

Int[(x^(7/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

((5*A*b - 9*a*B)*Sqrt[x])/(2*b^3) - ((5*A*b - 9*a*B)*x^(5/2))/(10*a*b^2) + ((A*b - a*B)*x^(9/2))/(2*a*b*(a + b
*x^2)) + (a^(1/4)*(5*A*b - 9*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*b^(13/4)) - (a^(1/
4)*(5*A*b - 9*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*b^(13/4)) + (a^(1/4)*(5*A*b - 9*a
*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*b^(13/4)) - (a^(1/4)*(5*A*b - 9*a*B
)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*b^(13/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}+\frac {\left (-\frac {5 A b}{2}+\frac {9 a B}{2}\right ) \int \frac {x^{7/2}}{a+b x^2} \, dx}{2 a b} \\ & = -\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}+\frac {(5 A b-9 a B) \int \frac {x^{3/2}}{a+b x^2} \, dx}{4 b^2} \\ & = \frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}-\frac {(a (5 A b-9 a B)) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{4 b^3} \\ & = \frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}-\frac {(a (5 A b-9 a B)) \text {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b^3} \\ & = \frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}-\frac {\left (\sqrt {a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^3}-\frac {\left (\sqrt {a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^3} \\ & = \frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}-\frac {\left (\sqrt {a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b^{7/2}}-\frac {\left (\sqrt {a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b^{7/2}}+\frac {\left (\sqrt [4]{a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} b^{13/4}}+\frac {\left (\sqrt [4]{a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} b^{13/4}} \\ & = \frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}+\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}}-\frac {\left (\sqrt [4]{a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {\left (\sqrt [4]{a} (5 A b-9 a B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}} \\ & = \frac {(5 A b-9 a B) \sqrt {x}}{2 b^3}-\frac {(5 A b-9 a B) x^{5/2}}{10 a b^2}+\frac {(A b-a B) x^{9/2}}{2 a b \left (a+b x^2\right )}+\frac {\sqrt [4]{a} (5 A b-9 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{13/4}}+\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}}-\frac {\sqrt [4]{a} (5 A b-9 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} b^{13/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.59 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{b} \sqrt {x} \left (-45 a^2 B+a b \left (25 A-36 B x^2\right )+4 b^2 x^2 \left (5 A+B x^2\right )\right )}{a+b x^2}-5 \sqrt {2} \sqrt [4]{a} (-5 A b+9 a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+5 \sqrt {2} \sqrt [4]{a} (-5 A b+9 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{40 b^{13/4}} \]

[In]

Integrate[(x^(7/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

((4*b^(1/4)*Sqrt[x]*(-45*a^2*B + a*b*(25*A - 36*B*x^2) + 4*b^2*x^2*(5*A + B*x^2)))/(a + b*x^2) - 5*Sqrt[2]*a^(
1/4)*(-5*A*b + 9*a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])] + 5*Sqrt[2]*a^(1/4)*(-5*
A*b + 9*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(40*b^(13/4))

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.55

method	result	size
risch	\(\frac {2 \left (b B \,x^{2}+5 A b -10 B a \right ) \sqrt {x}}{5 b^{3}}-\frac {a \left (\frac {2 \left (-\frac {A b}{4}+\frac {B a}{4}\right ) \sqrt {x}}{b \,x^{2}+a}+\frac {\left (5 A b -9 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 a}\right )}{b^{3}}\)	\(169\)
derivativedivides	\(\frac {\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 A b \sqrt {x}-4 B a \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {\left (-\frac {A b}{4}+\frac {B a}{4}\right ) \sqrt {x}}{b \,x^{2}+a}+\frac {\left (5 A b -9 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a}\right )}{b^{3}}\)	\(171\)
default	\(\frac {\frac {2 b B \,x^{\frac {5}{2}}}{5}+2 A b \sqrt {x}-4 B a \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {\left (-\frac {A b}{4}+\frac {B a}{4}\right ) \sqrt {x}}{b \,x^{2}+a}+\frac {\left (5 A b -9 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a}\right )}{b^{3}}\)	\(171\)

[In]

int(x^(7/2)*(B*x^2+A)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

2/5*(B*b*x^2+5*A*b-10*B*a)*x^(1/2)/b^3-a/b^3*(2*(-1/4*A*b+1/4*B*a)*x^(1/2)/(b*x^2+a)+1/16*(5*A*b-9*B*a)*(a/b)^
(1/4)/a*2^(1/2)*(ln((x+(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/(x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+2
*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 696, normalized size of antiderivative = 2.25 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {5 \, {\left (b^{4} x^{2} + a b^{3}\right )} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} \log \left (b^{3} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} - {\left (9 \, B a - 5 \, A b\right )} \sqrt {x}\right ) + 5 \, {\left (i \, b^{4} x^{2} + i \, a b^{3}\right )} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} \log \left (i \, b^{3} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} - {\left (9 \, B a - 5 \, A b\right )} \sqrt {x}\right ) + 5 \, {\left (-i \, b^{4} x^{2} - i \, a b^{3}\right )} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-i \, b^{3} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} - {\left (9 \, B a - 5 \, A b\right )} \sqrt {x}\right ) - 5 \, {\left (b^{4} x^{2} + a b^{3}\right )} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-b^{3} \left (-\frac {6561 \, B^{4} a^{5} - 14580 \, A B^{3} a^{4} b + 12150 \, A^{2} B^{2} a^{3} b^{2} - 4500 \, A^{3} B a^{2} b^{3} + 625 \, A^{4} a b^{4}}{b^{13}}\right )^{\frac {1}{4}} - {\left (9 \, B a - 5 \, A b\right )} \sqrt {x}\right ) - 4 \, {\left (4 \, B b^{2} x^{4} - 45 \, B a^{2} + 25 \, A a b - 4 \, {\left (9 \, B a b - 5 \, A b^{2}\right )} x^{2}\right )} \sqrt {x}}{40 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

[In]

integrate(x^(7/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/40*(5*(b^4*x^2 + a*b^3)*(-(6561*B^4*a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A^3*B*a^2*b^3 +
625*A^4*a*b^4)/b^13)^(1/4)*log(b^3*(-(6561*B^4*a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A^3*B*a^
2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4) - (9*B*a - 5*A*b)*sqrt(x)) + 5*(I*b^4*x^2 + I*a*b^3)*(-(6561*B^4*a^5 - 1458
0*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A^3*B*a^2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4)*log(I*b^3*(-(6561*B^4*
a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A^3*B*a^2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4) - (9*B*a - 5
*A*b)*sqrt(x)) + 5*(-I*b^4*x^2 - I*a*b^3)*(-(6561*B^4*a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A
^3*B*a^2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4)*log(-I*b^3*(-(6561*B^4*a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b
^2 - 4500*A^3*B*a^2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4) - (9*B*a - 5*A*b)*sqrt(x)) - 5*(b^4*x^2 + a*b^3)*(-(6561*
B^4*a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A^3*B*a^2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4)*log(-b^3
*(-(6561*B^4*a^5 - 14580*A*B^3*a^4*b + 12150*A^2*B^2*a^3*b^2 - 4500*A^3*B*a^2*b^3 + 625*A^4*a*b^4)/b^13)^(1/4)
- (9*B*a - 5*A*b)*sqrt(x)) - 4*(4*B*b^2*x^4 - 45*B*a^2 + 25*A*a*b - 4*(9*B*a*b - 5*A*b^2)*x^2)*sqrt(x))/(b^4*
x^2 + a*b^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 770 vs. \(2 (292) = 584\).

Time = 155.90 (sec) , antiderivative size = 770, normalized size of antiderivative = 2.48 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {9}{2}}}{9} + \frac {2 B x^{\frac {13}{2}}}{13}}{a^{2}} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{b^{2}} & \text {for}\: a = 0 \\\frac {100 A a b \sqrt {x}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {25 A a b \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {25 A a b \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {50 A a b \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {80 A b^{2} x^{\frac {5}{2}}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {25 A b^{2} x^{2} \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {25 A b^{2} x^{2} \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {50 A b^{2} x^{2} \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {180 B a^{2} \sqrt {x}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {45 B a^{2} \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {45 B a^{2} \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {90 B a^{2} \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {144 B a b x^{\frac {5}{2}}}{40 a b^{3} + 40 b^{4} x^{2}} - \frac {45 B a b x^{2} \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {45 B a b x^{2} \sqrt [4]{- \frac {a}{b}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {90 B a b x^{2} \sqrt [4]{- \frac {a}{b}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{40 a b^{3} + 40 b^{4} x^{2}} + \frac {16 B b^{2} x^{\frac {9}{2}}}{40 a b^{3} + 40 b^{4} x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(7/2)*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(5/2)/5), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(9/2)/9 + 2*B*x**(13/2)/13)/a**2
, Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(5/2)/5)/b**2, Eq(a, 0)), (100*A*a*b*sqrt(x)/(40*a*b**3 + 40*b**4*x**2) +
25*A*a*b*(-a/b)**(1/4)*log(sqrt(x) - (-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) - 25*A*a*b*(-a/b)**(1/4)*log(sq
rt(x) + (-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) - 50*A*a*b*(-a/b)**(1/4)*atan(sqrt(x)/(-a/b)**(1/4))/(40*a*b
**3 + 40*b**4*x**2) + 80*A*b**2*x**(5/2)/(40*a*b**3 + 40*b**4*x**2) + 25*A*b**2*x**2*(-a/b)**(1/4)*log(sqrt(x)
- (-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) - 25*A*b**2*x**2*(-a/b)**(1/4)*log(sqrt(x) + (-a/b)**(1/4))/(40*a
*b**3 + 40*b**4*x**2) - 50*A*b**2*x**2*(-a/b)**(1/4)*atan(sqrt(x)/(-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) -
180*B*a**2*sqrt(x)/(40*a*b**3 + 40*b**4*x**2) - 45*B*a**2*(-a/b)**(1/4)*log(sqrt(x) - (-a/b)**(1/4))/(40*a*b**
3 + 40*b**4*x**2) + 45*B*a**2*(-a/b)**(1/4)*log(sqrt(x) + (-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) + 90*B*a**
2*(-a/b)**(1/4)*atan(sqrt(x)/(-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) - 144*B*a*b*x**(5/2)/(40*a*b**3 + 40*b*
*4*x**2) - 45*B*a*b*x**2*(-a/b)**(1/4)*log(sqrt(x) - (-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) + 45*B*a*b*x**2
*(-a/b)**(1/4)*log(sqrt(x) + (-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) + 90*B*a*b*x**2*(-a/b)**(1/4)*atan(sqrt
(x)/(-a/b)**(1/4))/(40*a*b**3 + 40*b**4*x**2) + 16*B*b**2*x**(9/2)/(40*a*b**3 + 40*b**4*x**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.87 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {{\left (B a^{2} - A a b\right )} \sqrt {x}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {{\left (\frac {2 \, \sqrt {2} {\left (9 \, B a - 5 \, A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (9 \, B a - 5 \, A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (9 \, B a - 5 \, A b\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (9 \, B a - 5 \, A b\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )} a}{16 \, b^{3}} + \frac {2 \, {\left (B b x^{\frac {5}{2}} - 5 \, {\left (2 \, B a - A b\right )} \sqrt {x}\right )}}{5 \, b^{3}} \]

[In]

integrate(x^(7/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

-1/2*(B*a^2 - A*a*b)*sqrt(x)/(b^4*x^2 + a*b^3) + 1/16*(2*sqrt(2)*(9*B*a - 5*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*a
^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(9*B*a
- 5*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqr
t(sqrt(a)*sqrt(b))) + sqrt(2)*(9*B*a - 5*A*b)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3
/4)*b^(1/4)) - sqrt(2)*(9*B*a - 5*A*b)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^
(1/4)))*a/b^3 + 2/5*(B*b*x^(5/2) - 5*(2*B*a - A*b)*sqrt(x))/b^3

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.96 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {\sqrt {2} {\left (9 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{4}} + \frac {\sqrt {2} {\left (9 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{4}} + \frac {\sqrt {2} {\left (9 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, b^{4}} - \frac {\sqrt {2} {\left (9 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, b^{4}} - \frac {B a^{2} \sqrt {x} - A a b \sqrt {x}}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {2 \, {\left (B b^{8} x^{\frac {5}{2}} - 10 \, B a b^{7} \sqrt {x} + 5 \, A b^{8} \sqrt {x}\right )}}{5 \, b^{10}} \]

[In]

integrate(x^(7/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

1/8*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - 5*(a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(
a/b)^(1/4))/b^4 + 1/8*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - 5*(a*b^3)^(1/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(
1/4) - 2*sqrt(x))/(a/b)^(1/4))/b^4 + 1/16*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - 5*(a*b^3)^(1/4)*A*b)*log(sqrt(2)*sqrt
(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^4 - 1/16*sqrt(2)*(9*(a*b^3)^(1/4)*B*a - 5*(a*b^3)^(1/4)*A*b)*log(-sqrt(2)*s
qrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^4 - 1/2*(B*a^2*sqrt(x) - A*a*b*sqrt(x))/((b*x^2 + a)*b^3) + 2/5*(B*b^8*x
^(5/2) - 10*B*a*b^7*sqrt(x) + 5*A*b^8*sqrt(x))/b^10

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 823, normalized size of antiderivative = 2.65 \[ \int \frac {x^{7/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\sqrt {x}\,\left (\frac {2\,A}{b^2}-\frac {4\,B\,a}{b^3}\right )+\frac {2\,B\,x^{5/2}}{5\,b^2}-\frac {\sqrt {x}\,\left (\frac {B\,a^2}{2}-\frac {A\,a\,b}{2}\right )}{b^4\,x^2+a\,b^3}+\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}-\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )\,1{}\mathrm {i}}{8\,b^{13/4}}+\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}+\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )\,1{}\mathrm {i}}{8\,b^{13/4}}}{\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}-\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{13/4}}-\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}+\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{13/4}}}\right )\,\left (5\,A\,b-9\,B\,a\right )\,1{}\mathrm {i}}{4\,b^{13/4}}+\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}-\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )\,1{}\mathrm {i}}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{13/4}}+\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}+\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )\,1{}\mathrm {i}}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{13/4}}}{\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}-\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )\,1{}\mathrm {i}}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )\,1{}\mathrm {i}}{8\,b^{13/4}}-\frac {{\left (-a\right )}^{1/4}\,\left (\frac {\sqrt {x}\,\left (25\,A^2\,a^2\,b^2-90\,A\,B\,a^3\,b+81\,B^2\,a^4\right )}{b^3}+\frac {{\left (-a\right )}^{1/4}\,\left (5\,A\,b-9\,B\,a\right )\,\left (72\,B\,a^3-40\,A\,a^2\,b\right )\,1{}\mathrm {i}}{8\,b^{13/4}}\right )\,\left (5\,A\,b-9\,B\,a\right )\,1{}\mathrm {i}}{8\,b^{13/4}}}\right )\,\left (5\,A\,b-9\,B\,a\right )}{4\,b^{13/4}} \]

[In]

int((x^(7/2)*(A + B*x^2))/(a + b*x^2)^2,x)

[Out]

x^(1/2)*((2*A)/b^2 - (4*B*a)/b^3) + (2*B*x^(5/2))/(5*b^2) - (x^(1/2)*((B*a^2)/2 - (A*a*b)/2))/(a*b^3 + b^4*x^2
) + ((-a)^(1/4)*atan((((-a)^(1/4)*((x^(1/2)*(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b))/b^3 - ((-a)^(1/4)*(5
*A*b - 9*B*a)*(72*B*a^3 - 40*A*a^2*b))/(8*b^(13/4)))*(5*A*b - 9*B*a)*1i)/(8*b^(13/4)) + ((-a)^(1/4)*((x^(1/2)*
(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b))/b^3 + ((-a)^(1/4)*(5*A*b - 9*B*a)*(72*B*a^3 - 40*A*a^2*b))/(8*b^
(13/4)))*(5*A*b - 9*B*a)*1i)/(8*b^(13/4)))/(((-a)^(1/4)*((x^(1/2)*(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b)
)/b^3 - ((-a)^(1/4)*(5*A*b - 9*B*a)*(72*B*a^3 - 40*A*a^2*b))/(8*b^(13/4)))*(5*A*b - 9*B*a))/(8*b^(13/4)) - ((-
a)^(1/4)*((x^(1/2)*(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b))/b^3 + ((-a)^(1/4)*(5*A*b - 9*B*a)*(72*B*a^3 -
40*A*a^2*b))/(8*b^(13/4)))*(5*A*b - 9*B*a))/(8*b^(13/4))))*(5*A*b - 9*B*a)*1i)/(4*b^(13/4)) + ((-a)^(1/4)*ata
n((((-a)^(1/4)*((x^(1/2)*(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b))/b^3 - ((-a)^(1/4)*(5*A*b - 9*B*a)*(72*B
*a^3 - 40*A*a^2*b)*1i)/(8*b^(13/4)))*(5*A*b - 9*B*a))/(8*b^(13/4)) + ((-a)^(1/4)*((x^(1/2)*(81*B^2*a^4 + 25*A^
2*a^2*b^2 - 90*A*B*a^3*b))/b^3 + ((-a)^(1/4)*(5*A*b - 9*B*a)*(72*B*a^3 - 40*A*a^2*b)*1i)/(8*b^(13/4)))*(5*A*b
- 9*B*a))/(8*b^(13/4)))/(((-a)^(1/4)*((x^(1/2)*(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b))/b^3 - ((-a)^(1/4)
*(5*A*b - 9*B*a)*(72*B*a^3 - 40*A*a^2*b)*1i)/(8*b^(13/4)))*(5*A*b - 9*B*a)*1i)/(8*b^(13/4)) - ((-a)^(1/4)*((x^
(1/2)*(81*B^2*a^4 + 25*A^2*a^2*b^2 - 90*A*B*a^3*b))/b^3 + ((-a)^(1/4)*(5*A*b - 9*B*a)*(72*B*a^3 - 40*A*a^2*b)*
1i)/(8*b^(13/4)))*(5*A*b - 9*B*a)*1i)/(8*b^(13/4))))*(5*A*b - 9*B*a))/(4*b^(13/4))