\(\int \frac {x^{5/2} (A+B x^2)}{a+b x^2} \, dx\) [368]

Optimal result

Integrand size = 22, antiderivative size = 257 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}+\frac {a^{3/4} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {a^{3/4} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}} \]

[Out]

2/3*(A*b-B*a)*x^(3/2)/b^2+2/7*B*x^(7/2)/b+1/2*a^(3/4)*(A*b-B*a)*arctan(1-b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(1
1/4)*2^(1/2)-1/2*a^(3/4)*(A*b-B*a)*arctan(1+b^(1/4)*2^(1/2)*x^(1/2)/a^(1/4))/b^(11/4)*2^(1/2)-1/4*a^(3/4)*(A*b
-B*a)*ln(a^(1/2)+x*b^(1/2)-a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(11/4)*2^(1/2)+1/4*a^(3/4)*(A*b-B*a)*ln(a^(1/2)+
x*b^(1/2)+a^(1/4)*b^(1/4)*2^(1/2)*x^(1/2))/b^(11/4)*2^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {470, 327, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {a^{3/4} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (A b-a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {a^{3/4} (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {2 x^{3/2} (A b-a B)}{3 b^2}+\frac {2 B x^{7/2}}{7 b} \]

[In]

Int[(x^(5/2)*(A + B*x^2))/(a + b*x^2),x]

[Out]

(2*(A*b - a*B)*x^(3/2))/(3*b^2) + (2*B*x^(7/2))/(7*b) + (a^(3/4)*(A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[
x])/a^(1/4)])/(Sqrt[2]*b^(11/4)) - (a^(3/4)*(A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2
]*b^(11/4)) - (a^(3/4)*(A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(1
1/4)) + (a^(3/4)*(A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(11/4))

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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 470

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

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 {2 B x^{7/2}}{7 b}-\frac {\left (2 \left (-\frac {7 A b}{2}+\frac {7 a B}{2}\right )\right ) \int \frac {x^{5/2}}{a+b x^2} \, dx}{7 b} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}-\frac {(a (A b-a B)) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{b^2} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}-\frac {(2 a (A b-a B)) \text {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{b^2} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}+\frac {(a (A b-a B)) \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{b^{5/2}}-\frac {(a (A b-a B)) \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{b^{5/2}} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}-\frac {(a (A b-a B)) \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 )}{2 b^3}-\frac {(a (A b-a B)) \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 )}{2 b^3}-\frac {\left (a^{3/4} (A b-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 )}{2 \sqrt {2} b^{11/4}}-\frac {\left (a^{3/4} (A b-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 )}{2 \sqrt {2} b^{11/4}} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}-\frac {a^{3/4} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {a^{3/4} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}}-\frac {\left (a^{3/4} (A b-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 )}{\sqrt {2} b^{11/4}}+\frac {\left (a^{3/4} (A b-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 )}{\sqrt {2} b^{11/4}} \\ & = \frac {2 (A b-a B) x^{3/2}}{3 b^2}+\frac {2 B x^{7/2}}{7 b}+\frac {a^{3/4} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (A b-a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}}+\frac {a^{3/4} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} b^{11/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.59 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 x^{3/2} \left (7 A b-7 a B+3 b B x^2\right )}{21 b^2}-\frac {a^{3/4} (-A b+a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt {2} b^{11/4}}-\frac {a^{3/4} (-A b+a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} b^{11/4}} \]

[In]

Integrate[(x^(5/2)*(A + B*x^2))/(a + b*x^2),x]

[Out]

(2*x^(3/2)*(7*A*b - 7*a*B + 3*b*B*x^2))/(21*b^2) - (a^(3/4)*(-(A*b) + a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[
2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(Sqrt[2]*b^(11/4)) - (a^(3/4)*(-(A*b) + a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[x])/(Sqrt[a] + Sqrt[b]*x)])/(Sqrt[2]*b^(11/4))

Maple [A] (verified)

Time = 2.78 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.54

method	result	size
risch	\(\frac {2 x^{\frac {3}{2}} \left (3 b B \,x^{2}+7 A b -7 B a \right )}{21 b^{2}}-\frac {a \left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{3} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(140\)
derivativedivides	\(\frac {\frac {2 b B \,x^{\frac {7}{2}}}{7}+\frac {2 \left (A b -B a \right ) x^{\frac {3}{2}}}{3}}{b^{2}}-\frac {a \left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{3} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(142\)
default	\(\frac {\frac {2 b B \,x^{\frac {7}{2}}}{7}+\frac {2 \left (A b -B a \right ) x^{\frac {3}{2}}}{3}}{b^{2}}-\frac {a \left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{3} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(142\)

[In]

int(x^(5/2)*(B*x^2+A)/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

2/21*x^(3/2)*(3*B*b*x^2+7*A*b-7*B*a)/b^2-1/4*a*(A*b-B*a)/b^3/(a/b)^(1/4)*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*arct
an(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 = 749, normalized size of antiderivative = 2.91 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=-\frac {21 \, b^{2} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {1}{4}} \log \left (b^{8} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{5} - 3 \, A B^{2} a^{4} b + 3 \, A^{2} B a^{3} b^{2} - A^{3} a^{2} b^{3}\right )} \sqrt {x}\right ) - 21 i \, b^{2} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {1}{4}} \log \left (i \, b^{8} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{5} - 3 \, A B^{2} a^{4} b + 3 \, A^{2} B a^{3} b^{2} - A^{3} a^{2} b^{3}\right )} \sqrt {x}\right ) + 21 i \, b^{2} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-i \, b^{8} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{5} - 3 \, A B^{2} a^{4} b + 3 \, A^{2} B a^{3} b^{2} - A^{3} a^{2} b^{3}\right )} \sqrt {x}\right ) - 21 \, b^{2} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-b^{8} \left (-\frac {B^{4} a^{7} - 4 \, A B^{3} a^{6} b + 6 \, A^{2} B^{2} a^{5} b^{2} - 4 \, A^{3} B a^{4} b^{3} + A^{4} a^{3} b^{4}}{b^{11}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{5} - 3 \, A B^{2} a^{4} b + 3 \, A^{2} B a^{3} b^{2} - A^{3} a^{2} b^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (3 \, B b x^{3} - 7 \, {\left (B a - A b\right )} x\right )} \sqrt {x}}{42 \, b^{2}} \]

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a),x, algorithm="fricas")

[Out]

-1/42*(21*b^2*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*log(
b^8*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(3/4) - (B^3*a^5 - 3
*A*B^2*a^4*b + 3*A^2*B*a^3*b^2 - A^3*a^2*b^3)*sqrt(x)) - 21*I*b^2*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b
^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*log(I*b^8*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A
^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(3/4) - (B^3*a^5 - 3*A*B^2*a^4*b + 3*A^2*B*a^3*b^2 - A^3*a^2*b^3)*sqrt(x)) +
21*I*b^2*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*log(-I*b
^8*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(3/4) - (B^3*a^5 - 3*
A*B^2*a^4*b + 3*A^2*B*a^3*b^2 - A^3*a^2*b^3)*sqrt(x)) - 21*b^2*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2
- 4*A^3*B*a^4*b^3 + A^4*a^3*b^4)/b^11)^(1/4)*log(-b^8*(-(B^4*a^7 - 4*A*B^3*a^6*b + 6*A^2*B^2*a^5*b^2 - 4*A^3*B
*a^4*b^3 + A^4*a^3*b^4)/b^11)^(3/4) - (B^3*a^5 - 3*A*B^2*a^4*b + 3*A^2*B*a^3*b^2 - A^3*a^2*b^3)*sqrt(x)) - 4*(
3*B*b*x^3 - 7*(B*a - A*b)*x)*sqrt(x))/b^2

Sympy [A] (verification not implemented)

Time = 16.09 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.35 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\begin {cases} \tilde {\infty } \left (\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {7}{2}}}{7}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {7}{2}}}{7} + \frac {2 B x^{\frac {11}{2}}}{11}}{a} & \text {for}\: b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {7}{2}}}{7}}{b} & \text {for}\: a = 0 \\- \frac {2 A a \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {2 A x^{\frac {3}{2}}}{3 b} + \frac {A \left (- \frac {a}{b}\right )^{\frac {3}{4}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b} - \frac {A \left (- \frac {a}{b}\right )^{\frac {3}{4}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b} - \frac {A \left (- \frac {a}{b}\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b} + \frac {2 B a^{2} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b^{3} \sqrt [4]{- \frac {a}{b}}} - \frac {2 B a x^{\frac {3}{2}}}{3 b^{2}} - \frac {B a \left (- \frac {a}{b}\right )^{\frac {3}{4}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {B a \left (- \frac {a}{b}\right )^{\frac {3}{4}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {B a \left (- \frac {a}{b}\right )^{\frac {3}{4}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b^{2}} + \frac {2 B x^{\frac {7}{2}}}{7 b} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(5/2)*(B*x**2+A)/(b*x**2+a),x)

[Out]

Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(7/2)/7), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(7/2)/7 + 2*B*x**(11/2)/11)/a
, Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(7/2)/7)/b, Eq(a, 0)), (-2*A*a*atan(sqrt(x)/(-a/b)**(1/4))/(b**2*(-a/b)
**(1/4)) + 2*A*x**(3/2)/(3*b) + A*(-a/b)**(3/4)*log(sqrt(x) - (-a/b)**(1/4))/(2*b) - A*(-a/b)**(3/4)*log(sqrt(
x) + (-a/b)**(1/4))/(2*b) - A*(-a/b)**(3/4)*atan(sqrt(x)/(-a/b)**(1/4))/b + 2*B*a**2*atan(sqrt(x)/(-a/b)**(1/4
))/(b**3*(-a/b)**(1/4)) - 2*B*a*x**(3/2)/(3*b**2) - B*a*(-a/b)**(3/4)*log(sqrt(x) - (-a/b)**(1/4))/(2*b**2) +
B*a*(-a/b)**(3/4)*log(sqrt(x) + (-a/b)**(1/4))/(2*b**2) + B*a*(-a/b)**(3/4)*atan(sqrt(x)/(-a/b)**(1/4))/b**2 +
2*B*x**(7/2)/(7*b), True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.83 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {{\left (B a^{2} - A a b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, b^{2}} + \frac {2 \, {\left (3 \, B b x^{\frac {7}{2}} - 7 \, {\left (B a - A b\right )} x^{\frac {3}{2}}\right )}}{21 \, b^{2}} \]

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a),x, algorithm="maxima")

[Out]

1/4*(B*a^2 - A*a*b)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*s
qrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*
sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x)
+ sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))
/(a^(1/4)*b^(3/4)))/b^2 + 2/21*(3*B*b*x^(7/2) - 7*(B*a - A*b)*x^(3/2))/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.03 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{5}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{5}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{5}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{5}} + \frac {2 \, {\left (3 \, B b^{6} x^{\frac {7}{2}} - 7 \, B a b^{5} x^{\frac {3}{2}} + 7 \, A b^{6} x^{\frac {3}{2}}\right )}}{21 \, b^{7}} \]

[In]

integrate(x^(5/2)*(B*x^2+A)/(b*x^2+a),x, algorithm="giac")

[Out]

1/2*sqrt(2)*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)
^(1/4))/b^5 + 1/2*sqrt(2)*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2
*sqrt(x))/(a/b)^(1/4))/b^5 - 1/4*sqrt(2)*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/
4) + x + sqrt(a/b))/b^5 + 1/4*sqrt(2)*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4)
+ x + sqrt(a/b))/b^5 + 2/21*(3*B*b^6*x^(7/2) - 7*B*a*b^5*x^(3/2) + 7*A*b^6*x^(3/2))/b^7

Mupad [B] (verification not implemented)

Time = 5.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.36 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{a+b x^2} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )+\frac {2\,B\,x^{7/2}}{7\,b}+\frac {{\left (-a\right )}^{3/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b-B\,a\right )}{b^{11/4}}+\frac {{\left (-a\right )}^{3/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{b^{11/4}} \]

[In]

int((x^(5/2)*(A + B*x^2))/(a + b*x^2),x)

[Out]

x^(3/2)*((2*A)/(3*b) - (2*B*a)/(3*b^2)) + (2*B*x^(7/2))/(7*b) + ((-a)^(3/4)*atan((b^(1/4)*x^(1/2))/(-a)^(1/4))
*(A*b - B*a))/b^(11/4) + ((-a)^(3/4)*atan((b^(1/4)*x^(1/2)*1i)/(-a)^(1/4))*(A*b - B*a)*1i)/b^(11/4)