\(\int \frac {A+B x^2}{x^{7/2} (a+b x^2)} \, dx\) [374]

Optimal result

Integrand size = 22, antiderivative size = 255 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 255, 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 = {464, 331, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=-\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} a^{9/4}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {2 A}{5 a x^{5/2}} \]

[In]

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

[Out]

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

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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 {2 A}{5 a x^{5/2}}-\frac {\left (2 \left (\frac {5 A b}{2}-\frac {5 a B}{2}\right )\right ) \int \frac {1}{x^{3/2} \left (a+b x^2\right )} \, dx}{5 a} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {(b (A b-a B)) \int \frac {\sqrt {x}}{a+b x^2} \, dx}{a^2} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {(2 b (A b-a B)) \text {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {\left (\sqrt {b} (A b-a B)\right ) \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2}+\frac {\left (\sqrt {b} (A b-a B)\right ) \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{a^2} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {(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 a^2}+\frac {(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 a^2}+\frac {\left (\sqrt [4]{b} (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} a^{9/4}}+\frac {\left (\sqrt [4]{b} (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} a^{9/4}} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {\sqrt [4]{b} (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} a^{9/4}}-\frac {\sqrt [4]{b} (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} a^{9/4}}+\frac {\left (\sqrt [4]{b} (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} a^{9/4}}-\frac {\left (\sqrt [4]{b} (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} a^{9/4}} \\ & = -\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (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} a^{9/4}}-\frac {\sqrt [4]{b} (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} a^{9/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.60 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=-\frac {2 \left (a A-5 A b x^2+5 a B x^2\right )}{5 a^2 x^{5/2}}+\frac {\sqrt [4]{b} (-A b+a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (-A b+a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} a^{9/4}} \]

[In]

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

[Out]

(-2*(a*A - 5*A*b*x^2 + 5*a*B*x^2))/(5*a^2*x^(5/2)) + (b^(1/4)*(-(A*b) + a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqr
t[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(Sqrt[2]*a^(9/4)) + (b^(1/4)*(-(A*b) + a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*S
qrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(Sqrt[2]*a^(9/4))

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.55

method	result	size
derivativedivides	\(-\frac {2 A}{5 a \,x^{\frac {5}{2}}}-\frac {2 \left (-A b +B a \right )}{a^{2} \sqrt {x}}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 a^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(140\)
default	\(-\frac {2 A}{5 a \,x^{\frac {5}{2}}}-\frac {2 \left (-A b +B a \right )}{a^{2} \sqrt {x}}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 a^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(140\)
risch	\(-\frac {2 \left (-5 A b \,x^{2}+5 B a \,x^{2}+A a \right )}{5 a^{2} x^{\frac {5}{2}}}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 a^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(141\)

[In]

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

[Out]

-2/5*A/a/x^(5/2)-2/a^2*(-A*b+B*a)/x^(1/2)+1/4*(A*b-B*a)/a^2/(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*arcta
n(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 = 738, normalized size of antiderivative = 2.89 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=\frac {5 \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} \sqrt {x}\right ) - 5 i \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (i \, a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} \sqrt {x}\right ) + 5 i \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-i \, a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} \sqrt {x}\right ) - 5 \, a^{2} x^{3} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} \sqrt {x}\right ) - 4 \, {\left (5 \, {\left (B a - A b\right )} x^{2} + A a\right )} \sqrt {x}}{10 \, a^{2} x^{3}} \]

[In]

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

[Out]

1/10*(5*a^2*x^3*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(a
^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A
*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*sqrt(x)) - 5*I*a^2*x^3*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*
b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(I*a^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3
*B*a*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*sqrt(x)) + 5*I*a^2*x^
3*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(-I*a^7*(-(B^4*a
^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B^2*a^2*b^
2 + 3*A^2*B*a*b^3 - A^3*b^4)*sqrt(x)) - 5*a^2*x^3*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B
*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(-a^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4
*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*sqrt(x)) - 4*(5*(B*a - A*b)*x^2 + A
*a)*sqrt(x))/(a^2*x^3)

Sympy [A] (verification not implemented)

Time = 44.11 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=A \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{9 b x^{\frac {9}{2}}} & \text {for}\: a = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} + \frac {b \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 a^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {b \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 a^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {b \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{a^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {2 b}{a^{2} \sqrt {x}} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {2}{a \sqrt {x}} & \text {for}\: b = 0 \\- \frac {\log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 a \sqrt [4]{- \frac {a}{b}}} + \frac {\log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 a \sqrt [4]{- \frac {a}{b}}} - \frac {\operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{a \sqrt [4]{- \frac {a}{b}}} - \frac {2}{a \sqrt {x}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

A*Piecewise((zoo/x**(9/2), Eq(a, 0) & Eq(b, 0)), (-2/(9*b*x**(9/2)), Eq(a, 0)), (-2/(5*a*x**(5/2)), Eq(b, 0)),
(-2/(5*a*x**(5/2)) + b*log(sqrt(x) - (-a/b)**(1/4))/(2*a**2*(-a/b)**(1/4)) - b*log(sqrt(x) + (-a/b)**(1/4))/(
2*a**2*(-a/b)**(1/4)) + b*atan(sqrt(x)/(-a/b)**(1/4))/(a**2*(-a/b)**(1/4)) + 2*b/(a**2*sqrt(x)), True)) + B*Pi
ecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(a, 0)), (-2/(a*sqrt(x)), Eq(b, 0)), (-log(
sqrt(x) - (-a/b)**(1/4))/(2*a*(-a/b)**(1/4)) + log(sqrt(x) + (-a/b)**(1/4))/(2*a*(-a/b)**(1/4)) - atan(sqrt(x)
/(-a/b)**(1/4))/(a*(-a/b)**(1/4)) - 2/(a*sqrt(x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=-\frac {{\left (B a b - A b^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, a^{2}} - \frac {2 \, {\left (5 \, {\left (B a - A b\right )} x^{2} + A a\right )}}{5 \, a^{2} x^{\frac {5}{2}}} \]

[In]

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

[Out]

-1/4*(B*a*b - A*b^2)*(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)) + 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)))/a^2 - 2/5*(5*(B*a - A*b)*x^2 + A*a)/(a^2*x^(5/2))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=-\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{3} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{3} b^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{3} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{3} b^{2}} - \frac {2 \, {\left (5 \, B a x^{2} - 5 \, A b x^{2} + A a\right )}}{5 \, a^{2} x^{\frac {5}{2}}} \]

[In]

integrate((B*x^2+A)/x^(7/2)/(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))/(a^3*b^2) - 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))/(a^3*b^2) + 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))/(a^3*b^2) - 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))/(a^3*b^2) - 2/5*(5*B*a*x^2 - 5*A*b*x^2 + A*a)/(a^2*x^(5/2))

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.35 \[ \int \frac {A+B x^2}{x^{7/2} \left (a+b x^2\right )} \, dx=\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (A\,b-B\,a\right )}{a^{9/4}}-\frac {\frac {2\,A}{5\,a}-\frac {2\,x^2\,\left (A\,b-B\,a\right )}{a^2}}{x^{5/2}}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,\sqrt {x}}{a^{1/4}}\right )\,\left (A\,b-B\,a\right )}{a^{9/4}} \]

[In]

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

[Out]

((-b)^(1/4)*atan(((-b)^(1/4)*x^(1/2))/a^(1/4))*(A*b - B*a))/a^(9/4) - ((2*A)/(5*a) - (2*x^2*(A*b - B*a))/a^2)/
x^(5/2) - ((-b)^(1/4)*atanh(((-b)^(1/4)*x^(1/2))/a^(1/4))*(A*b - B*a))/a^(9/4)