\(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^6} \, dx\) [549]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

[Out]

-1/3*b*(2*A*b+5*B*a)*(b*x^2+a)^(3/2)/a/x-1/15*(2*A*b+5*B*a)*(b*x^2+a)^(5/2)/a/x^3-1/5*A*(b*x^2+a)^(7/2)/a/x^5+
1/2*b^(3/2)*(2*A*b+5*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))+1/2*b^2*(2*A*b+5*B*a)*x*(b*x^2+a)^(1/2)/a

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {464, 283, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\frac {1}{2} b^{3/2} (5 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {b^2 x \sqrt {a+b x^2} (5 a B+2 A b)}{2 a}-\frac {b \left (a+b x^2\right )^{3/2} (5 a B+2 A b)}{3 a x}-\frac {\left (a+b x^2\right )^{5/2} (5 a B+2 A b)}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5} \]

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

(b^2*(2*A*b + 5*a*B)*x*Sqrt[a + b*x^2])/(2*a) - (b*(2*A*b + 5*a*B)*(a + b*x^2)^(3/2))/(3*a*x) - ((2*A*b + 5*a*
B)*(a + b*x^2)^(5/2))/(15*a*x^3) - (A*(a + b*x^2)^(7/2))/(5*a*x^5) + (b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/2

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 283

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

Rubi steps \begin{align*} \text {integral}& = -\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}-\frac {(-2 A b-5 a B) \int \frac {\left (a+b x^2\right )^{5/2}}{x^4} \, dx}{5 a} \\ & = -\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {(b (2 A b+5 a B)) \int \frac {\left (a+b x^2\right )^{3/2}}{x^2} \, dx}{3 a} \\ & = -\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {\left (b^2 (2 A b+5 a B)\right ) \int \sqrt {a+b x^2} \, dx}{a} \\ & = \frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} \left (b^2 (2 A b+5 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx \\ & = \frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} \left (b^2 (2 A b+5 a B)\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right ) \\ & = \frac {b^2 (2 A b+5 a B) x \sqrt {a+b x^2}}{2 a}-\frac {b (2 A b+5 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {(2 A b+5 a B) \left (a+b x^2\right )^{5/2}}{15 a x^3}-\frac {A \left (a+b x^2\right )^{7/2}}{5 a x^5}+\frac {1}{2} b^{3/2} (2 A b+5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\frac {\sqrt {a+b x^2} \left (-6 a^2 A-22 a A b x^2-10 a^2 B x^2-46 A b^2 x^4-70 a b B x^4+15 b^2 B x^6\right )}{30 x^5}+b^{3/2} (2 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right ) \]

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^6,x]

[Out]

(Sqrt[a + b*x^2]*(-6*a^2*A - 22*a*A*b*x^2 - 10*a^2*B*x^2 - 46*A*b^2*x^4 - 70*a*b*B*x^4 + 15*b^2*B*x^6))/(30*x^
5) + b^(3/2)*(2*A*b + 5*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])]

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-15 b^{2} B \,x^{6}+46 A \,b^{2} x^{4}+70 B a b \,x^{4}+22 a A b \,x^{2}+10 a^{2} B \,x^{2}+6 a^{2} A \right )}{30 x^{5}}+\frac {\left (2 A b +5 B a \right ) b^{\frac {3}{2}} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2}\)	\(96\)
pseudoelliptic	\(-\frac {-5 x^{5} b^{2} \left (A b +\frac {5 B a}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\sqrt {b \,x^{2}+a}\, \left (\frac {11 x^{2} \left (\frac {35 x^{2} B}{11}+A \right ) a \,b^{\frac {3}{2}}}{3}+\left (-\frac {5}{2} B \,x^{6}+\frac {23}{3} A \,x^{4}\right ) b^{\frac {5}{2}}+a^{2} \sqrt {b}\, \left (\frac {5 x^{2} B}{3}+A \right )\right )}{5 \sqrt {b}\, x^{5}}\)	\(103\)
default	\(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{5 a \,x^{5}}+\frac {2 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )}{5 a}\right )\)	\(260\)

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/30*(b*x^2+a)^(1/2)*(-15*B*b^2*x^6+46*A*b^2*x^4+70*B*a*b*x^4+22*A*a*b*x^2+10*B*a^2*x^2+6*A*a^2)/x^5+1/2*(2*A
*b+5*B*a)*b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\left [\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {b} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (15 \, B b^{2} x^{6} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{60 \, x^{5}}, -\frac {15 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (15 \, B b^{2} x^{6} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{4} - 6 \, A a^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, x^{5}}\right ] \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="fricas")

[Out]

[1/60*(15*(5*B*a*b + 2*A*b^2)*sqrt(b)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(15*B*b^2*x^6 -
2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5, -1/30*(15*(5*B*a*b +
2*A*b^2)*sqrt(-b)*x^5*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (15*B*b^2*x^6 - 2*(35*B*a*b + 23*A*b^2)*x^4 - 6*A*
a^2 - 2*(5*B*a^2 + 11*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^5]

Sympy [A] (verification not implemented)

Time = 3.15 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.22 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=- \frac {A \sqrt {a} b^{2}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {8 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} + A b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b^{3} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {2 B a^{\frac {3}{2}} b}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {2 B \sqrt {a} b^{2} x}{\sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3} + 2 B a b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + B b^{2} \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**6,x)

[Out]

-A*sqrt(a)*b**2/(x*sqrt(1 + b*x**2/a)) - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 11*A*a*b**(3/2)*sqrt(a
/(b*x**2) + 1)/(15*x**2) - 8*A*b**(5/2)*sqrt(a/(b*x**2) + 1)/15 + A*b**(5/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**3
*x/(sqrt(a)*sqrt(1 + b*x**2/a)) - 2*B*a**(3/2)*b/(x*sqrt(1 + b*x**2/a)) - 2*B*sqrt(a)*b**2*x/sqrt(1 + b*x**2/a
) - B*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - B*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/3 + 2*B*a*b**(3/2)*asinh(
sqrt(b)*x/sqrt(a)) + B*b**2*Piecewise((a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0))
, (x*log(x)/sqrt(b*x**2), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\frac {5}{2} \, \sqrt {b x^{2} + a} B b^{2} x + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2} x}{3 \, a} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} A b^{3} x}{a} + \frac {5}{2} \, B a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) + A b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{3 \, a x} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{15 \, a^{2} x} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{3 \, a x^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{5 \, a x^{5}} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="maxima")

[Out]

5/2*sqrt(b*x^2 + a)*B*b^2*x + 5/3*(b*x^2 + a)^(3/2)*B*b^2*x/a + 2/3*(b*x^2 + a)^(3/2)*A*b^3*x/a^2 + sqrt(b*x^2
+ a)*A*b^3*x/a + 5/2*B*a*b^(3/2)*arcsinh(b*x/sqrt(a*b)) + A*b^(5/2)*arcsinh(b*x/sqrt(a*b)) - 4/3*(b*x^2 + a)^
(5/2)*B*b/(a*x) - 8/15*(b*x^2 + a)^(5/2)*A*b^2/(a^2*x) - 1/3*(b*x^2 + a)^(7/2)*B/(a*x^3) - 2/15*(b*x^2 + a)^(7
/2)*A*b/(a^2*x^3) - 1/5*(b*x^2 + a)^(7/2)*A/(a*x^5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (128) = 256\).

Time = 0.32 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.11 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a} B b^{2} x - \frac {1}{4} \, {\left (5 \, B a b^{\frac {3}{2}} + 2 \, A b^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{2} b^{\frac {3}{2}} + 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a b^{\frac {5}{2}} - 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{3} b^{\frac {3}{2}} - 90 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A a^{2} b^{\frac {5}{2}} + 200 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{4} b^{\frac {3}{2}} + 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{3} b^{\frac {5}{2}} - 130 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{5} b^{\frac {3}{2}} - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{4} b^{\frac {5}{2}} + 35 \, B a^{6} b^{\frac {3}{2}} + 23 \, A a^{5} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^6,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*B*b^2*x - 1/4*(5*B*a*b^(3/2) + 2*A*b^(5/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/15*(4
5*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^2*b^(3/2) + 45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*a*b^(5/2) - 150*(sqrt(b
)*x - sqrt(b*x^2 + a))^6*B*a^3*b^(3/2) - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a^2*b^(5/2) + 200*(sqrt(b)*x - s
qrt(b*x^2 + a))^4*B*a^4*b^(3/2) + 140*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^3*b^(5/2) - 130*(sqrt(b)*x - sqrt(b*
x^2 + a))^2*B*a^5*b^(3/2) - 70*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^4*b^(5/2) + 35*B*a^6*b^(3/2) + 23*A*a^5*b^(
5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^6} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2}}{x^6} \,d x \]

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^6,x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^6, x)