\(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^{11}} \, dx\) [554]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {b^2 (3 A b-10 a B) \sqrt {a+b x^2}}{128 a x^4}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x^2}}{256 a^2 x^2}+\frac {b (3 A b-10 a B) \left (a+b x^2\right )^{3/2}}{96 a x^6}+\frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}-\frac {b^4 (3 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{5/2}} \]

[Out]

1/96*b*(3*A*b-10*B*a)*(b*x^2+a)^(3/2)/a/x^6+1/80*(3*A*b-10*B*a)*(b*x^2+a)^(5/2)/a/x^8-1/10*A*(b*x^2+a)^(7/2)/a
/x^10-1/256*b^4*(3*A*b-10*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+1/128*b^2*(3*A*b-10*B*a)*(b*x^2+a)^(1/
2)/a/x^4+1/256*b^3*(3*A*b-10*B*a)*(b*x^2+a)^(1/2)/a^2/x^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {457, 79, 43, 44, 65, 214} \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {b^4 (3 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{5/2}}+\frac {b^3 \sqrt {a+b x^2} (3 A b-10 a B)}{256 a^2 x^2}+\frac {b^2 \sqrt {a+b x^2} (3 A b-10 a B)}{128 a x^4}+\frac {\left (a+b x^2\right )^{5/2} (3 A b-10 a B)}{80 a x^8}+\frac {b \left (a+b x^2\right )^{3/2} (3 A b-10 a B)}{96 a x^6}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}} \]

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^11,x]

[Out]

(b^2*(3*A*b - 10*a*B)*Sqrt[a + b*x^2])/(128*a*x^4) + (b^3*(3*A*b - 10*a*B)*Sqrt[a + b*x^2])/(256*a^2*x^2) + (b
*(3*A*b - 10*a*B)*(a + b*x^2)^(3/2))/(96*a*x^6) + ((3*A*b - 10*a*B)*(a + b*x^2)^(5/2))/(80*a*x^8) - (A*(a + b*
x^2)^(7/2))/(10*a*x^10) - (b^4*(3*A*b - 10*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a^(5/2))

Rule 43

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

Rule 44

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 457

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{5/2} (A+B x)}{x^6} \, dx,x,x^2\right ) \\ & = -\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}+\frac {\left (-\frac {3 A b}{2}+5 a B\right ) \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^5} \, dx,x,x^2\right )}{10 a} \\ & = \frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}-\frac {(b (3 A b-10 a B)) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^4} \, dx,x,x^2\right )}{32 a} \\ & = \frac {b (3 A b-10 a B) \left (a+b x^2\right )^{3/2}}{96 a x^6}+\frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}-\frac {\left (b^2 (3 A b-10 a B)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,x^2\right )}{64 a} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x^2}}{128 a x^4}+\frac {b (3 A b-10 a B) \left (a+b x^2\right )^{3/2}}{96 a x^6}+\frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}-\frac {\left (b^3 (3 A b-10 a B)\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{256 a} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x^2}}{128 a x^4}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x^2}}{256 a^2 x^2}+\frac {b (3 A b-10 a B) \left (a+b x^2\right )^{3/2}}{96 a x^6}+\frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}+\frac {\left (b^4 (3 A b-10 a B)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{512 a^2} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x^2}}{128 a x^4}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x^2}}{256 a^2 x^2}+\frac {b (3 A b-10 a B) \left (a+b x^2\right )^{3/2}}{96 a x^6}+\frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}+\frac {\left (b^3 (3 A b-10 a B)\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{256 a^2} \\ & = \frac {b^2 (3 A b-10 a B) \sqrt {a+b x^2}}{128 a x^4}+\frac {b^3 (3 A b-10 a B) \sqrt {a+b x^2}}{256 a^2 x^2}+\frac {b (3 A b-10 a B) \left (a+b x^2\right )^{3/2}}{96 a x^6}+\frac {(3 A b-10 a B) \left (a+b x^2\right )^{5/2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{7/2}}{10 a x^{10}}-\frac {b^4 (3 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {\sqrt {a+b x^2} \left (-45 A b^4 x^8+30 a b^3 x^6 \left (A+5 B x^2\right )+96 a^4 \left (4 A+5 B x^2\right )+16 a^3 b x^2 \left (63 A+85 B x^2\right )+4 a^2 b^2 x^4 \left (186 A+295 B x^2\right )\right )}{3840 a^2 x^{10}}+\frac {b^4 (-3 A b+10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{5/2}} \]

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^11,x]

[Out]

-1/3840*(Sqrt[a + b*x^2]*(-45*A*b^4*x^8 + 30*a*b^3*x^6*(A + 5*B*x^2) + 96*a^4*(4*A + 5*B*x^2) + 16*a^3*b*x^2*(
63*A + 85*B*x^2) + 4*a^2*b^2*x^4*(186*A + 295*B*x^2)))/(a^2*x^10) + (b^4*(-3*A*b + 10*a*B)*ArcTanh[Sqrt[a + b*
x^2]/Sqrt[a]])/(256*a^(5/2))

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(-\frac {31 \left (\frac {15 x^{10} b^{4} \left (A b -\frac {10 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{248}+\left (\frac {5 b^{3} x^{6} \left (5 x^{2} B +A \right ) a^{\frac {3}{2}}}{124}+b^{2} x^{4} \left (\frac {295 x^{2} B}{186}+A \right ) a^{\frac {5}{2}}+\frac {42 x^{2} b \left (\frac {85 x^{2} B}{63}+A \right ) a^{\frac {7}{2}}}{31}+\frac {4 \left (5 x^{2} B +4 A \right ) a^{\frac {9}{2}}}{31}-\frac {15 A \sqrt {a}\, b^{4} x^{8}}{248}\right ) \sqrt {b \,x^{2}+a}\right )}{160 a^{\frac {5}{2}} x^{10}}\)	\(132\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-45 A \,b^{4} x^{8}+150 B a \,b^{3} x^{8}+30 A a \,b^{3} x^{6}+1180 B \,a^{2} b^{2} x^{6}+744 A \,a^{2} b^{2} x^{4}+1360 B \,a^{3} b \,x^{4}+1008 A \,a^{3} b \,x^{2}+480 B \,a^{4} x^{2}+384 A \,a^{4}\right )}{3840 x^{10} a^{2}}-\frac {\left (3 A b -10 B a \right ) b^{4} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{256 a^{\frac {5}{2}}}\)	\(148\)
default	\(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 a \,x^{8}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{10 a \,x^{10}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 a \,x^{8}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )}{10 a}\right )\)	\(354\)

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x,method=_RETURNVERBOSE)

[Out]

-31/160/a^(5/2)*(15/248*x^10*b^4*(A*b-10/3*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(5/124*b^3*x^6*(5*B*x^2+A)*a^
(3/2)+b^2*x^4*(295/186*x^2*B+A)*a^(5/2)+42/31*x^2*b*(85/63*x^2*B+A)*a^(7/2)+4/31*(5*B*x^2+4*A)*a^(9/2)-15/248*
A*a^(1/2)*b^4*x^8)*(b*x^2+a)^(1/2))/x^10

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\left [-\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {a} x^{10} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{8} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{7680 \, a^{3} x^{10}}, -\frac {15 \, {\left (10 \, B a b^{4} - 3 \, A b^{5}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, {\left (10 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{8} + 10 \, {\left (118 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (170 \, B a^{4} b + 93 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + 21 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3840 \, a^{3} x^{10}}\right ] \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x, algorithm="fricas")

[Out]

[-1/7680*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt(a)*x^10*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(15*(
10*B*a^2*b^3 - 3*A*a*b^4)*x^8 + 10*(118*B*a^3*b^2 + 3*A*a^2*b^3)*x^6 + 384*A*a^5 + 8*(170*B*a^4*b + 93*A*a^3*b
^2)*x^4 + 48*(10*B*a^5 + 21*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^10), -1/3840*(15*(10*B*a*b^4 - 3*A*b^5)*sqrt
(-a)*x^10*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*(10*B*a^2*b^3 - 3*A*a*b^4)*x^8 + 10*(118*B*a^3*b^2 + 3*A*a^2*
b^3)*x^6 + 384*A*a^5 + 8*(170*B*a^4*b + 93*A*a^3*b^2)*x^4 + 48*(10*B*a^5 + 21*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(
a^3*x^10)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\text {Timed out} \]

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**11,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (161) = 322\).

Time = 0.22 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {5 \, B b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {3 \, A b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {5}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{4}}{128 \, a^{4}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{4}}{384 \, a^{3}} - \frac {5 \, \sqrt {b x^{2} + a} B b^{4}}{128 \, a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{5}}{1280 \, a^{5}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{5}}{256 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} A b^{5}}{256 \, a^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B b^{3}}{128 \, a^{4} x^{2}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{4}}{1280 \, a^{5} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B b^{2}}{192 \, a^{3} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{3}}{640 \, a^{4} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B b}{48 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{2}}{160 \, a^{3} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{8 \, a x^{8}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{80 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{10 \, a x^{10}} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x, algorithm="maxima")

[Out]

5/128*B*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 3/256*A*b^5*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) - 1/128*
(b*x^2 + a)^(5/2)*B*b^4/a^4 - 5/384*(b*x^2 + a)^(3/2)*B*b^4/a^3 - 5/128*sqrt(b*x^2 + a)*B*b^4/a^2 + 3/1280*(b*
x^2 + a)^(5/2)*A*b^5/a^5 + 1/256*(b*x^2 + a)^(3/2)*A*b^5/a^4 + 3/256*sqrt(b*x^2 + a)*A*b^5/a^3 + 1/128*(b*x^2
+ a)^(7/2)*B*b^3/(a^4*x^2) - 3/1280*(b*x^2 + a)^(7/2)*A*b^4/(a^5*x^2) + 1/192*(b*x^2 + a)^(7/2)*B*b^2/(a^3*x^4
) - 1/640*(b*x^2 + a)^(7/2)*A*b^3/(a^4*x^4) + 1/48*(b*x^2 + a)^(7/2)*B*b/(a^2*x^6) - 1/160*(b*x^2 + a)^(7/2)*A
*b^2/(a^3*x^6) - 1/8*(b*x^2 + a)^(7/2)*B/(a*x^8) + 3/80*(b*x^2 + a)^(7/2)*A*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(7/
2)*A/(a*x^10)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {\frac {15 \, {\left (10 \, B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {150 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a b^{5} + 580 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} b^{5} - 1280 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{3} b^{5} + 700 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 150 \, \sqrt {b x^{2} + a} B a^{5} b^{5} - 45 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b^{6} + 210 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b^{6} + 384 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 210 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 45 \, \sqrt {b x^{2} + a} A a^{4} b^{6}}{a^{2} b^{5} x^{10}}}{3840 \, b} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^11,x, algorithm="giac")

[Out]

-1/3840*(15*(10*B*a*b^5 - 3*A*b^6)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (150*(b*x^2 + a)^(9/2)*B*
a*b^5 + 580*(b*x^2 + a)^(7/2)*B*a^2*b^5 - 1280*(b*x^2 + a)^(5/2)*B*a^3*b^5 + 700*(b*x^2 + a)^(3/2)*B*a^4*b^5 -
150*sqrt(b*x^2 + a)*B*a^5*b^5 - 45*(b*x^2 + a)^(9/2)*A*b^6 + 210*(b*x^2 + a)^(7/2)*A*a*b^6 + 384*(b*x^2 + a)^
(5/2)*A*a^2*b^6 - 210*(b*x^2 + a)^(3/2)*A*a^3*b^6 + 45*sqrt(b*x^2 + a)*A*a^4*b^6)/(a^2*b^5*x^10))/b

Mupad [B] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {7\,A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^{10}}-\frac {73\,B\,{\left (b\,x^2+a\right )}^{5/2}}{384\,x^8}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{10\,x^{10}}+\frac {55\,B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{384\,x^8}-\frac {3\,A\,a^2\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {7\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^{10}}+\frac {3\,A\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^2\,x^{10}}-\frac {5\,B\,a^2\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {5\,B\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a\,x^8}+\frac {A\,b^5\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{256\,a^{5/2}}-\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{3/2}} \]

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^11,x)

[Out]

(A*b^5*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(256*a^(5/2)) - (73*B*(a + b*x^2)^(5/2))/(384*x^8) - (A*(a + b
*x^2)^(5/2))/(10*x^10) - (B*b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128*a^(3/2)) + (7*A*a*(a + b*x^2)^(3
/2))/(128*x^10) + (55*B*a*(a + b*x^2)^(3/2))/(384*x^8) - (3*A*a^2*(a + b*x^2)^(1/2))/(256*x^10) - (7*A*(a + b*
x^2)^(7/2))/(128*a*x^10) + (3*A*(a + b*x^2)^(9/2))/(256*a^2*x^10) - (5*B*a^2*(a + b*x^2)^(1/2))/(128*x^8) - (5
*B*(a + b*x^2)^(7/2))/(128*a*x^8)