3.6.19 \(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^{11}} \, dx\)

Optimal. Leaf size=184 \[ \frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{7/2}}-\frac {3 b^3 \sqrt {a+b x^2} (A b-2 a B)}{256 a^3 x^2}+\frac {b^2 \sqrt {a+b x^2} (A b-2 a B)}{128 a^2 x^4}+\frac {\left (a+b x^2\right )^{3/2} (A b-2 a B)}{16 a x^8}+\frac {b \sqrt {a+b x^2} (A b-2 a B)}{32 a x^6}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}} \]

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Rubi [A]  time = 0.15, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {446, 78, 47, 51, 63, 208} \begin {gather*} -\frac {3 b^3 \sqrt {a+b x^2} (A b-2 a B)}{256 a^3 x^2}+\frac {b^2 \sqrt {a+b x^2} (A b-2 a B)}{128 a^2 x^4}+\frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{7/2}}+\frac {b \sqrt {a+b x^2} (A b-2 a B)}{32 a x^6}+\frac {\left (a+b x^2\right )^{3/2} (A b-2 a B)}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]

[Out]

(b*(A*b - 2*a*B)*Sqrt[a + b*x^2])/(32*a*x^6) + (b^2*(A*b - 2*a*B)*Sqrt[a + b*x^2])/(128*a^2*x^4) - (3*b^3*(A*b
- 2*a*B)*Sqrt[a + b*x^2])/(256*a^3*x^2) + ((A*b - 2*a*B)*(a + b*x^2)^(3/2))/(16*a*x^8) - (A*(a + b*x^2)^(5/2)
)/(10*a*x^10) + (3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a^(7/2))

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 78

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] || LtQ
[p, n]))))

Rule 208

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

Rule 446

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*} \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2} (A+B x)}{x^6} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}+\frac {\left (-\frac {5 A b}{2}+5 a B\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^5} \, dx,x,x^2\right )}{10 a}\\ &=\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac {(3 b (A b-2 a B)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^4} \, dx,x,x^2\right )}{32 a}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x^2}}{32 a x^6}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac {\left (b^2 (A b-2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,x^2\right )}{64 a}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x^2}}{32 a x^6}+\frac {b^2 (A b-2 a B) \sqrt {a+b x^2}}{128 a^2 x^4}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}+\frac {\left (3 b^3 (A b-2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{256 a^2}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x^2}}{32 a x^6}+\frac {b^2 (A b-2 a B) \sqrt {a+b x^2}}{128 a^2 x^4}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x^2}}{256 a^3 x^2}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac {\left (3 b^4 (A b-2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{512 a^3}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x^2}}{32 a x^6}+\frac {b^2 (A b-2 a B) \sqrt {a+b x^2}}{128 a^2 x^4}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x^2}}{256 a^3 x^2}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}-\frac {\left (3 b^3 (A b-2 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{256 a^3}\\ &=\frac {b (A b-2 a B) \sqrt {a+b x^2}}{32 a x^6}+\frac {b^2 (A b-2 a B) \sqrt {a+b x^2}}{128 a^2 x^4}-\frac {3 b^3 (A b-2 a B) \sqrt {a+b x^2}}{256 a^3 x^2}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{16 a x^8}-\frac {A \left (a+b x^2\right )^{5/2}}{10 a x^{10}}+\frac {3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 61, normalized size = 0.33 \begin {gather*} -\frac {\left (a+b x^2\right )^{5/2} \left (a^5 A+b^4 x^{10} (2 a B-A b) \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {b x^2}{a}+1\right )\right )}{10 a^6 x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]

[Out]

-1/10*((a + b*x^2)^(5/2)*(a^5*A + b^4*(-(A*b) + 2*a*B)*x^10*Hypergeometric2F1[5/2, 5, 7/2, 1 + (b*x^2)/a]))/(a
^6*x^10)

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IntegrateAlgebraic [A]  time = 0.26, size = 152, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-128 a^4 A-160 a^4 B x^2-176 a^3 A b x^2-240 a^3 b B x^4-8 a^2 A b^2 x^4-20 a^2 b^2 B x^6+10 a A b^3 x^6+30 a b^3 B x^8-15 A b^4 x^8\right )}{1280 a^3 x^{10}}-\frac {3 \left (2 a b^4 B-A b^5\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^2)^(3/2)*(A + B*x^2))/x^11,x]

[Out]

(Sqrt[a + b*x^2]*(-128*a^4*A - 176*a^3*A*b*x^2 - 160*a^4*B*x^2 - 8*a^2*A*b^2*x^4 - 240*a^3*b*B*x^4 + 10*a*A*b^
3*x^6 - 20*a^2*b^2*B*x^6 - 15*A*b^4*x^8 + 30*a*b^3*B*x^8))/(1280*a^3*x^10) - (3*(-(A*b^5) + 2*a*b^4*B)*ArcTanh
[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a^(7/2))

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fricas [A]  time = 1.08, size = 317, normalized size = 1.72 \begin {gather*} \left [-\frac {15 \, {\left (2 \, B a b^{4} - 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 (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{8} - 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{6} - 128 \, A a^{5} - 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{4} - 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{2560 \, a^{4} x^{10}}, \frac {15 \, {\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, {\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{8} - 10 \, {\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{6} - 128 \, A a^{5} - 8 \, {\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{4} - 16 \, {\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{1280 \, a^{4} x^{10}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="fricas")

[Out]

[-1/2560*(15*(2*B*a*b^4 - 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*(2*B
*a^2*b^3 - A*a*b^4)*x^8 - 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^6 - 128*A*a^5 - 8*(30*B*a^4*b + A*a^3*b^2)*x^4 - 16*(
10*B*a^5 + 11*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^4*x^10), 1/1280*(15*(2*B*a*b^4 - A*b^5)*sqrt(-a)*x^10*arctan(s
qrt(-a)/sqrt(b*x^2 + a)) + (15*(2*B*a^2*b^3 - A*a*b^4)*x^8 - 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^6 - 128*A*a^5 - 8*
(30*B*a^4*b + A*a^3*b^2)*x^4 - 16*(10*B*a^5 + 11*A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^4*x^10)]

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giac [A]  time = 0.48, size = 212, normalized size = 1.15 \begin {gather*} \frac {\frac {15 \, {\left (2 \, B a b^{5} - A b^{6}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {30 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a b^{5} - 140 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} b^{5} + 140 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 30 \, \sqrt {b x^{2} + a} B a^{5} b^{5} - 15 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b^{6} + 70 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b^{6} - 128 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a^{2} b^{6} - 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 15 \, \sqrt {b x^{2} + a} A a^{4} b^{6}}{a^{3} b^{5} x^{10}}}{1280 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="giac")

[Out]

1/1280*(15*(2*B*a*b^5 - A*b^6)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + (30*(b*x^2 + a)^(9/2)*B*a*b^5
- 140*(b*x^2 + a)^(7/2)*B*a^2*b^5 + 140*(b*x^2 + a)^(3/2)*B*a^4*b^5 - 30*sqrt(b*x^2 + a)*B*a^5*b^5 - 15*(b*x^
2 + a)^(9/2)*A*b^6 + 70*(b*x^2 + a)^(7/2)*A*a*b^6 - 128*(b*x^2 + a)^(5/2)*A*a^2*b^6 - 70*(b*x^2 + a)^(3/2)*A*a
^3*b^6 + 15*sqrt(b*x^2 + a)*A*a^4*b^6)/(a^3*b^5*x^10))/b

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maple [B]  time = 0.04, size = 317, normalized size = 1.72 \begin {gather*} \frac {3 A \,b^{5} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{256 a^{\frac {7}{2}}}-\frac {3 B \,b^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {5}{2}}}-\frac {3 \sqrt {b \,x^{2}+a}\, A \,b^{5}}{256 a^{4}}+\frac {3 \sqrt {b \,x^{2}+a}\, B \,b^{4}}{128 a^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{5}}{256 a^{5}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{4}}{128 a^{4}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{4}}{256 a^{5} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,b^{3}}{128 a^{4} x^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{3}}{128 a^{4} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,b^{2}}{64 a^{3} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,b^{2}}{32 a^{3} x^{6}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B b}{16 a^{2} x^{6}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A b}{16 a^{2} x^{8}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B}{8 a \,x^{8}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{10 a \,x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x)

[Out]

-1/10*A*(b*x^2+a)^(5/2)/a/x^10+1/16*A*b/a^2/x^8*(b*x^2+a)^(5/2)-1/32*A*b^2/a^3/x^6*(b*x^2+a)^(5/2)+1/128*A*b^3
/a^4/x^4*(b*x^2+a)^(5/2)+1/256*A*b^4/a^5/x^2*(b*x^2+a)^(5/2)-1/256*A*b^5/a^5*(b*x^2+a)^(3/2)+3/256*A*b^5/a^(7/
2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)-3/256*A*b^5/a^4*(b*x^2+a)^(1/2)-1/8*B/a/x^8*(b*x^2+a)^(5/2)+1/16*B*b/
a^2/x^6*(b*x^2+a)^(5/2)-1/64*B*b^2/a^3/x^4*(b*x^2+a)^(5/2)-1/128*B*b^3/a^4/x^2*(b*x^2+a)^(5/2)+1/128*B*b^4/a^4
*(b*x^2+a)^(3/2)-3/128*B*b^4/a^(5/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+3/128*B*b^4/a^3*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.20, size = 294, normalized size = 1.60 \begin {gather*} -\frac {3 \, B b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {5}{2}}} + \frac {3 \, A b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {7}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{4}}{128 \, a^{4}} + \frac {3 \, \sqrt {b x^{2} + a} B b^{4}}{128 \, a^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{5}}{256 \, a^{5}} - \frac {3 \, \sqrt {b x^{2} + a} A b^{5}}{256 \, a^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{4}}{256 \, a^{5} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{64 \, a^{3} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{3}}{128 \, a^{4} x^{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{32 \, a^{3} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{8 \, a x^{8}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{16 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{10 \, a x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^11,x, algorithm="maxima")

[Out]

-3/128*B*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 3/256*A*b^5*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) + 1/128
*(b*x^2 + a)^(3/2)*B*b^4/a^4 + 3/128*sqrt(b*x^2 + a)*B*b^4/a^3 - 1/256*(b*x^2 + a)^(3/2)*A*b^5/a^5 - 3/256*sqr
t(b*x^2 + a)*A*b^5/a^4 - 1/128*(b*x^2 + a)^(5/2)*B*b^3/(a^4*x^2) + 1/256*(b*x^2 + a)^(5/2)*A*b^4/(a^5*x^2) - 1
/64*(b*x^2 + a)^(5/2)*B*b^2/(a^3*x^4) + 1/128*(b*x^2 + a)^(5/2)*A*b^3/(a^4*x^4) + 1/16*(b*x^2 + a)^(5/2)*B*b/(
a^2*x^6) - 1/32*(b*x^2 + a)^(5/2)*A*b^2/(a^3*x^6) - 1/8*(b*x^2 + a)^(5/2)*B/(a*x^8) + 1/16*(b*x^2 + a)^(5/2)*A
*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(5/2)*A/(a*x^10)

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mupad [B]  time = 4.89, size = 205, normalized size = 1.11 \begin {gather*} \frac {3\,A\,a\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {11\,B\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^8}-\frac {7\,A\,{\left (b\,x^2+a\right )}^{3/2}}{128\,x^{10}}+\frac {3\,B\,a\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{10\,a\,x^{10}}+\frac {7\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,x^{10}}-\frac {3\,A\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^3\,x^{10}}-\frac {11\,B\,{\left (b\,x^2+a\right )}^{5/2}}{128\,a\,x^8}+\frac {3\,B\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^2\,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^{7/2}}+\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,3{}\mathrm {i}}{128\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^11,x)

[Out]

(B*b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(128*a^(5/2)) - (11*B*(a + b*x^2)^(3/2))/(128*x^8) - (A*b^5*at
an(((a + b*x^2)^(1/2)*1i)/a^(1/2))*3i)/(256*a^(7/2)) - (7*A*(a + b*x^2)^(3/2))/(128*x^10) + (3*A*a*(a + b*x^2)
^(1/2))/(256*x^10) + (3*B*a*(a + b*x^2)^(1/2))/(128*x^8) - (A*(a + b*x^2)^(5/2))/(10*a*x^10) + (7*A*(a + b*x^2
)^(7/2))/(128*a^2*x^10) - (3*A*(a + b*x^2)^(9/2))/(256*a^3*x^10) - (11*B*(a + b*x^2)^(5/2))/(128*a*x^8) + (3*B
*(a + b*x^2)^(7/2))/(128*a^2*x^8)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**11,x)

[Out]

Timed out

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