3.520 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^{11}} \, dx\)

Optimal. Leaf size=189 \[ -\frac {b^4 (7 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{9/2}}+\frac {b^3 \sqrt {a+b x^2} (7 A b-10 a B)}{256 a^4 x^2}-\frac {b^2 \sqrt {a+b x^2} (7 A b-10 a B)}{384 a^3 x^4}+\frac {b \sqrt {a+b x^2} (7 A b-10 a B)}{480 a^2 x^6}+\frac {\sqrt {a+b x^2} (7 A b-10 a B)}{80 a x^8}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}} \]

[Out]

-1/10*A*(b*x^2+a)^(3/2)/a/x^10-1/256*b^4*(7*A*b-10*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(9/2)+1/80*(7*A*b-1
0*B*a)*(b*x^2+a)^(1/2)/a/x^8+1/480*b*(7*A*b-10*B*a)*(b*x^2+a)^(1/2)/a^2/x^6-1/384*b^2*(7*A*b-10*B*a)*(b*x^2+a)
^(1/2)/a^3/x^4+1/256*b^3*(7*A*b-10*B*a)*(b*x^2+a)^(1/2)/a^4/x^2

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Rubi [A]  time = 0.15, antiderivative size = 189, 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} \[ \frac {b^3 \sqrt {a+b x^2} (7 A b-10 a B)}{256 a^4 x^2}-\frac {b^2 \sqrt {a+b x^2} (7 A b-10 a B)}{384 a^3 x^4}-\frac {b^4 (7 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{9/2}}+\frac {b \sqrt {a+b x^2} (7 A b-10 a B)}{480 a^2 x^6}+\frac {\sqrt {a+b x^2} (7 A b-10 a B)}{80 a x^8}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*A*b - 10*a*B)*Sqrt[a + b*x^2])/(80*a*x^8) + (b*(7*A*b - 10*a*B)*Sqrt[a + b*x^2])/(480*a^2*x^6) - (b^2*(7*A
*b - 10*a*B)*Sqrt[a + b*x^2])/(384*a^3*x^4) + (b^3*(7*A*b - 10*a*B)*Sqrt[a + b*x^2])/(256*a^4*x^2) - (A*(a + b
*x^2)^(3/2))/(10*a*x^10) - (b^4*(7*A*b - 10*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(256*a^(9/2))

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 {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x} (A+B x)}{x^6} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}+\frac {\left (-\frac {7 A b}{2}+5 a B\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^5} \, dx,x,x^2\right )}{10 a}\\ &=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}-\frac {(b (7 A b-10 a B)) \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x}} \, dx,x,x^2\right )}{160 a}\\ &=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}+\frac {b (7 A b-10 a B) \sqrt {a+b x^2}}{480 a^2 x^6}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}+\frac {\left (b^2 (7 A b-10 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,x^2\right )}{192 a^2}\\ &=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}+\frac {b (7 A b-10 a B) \sqrt {a+b x^2}}{480 a^2 x^6}-\frac {b^2 (7 A b-10 a B) \sqrt {a+b x^2}}{384 a^3 x^4}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}-\frac {\left (b^3 (7 A b-10 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )}{256 a^3}\\ &=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}+\frac {b (7 A b-10 a B) \sqrt {a+b x^2}}{480 a^2 x^6}-\frac {b^2 (7 A b-10 a B) \sqrt {a+b x^2}}{384 a^3 x^4}+\frac {b^3 (7 A b-10 a B) \sqrt {a+b x^2}}{256 a^4 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}+\frac {\left (b^4 (7 A b-10 a B)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{512 a^4}\\ &=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}+\frac {b (7 A b-10 a B) \sqrt {a+b x^2}}{480 a^2 x^6}-\frac {b^2 (7 A b-10 a B) \sqrt {a+b x^2}}{384 a^3 x^4}+\frac {b^3 (7 A b-10 a B) \sqrt {a+b x^2}}{256 a^4 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}+\frac {\left (b^3 (7 A b-10 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^4}\\ &=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}+\frac {b (7 A b-10 a B) \sqrt {a+b x^2}}{480 a^2 x^6}-\frac {b^2 (7 A b-10 a B) \sqrt {a+b x^2}}{384 a^3 x^4}+\frac {b^3 (7 A b-10 a B) \sqrt {a+b x^2}}{256 a^4 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}-\frac {b^4 (7 A b-10 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 62, normalized size = 0.33 \[ -\frac {\left (a+b x^2\right )^{3/2} \left (3 a^5 A+b^4 x^{10} (10 a B-7 A b) \, _2F_1\left (\frac {3}{2},5;\frac {5}{2};\frac {b x^2}{a}+1\right )\right )}{30 a^6 x^{10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*((a + b*x^2)^(3/2)*(3*a^5*A + b^4*(-7*A*b + 10*a*B)*x^10*Hypergeometric2F1[3/2, 5, 5/2, 1 + (b*x^2)/a]))
/(a^6*x^10)

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fricas [A]  time = 1.08, size = 317, normalized size = 1.68 \[ \left [-\frac {15 \, {\left (10 \, B a b^{4} - 7 \, A b^{5}\right )} \sqrt {a} x^{10} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, {\left (10 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{8} - 10 \, {\left (10 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (10 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{7680 \, a^{5} x^{10}}, -\frac {15 \, {\left (10 \, B a b^{4} - 7 \, A b^{5}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, {\left (10 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{8} - 10 \, {\left (10 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (10 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3840 \, a^{5} x^{10}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^11,x, algorithm="fricas")

[Out]

[-1/7680*(15*(10*B*a*b^4 - 7*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 - 7*A*a*b^4)*x^8 - 10*(10*B*a^3*b^2 - 7*A*a^2*b^3)*x^6 + 384*A*a^5 + 8*(10*B*a^4*b - 7*A*a^3*b^2)
*x^4 + 48*(10*B*a^5 + A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^5*x^10), -1/3840*(15*(10*B*a*b^4 - 7*A*b^5)*sqrt(-a)*x
^10*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*(10*B*a^2*b^3 - 7*A*a*b^4)*x^8 - 10*(10*B*a^3*b^2 - 7*A*a^2*b^3)*x^
6 + 384*A*a^5 + 8*(10*B*a^4*b - 7*A*a^3*b^2)*x^4 + 48*(10*B*a^5 + A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^5*x^10)]

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giac [A]  time = 0.34, size = 230, normalized size = 1.22 \[ -\frac {\frac {15 \, {\left (10 \, B a b^{5} - 7 \, A b^{6}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {150 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a b^{5} - 700 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} b^{5} + 1280 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{3} b^{5} - 580 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 150 \, \sqrt {b x^{2} + a} B a^{5} b^{5} - 105 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b^{6} + 490 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b^{6} - 896 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a^{2} b^{6} + 790 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 105 \, \sqrt {b x^{2} + a} A a^{4} b^{6}}{a^{4} b^{5} x^{10}}}{3840 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^11,x, algorithm="giac")

[Out]

-1/3840*(15*(10*B*a*b^5 - 7*A*b^6)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4) + (150*(b*x^2 + a)^(9/2)*B*
a*b^5 - 700*(b*x^2 + a)^(7/2)*B*a^2*b^5 + 1280*(b*x^2 + a)^(5/2)*B*a^3*b^5 - 580*(b*x^2 + a)^(3/2)*B*a^4*b^5 -
150*sqrt(b*x^2 + a)*B*a^5*b^5 - 105*(b*x^2 + a)^(9/2)*A*b^6 + 490*(b*x^2 + a)^(7/2)*A*a*b^6 - 896*(b*x^2 + a)
^(5/2)*A*a^2*b^6 + 790*(b*x^2 + a)^(3/2)*A*a^3*b^6 + 105*sqrt(b*x^2 + a)*A*a^4*b^6)/(a^4*b^5*x^10))/b

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maple [A]  time = 0.03, size = 281, normalized size = 1.49 \[ -\frac {7 A \,b^{5} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{256 a^{\frac {9}{2}}}+\frac {5 B \,b^{4} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {7}{2}}}+\frac {7 \sqrt {b \,x^{2}+a}\, A \,b^{5}}{256 a^{5}}-\frac {5 \sqrt {b \,x^{2}+a}\, B \,b^{4}}{128 a^{4}}-\frac {7 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{4}}{256 a^{5} x^{2}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{3}}{128 a^{4} x^{2}}+\frac {7 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{3}}{128 a^{4} x^{4}}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{2}}{64 a^{3} x^{4}}-\frac {7 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,b^{2}}{96 a^{3} x^{6}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B b}{48 a^{2} x^{6}}+\frac {7 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A b}{80 a^{2} x^{8}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{8 a \,x^{8}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{10 a \,x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/x^11,x)

[Out]

-1/10*A*(b*x^2+a)^(3/2)/a/x^10+7/80*A*b/a^2/x^8*(b*x^2+a)^(3/2)-7/96*A*b^2/a^3/x^6*(b*x^2+a)^(3/2)+7/128*A*b^3
/a^4/x^4*(b*x^2+a)^(3/2)-7/256*A*b^4/a^5/x^2*(b*x^2+a)^(3/2)-7/256*A*b^5/a^(9/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(
1/2))/x)+7/256*A*b^5/a^5*(b*x^2+a)^(1/2)-1/8*B/a/x^8*(b*x^2+a)^(3/2)+5/48*B*b/a^2/x^6*(b*x^2+a)^(3/2)-5/64*B*b
^2/a^3/x^4*(b*x^2+a)^(3/2)+5/128*B*b^3/a^4/x^2*(b*x^2+a)^(3/2)+5/128*B*b^4/a^(7/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a
^(1/2))/x)-5/128*B*b^4/a^4*(b*x^2+a)^(1/2)

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maxima [A]  time = 1.17, size = 258, normalized size = 1.37 \[ \frac {5 \, B b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {7}{2}}} - \frac {7 \, A b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {9}{2}}} - \frac {5 \, \sqrt {b x^{2} + a} B b^{4}}{128 \, a^{4}} + \frac {7 \, \sqrt {b x^{2} + a} A b^{5}}{256 \, a^{5}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{3}}{128 \, a^{4} x^{2}} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{4}}{256 \, a^{5} x^{2}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{64 \, a^{3} x^{4}} + \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3}}{128 \, a^{4} x^{4}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{48 \, a^{2} x^{6}} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{96 \, a^{3} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{8 \, a x^{8}} + \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{80 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{10 \, a x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^11,x, algorithm="maxima")

[Out]

5/128*B*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) - 7/256*A*b^5*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(9/2) - 5/128*
sqrt(b*x^2 + a)*B*b^4/a^4 + 7/256*sqrt(b*x^2 + a)*A*b^5/a^5 + 5/128*(b*x^2 + a)^(3/2)*B*b^3/(a^4*x^2) - 7/256*
(b*x^2 + a)^(3/2)*A*b^4/(a^5*x^2) - 5/64*(b*x^2 + a)^(3/2)*B*b^2/(a^3*x^4) + 7/128*(b*x^2 + a)^(3/2)*A*b^3/(a^
4*x^4) + 5/48*(b*x^2 + a)^(3/2)*B*b/(a^2*x^6) - 7/96*(b*x^2 + a)^(3/2)*A*b^2/(a^3*x^6) - 1/8*(b*x^2 + a)^(3/2)
*B/(a*x^8) + 7/80*(b*x^2 + a)^(3/2)*A*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(3/2)*A/(a*x^10)

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mupad [B]  time = 3.50, size = 209, normalized size = 1.11 \[ \frac {7\,A\,{\left (b\,x^2+a\right )}^{5/2}}{30\,a^2\,x^{10}}-\frac {5\,B\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {79\,A\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a\,x^{10}}-\frac {7\,A\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {49\,A\,{\left (b\,x^2+a\right )}^{7/2}}{384\,a^3\,x^{10}}+\frac {7\,A\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^4\,x^{10}}-\frac {73\,B\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a\,x^8}+\frac {55\,B\,{\left (b\,x^2+a\right )}^{5/2}}{384\,a^2\,x^8}-\frac {5\,B\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^3\,x^8}+\frac {A\,b^5\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,7{}\mathrm {i}}{256\,a^{9/2}}-\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/x^11,x)

[Out]

(A*b^5*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*7i)/(256*a^(9/2)) - (5*B*(a + b*x^2)^(1/2))/(128*x^8) - (7*A*(a +
b*x^2)^(1/2))/(256*x^10) - (B*b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128*a^(7/2)) - (79*A*(a + b*x^2)^(
3/2))/(384*a*x^10) + (7*A*(a + b*x^2)^(5/2))/(30*a^2*x^10) - (49*A*(a + b*x^2)^(7/2))/(384*a^3*x^10) + (7*A*(a
+ b*x^2)^(9/2))/(256*a^4*x^10) - (73*B*(a + b*x^2)^(3/2))/(384*a*x^8) + (55*B*(a + b*x^2)^(5/2))/(384*a^2*x^8
) - (5*B*(a + b*x^2)^(7/2))/(128*a^3*x^8)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/x**11,x)

[Out]

Timed out

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