3.645 \(\int \frac{x^{11}}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=238 \[ \frac{a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a^2}{b^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x^2 \left (a+b x^2\right )}{2 b^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

(-5*a^2)/(b^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + a^5/(8*b^6*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (
5*a^4)/(6*b^6*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*a^3)/(2*b^6*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]) + (x^2*(a + b*x^2))/(2*b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (5*a*(a + b*x^2)*Log[a + b*x^2])/(2
*b^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.189469, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \[ \frac{a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a^2}{b^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x^2 \left (a+b x^2\right )}{2 b^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(-5*a^2)/(b^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + a^5/(8*b^6*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (
5*a^4)/(6*b^6*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*a^3)/(2*b^6*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]) + (x^2*(a + b*x^2))/(2*b^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (5*a*(a + b*x^2)*Log[a + b*x^2])/(2
*b^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,x^2\right )\\ &=\frac{\left (b^4 \left (a b+b^2 x^2\right )\right ) \operatorname{Subst}\left (\int \frac{x^5}{\left (a b+b^2 x\right )^5} \, dx,x,x^2\right )}{2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{\left (b^4 \left (a b+b^2 x^2\right )\right ) \operatorname{Subst}\left (\int \left (\frac{1}{b^{10}}-\frac{a^5}{b^{10} (a+b x)^5}+\frac{5 a^4}{b^{10} (a+b x)^4}-\frac{10 a^3}{b^{10} (a+b x)^3}+\frac{10 a^2}{b^{10} (a+b x)^2}-\frac{5 a}{b^{10} (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{5 a^2}{b^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x^2 \left (a+b x^2\right )}{2 b^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0382, size = 103, normalized size = 0.43 \[ \frac{-48 a^2 b^3 x^6-252 a^3 b^2 x^4-248 a^4 b x^2-77 a^5+48 a b^4 x^8-60 a \left (a+b x^2\right )^4 \log \left (a+b x^2\right )+12 b^5 x^{10}}{24 b^6 \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(-77*a^5 - 248*a^4*b*x^2 - 252*a^3*b^2*x^4 - 48*a^2*b^3*x^6 + 48*a*b^4*x^8 + 12*b^5*x^10 - 60*a*(a + b*x^2)^4*
Log[a + b*x^2])/(24*b^6*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.238, size = 163, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -12\,{b}^{5}{x}^{10}+60\,\ln \left ( b{x}^{2}+a \right ){x}^{8}a{b}^{4}-48\,a{b}^{4}{x}^{8}+240\,\ln \left ( b{x}^{2}+a \right ){x}^{6}{a}^{2}{b}^{3}+48\,{a}^{2}{b}^{3}{x}^{6}+360\,\ln \left ( b{x}^{2}+a \right ){x}^{4}{a}^{3}{b}^{2}+252\,{b}^{2}{a}^{3}{x}^{4}+240\,\ln \left ( b{x}^{2}+a \right ){x}^{2}{a}^{4}b+248\,{a}^{4}b{x}^{2}+60\,\ln \left ( b{x}^{2}+a \right ){a}^{5}+77\,{a}^{5} \right ) \left ( b{x}^{2}+a \right ) }{24\,{b}^{6}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

-1/24*(-12*b^5*x^10+60*ln(b*x^2+a)*x^8*a*b^4-48*a*b^4*x^8+240*ln(b*x^2+a)*x^6*a^2*b^3+48*a^2*b^3*x^6+360*ln(b*
x^2+a)*x^4*a^3*b^2+252*b^2*a^3*x^4+240*ln(b*x^2+a)*x^2*a^4*b+248*a^4*b*x^2+60*ln(b*x^2+a)*a^5+77*a^5)*(b*x^2+a
)/b^6/((b*x^2+a)^2)^(5/2)

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Maxima [A]  time = 1.02443, size = 161, normalized size = 0.68 \begin{align*} \frac{12 \, b^{5} x^{10} + 48 \, a b^{4} x^{8} - 48 \, a^{2} b^{3} x^{6} - 252 \, a^{3} b^{2} x^{4} - 248 \, a^{4} b x^{2} - 77 \, a^{5}}{24 \,{\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}} - \frac{5 \, a \log \left (b x^{2} + a\right )}{2 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/24*(12*b^5*x^10 + 48*a*b^4*x^8 - 48*a^2*b^3*x^6 - 252*a^3*b^2*x^4 - 248*a^4*b*x^2 - 77*a^5)/(b^10*x^8 + 4*a*
b^9*x^6 + 6*a^2*b^8*x^4 + 4*a^3*b^7*x^2 + a^4*b^6) - 5/2*a*log(b*x^2 + a)/b^6

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Fricas [A]  time = 1.31697, size = 332, normalized size = 1.39 \begin{align*} \frac{12 \, b^{5} x^{10} + 48 \, a b^{4} x^{8} - 48 \, a^{2} b^{3} x^{6} - 252 \, a^{3} b^{2} x^{4} - 248 \, a^{4} b x^{2} - 77 \, a^{5} - 60 \,{\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \log \left (b x^{2} + a\right )}{24 \,{\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/24*(12*b^5*x^10 + 48*a*b^4*x^8 - 48*a^2*b^3*x^6 - 252*a^3*b^2*x^4 - 248*a^4*b*x^2 - 77*a^5 - 60*(a*b^4*x^8 +
4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4*b*x^2 + a^5)*log(b*x^2 + a))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 +
4*a^3*b^7*x^2 + a^4*b^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{11}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x**11/((a + b*x**2)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x