3.651 \(\int \frac{1}{x (a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=223 \[ \frac{1}{4 a^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{6 a^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{2 a^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\log (x) \left (a+b x^2\right )}{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

1/(2*a^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(6*a^2*(
a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a +
b*x^2)*Log[x])/(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((a + b*x^2)*Log[a + b*x^2])/(2*a^5*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4])

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Rubi [A]  time = 0.120767, antiderivative size = 223, 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 = {1112, 266, 44} \[ \frac{1}{4 a^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{6 a^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{8 a \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{1}{2 a^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\log (x) \left (a+b x^2\right )}{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

1/(2*a^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(8*a*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(6*a^2*(
a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a +
b*x^2)*Log[x])/(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((a + b*x^2)*Log[a + b*x^2])/(2*a^5*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4])

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 266

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.039439, size = 96, normalized size = 0.43 \[ \frac{a \left (52 a^2 b x^2+25 a^3+42 a b^2 x^4+12 b^3 x^6\right )+24 \log (x) \left (a+b x^2\right )^4-12 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 a^5 \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]

[Out]

(a*(25*a^3 + 52*a^2*b*x^2 + 42*a*b^2*x^4 + 12*b^3*x^6) + 24*(a + b*x^2)^4*Log[x] - 12*(a + b*x^2)^4*Log[a + b*
x^2])/(24*a^5*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.23, size = 193, normalized size = 0.9 \begin{align*}{\frac{ \left ( 24\,\ln \left ( x \right ){x}^{8}{b}^{4}-12\,\ln \left ( b{x}^{2}+a \right ){x}^{8}{b}^{4}+96\,\ln \left ( x \right ){x}^{6}a{b}^{3}-48\,\ln \left ( b{x}^{2}+a \right ){x}^{6}a{b}^{3}+12\,a{b}^{3}{x}^{6}+144\,\ln \left ( x \right ){x}^{4}{a}^{2}{b}^{2}-72\,\ln \left ( b{x}^{2}+a \right ){x}^{4}{a}^{2}{b}^{2}+42\,{a}^{2}{b}^{2}{x}^{4}+96\,\ln \left ( x \right ){x}^{2}{a}^{3}b-48\,\ln \left ( b{x}^{2}+a \right ){x}^{2}{a}^{3}b+52\,{a}^{3}b{x}^{2}+24\,{a}^{4}\ln \left ( x \right ) -12\,\ln \left ( b{x}^{2}+a \right ){a}^{4}+25\,{a}^{4} \right ) \left ( b{x}^{2}+a \right ) }{24\,{a}^{5}} \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(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/24*(24*ln(x)*x^8*b^4-12*ln(b*x^2+a)*x^8*b^4+96*ln(x)*x^6*a*b^3-48*ln(b*x^2+a)*x^6*a*b^3+12*a*b^3*x^6+144*ln(
x)*x^4*a^2*b^2-72*ln(b*x^2+a)*x^4*a^2*b^2+42*a^2*b^2*x^4+96*ln(x)*x^2*a^3*b-48*ln(b*x^2+a)*x^2*a^3*b+52*a^3*b*
x^2+24*a^4*ln(x)-12*ln(b*x^2+a)*a^4+25*a^4)*(b*x^2+a)/a^5/((b*x^2+a)^2)^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33906, size = 378, normalized size = 1.7 \begin{align*} \frac{12 \, a b^{3} x^{6} + 42 \, a^{2} b^{2} x^{4} + 52 \, a^{3} b x^{2} + 25 \, a^{4} - 12 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right ) + 24 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (x\right )}{24 \,{\left (a^{5} b^{4} x^{8} + 4 \, a^{6} b^{3} x^{6} + 6 \, a^{7} b^{2} x^{4} + 4 \, a^{8} b x^{2} + a^{9}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/24*(12*a*b^3*x^6 + 42*a^2*b^2*x^4 + 52*a^3*b*x^2 + 25*a^4 - 12*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^
3*b*x^2 + a^4)*log(b*x^2 + a) + 24*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*log(x))/(a^5*b^
4*x^8 + 4*a^6*b^3*x^6 + 6*a^7*b^2*x^4 + 4*a^8*b*x^2 + a^9)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \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(1/x/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(1/(x*((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(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x