3.635 \(\int \frac{x^5}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=113 \[ -\frac{a^2}{4 b^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{a}{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0983267, antiderivative size = 113, 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} \[ -\frac{a^2}{4 b^3 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{a}{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

a/(b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - a^2/(4*b^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a + b*x^
2)*Log[a + b*x^2])/(2*b^3*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^5}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a b+b^2 x\right )^3} \, dx,x,x^2\right )}{2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \operatorname{Subst}\left (\int \left (\frac{a^2}{b^5 (a+b x)^3}-\frac{2 a}{b^5 (a+b x)^2}+\frac{1}{b^5 (a+b x)}\right ) \, dx,x,x^2\right )}{2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{a}{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{a^2}{4 b^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 \left (a+b x^2\right )}{2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0216094, size = 61, normalized size = 0.54 \[ \frac{a \left (3 a+4 b x^2\right )+2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 b^3 \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.237, size = 81, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2\,\ln \left ( b{x}^{2}+a \right ){x}^{4}{b}^{2}+4\,\ln \left ( b{x}^{2}+a \right ){x}^{2}ab+4\,ab{x}^{2}+2\,{a}^{2}\ln \left ( b{x}^{2}+a \right ) +3\,{a}^{2} \right ) \left ( b{x}^{2}+a \right ) }{4\,{b}^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/4*(2*ln(b*x^2+a)*x^4*b^2+4*ln(b*x^2+a)*x^2*a*b+4*a*b*x^2+2*a^2*ln(b*x^2+a)+3*a^2)*(b*x^2+a)/b^3/((b*x^2+a)^2
)^(3/2)

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Maxima [A]  time = 1.22592, size = 86, normalized size = 0.76 \begin{align*} \frac{a b x^{2}}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{2}} + \frac{\log \left (x^{2} + \frac{a}{b}\right )}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{3 \, a^{2} b^{2}}{4 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x^{2} + \frac{a}{b}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

a*b*x^2/((x^2 + a/b)^2*(b^2)^(5/2)) + 1/2*log(x^2 + a/b)/(b^2)^(3/2) + 3/4*a^2*b^2/((b^2)^(7/2)*(x^2 + a/b)^2)

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Fricas [A]  time = 1.19541, size = 143, normalized size = 1.27 \begin{align*} \frac{4 \, a b x^{2} + 3 \, a^{2} + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right )}{4 \,{\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/4*(4*a*b*x^2 + 3*a^2 + 2*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**5/((a + b*x**2)**2)**(3/2), 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^5/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x