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

Optimal. Leaf size=250 \[ \frac{b^5 x^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac{5 a b^4 x^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac{5 a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}-\frac{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}+\frac{10 a^3 b^2 \log (x) \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^4*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*
(a + b*x^2)) + (5*a^2*b^3*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b*x^2) + (5*a*b^4*x^4*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(4*(a + b*x^2)) + (b^5*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (10*a^3*b^2*Sqrt[a^
2 + 2*a*b*x^2 + b^2*x^4]*Log[x])/(a + b*x^2)

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Rubi [A]  time = 0.0700548, antiderivative size = 250, 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, 43} \[ \frac{b^5 x^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac{5 a b^4 x^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac{5 a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}-\frac{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}+\frac{10 a^3 b^2 \log (x) \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^4*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*
(a + b*x^2)) + (5*a^2*b^3*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b*x^2) + (5*a*b^4*x^4*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(4*(a + b*x^2)) + (b^5*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (10*a^3*b^2*Sqrt[a^
2 + 2*a*b*x^2 + b^2*x^4]*Log[x])/(a + b*x^2)

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 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{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^5} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \frac{\left (a b+b^2 x^2\right )^5}{x^5} \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^5}{x^3} \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \operatorname{Subst}\left (\int \left (10 a^2 b^8+\frac{a^5 b^5}{x^3}+\frac{5 a^4 b^6}{x^2}+\frac{10 a^3 b^7}{x}+5 a b^9 x+b^{10} x^2\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac{a^5 \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac{5 a^4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac{5 a b^4 x^4 \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac{b^5 x^6 \sqrt{a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac{10 a^3 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2}\\ \end{align*}

Mathematica [A]  time = 0.0264782, size = 85, normalized size = 0.34 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (60 a^2 b^3 x^6+120 a^3 b^2 x^4 \log (x)-30 a^4 b x^2-3 a^5+15 a b^4 x^8+2 b^5 x^{10}\right )}{12 x^4 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(-3*a^5 - 30*a^4*b*x^2 + 60*a^2*b^3*x^6 + 15*a*b^4*x^8 + 2*b^5*x^10 + 120*a^3*b^2*x^4*Log
[x]))/(12*x^4*(a + b*x^2))

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Maple [A]  time = 0.213, size = 82, normalized size = 0.3 \begin{align*}{\frac{2\,{b}^{5}{x}^{10}+15\,a{b}^{4}{x}^{8}+60\,{a}^{2}{b}^{3}{x}^{6}+120\,{b}^{2}{a}^{3}\ln \left ( x \right ){x}^{4}-30\,{a}^{4}b{x}^{2}-3\,{a}^{5}}{12\, \left ( b{x}^{2}+a \right ) ^{5}{x}^{4}} \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((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^5,x)

[Out]

1/12*((b*x^2+a)^2)^(5/2)*(2*b^5*x^10+15*a*b^4*x^8+60*a^2*b^3*x^6+120*b^2*a^3*ln(x)*x^4-30*a^4*b*x^2-3*a^5)/(b*
x^2+a)^5/x^4

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46848, size = 139, normalized size = 0.56 \begin{align*} \frac{2 \, b^{5} x^{10} + 15 \, a b^{4} x^{8} + 60 \, a^{2} b^{3} x^{6} + 120 \, a^{3} b^{2} x^{4} \log \left (x\right ) - 30 \, a^{4} b x^{2} - 3 \, a^{5}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*b^5*x^10 + 15*a*b^4*x^8 + 60*a^2*b^3*x^6 + 120*a^3*b^2*x^4*log(x) - 30*a^4*b*x^2 - 3*a^5)/x^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13494, size = 171, normalized size = 0.68 \begin{align*} \frac{1}{6} \, b^{5} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{5}{4} \, a b^{4} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 5 \, a^{2} b^{3} x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 5 \, a^{3} b^{2} \log \left (x^{2}\right ) \mathrm{sgn}\left (b x^{2} + a\right ) - \frac{30 \, a^{3} b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 10 \, a^{4} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{5} \mathrm{sgn}\left (b x^{2} + a\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^5*x^6*sgn(b*x^2 + a) + 5/4*a*b^4*x^4*sgn(b*x^2 + a) + 5*a^2*b^3*x^2*sgn(b*x^2 + a) + 5*a^3*b^2*log(x^2)*
sgn(b*x^2 + a) - 1/4*(30*a^3*b^2*x^4*sgn(b*x^2 + a) + 10*a^4*b*x^2*sgn(b*x^2 + a) + a^5*sgn(b*x^2 + a))/x^4