3.42 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{10}} \, dx\)

Optimal. Leaf size=161 \[ -\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac{a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac{a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac{b^3 \log (x) \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*x^9*(a + b*x^3)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*x^6*(a
+ b*x^3)) - (a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^3*(a + b*x^3)) + (b^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*
Log[x])/(a + b*x^3)

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Rubi [A]  time = 0.0464384, antiderivative size = 161, 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 = {1355, 266, 43} \[ -\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac{a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac{a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac{b^3 \log (x) \sqrt{a^2+2 a b x^3+b^2 x^6}}{a+b x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^10,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(9*x^9*(a + b*x^3)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(2*x^6*(a
+ b*x^3)) - (a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^3*(a + b*x^3)) + (b^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*
Log[x])/(a + b*x^3)

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && 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^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int \frac{\left (a b+b^2 x^3\right )^3}{x^{10}} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^3}{x^4} \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \operatorname{Subst}\left (\int \left (\frac{a^3 b^3}{x^4}+\frac{3 a^2 b^4}{x^3}+\frac{3 a b^5}{x^2}+\frac{b^6}{x}\right ) \, dx,x,x^3\right )}{3 b^2 \left (a b+b^2 x^3\right )}\\ &=-\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac{a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac{a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac{b^3 \sqrt{a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3}\\ \end{align*}

Mathematica [A]  time = 0.0220902, size = 63, normalized size = 0.39 \[ -\frac{\sqrt{\left (a+b x^3\right )^2} \left (a \left (2 a^2+9 a b x^3+18 b^2 x^6\right )-18 b^3 x^9 \log (x)\right )}{18 x^9 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^10,x]

[Out]

-(Sqrt[(a + b*x^3)^2]*(a*(2*a^2 + 9*a*b*x^3 + 18*b^2*x^6) - 18*b^3*x^9*Log[x]))/(18*x^9*(a + b*x^3))

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Maple [A]  time = 0.011, size = 60, normalized size = 0.4 \begin{align*}{\frac{18\,{b}^{3}\ln \left ( x \right ){x}^{9}-18\,a{b}^{2}{x}^{6}-9\,{a}^{2}b{x}^{3}-2\,{a}^{3}}{18\, \left ( b{x}^{3}+a \right ) ^{3}{x}^{9}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x)

[Out]

1/18*((b*x^3+a)^2)^(3/2)*(18*b^3*ln(x)*x^9-18*a*b^2*x^6-9*a^2*b*x^3-2*a^3)/(b*x^3+a)^3/x^9

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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^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7255, size = 90, normalized size = 0.56 \begin{align*} \frac{18 \, b^{3} x^{9} \log \left (x\right ) - 18 \, a b^{2} x^{6} - 9 \, a^{2} b x^{3} - 2 \, a^{3}}{18 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="fricas")

[Out]

1/18*(18*b^3*x^9*log(x) - 18*a*b^2*x^6 - 9*a^2*b*x^3 - 2*a^3)/x^9

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac{3}{2}}}{x^{10}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**10,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**10, x)

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Giac [A]  time = 1.11498, size = 115, normalized size = 0.71 \begin{align*} b^{3} \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x^{3} + a\right ) - \frac{11 \, b^{3} x^{9} \mathrm{sgn}\left (b x^{3} + a\right ) + 18 \, a b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 9 \, a^{2} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + 2 \, a^{3} \mathrm{sgn}\left (b x^{3} + a\right )}{18 \, x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="giac")

[Out]

b^3*log(abs(x))*sgn(b*x^3 + a) - 1/18*(11*b^3*x^9*sgn(b*x^3 + a) + 18*a*b^2*x^6*sgn(b*x^3 + a) + 9*a^2*b*x^3*s
gn(b*x^3 + a) + 2*a^3*sgn(b*x^3 + a))/x^9