3.43 \(\int \frac{(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{11}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac{b^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(10*x^10*(a + b*x^3)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^
7*(a + b*x^3)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^4*(a + b*x^3)) - (b^3*Sqrt[a^2 + 2*a*b*x^3 + b
^2*x^6])/(x*(a + b*x^3))

________________________________________________________________________________________

Rubi [A]  time = 0.0402338, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {1355, 270} \[ -\frac{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac{b^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^11,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(10*x^10*(a + b*x^3)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(7*x^
7*(a + b*x^3)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(4*x^4*(a + b*x^3)) - (b^3*Sqrt[a^2 + 2*a*b*x^3 + b
^2*x^6])/(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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, 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^{11}} \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac{\sqrt{a^2+2 a b x^3+b^2 x^6} \int \left (\frac{a^3 b^3}{x^{11}}+\frac{3 a^2 b^4}{x^8}+\frac{3 a b^5}{x^5}+\frac{b^6}{x^2}\right ) \, dx}{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}}{10 x^{10} \left (a+b x^3\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac{b^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}\\ \end{align*}

Mathematica [A]  time = 0.0128461, size = 61, normalized size = 0.37 \[ -\frac{\sqrt{\left (a+b x^3\right )^2} \left (60 a^2 b x^3+14 a^3+105 a b^2 x^6+140 b^3 x^9\right )}{140 x^{10} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^11,x]

[Out]

-(Sqrt[(a + b*x^3)^2]*(14*a^3 + 60*a^2*b*x^3 + 105*a*b^2*x^6 + 140*b^3*x^9))/(140*x^10*(a + b*x^3))

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 58, normalized size = 0.4 \begin{align*} -{\frac{140\,{b}^{3}{x}^{9}+105\,a{b}^{2}{x}^{6}+60\,{a}^{2}b{x}^{3}+14\,{a}^{3}}{140\,{x}^{10} \left ( b{x}^{3}+a \right ) ^{3}} \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^11,x)

[Out]

-1/140*(140*b^3*x^9+105*a*b^2*x^6+60*a^2*b*x^3+14*a^3)*((b*x^3+a)^2)^(3/2)/x^10/(b*x^3+a)^3

________________________________________________________________________________________

Maxima [A]  time = 1.03771, size = 50, normalized size = 0.3 \begin{align*} -\frac{140 \, b^{3} x^{9} + 105 \, a b^{2} x^{6} + 60 \, a^{2} b x^{3} + 14 \, a^{3}}{140 \, x^{10}} \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^11,x, algorithm="maxima")

[Out]

-1/140*(140*b^3*x^9 + 105*a*b^2*x^6 + 60*a^2*b*x^3 + 14*a^3)/x^10

________________________________________________________________________________________

Fricas [A]  time = 1.64957, size = 90, normalized size = 0.55 \begin{align*} -\frac{140 \, b^{3} x^{9} + 105 \, a b^{2} x^{6} + 60 \, a^{2} b x^{3} + 14 \, a^{3}}{140 \, x^{10}} \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^11,x, algorithm="fricas")

[Out]

-1/140*(140*b^3*x^9 + 105*a*b^2*x^6 + 60*a^2*b*x^3 + 14*a^3)/x^10

________________________________________________________________________________________

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^{11}}\, 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**11,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**11, x)

________________________________________________________________________________________

Giac [A]  time = 1.14586, size = 93, normalized size = 0.56 \begin{align*} -\frac{140 \, b^{3} x^{9} \mathrm{sgn}\left (b x^{3} + a\right ) + 105 \, a b^{2} x^{6} \mathrm{sgn}\left (b x^{3} + a\right ) + 60 \, a^{2} b x^{3} \mathrm{sgn}\left (b x^{3} + a\right ) + 14 \, a^{3} \mathrm{sgn}\left (b x^{3} + a\right )}{140 \, x^{10}} \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^11,x, algorithm="giac")

[Out]

-1/140*(140*b^3*x^9*sgn(b*x^3 + a) + 105*a*b^2*x^6*sgn(b*x^3 + a) + 60*a^2*b*x^3*sgn(b*x^3 + a) + 14*a^3*sgn(b
*x^3 + a))/x^10