\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{11}} \, dx\) [43]

Optimal result

Integrand size = 26, antiderivative size = 165 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )} \]

[Out]

-1/10*a^3*((b*x^3+a)^2)^(1/2)/x^10/(b*x^3+a)-3/7*a^2*b*((b*x^3+a)^2)^(1/2)/x^7/(b*x^3+a)-3/4*a*b^2*((b*x^3+a)^
2)^(1/2)/x^4/(b*x^3+a)-b^3*((b*x^3+a)^2)^(1/2)/x/(b*x^3+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 276} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {3 a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^4 \left (a+b x^3\right )}-\frac {b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )} \]

[In]

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

[Out]

-1/10*(a^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(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 276

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]

Rule 1369

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (14 a^3+60 a^2 b x^3+105 a b^2 x^6+140 b^3 x^9\right )}{140 x^{10} \left (a+b x^3\right )} \]

[In]

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

[Out]

-1/140*(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))/(x^10*(a + b*x^3))

Maple [A] (verified)

Time = 12.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.35

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-b^{3} x^{9}-\frac {3}{4} b^{2} x^{6} a -\frac {3}{7} a^{2} b \,x^{3}-\frac {1}{10} a^{3}\right )}{\left (b \,x^{3}+a \right ) x^{10}}\)	\(57\)
gosper	\(-\frac {\left (140 b^{3} x^{9}+105 b^{2} x^{6} a +60 a^{2} b \,x^{3}+14 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{140 x^{10} \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
default	\(-\frac {\left (140 b^{3} x^{9}+105 b^{2} x^{6} a +60 a^{2} b \,x^{3}+14 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{140 x^{10} \left (b \,x^{3}+a \right )^{3}}\)	\(58\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^11,x,method=_RETURNVERBOSE)

[Out]

((b*x^3+a)^2)^(1/2)/(b*x^3+a)*(-b^3*x^9-3/4*b^2*x^6*a-3/7*a^2*b*x^3-1/10*a^3)/x^10

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {140 \, b^{3} x^{9} + 105 \, a b^{2} x^{6} + 60 \, a^{2} b x^{3} + 14 \, a^{3}}{140 \, x^{10}} \]

[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]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{11}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {140 \, b^{3} x^{9} + 105 \, a b^{2} x^{6} + 60 \, a^{2} b x^{3} + 14 \, a^{3}}{140 \, x^{10}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {140 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 105 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 60 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 14 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{140 \, x^{10}} \]

[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

Mupad [B] (verification not implemented)

Time = 8.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{11}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{10\,x^{10}\,\left (b\,x^3+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{x\,\left (b\,x^3+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{4\,x^4\,\left (b\,x^3+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{7\,x^7\,\left (b\,x^3+a\right )} \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^11,x)

[Out]

- (a^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(10*x^10*(a + b*x^3)) - (b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(x*(
a + b*x^3)) - (3*a*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(4*x^4*(a + b*x^3)) - (3*a^2*b*(a^2 + b^2*x^6 + 2*a*
b*x^3)^(1/2))/(7*x^7*(a + b*x^3))