\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{12}} \, dx\) [75]

Optimal result

Integrand size = 26, antiderivative size = 247 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5 \left (a+b x^3\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^2 \left (a+b x^3\right )}+\frac {5 a b^4 x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^5 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )} \]

[Out]

-1/11*a^5*((b*x^3+a)^2)^(1/2)/x^11/(b*x^3+a)-5/8*a^4*b*((b*x^3+a)^2)^(1/2)/x^8/(b*x^3+a)-2*a^3*b^2*((b*x^3+a)^
2)^(1/2)/x^5/(b*x^3+a)-5*a^2*b^3*((b*x^3+a)^2)^(1/2)/x^2/(b*x^3+a)+5*a*b^4*x*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/4
*b^5*x^4*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 247, 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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=\frac {b^5 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )}+\frac {5 a b^4 x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^2 \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5 \left (a+b x^3\right )} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^12,x]

[Out]

-1/11*(a^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^11*(a + b*x^3)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*
x^8*(a + b*x^3)) - (2*a^3*b^2*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(x^5*(a + b*x^3)) - (5*a^2*b^3*Sqrt[a^2 + 2*a*b
*x^3 + b^2*x^6])/(x^2*(a + b*x^3)) + (5*a*b^4*x*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a + b*x^3) + (b^5*x^4*Sqrt[a
^2 + 2*a*b*x^3 + b^2*x^6])/(4*(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 )^5}{x^{12}} \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (5 a b^9+\frac {a^5 b^5}{x^{12}}+\frac {5 a^4 b^6}{x^9}+\frac {10 a^3 b^7}{x^6}+\frac {10 a^2 b^8}{x^3}+b^{10} x^3\right ) \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = -\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 x^{11} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {2 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^5 \left (a+b x^3\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^2 \left (a+b x^3\right )}+\frac {5 a b^4 x \sqrt {a^2+2 a b x^3+b^2 x^6}}{a+b x^3}+\frac {b^5 x^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (8 a^5+55 a^4 b x^3+176 a^3 b^2 x^6+440 a^2 b^3 x^9-440 a b^4 x^{12}-22 b^5 x^{15}\right )}{88 x^{11} \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^12,x]

[Out]

-1/88*(Sqrt[(a + b*x^3)^2]*(8*a^5 + 55*a^4*b*x^3 + 176*a^3*b^2*x^6 + 440*a^2*b^3*x^9 - 440*a*b^4*x^12 - 22*b^5
*x^15))/(x^11*(a + b*x^3))

Maple [A] (verified)

Time = 12.87 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32

method	result	size
gosper	\(-\frac {\left (-22 b^{5} x^{15}-440 a \,b^{4} x^{12}+440 a^{2} b^{3} x^{9}+176 a^{3} b^{2} x^{6}+55 a^{4} b \,x^{3}+8 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{88 \left (b \,x^{3}+a \right )^{5} x^{11}}\)	\(80\)
default	\(-\frac {\left (-22 b^{5} x^{15}-440 a \,b^{4} x^{12}+440 a^{2} b^{3} x^{9}+176 a^{3} b^{2} x^{6}+55 a^{4} b \,x^{3}+8 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{88 \left (b \,x^{3}+a \right )^{5} x^{11}}\)	\(80\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4} \left (\frac {1}{4} b \,x^{4}+5 a x \right )}{b \,x^{3}+a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-5 a^{2} b^{3} x^{9}-2 a^{3} b^{2} x^{6}-\frac {5}{8} a^{4} b \,x^{3}-\frac {1}{11} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{11}}\)	\(98\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^12,x,method=_RETURNVERBOSE)

[Out]

-1/88*(-22*b^5*x^15-440*a*b^4*x^12+440*a^2*b^3*x^9+176*a^3*b^2*x^6+55*a^4*b*x^3+8*a^5)*((b*x^3+a)^2)^(5/2)/(b*
x^3+a)^5/x^11

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=\frac {22 \, b^{5} x^{15} + 440 \, a b^{4} x^{12} - 440 \, a^{2} b^{3} x^{9} - 176 \, a^{3} b^{2} x^{6} - 55 \, a^{4} b x^{3} - 8 \, a^{5}}{88 \, x^{11}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^12,x, algorithm="fricas")

[Out]

1/88*(22*b^5*x^15 + 440*a*b^4*x^12 - 440*a^2*b^3*x^9 - 176*a^3*b^2*x^6 - 55*a^4*b*x^3 - 8*a^5)/x^11

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{12}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**12,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**12, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=\frac {22 \, b^{5} x^{15} + 440 \, a b^{4} x^{12} - 440 \, a^{2} b^{3} x^{9} - 176 \, a^{3} b^{2} x^{6} - 55 \, a^{4} b x^{3} - 8 \, a^{5}}{88 \, x^{11}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^12,x, algorithm="maxima")

[Out]

1/88*(22*b^5*x^15 + 440*a*b^4*x^12 - 440*a^2*b^3*x^9 - 176*a^3*b^2*x^6 - 55*a^4*b*x^3 - 8*a^5)/x^11

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=\frac {1}{4} \, b^{5} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a b^{4} x \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {440 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 176 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 55 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 8 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{88 \, x^{11}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^12,x, algorithm="giac")

[Out]

1/4*b^5*x^4*sgn(b*x^3 + a) + 5*a*b^4*x*sgn(b*x^3 + a) - 1/88*(440*a^2*b^3*x^9*sgn(b*x^3 + a) + 176*a^3*b^2*x^6
*sgn(b*x^3 + a) + 55*a^4*b*x^3*sgn(b*x^3 + a) + 8*a^5*sgn(b*x^3 + a))/x^11

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{12}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^{12}} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^12,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^12, x)