\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{11}} \, dx\) [74]

Optimal result

Integrand size = 26, antiderivative size = 253 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^4 \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}+\frac {5 a b^4 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {b^5 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )} \]

[Out]

-1/10*a^5*((b*x^3+a)^2)^(1/2)/x^10/(b*x^3+a)-5/7*a^4*b*((b*x^3+a)^2)^(1/2)/x^7/(b*x^3+a)-5/2*a^3*b^2*((b*x^3+a
)^2)^(1/2)/x^4/(b*x^3+a)-10*a^2*b^3*((b*x^3+a)^2)^(1/2)/x/(b*x^3+a)+5/2*a*b^4*x^2*((b*x^3+a)^2)^(1/2)/(b*x^3+a
)+1/5*b^5*x^5*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 253, 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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=\frac {b^5 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {5 a b^4 x^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x \left (a+b x^3\right )}-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{10 x^{10} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 x^7 \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^4 \left (a+b x^3\right )} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^11,x]

[Out]

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

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (7 a^5+50 a^4 b x^3+175 a^3 b^2 x^6+700 a^2 b^3 x^9-175 a b^4 x^{12}-14 b^5 x^{15}\right )}{70 x^{10} \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^11,x]

[Out]

-1/70*(Sqrt[(a + b*x^3)^2]*(7*a^5 + 50*a^4*b*x^3 + 175*a^3*b^2*x^6 + 700*a^2*b^3*x^9 - 175*a*b^4*x^12 - 14*b^5
*x^15))/(x^10*(a + b*x^3))

Maple [A] (verified)

Time = 11.74 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32

method	result	size
gosper	\(-\frac {\left (-14 b^{5} x^{15}-175 a \,b^{4} x^{12}+700 a^{2} b^{3} x^{9}+175 a^{3} b^{2} x^{6}+50 a^{4} b \,x^{3}+7 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{70 \left (b \,x^{3}+a \right )^{5} x^{10}}\)	\(80\)
default	\(-\frac {\left (-14 b^{5} x^{15}-175 a \,b^{4} x^{12}+700 a^{2} b^{3} x^{9}+175 a^{3} b^{2} x^{6}+50 a^{4} b \,x^{3}+7 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{70 \left (b \,x^{3}+a \right )^{5} x^{10}}\)	\(80\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4} \left (\frac {1}{5} b \,x^{5}+\frac {5}{2} a \,x^{2}\right )}{b \,x^{3}+a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-10 a^{2} b^{3} x^{9}-\frac {5}{2} a^{3} b^{2} x^{6}-\frac {5}{7} a^{4} b \,x^{3}-\frac {1}{10} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{10}}\)	\(100\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^11,x,method=_RETURNVERBOSE)

[Out]

-1/70*(-14*b^5*x^15-175*a*b^4*x^12+700*a^2*b^3*x^9+175*a^3*b^2*x^6+50*a^4*b*x^3+7*a^5)*((b*x^3+a)^2)^(5/2)/(b*
x^3+a)^5/x^10

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=\frac {14 \, b^{5} x^{15} + 175 \, a b^{4} x^{12} - 700 \, a^{2} b^{3} x^{9} - 175 \, a^{3} b^{2} x^{6} - 50 \, a^{4} b x^{3} - 7 \, a^{5}}{70 \, x^{10}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^11,x, algorithm="fricas")

[Out]

1/70*(14*b^5*x^15 + 175*a*b^4*x^12 - 700*a^2*b^3*x^9 - 175*a^3*b^2*x^6 - 50*a^4*b*x^3 - 7*a^5)/x^10

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{11}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**11,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**11, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=\frac {14 \, b^{5} x^{15} + 175 \, a b^{4} x^{12} - 700 \, a^{2} b^{3} x^{9} - 175 \, a^{3} b^{2} x^{6} - 50 \, a^{4} b x^{3} - 7 \, a^{5}}{70 \, x^{10}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^11,x, algorithm="maxima")

[Out]

1/70*(14*b^5*x^15 + 175*a*b^4*x^12 - 700*a^2*b^3*x^9 - 175*a^3*b^2*x^6 - 50*a^4*b*x^3 - 7*a^5)/x^10

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=\frac {1}{5} \, b^{5} x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{2} \, a b^{4} x^{2} \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {700 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 175 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 50 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 7 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{70 \, x^{10}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^11,x, algorithm="giac")

[Out]

1/5*b^5*x^5*sgn(b*x^3 + a) + 5/2*a*b^4*x^2*sgn(b*x^3 + a) - 1/70*(700*a^2*b^3*x^9*sgn(b*x^3 + a) + 175*a^3*b^2
*x^6*sgn(b*x^3 + a) + 50*a^4*b*x^3*sgn(b*x^3 + a) + 7*a^5*sgn(b*x^3 + a))/x^10

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{11}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^{11}} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^11,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^11, x)