\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{3/2}}{x^{13}} \, dx\) [45]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 a x^{12}} \]

[Out]

-1/12*(b*x^3+a)^3*((b*x^3+a)^2)^(1/2)/a/x^12

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 270} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 a x^{12}} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^13,x]

[Out]

-1/12*((a + b*x^3)^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a*x^12)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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^{13}} \, dx}{b^2 \left (a b+b^2 x^3\right )} \\ & = -\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 a x^{12}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (a^3+4 a^2 b x^3+6 a b^2 x^6+4 b^3 x^9\right )}{12 x^{12} \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^13,x]

[Out]

-1/12*(Sqrt[(a + b*x^3)^2]*(a^3 + 4*a^2*b*x^3 + 6*a*b^2*x^6 + 4*b^3*x^9))/(x^12*(a + b*x^3))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (2 b \,x^{3}+a \right ) \left (2 b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )}{12 x^{12}}\)	\(41\)
gosper	\(-\frac {\left (4 b^{3} x^{9}+6 b^{2} x^{6} a +4 a^{2} b \,x^{3}+a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{12 x^{12} \left (b \,x^{3}+a \right )^{3}}\)	\(56\)
default	\(-\frac {\left (4 b^{3} x^{9}+6 b^{2} x^{6} a +4 a^{2} b \,x^{3}+a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{12 x^{12} \left (b \,x^{3}+a \right )^{3}}\)	\(56\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{3} b^{3} x^{9}-\frac {1}{2} b^{2} x^{6} a -\frac {1}{3} a^{2} b \,x^{3}-\frac {1}{12} a^{3}\right )}{\left (b \,x^{3}+a \right ) x^{12}}\)	\(57\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^13,x,method=_RETURNVERBOSE)

[Out]

-1/12*csgn(b*x^3+a)*(2*b*x^3+a)*(2*b^2*x^6+2*a*b*x^3+a^2)/x^12

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {4 \, b^{3} x^{9} + 6 \, a b^{2} x^{6} + 4 \, a^{2} b x^{3} + a^{3}}{12 \, x^{12}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^13,x, algorithm="fricas")

[Out]

-1/12*(4*b^3*x^9 + 6*a*b^2*x^6 + 4*a^2*b*x^3 + a^3)/x^12

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{13}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**13,x)

[Out]

Integral(((a + b*x**3)**2)**(3/2)/x**13, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (28) = 56\).

Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 3.61 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4}}{12 \, a^{4}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{3}}{12 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{12 \, a^{4} x^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{12 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{12 \, a^{2} x^{12}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^13,x, algorithm="maxima")

[Out]

1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^4/a^4 + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^3/(a^3*x^3) - 1/12*(
b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/(a^4*x^6) + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a^3*x^9) - 1/12*(b^
2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^12)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.66 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {4 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 6 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 4 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{12 \, x^{12}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^13,x, algorithm="giac")

[Out]

-1/12*(4*b^3*x^9*sgn(b*x^3 + a) + 6*a*b^2*x^6*sgn(b*x^3 + a) + 4*a^2*b*x^3*sgn(b*x^3 + a) + a^3*sgn(b*x^3 + a)
)/x^12

Mupad [B] (verification not implemented)

Time = 8.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.68 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{13}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{12\,x^{12}\,\left (b\,x^3+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^3\,\left (b\,x^3+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{2\,x^6\,\left (b\,x^3+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^9\,\left (b\,x^3+a\right )} \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^13,x)

[Out]

- (a^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(12*x^12*(a + b*x^3)) - (b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(3*x
^3*(a + b*x^3)) - (a*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(2*x^6*(a + b*x^3)) - (a^2*b*(a^2 + b^2*x^6 + 2*a*
b*x^3)^(1/2))/(3*x^9*(a + b*x^3))