\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^{5/2}}{x^{19}} \, dx\) [82]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=-\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 a x^{18}} \]

[Out]

-1/18*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/a/x^18

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} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=-\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 a x^{18}} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^19,x]

[Out]

-1/18*((a + b*x^3)^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(a*x^18)

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 )^5}{x^{19}} \, dx}{b^4 \left (a b+b^2 x^3\right )} \\ & = -\frac {\left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 a x^{18}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (a^5+6 a^4 b x^3+15 a^3 b^2 x^6+20 a^2 b^3 x^9+15 a b^4 x^{12}+6 b^5 x^{15}\right )}{18 x^{18} \left (a+b x^3\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)/x^19,x]

[Out]

-1/18*(Sqrt[(a + b*x^3)^2]*(a^5 + 6*a^4*b*x^3 + 15*a^3*b^2*x^6 + 20*a^2*b^3*x^9 + 15*a*b^4*x^12 + 6*b^5*x^15))
/(x^18*(a + b*x^3))

Maple [C] (warning: unable to verify)

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

Time = 1.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.41

method	result	size
pseudoelliptic	\(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (2 b \,x^{3}+a \right ) \left (b^{2} x^{6}+a b \,x^{3}+a^{2}\right ) \left (3 b^{2} x^{6}+3 a b \,x^{3}+a^{2}\right )}{18 x^{18}}\)	\(58\)
gosper	\(-\frac {\left (6 b^{5} x^{15}+15 a \,b^{4} x^{12}+20 a^{2} b^{3} x^{9}+15 a^{3} b^{2} x^{6}+6 a^{4} b \,x^{3}+a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{18 x^{18} \left (b \,x^{3}+a \right )^{5}}\)	\(78\)
default	\(-\frac {\left (6 b^{5} x^{15}+15 a \,b^{4} x^{12}+20 a^{2} b^{3} x^{9}+15 a^{3} b^{2} x^{6}+6 a^{4} b \,x^{3}+a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{18 x^{18} \left (b \,x^{3}+a \right )^{5}}\)	\(78\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{18} a^{5}-\frac {1}{3} a^{4} b \,x^{3}-\frac {5}{6} a^{3} b^{2} x^{6}-\frac {10}{9} a^{2} b^{3} x^{9}-\frac {5}{6} a \,b^{4} x^{12}-\frac {1}{3} b^{5} x^{15}\right )}{\left (b \,x^{3}+a \right ) x^{18}}\)	\(79\)

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^19,x,method=_RETURNVERBOSE)

[Out]

-1/18*csgn(b*x^3+a)*(2*b*x^3+a)*(b^2*x^6+a*b*x^3+a^2)*(3*b^2*x^6+3*a*b*x^3+a^2)/x^18

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=-\frac {6 \, b^{5} x^{15} + 15 \, a b^{4} x^{12} + 20 \, a^{2} b^{3} x^{9} + 15 \, a^{3} b^{2} x^{6} + 6 \, a^{4} b x^{3} + a^{5}}{18 \, x^{18}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^19,x, algorithm="fricas")

[Out]

-1/18*(6*b^5*x^15 + 15*a*b^4*x^12 + 20*a^2*b^3*x^9 + 15*a^3*b^2*x^6 + 6*a^4*b*x^3 + a^5)/x^18

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{19}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**(5/2)/x**19,x)

[Out]

Integral(((a + b*x**3)**2)**(5/2)/x**19, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (28) = 56\).

Time = 0.22 (sec) , antiderivative size = 210, normalized size of antiderivative = 5.12 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{6}}{18 \, a^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{5}}{18 \, a^{5} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{4}}{18 \, a^{6} x^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{3}}{18 \, a^{5} x^{9}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{2}}{18 \, a^{4} x^{12}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{18 \, a^{3} x^{15}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{18 \, a^{2} x^{18}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^19,x, algorithm="maxima")

[Out]

1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^6/a^6 + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^5/(a^5*x^3) - 1/18*(
b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^4/(a^6*x^6) + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^3/(a^5*x^9) - 1/18*(
b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^2/(a^4*x^12) + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b/(a^3*x^15) - 1/18*(
b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^18)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (28) = 56\).

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.59 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=-\frac {6 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 15 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 20 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 15 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 6 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{18 \, x^{18}} \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^(5/2)/x^19,x, algorithm="giac")

[Out]

-1/18*(6*b^5*x^15*sgn(b*x^3 + a) + 15*a*b^4*x^12*sgn(b*x^3 + a) + 20*a^2*b^3*x^9*sgn(b*x^3 + a) + 15*a^3*b^2*x
^6*sgn(b*x^3 + a) + 6*a^4*b*x^3*sgn(b*x^3 + a) + a^5*sgn(b*x^3 + a))/x^18

Mupad [B] (verification not implemented)

Time = 8.37 (sec) , antiderivative size = 231, normalized size of antiderivative = 5.63 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{19}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{18\,x^{18}\,\left (b\,x^3+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^3\,\left (b\,x^3+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^6\,\left (b\,x^3+a\right )}-\frac {a^4\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^{15}\,\left (b\,x^3+a\right )}-\frac {10\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{9\,x^9\,\left (b\,x^3+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^{12}\,\left (b\,x^3+a\right )} \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)/x^19,x)

[Out]

- (a^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(18*x^18*(a + b*x^3)) - (b^5*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(3*x
^3*(a + b*x^3)) - (5*a*b^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x^6*(a + b*x^3)) - (a^4*b*(a^2 + b^2*x^6 + 2*
a*b*x^3)^(1/2))/(3*x^15*(a + b*x^3)) - (10*a^2*b^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(9*x^9*(a + b*x^3)) - (5
*a^3*b^2*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x^12*(a + b*x^3))