\(\int \frac {1}{x^4 (a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\) [117]

Optimal result

Integrand size = 26, antiderivative size = 269 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {4 b}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{12 a^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {2 b}{9 a^3 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{2 a^4 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {5 b \left (a+b x^3\right ) \log (x)}{a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {5 b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1369, 272, 46} \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {b}{12 a^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {5 b \log (x) \left (a+b x^3\right )}{a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {5 b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {4 b}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{2 a^4 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {2 b}{9 a^3 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

Int[1/(x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 {\left (b^4 \left (a b+b^2 x^3\right )\right ) \int \frac {1}{x^4 \left (a b+b^2 x^3\right )^5} \, dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^3\right )\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (a b+b^2 x\right )^5} \, dx,x,x^3\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (b^4 \left (a b+b^2 x^3\right )\right ) \text {Subst}\left (\int \left (\frac {1}{a^5 b^5 x^2}-\frac {5}{a^6 b^4 x}+\frac {1}{a^2 b^3 (a+b x)^5}+\frac {2}{a^3 b^3 (a+b x)^4}+\frac {3}{a^4 b^3 (a+b x)^3}+\frac {4}{a^5 b^3 (a+b x)^2}+\frac {5}{a^6 b^3 (a+b x)}\right ) \, dx,x,x^3\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = -\frac {4 b}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{12 a^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {2 b}{9 a^3 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{2 a^4 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {5 b \left (a+b x^3\right ) \log (x)}{a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {5 b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {-a \left (12 a^4+125 a^3 b x^3+260 a^2 b^2 x^6+210 a b^3 x^9+60 b^4 x^{12}\right )-180 b x^3 \left (a+b x^3\right )^4 \log (x)+60 b x^3 \left (a+b x^3\right )^4 \log \left (a+b x^3\right )}{36 a^6 x^3 \left (a+b x^3\right )^3 \sqrt {\left (a+b x^3\right )^2}} \]

[In]

Integrate[1/(x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)),x]

[Out]

(-(a*(12*a^4 + 125*a^3*b*x^3 + 260*a^2*b^2*x^6 + 210*a*b^3*x^9 + 60*b^4*x^12)) - 180*b*x^3*(a + b*x^3)^4*Log[x
] + 60*b*x^3*(a + b*x^3)^4*Log[a + b*x^3])/(36*a^6*x^3*(a + b*x^3)^3*Sqrt[(a + b*x^3)^2])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.42

method	result	size
pseudoelliptic	\(-\frac {\left (-5 b \,x^{3} \left (b \,x^{3}+a \right )^{4} \ln \left (b \,x^{3}+a \right )+5 b \,x^{3} \left (b \,x^{3}+a \right )^{4} \ln \left (b \,x^{3}\right )+a \left (5 b^{4} x^{12}+\frac {35}{2} a \,b^{3} x^{9}+\frac {65}{3} a^{2} b^{2} x^{6}+\frac {125}{12} a^{3} b \,x^{3}+a^{4}\right )\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right )^{4} a^{6} x^{3}}\)	\(114\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{3 a}-\frac {125 b \,x^{3}}{36 a^{2}}-\frac {65 b^{2} x^{6}}{9 a^{3}}-\frac {35 b^{3} x^{9}}{6 a^{4}}-\frac {5 b^{4} x^{12}}{3 a^{5}}\right )}{\left (b \,x^{3}+a \right )^{5} x^{3}}-\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{6}}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (-b \,x^{3}-a \right )}{3 \left (b \,x^{3}+a \right ) a^{6}}\)	\(139\)
default	\(\frac {\left (60 \ln \left (b \,x^{3}+a \right ) b^{5} x^{15}-180 b^{5} \ln \left (x \right ) x^{15}+240 \ln \left (b \,x^{3}+a \right ) a \,b^{4} x^{12}-720 b^{4} a \ln \left (x \right ) x^{12}-60 a \,b^{4} x^{12}+360 \ln \left (b \,x^{3}+a \right ) a^{2} b^{3} x^{9}-1080 a^{2} b^{3} \ln \left (x \right ) x^{9}-210 a^{2} b^{3} x^{9}+240 \ln \left (b \,x^{3}+a \right ) a^{3} b^{2} x^{6}-720 a^{3} b^{2} \ln \left (x \right ) x^{6}-260 a^{3} b^{2} x^{6}+60 \ln \left (b \,x^{3}+a \right ) a^{4} b \,x^{3}-180 b \,a^{4} \ln \left (x \right ) x^{3}-125 a^{4} b \,x^{3}-12 a^{5}\right ) \left (b \,x^{3}+a \right )}{36 x^{3} a^{6} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(219\)

[In]

int(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(-5*b*x^3*(b*x^3+a)^4*ln(b*x^3+a)+5*b*x^3*(b*x^3+a)^4*ln(b*x^3)+a*(5*b^4*x^12+35/2*a*b^3*x^9+65/3*a^2*b^2
*x^6+125/12*a^3*b*x^3+a^4))*csgn(b*x^3+a)/(b*x^3+a)^4/a^6/x^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {60 \, a b^{4} x^{12} + 210 \, a^{2} b^{3} x^{9} + 260 \, a^{3} b^{2} x^{6} + 125 \, a^{4} b x^{3} + 12 \, a^{5} - 60 \, {\left (b^{5} x^{15} + 4 \, a b^{4} x^{12} + 6 \, a^{2} b^{3} x^{9} + 4 \, a^{3} b^{2} x^{6} + a^{4} b x^{3}\right )} \log \left (b x^{3} + a\right ) + 180 \, {\left (b^{5} x^{15} + 4 \, a b^{4} x^{12} + 6 \, a^{2} b^{3} x^{9} + 4 \, a^{3} b^{2} x^{6} + a^{4} b x^{3}\right )} \log \left (x\right )}{36 \, {\left (a^{6} b^{4} x^{15} + 4 \, a^{7} b^{3} x^{12} + 6 \, a^{8} b^{2} x^{9} + 4 \, a^{9} b x^{6} + a^{10} x^{3}\right )}} \]

[In]

integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/36*(60*a*b^4*x^12 + 210*a^2*b^3*x^9 + 260*a^3*b^2*x^6 + 125*a^4*b*x^3 + 12*a^5 - 60*(b^5*x^15 + 4*a*b^4*x^1
2 + 6*a^2*b^3*x^9 + 4*a^3*b^2*x^6 + a^4*b*x^3)*log(b*x^3 + a) + 180*(b^5*x^15 + 4*a*b^4*x^12 + 6*a^2*b^3*x^9 +
4*a^3*b^2*x^6 + a^4*b*x^3)*log(x))/(a^6*b^4*x^15 + 4*a^7*b^3*x^12 + 6*a^8*b^2*x^9 + 4*a^9*b*x^6 + a^10*x^3)

Sympy [F]

\[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {1}{x^{4} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/x**4/(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)

[Out]

Integral(1/(x**4*((a + b*x**3)**2)**(5/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {5 \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{6}} - \frac {5 \, b}{9 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{3}} - \frac {5 \, b}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{5}} - \frac {1}{3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{3}} - \frac {5}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{4} b} - \frac {1}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} a^{2} b^{3}} \]

[In]

integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {5 \, b \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{6} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {5 \, b \log \left ({\left | x \right |}\right )}{a^{6} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {5 \, b x^{3} - a}{3 \, a^{6} x^{3} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {125 \, b^{5} x^{12} + 548 \, a b^{4} x^{9} + 912 \, a^{2} b^{3} x^{6} + 688 \, a^{3} b^{2} x^{3} + 202 \, a^{4} b}{36 \, {\left (b x^{3} + a\right )}^{4} a^{6} \mathrm {sgn}\left (b x^{3} + a\right )} \]

[In]

integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")

[Out]

5/3*b*log(abs(b*x^3 + a))/(a^6*sgn(b*x^3 + a)) - 5*b*log(abs(x))/(a^6*sgn(b*x^3 + a)) + 1/3*(5*b*x^3 - a)/(a^6
*x^3*sgn(b*x^3 + a)) - 1/36*(125*b^5*x^12 + 548*a*b^4*x^9 + 912*a^2*b^3*x^6 + 688*a^3*b^2*x^3 + 202*a^4*b)/((b
*x^3 + a)^4*a^6*sgn(b*x^3 + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {1}{x^4\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}} \,d x \]

[In]

int(1/(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)),x)

[Out]

int(1/(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)), x)