\(\int \frac {1}{x (a^2+2 a b x^3+b^2 x^6)^{5/2}} \, dx\) [114]

Optimal result

Integrand size = 26, antiderivative size = 223 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {1}{3 a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{12 a \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{9 a^2 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{6 a^3 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log (x)}{a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 223, 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} \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {1}{9 a^2 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{12 a \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\log (x) \left (a+b x^3\right )}{a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{3 a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{6 a^3 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {a \left (25 a^3+52 a^2 b x^3+42 a b^2 x^6+12 b^3 x^9\right )+36 \left (a+b x^3\right )^4 \log (x)-12 \left (a+b x^3\right )^4 \log \left (a+b x^3\right )}{36 a^5 \left (a+b x^3\right )^3 \sqrt {\left (a+b x^3\right )^2}} \]

[In]

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

[Out]

(a*(25*a^3 + 52*a^2*b*x^3 + 42*a*b^2*x^6 + 12*b^3*x^9) + 36*(a + b*x^3)^4*Log[x] - 12*(a + b*x^3)^4*Log[a + b*
x^3])/(36*a^5*(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 = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.41

method	result	size
pseudoelliptic	\(\frac {\left (-\ln \left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{4}+\ln \left (b \,x^{3}\right ) \left (b \,x^{3}+a \right )^{4}+a \,b^{3} x^{9}+\frac {7 a^{2} b^{2} x^{6}}{2}+\frac {13 a^{3} b \,x^{3}}{3}+\frac {25 a^{4}}{12}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right )^{4} a^{5}}\)	\(92\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {b^{3} x^{9}}{3 a^{4}}+\frac {7 b^{2} x^{6}}{6 a^{3}}+\frac {13 b \,x^{3}}{9 a^{2}}+\frac {25}{36 a}\right )}{\left (b \,x^{3}+a \right )^{5}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{5}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) a^{5}}\)	\(119\)
default	\(\frac {\left (36 \ln \left (x \right ) b^{4} x^{12}-12 \ln \left (b \,x^{3}+a \right ) b^{4} x^{12}+144 \ln \left (x \right ) a \,b^{3} x^{9}-48 \ln \left (b \,x^{3}+a \right ) a \,b^{3} x^{9}+12 a \,b^{3} x^{9}+216 \ln \left (x \right ) a^{2} b^{2} x^{6}-72 \ln \left (b \,x^{3}+a \right ) a^{2} b^{2} x^{6}+42 a^{2} b^{2} x^{6}+144 \ln \left (x \right ) a^{3} b \,x^{3}-48 \ln \left (b \,x^{3}+a \right ) a^{3} b \,x^{3}+52 a^{3} b \,x^{3}+36 a^{4} \ln \left (x \right )-12 \ln \left (b \,x^{3}+a \right ) a^{4}+25 a^{4}\right ) \left (b \,x^{3}+a \right )}{36 a^{5} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(193\)

[In]

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

[Out]

1/3*(-ln(b*x^3+a)*(b*x^3+a)^4+ln(b*x^3)*(b*x^3+a)^4+a*b^3*x^9+7/2*a^2*b^2*x^6+13/3*a^3*b*x^3+25/12*a^4)*csgn(b
*x^3+a)/(b*x^3+a)^4/a^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {12 \, a b^{3} x^{9} + 42 \, a^{2} b^{2} x^{6} + 52 \, a^{3} b x^{3} + 25 \, a^{4} - 12 \, {\left (b^{4} x^{12} + 4 \, a b^{3} x^{9} + 6 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} + a^{4}\right )} \log \left (b x^{3} + a\right ) + 36 \, {\left (b^{4} x^{12} + 4 \, a b^{3} x^{9} + 6 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} + a^{4}\right )} \log \left (x\right )}{36 \, {\left (a^{5} b^{4} x^{12} + 4 \, a^{6} b^{3} x^{9} + 6 \, a^{7} b^{2} x^{6} + 4 \, a^{8} b x^{3} + a^{9}\right )}} \]

[In]

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

[Out]

1/36*(12*a*b^3*x^9 + 42*a^2*b^2*x^6 + 52*a^3*b*x^3 + 25*a^4 - 12*(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a
^3*b*x^3 + a^4)*log(b*x^3 + a) + 36*(b^4*x^12 + 4*a*b^3*x^9 + 6*a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4)*log(x))/(a^5*
b^4*x^12 + 4*a^6*b^3*x^9 + 6*a^7*b^2*x^6 + 4*a^8*b*x^3 + a^9)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{5}} + \frac {1}{9 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2}} + \frac {1}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{4}} + \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{3} b^{2}} + \frac {1}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} a b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {25 \, b^{4} x^{12} + 112 \, a b^{3} x^{9} + 192 \, a^{2} b^{2} x^{6} + 152 \, a^{3} b x^{3} + 50 \, a^{4}}{36 \, {\left (b x^{3} + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} \]

[In]

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

[Out]

-1/3*log(abs(b*x^3 + a))/(a^5*sgn(b*x^3 + a)) + log(abs(x))/(a^5*sgn(b*x^3 + a)) + 1/36*(25*b^4*x^12 + 112*a*b
^3*x^9 + 192*a^2*b^2*x^6 + 152*a^3*b*x^3 + 50*a^4)/((b*x^3 + a)^4*a^5*sgn(b*x^3 + a))

Mupad [F(-1)]

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

[In]

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

[Out]

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