\(\int \frac {1}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\) [90]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {b x}{a (b c-a d) \sqrt [3]{a+b x^3}}-\frac {d \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{4/3}}-\frac {d \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{4/3}}+\frac {d \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} (b c-a d)^{4/3}} \]

[Out]

b*x/a/(-a*d+b*c)/(b*x^3+a)^(1/3)-1/6*d*ln(d*x^3+c)/c^(2/3)/(-a*d+b*c)^(4/3)+1/2*d*ln((-a*d+b*c)^(1/3)*x/c^(1/3
)-(b*x^3+a)^(1/3))/c^(2/3)/(-a*d+b*c)^(4/3)-1/3*d*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*
3^(1/2))/c^(2/3)/(-a*d+b*c)^(4/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {390, 384} \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=-\frac {d \arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{4/3}}-\frac {d \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{4/3}}+\frac {d \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} (b c-a d)^{4/3}}+\frac {b x}{a \sqrt [3]{a+b x^3} (b c-a d)} \]

[In]

Int[1/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(b*x)/(a*(b*c - a*d)*(a + b*x^3)^(1/3)) - (d*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/
Sqrt[3]])/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(4/3)) - (d*Log[c + d*x^3])/(6*c^(2/3)*(b*c - a*d)^(4/3)) + (d*Log[((b*
c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*c^(2/3)*(b*c - a*d)^(4/3))

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {b x}{a (b c-a d) \sqrt [3]{a+b x^3}}-\frac {d \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{b c-a d} \\ & = \frac {b x}{a (b c-a d) \sqrt [3]{a+b x^3}}-\frac {d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{4/3}}-\frac {d \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{4/3}}+\frac {d \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} (b c-a d)^{4/3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.23 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.83 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {1}{12} \left (\frac {12 b x}{\left (a b c-a^2 d\right ) \sqrt [3]{a+b x^3}}+\frac {2 \sqrt {-6+6 i \sqrt {3}} d \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{2/3} (b c-a d)^{4/3}}-\frac {2 i \left (-i+\sqrt {3}\right ) d \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{c^{2/3} (b c-a d)^{4/3}}+\frac {\left (d+i \sqrt {3} d\right ) \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{c^{2/3} (b c-a d)^{4/3}}\right ) \]

[In]

Integrate[1/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

((12*b*x)/((a*b*c - a^2*d)*(a + b*x^3)^(1/3)) + (2*Sqrt[-6 + (6*I)*Sqrt[3]]*d*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(
Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))])/(c^(2/3)*(b*c - a*d)^(4/3)) - ((2*I
)*(-I + Sqrt[3])*d*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(c^(2/3)*(b*c - a*d
)^(4/3)) + ((d + I*Sqrt[3]*d)*Log[2*(b*c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a +
b*x^3)^(1/3) + I*(I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(c^(2/3)*(b*c - a*d)^(4/3)))/12

Maple [A] (verified)

Time = 4.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.36

method	result	size
pseudoelliptic	\(\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) a d \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a d \left (b \,x^{3}+a \right )^{\frac {1}{3}}-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a d \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{2}-3 b x c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (a d -b c \right ) c a}\)	\(243\)

[In]

int(1/(b*x^3+a)^(4/3)/(d*x^3+c),x,method=_RETURNVERBOSE)

[Out]

1/3/((a*d-b*c)/c)^(1/3)/(b*x^3+a)^(1/3)*(3^(1/2)*arctan(1/3*3^(1/2)*(((a*d-b*c)/c)^(1/3)*x-2*(b*x^3+a)^(1/3))/
((a*d-b*c)/c)^(1/3)/x)*a*d*(b*x^3+a)^(1/3)+ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)*a*d*(b*x^3+a)^(1/3)-1
/2*ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)*a*d*(b*x^3+a)^(1/3)
-3*b*x*c*((a*d-b*c)/c)^(1/3))/(a*d-b*c)/c/a

Fricas [F(-1)]

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

[In]

integrate(1/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

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

[In]

integrate(1/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(1/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

Maxima [F]

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

[In]

integrate(1/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)

Giac [F]

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

[In]

integrate(1/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)

Mupad [F(-1)]

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

[In]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)), x)