\(\int \frac {1}{(a+b x^3)^{4/3} (c+d x^3)^2} \, dx\) [102]

Optimal result

Integrand size = 21, antiderivative size = 261 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2} \, dx=\frac {b (3 b c+a d) x}{3 a c (b c-a d)^2 \sqrt [3]{a+b x^3}}-\frac {d x}{3 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )}-\frac {2 d (3 b c-a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{7/3}}-\frac {d (3 b c-a d) \log \left (c+d x^3\right )}{9 c^{5/3} (b c-a d)^{7/3}}+\frac {d (3 b c-a d) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{3 c^{5/3} (b c-a d)^{7/3}} \]

[Out]

1/3*b*(a*d+3*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^3+a)^(1/3)-1/3*d*x/c/(-a*d+b*c)/(b*x^3+a)^(1/3)/(d*x^3+c)-1/9*d*(-a*
d+3*b*c)*ln(d*x^3+c)/c^(5/3)/(-a*d+b*c)^(7/3)+1/3*d*(-a*d+3*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3)
)/c^(5/3)/(-a*d+b*c)^(7/3)-2/9*d*(-a*d+3*b*c)*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^(5/3)/(-a*d+b*c)^(7/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {425, 541, 12, 384} \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2} \, dx=-\frac {2 d (3 b c-a d) \arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{7/3}}-\frac {d (3 b c-a d) \log \left (c+d x^3\right )}{9 c^{5/3} (b c-a d)^{7/3}}+\frac {d (3 b c-a d) \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{3 c^{5/3} (b c-a d)^{7/3}}+\frac {b x (a d+3 b c)}{3 a c \sqrt [3]{a+b x^3} (b c-a d)^2}-\frac {d x}{3 c \sqrt [3]{a+b x^3} \left (c+d x^3\right ) (b c-a d)} \]

[In]

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

[Out]

(b*(3*b*c + a*d)*x)/(3*a*c*(b*c - a*d)^2*(a + b*x^3)^(1/3)) - (d*x)/(3*c*(b*c - a*d)*(a + b*x^3)^(1/3)*(c + d*
x^3)) - (2*d*(3*b*c - a*d)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[
3]*c^(5/3)*(b*c - a*d)^(7/3)) - (d*(3*b*c - a*d)*Log[c + d*x^3])/(9*c^(5/3)*(b*c - a*d)^(7/3)) + (d*(3*b*c - a
*d)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(3*c^(5/3)*(b*c - a*d)^(7/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 425

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 541

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

Rubi steps \begin{align*} \text {integral}& = -\frac {d x}{3 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )}+\frac {\int \frac {3 b c-2 a d-3 b d x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx}{3 c (b c-a d)} \\ & = \frac {b (3 b c+a d) x}{3 a c (b c-a d)^2 \sqrt [3]{a+b x^3}}-\frac {d x}{3 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )}-\frac {\int \frac {2 a d (3 b c-a d)}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{3 a c (b c-a d)^2} \\ & = \frac {b (3 b c+a d) x}{3 a c (b c-a d)^2 \sqrt [3]{a+b x^3}}-\frac {d x}{3 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )}-\frac {(2 d (3 b c-a d)) \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{3 c (b c-a d)^2} \\ & = \frac {b (3 b c+a d) x}{3 a c (b c-a d)^2 \sqrt [3]{a+b x^3}}-\frac {d x}{3 c (b c-a d) \sqrt [3]{a+b x^3} \left (c+d x^3\right )}-\frac {2 d (3 b c-a 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 )}{3 \sqrt {3} c^{5/3} (b c-a d)^{7/3}}-\frac {d (3 b c-a d) \log \left (c+d x^3\right )}{9 c^{5/3} (b c-a d)^{7/3}}+\frac {d (3 b c-a d) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{3 c^{5/3} (b c-a d)^{7/3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.59 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2} \, dx=\frac {\frac {6 c^{2/3} x \left (a^2 d^2+a b d^2 x^3+3 b^2 c \left (c+d x^3\right )\right )}{a (b c-a d)^2 \sqrt [3]{a+b x^3} \left (c+d x^3\right )}+\frac {2 i \left (3 i+\sqrt {3}\right ) d (3 b c-a d) \text {arctanh}\left (\frac {i+\frac {\left (-i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d} x}}{\sqrt {3}}\right )}{(b c-a d)^{7/3}}+\frac {2 \left (1+i \sqrt {3}\right ) d (-3 b c+a d) \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{7/3}}+\frac {\left (1+i \sqrt {3}\right ) d (3 b c-a d) \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{7/3}}}{18 c^{5/3}} \]

[In]

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

[Out]

((6*c^(2/3)*x*(a^2*d^2 + a*b*d^2*x^3 + 3*b^2*c*(c + d*x^3)))/(a*(b*c - a*d)^2*(a + b*x^3)^(1/3)*(c + d*x^3)) +
((2*I)*(3*I + Sqrt[3])*d*(3*b*c - a*d)*ArcTanh[(I + ((-I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))/((b*c - a*d)^(
1/3)*x))/Sqrt[3]])/(b*c - a*d)^(7/3) + (2*(1 + I*Sqrt[3])*d*(-3*b*c + a*d)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*
Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(7/3) + ((1 + I*Sqrt[3])*d*(3*b*c - a*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)])/(b*c - a*d)^(7/3))/(18*c^(5/3))

Maple [A] (verified)

Time = 4.46 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.16

method	result	size
pseudoelliptic	\(\frac {\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a d \left (d \,x^{3}+c \right ) \left (a d -3 b c \right ) \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )}{9}+\frac {x c \left (a \left (b \,x^{3}+a \right ) d^{2}+3 x^{3} b^{2} c d +3 b^{2} c^{2}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{3}+\frac {2 \left (a d -3 b c \right ) d a \left (d \,x^{3}+c \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (\arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )}{9}}{c^{2} \left (d \,x^{3}+c \right ) \left (a d -b c \right )^{2} \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a}\)	\(303\)

[In]

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

[Out]

2/9/((a*d-b*c)/c)^(1/3)/(b*x^3+a)^(1/3)*((b*x^3+a)^(1/3)*a*d*(d*x^3+c)*(a*d-3*b*c)*ln((((a*d-b*c)/c)^(1/3)*x+(
b*x^3+a)^(1/3))/x)+3/2*x*c*(a*(b*x^3+a)*d^2+3*x^3*b^2*c*d+3*b^2*c^2)*((a*d-b*c)/c)^(1/3)+(a*d-3*b*c)*d*a*(d*x^
3+c)*(b*x^3+a)^(1/3)*(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)*3^(1
/2)-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)))/c^2/(d*x^3+c
)/(a*d-b*c)^2/a

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

integrate(1/(b*x**3+a)**(4/3)/(d*x**3+c)**2,x)

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)^2), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)^2), x)

Mupad [F(-1)]

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

[In]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)^2),x)

[Out]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)^2), x)