\(\int \frac {1}{(a+b x^3)^{7/3} (c+d x^3)} \, dx\) [91]

Optimal result

Integrand size = 21, antiderivative size = 226 \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\frac {b x}{4 a (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {b (3 b c-7 a d) x}{4 a^2 (b c-a d)^2 \sqrt [3]{a+b x^3}}+\frac {d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{7/3}}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{7/3}}-\frac {d^2 \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} (b c-a d)^{7/3}} \]

[Out]

1/4*b*x/a/(-a*d+b*c)/(b*x^3+a)^(4/3)+1/4*b*(-7*a*d+3*b*c)*x/a^2/(-a*d+b*c)^2/(b*x^3+a)^(1/3)+1/6*d^2*ln(d*x^3+
c)/c^(2/3)/(-a*d+b*c)^(7/3)-1/2*d^2*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(2/3)/(-a*d+b*c)^(7/3)+1/
3*d^2*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)^(7/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 226, 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} \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\frac {b x (3 b c-7 a d)}{4 a^2 \sqrt [3]{a+b x^3} (b c-a d)^2}+\frac {d^2 \arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{7/3}}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{7/3}}-\frac {d^2 \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} (b c-a d)^{7/3}}+\frac {b x}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)} \]

[In]

Int[1/((a + b*x^3)^(7/3)*(c + d*x^3)),x]

[Out]

(b*x)/(4*a*(b*c - a*d)*(a + b*x^3)^(4/3)) + (b*(3*b*c - 7*a*d)*x)/(4*a^2*(b*c - a*d)^2*(a + b*x^3)^(1/3)) + (d
^2*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)^(7/
3)) + (d^2*Log[c + d*x^3])/(6*c^(2/3)*(b*c - a*d)^(7/3)) - (d^2*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)^(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 {b x}{4 a (b c-a d) \left (a+b x^3\right )^{4/3}}-\frac {\int \frac {-3 b c+4 a d-3 b d x^3}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx}{4 a (b c-a d)} \\ & = \frac {b x}{4 a (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {b (3 b c-7 a d) x}{4 a^2 (b c-a d)^2 \sqrt [3]{a+b x^3}}+\frac {\int \frac {4 a^2 d^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{4 a^2 (b c-a d)^2} \\ & = \frac {b x}{4 a (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {b (3 b c-7 a d) x}{4 a^2 (b c-a d)^2 \sqrt [3]{a+b x^3}}+\frac {d^2 \int \frac {1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{(b c-a d)^2} \\ & = \frac {b x}{4 a (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {b (3 b c-7 a d) x}{4 a^2 (b c-a d)^2 \sqrt [3]{a+b x^3}}+\frac {d^2 \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)^{7/3}}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{7/3}}-\frac {d^2 \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)^{7/3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.09 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\frac {1}{12} \left (\frac {3 b x \left (-8 a^2 d+3 b^2 c x^3+a b \left (4 c-7 d x^3\right )\right )}{a^2 (b c-a d)^2 \left (a+b x^3\right )^{4/3}}-\frac {2 \sqrt {-6+6 i \sqrt {3}} d^2 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{2/3} (b c-a d)^{7/3}}+\frac {2 \left (1+i \sqrt {3}\right ) d^2 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{c^{2/3} (b c-a d)^{7/3}}-\frac {i \left (-i+\sqrt {3}\right ) d^2 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{c^{2/3} (b c-a d)^{7/3}}\right ) \]

[In]

Integrate[1/((a + b*x^3)^(7/3)*(c + d*x^3)),x]

[Out]

((3*b*x*(-8*a^2*d + 3*b^2*c*x^3 + a*b*(4*c - 7*d*x^3)))/(a^2*(b*c - a*d)^2*(a + b*x^3)^(4/3)) - (2*Sqrt[-6 + (
6*I)*Sqrt[3]]*d^2*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)^(7/3)) + (2*(1 + I*Sqrt[3])*d^2*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)^(7/3)) - (I*(-I + Sqrt[3])*d^2*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)^(7/3)))/12

Maple [A] (verified)

Time = 5.06 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.19

method	result	size
pseudoelliptic	\(-\frac {-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2} d^{2} \left (b \,x^{3}+a \right )^{\frac {4}{3}}+12 x b c \left (a^{2} d -\frac {\left (-\frac {7 d \,x^{3}}{4}+c \right ) b a}{2}-\frac {3 b^{2} c \,x^{3}}{8}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+\left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right ) d^{2} \left (b \,x^{3}+a \right )^{\frac {4}{3}} a^{2}}{6 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (a d -b c \right )^{2} c \,a^{2}}\)	\(269\)

[In]

int(1/(b*x^3+a)^(7/3)/(d*x^3+c),x,method=_RETURNVERBOSE)

[Out]

-1/6/((a*d-b*c)/c)^(1/3)*(-2*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)*a^2*d^2*(b*x^3+a)^(4/3)+12*x*b*c*(a
^2*d-1/2*(-7/4*d*x^3+c)*b*a-3/8*b^2*c*x^3)*((a*d-b*c)/c)^(1/3)+(-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)*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))*d^2*(b*x^3+a)^(4/3)*a^2)/(b*x^3+a)^(4/3)/(a*d-b*c)^2/c/a^2

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x^3+a)^(7/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {7}{3}} \left (c + d x^{3}\right )}\, dx \]

[In]

integrate(1/(b*x**3+a)**(7/3)/(d*x**3+c),x)

[Out]

Integral(1/((a + b*x**3)**(7/3)*(c + d*x**3)), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {7}{3}} {\left (d x^{3} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(7/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(7/3)*(d*x^3 + c)), x)

Giac [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {7}{3}} {\left (d x^{3} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(7/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(7/3)*(d*x^3 + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{7/3}\,\left (d\,x^3+c\right )} \,d x \]

[In]

int(1/((a + b*x^3)^(7/3)*(c + d*x^3)),x)

[Out]

int(1/((a + b*x^3)^(7/3)*(c + d*x^3)), x)