\(\int \frac {1}{(a+b x^3)^{7/3} (c+d x^3)} \, dx\) [143]

Optimal result

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

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)+1/3*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))*3^(1/2)/c^(2/3)/(-a*d+b*c)^(7/3)+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)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.36 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\frac {1}{12} \left (\frac {3 b x \left (-8 a^2 d+3 b^2 c x^3+a b \left (4 c-7 d x^3\right )\right )}{a^2 (b c-a d)^2 \left (a+b x^3\right )^{4/3}}-\frac {2 \sqrt {-6+6 i \sqrt {3}} d^2 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{2/3} (b c-a d)^{7/3}}+\frac {2 \left (1+i \sqrt {3}\right ) d^2 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{c^{2/3} (b c-a d)^{7/3}}-\frac {i \left (-i+\sqrt {3}\right ) d^2 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{c^{2/3} (b c-a d)^{7/3}}\right ) \] Input:

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

Output:

((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
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {931, 25, 1024, 27, 901}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 931

\(\displaystyle \frac {b x}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}-\frac {\int -\frac {3 b d x^3+3 b c-4 a d}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx}{4 a (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {3 b d x^3+3 b c-4 a d}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}\)

\(\Big \downarrow \) 1024

\(\displaystyle \frac {\frac {b x (3 b c-7 a d)}{a \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\int -\frac {4 a^2 d^2}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{a (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 a d^2 \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{b c-a d}+\frac {b x (3 b c-7 a d)}{a \sqrt [3]{a+b x^3} (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {\frac {4 a d^2 \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{b c-a d}+\frac {b x (3 b c-7 a d)}{a \sqrt [3]{a+b x^3} (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[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 931

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] + Simp[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]) && IntBinomialQ[a, b, 
 c, d, n, p, q, x]
 

rule 1024

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] + Simp[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] /; Fr 
eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {-\frac {3 x b c \left (7 a b d \,x^{3}-3 b^{2} c \,x^{3}+8 d \,a^{2}-4 a b c \right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{2}+\left (b \,x^{3}+a \right )^{\frac {4}{3}} a^{2} d^{2} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{6 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (a d -b c \right )^{2} c \,a^{2}}\)	\(243\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d \,x^{3}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c \,x^{3}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b d \,x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c \,x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} d \,x^{9}}d x \] Input:

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

Output:

int(1/((a + b*x**3)**(1/3)*a**2*c + (a + b*x**3)**(1/3)*a**2*d*x**3 + 2*(a 
 + b*x**3)**(1/3)*a*b*c*x**3 + 2*(a + b*x**3)**(1/3)*a*b*d*x**6 + (a + b*x 
**3)**(1/3)*b**2*c*x**6 + (a + b*x**3)**(1/3)*b**2*d*x**9),x)