\(\int \frac {(a+b x^3)^{5/3}}{c+d x^3} \, dx\) [139]

Optimal result

Integrand size = 21, antiderivative size = 273 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\frac {b x \left (a+b x^3\right )^{2/3}}{3 d}-\frac {b^{2/3} (3 b c-5 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} d^2}+\frac {(b c-a d)^{5/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} d^2}+\frac {(b c-a d)^{5/3} \log \left (c+d x^3\right )}{6 c^{2/3} d^2}-\frac {(b c-a d)^{5/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} d^2}+\frac {b^{2/3} (3 b c-5 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 d^2} \] Output:

1/3*b*x*(b*x^3+a)^(2/3)/d-1/9*b^(2/3)*(-5*a*d+3*b*c)*arctan(1/3*(1+2*b^(1/ 
3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/d^2+1/3*(-a*d+b*c)^(5/3)*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) 
/d^2+1/6*(-a*d+b*c)^(5/3)*ln(d*x^3+c)/c^(2/3)/d^2-1/2*(-a*d+b*c)^(5/3)*ln( 
(-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(2/3)/d^2+1/6*b^(2/3)*(-5*a* 
d+3*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/d^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.91 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\frac {12 b d x \left (a+b x^3\right )^{2/3}-4 \sqrt {3} b^{2/3} (3 b c-5 a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-\frac {6 \sqrt {-6+6 i \sqrt {3}} (b c-a d)^{5/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{2/3}}+4 b^{2/3} (3 b c-5 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+\frac {6 \left (1+i \sqrt {3}\right ) (b c-a d)^{5/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{c^{2/3}}-2 b^{2/3} (3 b c-5 a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-\frac {3 i \left (-i+\sqrt {3}\right ) (b c-a d)^{5/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{c^{2/3}}}{36 d^2} \] Input:

Integrate[(a + b*x^3)^(5/3)/(c + d*x^3),x]
 

Output:

(12*b*d*x*(a + b*x^3)^(2/3) - 4*Sqrt[3]*b^(2/3)*(3*b*c - 5*a*d)*ArcTan[(Sq 
rt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] - (6*Sqrt[-6 + (6*I)*S 
qrt[3]]*(b*c - a*d)^(5/3)*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) + 4*b^( 
2/3)*(3*b*c - 5*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + (6*(1 + I*Sqr 
t[3])*(b*c - a*d)^(5/3)*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) - 2*b^(2/3)*(3*b*c - 5*a*d)*Log[b^(2/3)*x^2 
+ b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)] - ((3*I)*(-I + Sqrt[3]) 
*(b*c - a*d)^(5/3)*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))/(36*d^2)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {933, 25, 1026, 769, 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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx\)

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\frac {b (3 b c-5 a d) x^3+a (b c-3 a d)}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 d}+\frac {b x \left (a+b x^3\right )^{2/3}}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b x \left (a+b x^3\right )^{2/3}}{3 d}-\frac {\int \frac {b (3 b c-5 a d) x^3+a (b c-3 a d)}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 d}\)

\(\Big \downarrow \) 1026

\(\displaystyle \frac {b x \left (a+b x^3\right )^{2/3}}{3 d}-\frac {\frac {b (3 b c-5 a d) \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{d}-\frac {3 (b c-a d)^2 \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{d}}{3 d}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {b x \left (a+b x^3\right )^{2/3}}{3 d}-\frac {\frac {b (3 b c-5 a d) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{d}-\frac {3 (b c-a d)^2 \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{d}}{3 d}\)

\(\Big \downarrow \) 901

\(\displaystyle \frac {b x \left (a+b x^3\right )^{2/3}}{3 d}-\frac {\frac {b (3 b c-5 a d) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{d}-\frac {3 (b c-a d)^2 \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{d}}{3 d}\)

Input:

Int[(a + b*x^3)^(5/3)/(c + d*x^3),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, 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 933

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

rule 1026

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

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.19

method	result	size
pseudoelliptic	\(\frac {-\frac {5 c \left (-\frac {3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} x b d}{5}+\left (a d \,b^{\frac {2}{3}}-\frac {3 b^{\frac {5}{3}} c}{5}\right ) \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{3}+\frac {\left (a d -b c \right )^{2} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{2}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c \,d^{2}}\)	\(326\)

Input:

int((b*x^3+a)^(5/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 535 vs. \(2 (220) = 440\).

Time = 0.52 (sec) , antiderivative size = 535, normalized size of antiderivative = 1.96 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b d x + 6 \, \sqrt {3} {\left (b c - a d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) + 2 \, \sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} {\left (3 \, b c - 5 \, a d\right )} \arctan \left (-\frac {\sqrt {3} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}}}{3 \, b x}\right ) - 6 \, {\left (b c - a d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (\frac {c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}}{x}\right ) - 2 \, \left (-b^{2}\right )^{\frac {1}{3}} {\left (3 \, b c - 5 \, a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac {1}{3}} {\left (3 \, b c - 5 \, a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (b c - a d\right )} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (-\frac {{\left (b c - a d\right )} x^{2} \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x \left (\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{c^{2}}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (b c - a d\right )}}{x^{2}}\right )}{18 \, d^{2}} \] Input:

integrate((b*x^3+a)^(5/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

1/18*(6*(b*x^3 + a)^(2/3)*b*d*x + 6*sqrt(3)*(b*c - a*d)*((b^2*c^2 - 2*a*b* 
c*d + a^2*d^2)/c^2)^(1/3)*arctan(-1/3*(sqrt(3)*(b*c - a*d)*x + 2*sqrt(3)*( 
b*x^3 + a)^(1/3)*c*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3))/((b*c - a* 
d)*x)) + 2*sqrt(3)*(-b^2)^(1/3)*(3*b*c - 5*a*d)*arctan(-1/3*(sqrt(3)*b*x - 
 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b^2)^(1/3))/(b*x)) - 6*(b*c - a*d)*((b^2*c^ 
2 - 2*a*b*c*d + a^2*d^2)/c^2)^(1/3)*log((c*x*((b^2*c^2 - 2*a*b*c*d + a^2*d 
^2)/c^2)^(2/3) - (b*x^3 + a)^(1/3)*(b*c - a*d))/x) - 2*(-b^2)^(1/3)*(3*b*c 
 - 5*a*d)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + (-b^2)^(1/3)*(3 
*b*c - 5*a*d)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x 
- (b*x^3 + a)^(2/3)*b)/x^2) + 3*(b*c - a*d)*((b^2*c^2 - 2*a*b*c*d + a^2*d^ 
2)/c^2)^(1/3)*log(-((b*c - a*d)*x^2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^ 
(1/3) + (b*x^3 + a)^(1/3)*c*x*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)/c^2)^(2/3) 
+ (b*x^3 + a)^(2/3)*(b*c - a*d))/x^2))/d^2
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\int \frac {\left (a + b x^{3}\right )^{\frac {5}{3}}}{c + d x^{3}}\, dx \] Input:

integrate((b*x**3+a)**(5/3)/(d*x**3+c),x)
 

Output:

Integral((a + b*x**3)**(5/3)/(c + d*x**3), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{d x^{3} + c} \,d x } \] Input:

integrate((b*x^3+a)^(5/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{d x^{3} + c} \,d x } \] Input:

integrate((b*x^3+a)^(5/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(5/3)/(d*x^3 + c), x)
 

Mupad [F(-1)]

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

int((a + b*x^3)^(5/3)/(c + d*x^3),x)
 

Output:

int((a + b*x^3)^(5/3)/(c + d*x^3), x)
 

Reduce [F]

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

int((b*x^3+a)^(5/3)/(d*x^3+c),x)
 

Output:

((a + b*x**3)**(2/3)*b*x + 3*int((a + b*x**3)**(2/3)/(a*c + a*d*x**3 + b*c 
*x**3 + b*d*x**6),x)*a**2*d - int((a + b*x**3)**(2/3)/(a*c + a*d*x**3 + b* 
c*x**3 + b*d*x**6),x)*a*b*c + 5*int(((a + b*x**3)**(2/3)*x**3)/(a*c + a*d* 
x**3 + b*c*x**3 + b*d*x**6),x)*a*b*d - 3*int(((a + b*x**3)**(2/3)*x**3)/(a 
*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b**2*c)/(3*d)