3.1.87 \(\int \frac {(a+b x^3)^{5/3}}{c+d x^3} \, dx\) [87]

3.1.87.1 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+1/6*(-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^2-1/9*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))/d^2* 
3^(1/2)+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))/c^(2/3)/d^2*3^(1/2)
 

3.1.87.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 10.51 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{c+d x^3} \, dx=\frac {3 b \sqrt [3]{b c-a d} (-3 b c+5 a d) x^4 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 \sqrt [3]{c} \left (6 a b c^{2/3} \sqrt [3]{b c-a d} x+6 b^2 c^{2/3} \sqrt [3]{b c-a d} x^4+2 \sqrt {3} a (-b c+3 a d) \sqrt [3]{a+b x^3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{b+a x^3}}}{\sqrt {3}}\right )+2 a (b c-3 a d) \sqrt [3]{a+b x^3} \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )-a b c \sqrt [3]{a+b x^3} \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (b+a x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )+3 a^2 d \sqrt [3]{a+b x^3} \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (b+a x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{b+a x^3}}\right )\right )}{36 c d \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}} \]

input

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

output

(3*b*(b*c - a*d)^(1/3)*(-3*b*c + 5*a*d)*x^4*(1 + (b*x^3)/a)^(1/3)*AppellF1 
[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*c^(1/3)*(6*a*b*c^(2/3)* 
(b*c - a*d)^(1/3)*x + 6*b^2*c^(2/3)*(b*c - a*d)^(1/3)*x^4 + 2*Sqrt[3]*a*(- 
(b*c) + 3*a*d)*(a + b*x^3)^(1/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1 
/3)*(b + a*x^3)^(1/3)))/Sqrt[3]] + 2*a*(b*c - 3*a*d)*(a + b*x^3)^(1/3)*Log 
[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)] - a*b*c*(a + b*x^3)^(1 
/3)*Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b* 
c - a*d)^(1/3)*x)/(b + a*x^3)^(1/3)] + 3*a^2*d*(a + b*x^3)^(1/3)*Log[c^(2/ 
3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3 
)*x)/(b + a*x^3)^(1/3)]))/(36*c*d*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3))
 

3.1.87.3 Rubi [A] (verified)

Time = 0.36 (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)
 

3.1.87.3.1 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]
 

3.1.87.4 Maple [A] (verified)

Time = 4.57 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.47

method	result	size
pseudoelliptic	\(-\frac {-\frac {5 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c \left (a d \,b^{\frac {2}{3}}-\frac {3 b^{\frac {5}{3}} c}{5}\right ) \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{3}-2 \left (a d -b c \right )^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )+\frac {10 \sqrt {3}\, \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c \left (a d \,b^{\frac {2}{3}}-\frac {3 b^{\frac {5}{3}} c}{5}\right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )}{3}+\frac {10 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c \left (a d \,b^{\frac {2}{3}}-\frac {3 b^{\frac {5}{3}} c}{5}\right ) \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )}{3}-2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} x b c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} d +\left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right ) \left (a d -b c \right )^{2}}{6 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} c \,d^{2}}\)	\(402\)

input

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

output

-1/6*(-5/3*((a*d-b*c)/c)^(1/3)*c*(a*d*b^(2/3)-3/5*b^(5/3)*c)*ln((b^(2/3)*x 
^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*(a*d-b*c)^2*ln((((a*d 
-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)+10/3*3^(1/2)*((a*d-b*c)/c)^(1/3)*c*(a 
*d*b^(2/3)-3/5*b^(5/3)*c)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3)) 
/b^(1/3)/x)+10/3*((a*d-b*c)/c)^(1/3)*c*(a*d*b^(2/3)-3/5*b^(5/3)*c)*ln((-b^ 
(1/3)*x+(b*x^3+a)^(1/3))/x)-2*(b*x^3+a)^(2/3)*x*b*c*((a*d-b*c)/c)^(1/3)*d+ 
(-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))*(a*d-b*c)^2)/((a*d-b*c)/c)^(1/3)/c/d^ 
2
 

3.1.87.5 Fricas [B] (verification not implemented)

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

Time = 0.76 (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
 

3.1.87.6 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)
 

3.1.87.7 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)
 

3.1.87.8 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)
 

3.1.87.9 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)