\(\int \frac {(a+b x^3)^{4/3}}{x (c+d x^3)} \, dx\) [714]

Optimal result

Integrand size = 24, antiderivative size = 261 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x \left (c+d x^3\right )} \, dx=\frac {b \sqrt [3]{a+b x^3}}{d}-\frac {a^{4/3} \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} c}+\frac {(b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c d^{4/3}}-\frac {a^{4/3} \log (x)}{2 c}+\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c d^{4/3}}+\frac {a^{4/3} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 c}-\frac {(b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c d^{4/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x \left (c+d x^3\right )} \, dx=\frac {6 b c \sqrt [3]{d} \sqrt [3]{a+b x^3}-2 \sqrt {3} a^{4/3} d^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 \sqrt {3} (b c-a d)^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+2 a^{4/3} d^{4/3} \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^3}\right )-2 (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-a^{4/3} d^{4/3} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+(b c-a d)^{4/3} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 c d^{4/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {948, 95, 174, 69, 16, 70, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {\left (b x^3+a\right )^{4/3}}{x^3 \left (d x^3+c\right )}dx^3\)

\(\Big \downarrow \) 95

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 70

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 70

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) 
, x] + (Simp[3/(2*b*q)   Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] + Simp[3/(2*b*q^2)   Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && NegQ[(b*c - a*d)/b]
 

rule 95

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

rule 174

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

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(\frac {3 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b c d +\frac {\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d^{2} \left (-2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )\right ) a^{\frac {4}{3}}}{2}+\frac {\left (a d -b c \right )^{2} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{2}}{3 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d^{2} c}\)	\(293\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x \left (c+d x^3\right )} \, dx=-\frac {2 \, \sqrt {3} a^{\frac {4}{3}} d \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + a^{\frac {4}{3}} d \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 2 \, a^{\frac {4}{3}} d \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 2 \, \sqrt {3} {\left (b c - a d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) - 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b c - {\left (b c - a d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) + 2 \, {\left (b c - a d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}{6 \, c d} \] Input:

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

Output:

-1/6*(2*sqrt(3)*a^(4/3)*d*arctan(1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*a^(2/3) 
+ sqrt(3)*a)/a) + a^(4/3)*d*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1 
/3) + a^(2/3)) - 2*a^(4/3)*d*log((b*x^3 + a)^(1/3) - a^(1/3)) - 2*sqrt(3)* 
(b*c - a*d)*((b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3) 
*d*((b*c - a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) - 6*(b*x^3 + 
a)^(1/3)*b*c - (b*c - a*d)*((b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(2/3) - ( 
b*x^3 + a)^(1/3)*((b*c - a*d)/d)^(1/3) + ((b*c - a*d)/d)^(2/3)) + 2*(b*c - 
 a*d)*((b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) + ((b*c - a*d)/d)^(1/3)) 
)/(c*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x \left (c+d x^3\right )} \, dx=-\frac {\sqrt {3} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, c} - \frac {a^{\frac {4}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, c} + \frac {a^{\frac {4}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, c} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c^{2} d - a c d^{2}\right )}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{d} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, c d^{2}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, c d^{2}} \] Input:

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

Output:

-1/3*sqrt(3)*a^(4/3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^ 
(1/3))/c - 1/6*a^(4/3)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + 
 a^(2/3))/c + 1/3*a^(4/3)*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/c + 1/3*(b 
^2*c^2 - 2*a*b*c*d + a^2*d^2)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^( 
1/3) - (-(b*c - a*d)/d)^(1/3)))/(b*c^2*d - a*c*d^2) + (b*x^3 + a)^(1/3)*b/ 
d - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*arctan(1/3*sqrt(3)*(2 
*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/(c*d^ 
2) - 1/6*(-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d)*log((b*x^3 + a)^(2/3) + (b*x 
^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(c*d^2)
 

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 796, normalized size of antiderivative = 3.05 \[ \int \frac {\left (a+b x^3\right )^{4/3}}{x \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

log(c*d*(-(a*d - b*c)^4/(c^3*d^4))^(1/3) + a*d*(a + b*x^3)^(1/3) - b*c*(a 
+ b*x^3)^(1/3))*(-(a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 
 4*a^3*b*c*d^3)/(27*c^3*d^4))^(1/3) + log(c*(a^4/c^3)^(1/3) - a*(a + b*x^3 
)^(1/3))*(a^4/(27*c^3))^(1/3) + (b*(a + b*x^3)^(1/3))/d - log(c*(a^4/c^3)^ 
(1/3) + 2*a*(a + b*x^3)^(1/3) + 3^(1/2)*c*(a^4/c^3)^(1/3)*1i)*((3^(1/2)*1i 
)/2 + 1/2)*(a^4/(27*c^3))^(1/3) + log(c*(a^4/c^3)^(1/3)*1i + a*(a + b*x^3) 
^(1/3)*2i + 3^(1/2)*c*(a^4/c^3)^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(a^4/(27*c^3 
))^(1/3) + log((3*a^2*b^4*(a + b*x^3)^(1/3)*(a*d - b*c)^2*(2*a^4*d^4 + b^4 
*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3))/d + 3*a^2*b^4*c 
*((3^(1/2)*1i)/2 - 1/2)*(-(a*d - b*c)^4/(c^3*d^4))^(1/3)*(2*a^5*d^5 - b^5* 
c^5 - 10*a^2*b^3*c^3*d^2 + 10*a^3*b^2*c^2*d^3 + 5*a*b^4*c^4*d - 6*a^4*b*c* 
d^4))*((3^(1/2)*1i)/2 - 1/2)*(-(a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4* 
a*b^3*c^3*d - 4*a^3*b*c*d^3)/(27*c^3*d^4))^(1/3) - log((3*a^2*b^4*(a + b*x 
^3)^(1/3)*(a*d - b*c)^2*(2*a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3 
*c^3*d - 4*a^3*b*c*d^3))/d - 3*a^2*b^4*c*((3^(1/2)*1i)/2 + 1/2)*(-(a*d - b 
*c)^4/(c^3*d^4))^(1/3)*(2*a^5*d^5 - b^5*c^5 - 10*a^2*b^3*c^3*d^2 + 10*a^3* 
b^2*c^2*d^3 + 5*a*b^4*c^4*d - 6*a^4*b*c*d^4))*((3^(1/2)*1i)/2 + 1/2)*(-(a^ 
4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3)/(27*c 
^3*d^4))^(1/3)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{4/3}}{x \left (c+d x^3\right )} \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b d \,x^{7}+a d \,x^{4}+b c \,x^{4}+a c x}d x \right ) a^{2} c -2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b d +\left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{2} c}{c} \] Input:

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

Output:

(2*(a + b*x**3)**(1/3)*a + int((a + b*x**3)**(1/3)/(a*c*x + a*d*x**4 + b*c 
*x**4 + b*d*x**7),x)*a**2*c - 2*int(((a + b*x**3)**(1/3)*x**5)/(a*c + a*d* 
x**3 + b*c*x**3 + b*d*x**6),x)*a*b*d + int(((a + b*x**3)**(1/3)*x**5)/(a*c 
 + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b**2*c)/c