\(\int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx\) [147]

Optimal result

Integrand size = 19, antiderivative size = 435 \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx=\frac {\sqrt [3]{a+b x^3}}{d}+\frac {x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c \sqrt [3]{1+\frac {b x^3}{a}}}+\frac {\sqrt [3]{b} c \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}-\frac {\sqrt [3]{b c^3-a d^3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {\sqrt [3]{b c^3-a d^3} \arctan \left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {\sqrt [3]{b c^3-a d^3} \log \left (c^3+d^3 x^3\right )}{3 d^2}+\frac {\sqrt [3]{b} c \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d^2}-\frac {\sqrt [3]{b c^3-a d^3} \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{2 d^2}-\frac {\sqrt [3]{b c^3-a d^3} \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 d^2} \] Output:

(b*x^3+a)^(1/3)/d+x*(b*x^3+a)^(1/3)*AppellF1(1/3,-1/3,1,4/3,-b*x^3/a,-d^3* 
x^3/c^3)/c/(1+b*x^3/a)^(1/3)+1/3*b^(1/3)*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^3+b*c^3)^(1/3)*arctan(1/3*(1+2* 
(-a*d^3+b*c^3)^(1/3)*x/c/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/d^2+1/3*(-a*d^3 
+b*c^3)^(1/3)*arctan(1/3*(1-2*d*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)^(1/3))*3^(1 
/2))*3^(1/2)/d^2+1/3*(-a*d^3+b*c^3)^(1/3)*ln(d^3*x^3+c^3)/d^2+1/2*b^(1/3)* 
c*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/d^2-1/2*(-a*d^3+b*c^3)^(1/3)*ln((-a*d^3+b* 
c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/d^2-1/2*(-a*d^3+b*c^3)^(1/3)*ln((-a*d^3+b* 
c^3)^(1/3)+d*(b*x^3+a)^(1/3))/d^2
 

Mathematica [F]

\[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx=\int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2581, 2009}

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.

\(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx\)

\(\Big \downarrow \) 2581

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x \sqrt [3]{a+b x^3} \operatorname {AppellF1}\left (\frac {1}{3},-\frac {1}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c \sqrt [3]{\frac {b x^3}{a}+1}}-\frac {\sqrt [3]{b c^3-a d^3} \arctan \left (\frac {\frac {2 x \sqrt [3]{b c^3-a d^3}}{c \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {\sqrt [3]{b c^3-a d^3} \arctan \left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {\sqrt [3]{b} c \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {\sqrt [3]{b c^3-a d^3} \log \left (c^3+d^3 x^3\right )}{3 d^2}-\frac {\sqrt [3]{b c^3-a d^3} \log \left (\frac {x \sqrt [3]{b c^3-a d^3}}{c}-\sqrt [3]{a+b x^3}\right )}{2 d^2}-\frac {\sqrt [3]{b c^3-a d^3} \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{2 d^2}+\frac {\sqrt [3]{b} c \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d^2}+\frac {\sqrt [3]{a+b x^3}}{d}\)

Input:

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

Output:

(a + b*x^3)^(1/3)/d + (x*(a + b*x^3)^(1/3)*AppellF1[1/3, -1/3, 1, 4/3, -(( 
b*x^3)/a), -((d^3*x^3)/c^3)])/(c*(1 + (b*x^3)/a)^(1/3)) + (b^(1/3)*c*ArcTa 
n[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^2) - ((b*c^3 
- a*d^3)^(1/3)*ArcTan[(1 + (2*(b*c^3 - a*d^3)^(1/3)*x)/(c*(a + b*x^3)^(1/3 
)))/Sqrt[3]])/(Sqrt[3]*d^2) + ((b*c^3 - a*d^3)^(1/3)*ArcTan[(1 - (2*d*(a + 
 b*x^3)^(1/3))/(b*c^3 - a*d^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^2) + ((b*c^3 - 
a*d^3)^(1/3)*Log[c^3 + d^3*x^3])/(3*d^2) + (b^(1/3)*c*Log[b^(1/3)*x - (a + 
 b*x^3)^(1/3)])/(2*d^2) - ((b*c^3 - a*d^3)^(1/3)*Log[((b*c^3 - a*d^3)^(1/3 
)*x)/c - (a + b*x^3)^(1/3)])/(2*d^2) - ((b*c^3 - a*d^3)^(1/3)*Log[(b*c^3 - 
 a*d^3)^(1/3) + d*(a + b*x^3)^(1/3)])/(2*d^2)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2581

Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(c^3 + d^3*x^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d 
^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ[q, 0 
] && RationalQ[p] && EqQ[Denominator[p], 3]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\sqrt [3]{a+b x^3}}{c+d x} \, dx=\int \frac {\sqrt [3]{a + b x^{3}}}{c + d x}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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