\(\int \frac {x^3 (a+b x^3)^{2/3}}{a d-b d x^3} \, dx\) [808]

Optimal result

Integrand size = 28, antiderivative size = 229 \[ \int \frac {x^3 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {x \left (a+b x^3\right )^{2/3}}{3 b d}-\frac {5 a \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3} d}+\frac {2^{2/3} a \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{4/3} d}+\frac {a \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} b^{4/3} d}-\frac {a \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} b^{4/3} d}+\frac {5 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{6 b^{4/3} d} \] Output:

-1/3*x*(b*x^3+a)^(2/3)/b/d-5/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3) 
)*3^(1/2))*3^(1/2)/b^(4/3)/d+1/3*2^(2/3)*a*arctan(1/3*(1+2*2^(1/3)*b^(1/3) 
*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(4/3)/d+1/6*a*ln(-b*d*x^3+a*d)*2^(2 
/3)/b^(4/3)/d-1/2*a*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(2/3)/b^(4/3)/ 
d+5/6*a*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)/d
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx=-\frac {6 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3}+10 \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-6\ 2^{2/3} \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )-10 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+6\ 2^{2/3} a \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+5 a \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-3\ 2^{2/3} a \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{18 b^{4/3} d} \] Input:

Integrate[(x^3*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]
 

Output:

-1/18*(6*b^(1/3)*x*(a + b*x^3)^(2/3) + 10*Sqrt[3]*a*ArcTan[(Sqrt[3]*b^(1/3 
)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] - 6*2^(2/3)*Sqrt[3]*a*ArcTan[(Sqrt 
[3]*b^(1/3)*x)/(b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))] - 10*a*Log[-(b^(1/ 
3)*x) + (a + b*x^3)^(1/3)] + 6*2^(2/3)*a*Log[-2*b^(1/3)*x + 2^(2/3)*(a + b 
*x^3)^(1/3)] + 5*a*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b* 
x^3)^(2/3)] - 3*2^(2/3)*a*Log[2*b^(2/3)*x^2 + 2^(2/3)*b^(1/3)*x*(a + b*x^3 
)^(1/3) + 2^(1/3)*(a + b*x^3)^(2/3)])/(b^(4/3)*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {978, 27, 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.

\(\displaystyle \int \frac {x^3 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx\)

\(\Big \downarrow \) 978

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1026

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

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

Input:

Int[(x^3*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]
 

Output:

-1/3*(x*(a + b*x^3)^(2/3))/(b*d) + (a*(6*a*(ArcTan[(1 + (2*2^(1/3)*b^(1/3) 
*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]*a*b^(1/3)) + Log[a - b*x^ 
3]/(6*2^(1/3)*a*b^(1/3)) - Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*2 
^(1/3)*a*b^(1/3))) - 5*(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[ 
3]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))) 
))/(3*b*d)
 

Defintions of rubi rules used

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 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 978

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* 
((c + d*x^n)^q/(b*(m + n*(p + q) + 1))), x] - Simp[e^n/(b*(m + n*(p + q) + 
1))   Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 
 1) + (a*d*(m - n + 1) - n*q*(b*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c 
, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && GtQ[m - n 
 + 1, 0] && IntBinomialQ[a, b, c, d, e, m, 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 = 1.88 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {-6 \sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) a -6 \,2^{\frac {2}{3}} \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a +3 \,2^{\frac {2}{3}} \ln \left (\frac {b^{\frac {2}{3}} 2^{\frac {2}{3}} x^{2}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {1}{3}} 2^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a -6 \left (b \,x^{3}+a \right )^{\frac {2}{3}} x \,b^{\frac {1}{3}}+10 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a +10 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a -5 \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a}{18 d \,b^{\frac {4}{3}}}\)	\(241\)

Input:

int(x^3*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
 

Output:

1/18*(-6*3^(1/2)*2^(2/3)*arctan(1/3*3^(1/2)*(2^(2/3)*(b*x^3+a)^(1/3)+b^(1/ 
3)*x)/b^(1/3)/x)*a-6*2^(2/3)*ln((-2^(1/3)*b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a+ 
3*2^(2/3)*ln((b^(2/3)*2^(2/3)*x^2+(b*x^3+a)^(1/3)*b^(1/3)*2^(1/3)*x+(b*x^3 
+a)^(2/3))/x^2)*a-6*(b*x^3+a)^(2/3)*x*b^(1/3)+10*3^(1/2)*arctan(1/3*3^(1/2 
)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)*a+10*ln((-b^(1/3)*x+(b*x^3+a)^( 
1/3))/x)*a-5*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^(4/3)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 653, normalized size of antiderivative = 2.85 \[ \int \frac {x^3 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx =\text {Too large to display} \] Input:

integrate(x^3*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="fricas")
 

Output:

[-1/18*(6*4^(1/3)*sqrt(3)*a*b*(-1/b)^(1/3)*arctan(-1/3*(sqrt(3)*x - 4^(1/3 
)*sqrt(3)*(b*x^3 + a)^(1/3)*(-1/b)^(1/3))/x) - 15*sqrt(1/3)*a*b*sqrt(-1/b^ 
(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/3 
)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-1/b 
^(2/3)) + 2*a) - 6*4^(1/3)*a*b*(-1/b)^(1/3)*log(-(4^(2/3)*b*x*(-1/b)^(2/3) 
 - 2*(b*x^3 + a)^(1/3))/x) + 3*4^(1/3)*a*b*(-1/b)^(1/3)*log(-(2*4^(1/3)*b* 
x^2*(-1/b)^(1/3) - 4^(2/3)*(b*x^3 + a)^(1/3)*b*x*(-1/b)^(2/3) - 2*(b*x^3 + 
 a)^(2/3))/x^2) + 6*(b*x^3 + a)^(2/3)*b*x - 10*a*b^(2/3)*log(-(b^(1/3)*x - 
 (b*x^3 + a)^(1/3))/x) + 5*a*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)* 
b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2))/(b^2*d), -1/18*(6*4^(1/3)*sqrt(3)*a*b 
*(-1/b)^(1/3)*arctan(-1/3*(sqrt(3)*x - 4^(1/3)*sqrt(3)*(b*x^3 + a)^(1/3)*( 
-1/b)^(1/3))/x) - 6*4^(1/3)*a*b*(-1/b)^(1/3)*log(-(4^(2/3)*b*x*(-1/b)^(2/3 
) - 2*(b*x^3 + a)^(1/3))/x) + 3*4^(1/3)*a*b*(-1/b)^(1/3)*log(-(2*4^(1/3)*b 
*x^2*(-1/b)^(1/3) - 4^(2/3)*(b*x^3 + a)^(1/3)*b*x*(-1/b)^(2/3) - 2*(b*x^3 
+ a)^(2/3))/x^2) - 30*sqrt(1/3)*a*b^(2/3)*arctan(sqrt(1/3)*(b^(1/3)*x + 2* 
(b*x^3 + a)^(1/3))/(b^(1/3)*x)) + 6*(b*x^3 + a)^(2/3)*b*x - 10*a*b^(2/3)*l 
og(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + 5*a*b^(2/3)*log((b^(2/3)*x^2 + (b 
*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2))/(b^2*d)]
 

Sympy [F]

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

integrate(x**3*(b*x**3+a)**(2/3)/(-b*d*x**3+a*d),x)
 

Output:

-Integral(x**3*(a + b*x**3)**(2/3)/(-a + b*x**3), x)/d
 

Maxima [F]

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

integrate(x^3*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="maxima")
 

Output:

-integrate((b*x^3 + a)^(2/3)*x^3/(b*d*x^3 - a*d), x)
 

Giac [F]

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

integrate(x^3*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="giac")
 

Output:

integrate(-(b*x^3 + a)^(2/3)*x^3/(b*d*x^3 - a*d), x)
 

Mupad [F(-1)]

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

int((x^3*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x)
 

Output:

int((x^3*(a + b*x^3)^(2/3))/(a*d - b*d*x^3), x)
 

Reduce [F]

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

int(x^3*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x)
 

Output:

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