3.8.3 \(\int \frac {x (a+b x^3)^{4/3}}{c+d x^3} \, dx\) [703]

3.8.3.1 Optimal result

Integrand size = 22, antiderivative size = 277 \[ \int \frac {x \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {b x^2 \sqrt [3]{a+b x^3}}{3 d}+\frac {\sqrt [3]{b} (3 b c-4 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} d^2}-\frac {(b c-a d)^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} d^2}+\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 \sqrt [3]{c} d^2}+\frac {\sqrt [3]{b} (3 b c-4 a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 d^2}-\frac {(b c-a d)^{4/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} d^2} \]

output

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

3.8.3.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.54 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.69 \[ \int \frac {x \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {12 b d x^2 \sqrt [3]{a+b x^3}+4 \sqrt {3} \sqrt [3]{b} (3 b c-4 a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+\frac {6 \sqrt {-6-6 i \sqrt {3}} (b c-a d)^{4/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{c}}+4 \sqrt [3]{b} (3 b c-4 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+\frac {6 \left (1-i \sqrt {3}\right ) (b c-a d)^{4/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{c}}-2 \sqrt [3]{b} (3 b c-4 a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+\frac {3 i \left (i+\sqrt {3}\right ) (b c-a d)^{4/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{c}}}{36 d^2} \]

input

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

output

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

3.8.3.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {977, 25, 1054, 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.

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

\(\Big \downarrow \) 977

\(\displaystyle \frac {\int -\frac {x \left (b (3 b c-4 a d) x^3+a (2 b c-3 a d)\right )}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{3 d}+\frac {b x^2 \sqrt [3]{a+b x^3}}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b x^2 \sqrt [3]{a+b x^3}}{3 d}-\frac {\int \frac {x \left (b (3 b c-4 a d) x^3+a (2 b c-3 a d)\right )}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{3 d}\)

\(\Big \downarrow \) 1054

\(\displaystyle \frac {b x^2 \sqrt [3]{a+b x^3}}{3 d}-\frac {\int \left (\frac {b (3 b c-4 a d) x}{d \left (b x^3+a\right )^{2/3}}-\frac {3 \left (b^2 c^2-2 a b d c+a^2 d^2\right ) x}{d \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}\right )dx}{3 d}\)

\(\Big \downarrow \) 2009

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

input

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

output

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

3.8.3.3.1 Defintions of rubi rules used

rule 25

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

rule 977

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

rule 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

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

3.8.3.4 Maple [A] (verified)

Time = 5.19 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.39

method	result	size
pseudoelliptic	\(-\frac {-\frac {\left (a d -b c \right )^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}-\sqrt {3}\, \left (a d -b c \right )^{2} \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )-\frac {2 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} c \left (a d \,b^{\frac {1}{3}}-\frac {3 b^{\frac {4}{3}} c}{4}\right ) \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{3}+\left (a d -b c \right )^{2} \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )+\frac {4 \left (-\sqrt {3}\, \left (a d \,b^{\frac {1}{3}}-\frac {3 b^{\frac {4}{3}} c}{4}\right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\left (a d \,b^{\frac {1}{3}}-\frac {3 b^{\frac {4}{3}} c}{4}\right ) \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {3 b \,x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} d}{4}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} c}{3}}{3 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} c \,d^{2}}\)	\(385\)

input

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

output

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

3.8.3.5 Fricas [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.43 \[ \int \frac {x \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx=\frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b d x^{2} - 6 \, \sqrt {3} {\left (b c - a d\right )} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b c - a d}{c}\right )^{\frac {2}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) + 2 \, \sqrt {3} {\left (3 \, b c - 4 \, a d\right )} \left (-b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}}}{3 \, b x}\right ) - 2 \, {\left (3 \, b c - 4 \, a d\right )} \left (-b\right )^{\frac {1}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 6 \, {\left (b c - a d\right )} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \log \left (-\frac {x \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + {\left (3 \, b c - 4 \, a d\right )} \left (-b\right )^{\frac {1}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (b c - a d\right )} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \log \left (\frac {x^{2} \left (\frac {b c - a d}{c}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} x \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, d^{2}} \]

input

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

output

1/18*(6*(b*x^3 + a)^(1/3)*b*d*x^2 - 6*sqrt(3)*(b*c - a*d)*((b*c - a*d)/c)^ 
(1/3)*arctan(-1/3*(sqrt(3)*(b*c - a*d)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*c*( 
(b*c - a*d)/c)^(2/3))/((b*c - a*d)*x)) + 2*sqrt(3)*(3*b*c - 4*a*d)*(-b)^(1 
/3)*arctan(1/3*(sqrt(3)*b*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b)^(2/3))/(b*x 
)) - 2*(3*b*c - 4*a*d)*(-b)^(1/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x 
) - 6*(b*c - a*d)*((b*c - a*d)/c)^(1/3)*log(-(x*((b*c - a*d)/c)^(1/3) - (b 
*x^3 + a)^(1/3))/x) + (3*b*c - 4*a*d)*(-b)^(1/3)*log(((-b)^(2/3)*x^2 - (b* 
x^3 + a)^(1/3)*(-b)^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) + 3*(b*c - a*d)*((b* 
c - a*d)/c)^(1/3)*log((x^2*((b*c - a*d)/c)^(2/3) + (b*x^3 + a)^(1/3)*x*((b 
*c - a*d)/c)^(1/3) + (b*x^3 + a)^(2/3))/x^2))/d^2
 

3.8.3.6 Sympy [F]

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

input

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

output

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

3.8.3.7 Maxima [F]

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

input

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

output

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

3.8.3.8 Giac [F]

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

input

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

output

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

3.8.3.9 Mupad [F(-1)]

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

input

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

output

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