Integrand size = 28, antiderivative size = 233 \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}+\frac {4 a \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d}-\frac {\sqrt [3]{2} a \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a \log \left (a d-b d x^3\right )}{3\ 2^{2/3} b^{5/3} d}+\frac {2 a \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{3 b^{5/3} d}-\frac {a \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{5/3} d} \] Output:
-1/3*x^2*(b*x^3+a)^(1/3)/b/d+4/9*a*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/ 3))*3^(1/2))*3^(1/2)/b^(5/3)/d-1/3*2^(1/3)*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^(5/3)/d+1/6*a*ln(-b*d*x^3+a*d)*2^ (1/3)/b^(5/3)/d+2/3*a*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)/d-1/2*a*ln(2^( 1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(1/3)/b^(5/3)/d
Time = 1.03 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.26 \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {6 b^{2/3} x^2 \sqrt [3]{a+b x^3}-8 \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+6 \sqrt [3]{2} \sqrt {3} a \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )-8 a \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+6 \sqrt [3]{2} a \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+4 a \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-3 \sqrt [3]{2} a \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{18 b^{5/3} d} \] Input:
Integrate[(x^4*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
Output:
-1/18*(6*b^(2/3)*x^2*(a + b*x^3)^(1/3) - 8*Sqrt[3]*a*ArcTan[(Sqrt[3]*b^(1/ 3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] + 6*2^(1/3)*Sqrt[3]*a*ArcTan[(Sqr t[3]*b^(1/3)*x)/(b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))] - 8*a*Log[-(b^(1/ 3)*x) + (a + b*x^3)^(1/3)] + 6*2^(1/3)*a*Log[-2*b^(1/3)*x + 2^(2/3)*(a + b *x^3)^(1/3)] + 4*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^(1/3)*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^(5/3)*d)
Time = 0.59 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {978, 27, 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.
| \(\displaystyle \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) |
| \(\Big \downarrow \) 978 |
| \(\displaystyle \frac {\int \frac {2 a x \left (2 b x^3+a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{3 b d}-\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \int \frac {x \left (2 b x^3+a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{3 b d}-\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {2 a \int \left (\frac {3 a x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}-\frac {2 x}{\left (b x^3+a\right )^{2/3}}\right )dx}{3 b d}-\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a \left (\frac {2 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} b^{2/3}}+\frac {\log \left (a-b x^3\right )}{2\ 2^{2/3} b^{2/3}}+\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{b^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2\ 2^{2/3} b^{2/3}}\right )}{3 b d}-\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}\) |
Input:
Int[(x^4*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
Output:
-1/3*(x^2*(a + b*x^3)^(1/3))/(b*d) + (2*a*((2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)) - (Sqrt[3]*ArcTan[(1 + (2*2^( 1/3)*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(2^(2/3)*b^(2/3)) + Log[a - b *x^3]/(2*2^(2/3)*b^(2/3)) + Log[b^(1/3)*x - (a + b*x^3)^(1/3)]/b^(2/3) - ( 3*Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*2^(2/3)*b^(2/3))))/(3*b*d )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
Time = 2.06 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.04
| method | result | size |
| pseudoelliptic | \(\frac {-6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2} b^{\frac {2}{3}}+6 \,2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) a -8 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a -6 \,2^{\frac {1}{3}} \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a +3 \,2^{\frac {1}{3}} \ln \left (\frac {b^{\frac {2}{3}} 2^{\frac {2}{3}} x^{2}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {1}{3}} 2^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a +8 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a -4 \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a}{18 b^{\frac {5}{3}} d}\) | \(243\) |
Input:
int(x^4*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
Output:
1/18*(-6*(b*x^3+a)^(1/3)*x^2*b^(2/3)+6*2^(1/3)*3^(1/2)*arctan(1/3*3^(1/2)* (2^(2/3)*(b*x^3+a)^(1/3)+b^(1/3)*x)/b^(1/3)/x)*a-8*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-6*2^(1/3)*ln((-2^(1/3)*b^(1 /3)*x+(b*x^3+a)^(1/3))/x)*a+3*2^(1/3)*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+8*ln((-b^(1/3)*x+(b*x^3+a)^( 1/3))/x)*a-4*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)/b^(5/3)/d
Time = 0.11 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.43 \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {6 \, \sqrt {3} 2^{\frac {1}{3}} a b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} + \sqrt {3} x}{3 \, x}\right ) - 6 \cdot 2^{\frac {1}{3}} a b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 3 \cdot 2^{\frac {1}{3}} a b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {2}{3}} b^{2} x^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} x^{2} + 24 \, \sqrt {\frac {1}{3}} a {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x}\right ) - 8 \, a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + 4 \, a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{3} d} \] Input:
integrate(x^4*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="fricas")
Output:
-1/18*(6*sqrt(3)*2^(1/3)*a*b^2*(-1/b^2)^(1/3)*arctan(1/3*(sqrt(3)*2^(2/3)* (b*x^3 + a)^(1/3)*b*(-1/b^2)^(2/3) + sqrt(3)*x)/x) - 6*2^(1/3)*a*b^2*(-1/b ^2)^(1/3)*log((2^(1/3)*b*x*(-1/b^2)^(1/3) + (b*x^3 + a)^(1/3))/x) + 3*2^(1 /3)*a*b^2*(-1/b^2)^(1/3)*log((2^(2/3)*b^2*x^2*(-1/b^2)^(2/3) - 2^(1/3)*(b* x^3 + a)^(1/3)*b*x*(-1/b^2)^(1/3) + (b*x^3 + a)^(2/3))/x^2) + 6*(b*x^3 + a )^(1/3)*b^2*x^2 + 24*sqrt(1/3)*a*(b^2)^(1/6)*b*arctan(sqrt(1/3)*((b^2)^(1/ 3)*b*x + 2*(b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x)) - 8*a*(b^2) ^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + 4*a*(b^2)^(2/3)*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))/(b^3*d)
\[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=- \frac {\int \frac {x^{4} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \] Input:
integrate(x**4*(b*x**3+a)**(1/3)/(-b*d*x**3+a*d),x)
Output:
-Integral(x**4*(a + b*x**3)**(1/3)/(-a + b*x**3), x)/d
\[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{4}}{b d x^{3} - a d} \,d x } \] Input:
integrate(x^4*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="maxima")
Output:
-integrate((b*x^3 + a)^(1/3)*x^4/(b*d*x^3 - a*d), x)
\[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{4}}{b d x^{3} - a d} \,d x } \] Input:
integrate(x^4*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="giac")
Output:
integrate(-(b*x^3 + a)^(1/3)*x^4/(b*d*x^3 - a*d), x)
Timed out. \[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int \frac {x^4\,{\left (b\,x^3+a\right )}^{1/3}}{a\,d-b\,d\,x^3} \,d x \] Input:
int((x^4*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x)
Output:
int((x^4*(a + b*x^3)^(1/3))/(a*d - b*d*x^3), x)
\[ \int \frac {x^4 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {-\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2}+4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{4}}{-b^{2} x^{6}+a^{2}}d x \right ) a b +2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x}{-b^{2} x^{6}+a^{2}}d x \right ) a^{2}}{3 b d} \] Input:
int(x^4*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x)
Output:
( - (a + b*x**3)**(1/3)*x**2 + 4*int(((a + b*x**3)**(1/3)*x**4)/(a**2 - b* *2*x**6),x)*a*b + 2*int(((a + b*x**3)**(1/3)*x)/(a**2 - b**2*x**6),x)*a**2 )/(3*b*d)