Integrand size = 21, antiderivative size = 217 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\frac {x \left (a+b x^3\right )^{5/3}}{6 c \left (c+d x^3\right )^2}+\frac {5 a x \left (a+b x^3\right )^{2/3}}{18 c^2 \left (c+d x^3\right )}+\frac {5 a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} c^{8/3} \sqrt [3]{b c-a d}}+\frac {5 a^2 \log \left (c+d x^3\right )}{54 c^{8/3} \sqrt [3]{b c-a d}}-\frac {5 a^2 \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} \sqrt [3]{b c-a d}} \] Output:
1/6*x*(b*x^3+a)^(5/3)/c/(d*x^3+c)^2+5/18*a*x*(b*x^3+a)^(2/3)/c^2/(d*x^3+c) +5/27*a^2*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1 /2))*3^(1/2)/c^(8/3)/(-a*d+b*c)^(1/3)+5/54*a^2*ln(d*x^3+c)/c^(8/3)/(-a*d+b *c)^(1/3)-5/18*a^2*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(8/3)/ (-a*d+b*c)^(1/3)
Result contains complex when optimal does not.
Time = 4.76 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\frac {\frac {6 c^{2/3} \left (a+b x^3\right )^{2/3} \left (8 a c x+3 b c x^4+5 a d x^4\right )}{\left (c+d x^3\right )^2}-\frac {10 \sqrt {-6+6 i \sqrt {3}} a^2 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt [3]{b c-a d}}+\frac {10 \left (1+i \sqrt {3}\right ) a^2 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{\sqrt [3]{b c-a d}}-\frac {5 i \left (-i+\sqrt {3}\right ) a^2 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{\sqrt [3]{b c-a d}}}{108 c^{8/3}} \] Input:
Integrate[(a + b*x^3)^(5/3)/(c + d*x^3)^3,x]
Output:
((6*c^(2/3)*(a + b*x^3)^(2/3)*(8*a*c*x + 3*b*c*x^4 + 5*a*d*x^4))/(c + d*x^ 3)^2 - (10*Sqrt[-6 + (6*I)*Sqrt[3]]*a^2*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sq rt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))])/( b*c - a*d)^(1/3) + (10*(1 + I*Sqrt[3])*a^2*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(1/3) - ((5*I)*(-I + Sqrt[3])*a^2*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 )])/(b*c - a*d)^(1/3))/(108*c^(8/3))
Time = 0.48 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {903, 903, 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.
| \(\displaystyle \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx\) |
| \(\Big \downarrow \) 903 |
| \(\displaystyle \frac {5 a \int \frac {\left (b x^3+a\right )^{2/3}}{\left (d x^3+c\right )^2}dx}{6 c}+\frac {x \left (a+b x^3\right )^{5/3}}{6 c \left (c+d x^3\right )^2}\) |
| \(\Big \downarrow \) 903 |
| \(\displaystyle \frac {5 a \left (\frac {2 a \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{3 c}+\frac {x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right )}\right )}{6 c}+\frac {x \left (a+b x^3\right )^{5/3}}{6 c \left (c+d x^3\right )^2}\) |
| \(\Big \downarrow \) 901 |
| \(\displaystyle \frac {5 a \left (\frac {2 a \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{3 c}+\frac {x \left (a+b x^3\right )^{2/3}}{3 c \left (c+d x^3\right )}\right )}{6 c}+\frac {x \left (a+b x^3\right )^{5/3}}{6 c \left (c+d x^3\right )^2}\) |
Input:
Int[(a + b*x^3)^(5/3)/(c + d*x^3)^3,x]
Output:
(x*(a + b*x^3)^(5/3))/(6*c*(c + d*x^3)^2) + (5*a*((x*(a + b*x^3)^(2/3))/(3 *c*(c + d*x^3)) + (2*a*(ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(1/3)) + Log[c + d*x^ 3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/(3*c)))/(6*c)
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]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*n*(p + 1))), x] - Simp[ c*(q/(a*(p + 1))) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]
Time = 1.31 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.12
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {5 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2} \left (d \,x^{3}+c \right )^{2}}{54}+\frac {5 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2} \left (d \,x^{3}+c \right )^{2}}{27}+\frac {4 \left (\frac {\left (5 a d +3 b c \right ) x^{3}}{8}+a c \right ) c \left (b \,x^{3}+a \right )^{\frac {2}{3}} x \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}{9}+\frac {5 a^{2} \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \left (d \,x^{3}+c \right )^{2} \sqrt {3}}{27}}{c^{3} \left (d \,x^{3}+c \right )^{2} \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}\) | \(244\) |
Input:
int((b*x^3+a)^(5/3)/(d*x^3+c)^3,x,method=_RETURNVERBOSE)
Output:
5/27/((a*d-b*c)/c)^(1/3)*(-1/2*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)*a^2*(d*x^3+c)^2+ln((((a*d-b*c )/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)*a^2*(d*x^3+c)^2+12/5*(1/8*(5*a*d+3*b*c)*x ^3+a*c)*c*(b*x^3+a)^(2/3)*x*((a*d-b*c)/c)^(1/3)+a^2*arctan(1/3*3^(1/2)*(-2 /((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/x)*(d*x^3+c)^2*3^(1/2))/c^3/(d*x^3 +c)^2
Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(5/3)/(d*x^3+c)^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\text {Timed out} \] Input:
integrate((b*x**3+a)**(5/3)/(d*x**3+c)**3,x)
Output:
Timed out
\[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{{\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:
integrate((b*x^3+a)^(5/3)/(d*x^3+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^3 + a)^(5/3)/(d*x^3 + c)^3, x)
\[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}}}{{\left (d x^{3} + c\right )}^{3}} \,d x } \] Input:
integrate((b*x^3+a)^(5/3)/(d*x^3+c)^3,x, algorithm="giac")
Output:
integrate((b*x^3 + a)^(5/3)/(d*x^3 + c)^3, x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{5/3}}{{\left (d\,x^3+c\right )}^3} \,d x \] Input:
int((a + b*x^3)^(5/3)/(c + d*x^3)^3,x)
Output:
int((a + b*x^3)^(5/3)/(c + d*x^3)^3, x)
\[ \int \frac {\left (a+b x^3\right )^{5/3}}{\left (c+d x^3\right )^3} \, dx=\text {too large to display} \] Input:
int((b*x^3+a)^(5/3)/(d*x^3+c)^3,x)
Output:
(2*(a + b*x**3)**(2/3)*a**2*d*x - 6*(a + b*x**3)**(2/3)*a*b*c*x - 3*(a + b *x**3)**(2/3)*a*b*d*x**4 - (a + b*x**3)**(2/3)*b**2*c*x**4 + 10*int(((a + b*x**3)**(2/3)*x**3)/(a**2*c**3*d + 3*a**2*c**2*d**2*x**3 + 3*a**2*c*d**3* x**6 + a**2*d**4*x**9 - 3*a*b*c**4 - 8*a*b*c**3*d*x**3 - 6*a*b*c**2*d**2*x **6 + a*b*d**4*x**12 - 3*b**2*c**4*x**3 - 9*b**2*c**3*d*x**6 - 9*b**2*c**2 *d**2*x**9 - 3*b**2*c*d**3*x**12),x)*a**4*c**2*d**3 + 20*int(((a + b*x**3) **(2/3)*x**3)/(a**2*c**3*d + 3*a**2*c**2*d**2*x**3 + 3*a**2*c*d**3*x**6 + a**2*d**4*x**9 - 3*a*b*c**4 - 8*a*b*c**3*d*x**3 - 6*a*b*c**2*d**2*x**6 + a *b*d**4*x**12 - 3*b**2*c**4*x**3 - 9*b**2*c**3*d*x**6 - 9*b**2*c**2*d**2*x **9 - 3*b**2*c*d**3*x**12),x)*a**4*c*d**4*x**3 + 10*int(((a + b*x**3)**(2/ 3)*x**3)/(a**2*c**3*d + 3*a**2*c**2*d**2*x**3 + 3*a**2*c*d**3*x**6 + a**2* d**4*x**9 - 3*a*b*c**4 - 8*a*b*c**3*d*x**3 - 6*a*b*c**2*d**2*x**6 + a*b*d* *4*x**12 - 3*b**2*c**4*x**3 - 9*b**2*c**3*d*x**6 - 9*b**2*c**2*d**2*x**9 - 3*b**2*c*d**3*x**12),x)*a**4*d**5*x**6 - 50*int(((a + b*x**3)**(2/3)*x**3 )/(a**2*c**3*d + 3*a**2*c**2*d**2*x**3 + 3*a**2*c*d**3*x**6 + a**2*d**4*x* *9 - 3*a*b*c**4 - 8*a*b*c**3*d*x**3 - 6*a*b*c**2*d**2*x**6 + a*b*d**4*x**1 2 - 3*b**2*c**4*x**3 - 9*b**2*c**3*d*x**6 - 9*b**2*c**2*d**2*x**9 - 3*b**2 *c*d**3*x**12),x)*a**3*b*c**3*d**2 - 100*int(((a + b*x**3)**(2/3)*x**3)/(a **2*c**3*d + 3*a**2*c**2*d**2*x**3 + 3*a**2*c*d**3*x**6 + a**2*d**4*x**9 - 3*a*b*c**4 - 8*a*b*c**3*d*x**3 - 6*a*b*c**2*d**2*x**6 + a*b*d**4*x**12...