Integrand size = 21, antiderivative size = 324 \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\frac {b (3 b c+4 a d) x}{12 a c (b c-a d)^2 \left (a+b x^3\right )^{4/3}}+\frac {b \left (9 b^2 c^2-33 a b c d-4 a^2 d^2\right ) x}{12 a^2 c (b c-a d)^3 \sqrt [3]{a+b x^3}}-\frac {d x}{3 c (b c-a d) \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}+\frac {d^2 (9 b c-2 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^{5/3} (b c-a d)^{10/3}}+\frac {d^2 (9 b c-2 a d) \log \left (c+d x^3\right )}{18 c^{5/3} (b c-a d)^{10/3}}-\frac {d^2 (9 b c-2 a d) \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{6 c^{5/3} (b c-a d)^{10/3}} \] Output:
1/12*b*(4*a*d+3*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^3+a)^(4/3)+1/12*b*(-4*a^2*d^2 -33*a*b*c*d+9*b^2*c^2)*x/a^2/c/(-a*d+b*c)^3/(b*x^3+a)^(1/3)-1/3*d*x/c/(-a* d+b*c)/(b*x^3+a)^(4/3)/(d*x^3+c)+1/9*d^2*(-2*a*d+9*b*c)*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^(5/3)/(-a*d+b *c)^(10/3)+1/18*d^2*(-2*a*d+9*b*c)*ln(d*x^3+c)/c^(5/3)/(-a*d+b*c)^(10/3)-1 /6*d^2*(-2*a*d+9*b*c)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(5/ 3)/(-a*d+b*c)^(10/3)
Result contains complex when optimal does not.
Time = 8.86 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\frac {\frac {3 c^{2/3} x \left (4 a^4 d^3+8 a^3 b d^3 x^3-9 b^4 c^2 x^3 \left (c+d x^3\right )+4 a^2 b^2 d \left (9 c^2+9 c d x^3+d^2 x^6\right )+3 a b^3 c \left (-4 c^2+7 c d x^3+11 d^2 x^6\right )\right )}{a^2 (-b c+a d)^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}+\frac {2 i \left (3 i+\sqrt {3}\right ) d^2 (-9 b c+2 a d) \text {arctanh}\left (\frac {i+\frac {\left (-i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d} x}}{\sqrt {3}}\right )}{(b c-a d)^{10/3}}+\frac {2 \left (1+i \sqrt {3}\right ) d^2 (9 b c-2 a d) \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{10/3}}+\frac {\left (1+i \sqrt {3}\right ) d^2 (-9 b c+2 a d) \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{10/3}}}{36 c^{5/3}} \] Input:
Integrate[1/((a + b*x^3)^(7/3)*(c + d*x^3)^2),x]
Output:
((3*c^(2/3)*x*(4*a^4*d^3 + 8*a^3*b*d^3*x^3 - 9*b^4*c^2*x^3*(c + d*x^3) + 4 *a^2*b^2*d*(9*c^2 + 9*c*d*x^3 + d^2*x^6) + 3*a*b^3*c*(-4*c^2 + 7*c*d*x^3 + 11*d^2*x^6)))/(a^2*(-(b*c) + a*d)^3*(a + b*x^3)^(4/3)*(c + d*x^3)) + ((2* I)*(3*I + Sqrt[3])*d^2*(-9*b*c + 2*a*d)*ArcTanh[(I + ((-I + Sqrt[3])*c^(1/ 3)*(a + b*x^3)^(1/3))/((b*c - a*d)^(1/3)*x))/Sqrt[3]])/(b*c - a*d)^(10/3) + (2*(1 + I*Sqrt[3])*d^2*(9*b*c - 2*a*d)*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)^(10/3) + ((1 + I*Sqrt[3 ])*d^2*(-9*b*c + 2*a*d)*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)^(10/3))/(36*c^(5/3))
Time = 0.75 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {931, 1024, 25, 1024, 27, 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.
\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx\) |
\(\Big \downarrow \) 931 |
\(\displaystyle \frac {\int \frac {-6 b d x^3+3 b c-2 a d}{\left (b x^3+a\right )^{7/3} \left (d x^3+c\right )}dx}{3 c (b c-a d)}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}\) |
\(\Big \downarrow \) 1024 |
\(\displaystyle \frac {\frac {b x (4 a d+3 b c)}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}-\frac {\int -\frac {3 b d (3 b c+4 a d) x^3+9 b^2 c^2+8 a^2 d^2-24 a b c d}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx}{4 a (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {3 b d (3 b c+4 a d) x^3+9 b^2 c^2+8 a^2 d^2-24 a b c d}{\left (b x^3+a\right )^{4/3} \left (d x^3+c\right )}dx}{4 a (b c-a d)}+\frac {b x (4 a d+3 b c)}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}\) |
\(\Big \downarrow \) 1024 |
\(\displaystyle \frac {\frac {\frac {b x \left (-4 a^2 d^2-33 a b c d+9 b^2 c^2\right )}{a \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\int -\frac {4 a^2 d^2 (9 b c-2 a d)}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{a (b c-a d)}}{4 a (b c-a d)}+\frac {b x (4 a d+3 b c)}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {4 a d^2 (9 b c-2 a d) \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{b c-a d}+\frac {b x \left (-4 a^2 d^2-33 a b c d+9 b^2 c^2\right )}{a \sqrt [3]{a+b x^3} (b c-a d)}}{4 a (b c-a d)}+\frac {b x (4 a d+3 b c)}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}\) |
\(\Big \downarrow \) 901 |
\(\displaystyle \frac {\frac {\frac {b x \left (-4 a^2 d^2-33 a b c d+9 b^2 c^2\right )}{a \sqrt [3]{a+b x^3} (b c-a d)}+\frac {4 a d^2 (9 b c-2 a d) \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{b c-a d}}{4 a (b c-a d)}+\frac {b x (4 a d+3 b c)}{4 a \left (a+b x^3\right )^{4/3} (b c-a d)}}{3 c (b c-a d)}-\frac {d x}{3 c \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) (b c-a d)}\) |
Input:
Int[1/((a + b*x^3)^(7/3)*(c + d*x^3)^2),x]
Output:
-1/3*(d*x)/(c*(b*c - a*d)*(a + b*x^3)^(4/3)*(c + d*x^3)) + ((b*(3*b*c + 4* a*d)*x)/(4*a*(b*c - a*d)*(a + b*x^3)^(4/3)) + ((b*(9*b^2*c^2 - 33*a*b*c*d - 4*a^2*d^2)*x)/(a*(b*c - a*d)*(a + b*x^3)^(1/3)) + (4*a*d^2*(9*b*c - 2*a* d)*(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))))/(b*c - a*d))/(4*a*(b*c - a*d)))/(3*c*(b*c - a* d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Simp[1/(a*n*(p + 1)*(b*c - a*d)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, n, p, q, x]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f _.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*( p + 1)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b *c - a*d)*(p + 1) + d*(b*e - a*f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; Fr eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Time = 1.51 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.01
method | result | size |
pseudoelliptic | \(-\frac {-3 c x \left (a^{4} d^{3}+2 a^{3} b \,d^{3} x^{3}+9 d \left (\frac {1}{9} d^{2} x^{6}+c d \,x^{3}+c^{2}\right ) b^{2} a^{2}-3 c \left (d \,x^{3}+c \right ) \left (-\frac {11 d \,x^{3}}{4}+c \right ) b^{3} a -\frac {9 b^{4} c^{2} x^{3} \left (d \,x^{3}+c \right )}{4}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}-\frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} a^{2} d^{2} \left (d \,x^{3}+c \right ) \left (2 a d -9 b c \right ) \left (2 \arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{2}}{9 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {4}{3}} c^{2} \left (d \,x^{3}+c \right ) \left (a d -b c \right )^{3} a^{2}}\) | \(326\) |
Input:
int(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/9/((a*d-b*c)/c)^(1/3)*(-3*c*x*(a^4*d^3+2*a^3*b*d^3*x^3+9*d*(1/9*d^2*x^6 +c*d*x^3+c^2)*b^2*a^2-3*c*(d*x^3+c)*(-11/4*d*x^3+c)*b^3*a-9/4*b^4*c^2*x^3* (d*x^3+c))*((a*d-b*c)/c)^(1/3)-1/2*(b*x^3+a)^(4/3)*a^2*d^2*(d*x^3+c)*(2*a* d-9*b*c)*(2*arctan(1/3*3^(1/2)*(-2/((a*d-b*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/ x)*3^(1/2)+2*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)-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))) /(b*x^3+a)^(4/3)/c^2/(d*x^3+c)/(a*d-b*c)^3/a^2
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {7}{3}} \left (c + d x^{3}\right )^{2}}\, dx \] Input:
integrate(1/(b*x**3+a)**(7/3)/(d*x**3+c)**2,x)
Output:
Integral(1/((a + b*x**3)**(7/3)*(c + d*x**3)**2), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {7}{3}} {\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^(7/3)*(d*x^3 + c)^2), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {7}{3}} {\left (d x^{3} + c\right )}^{2}} \,d x } \] Input:
integrate(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(7/3)*(d*x^3 + c)^2), x)
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{7/3}\,{\left (d\,x^3+c\right )}^2} \,d x \] Input:
int(1/((a + b*x^3)^(7/3)*(c + d*x^3)^2),x)
Output:
int(1/((a + b*x^3)^(7/3)*(c + d*x^3)^2), x)
\[ \int \frac {1}{\left (a+b x^3\right )^{7/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} c^{2}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} c d \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d^{2} x^{6}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,c^{2} x^{3}+4 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c d \,x^{6}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,d^{2} x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c^{2} x^{6}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c d \,x^{9}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} d^{2} x^{12}}d x \] Input:
int(1/(b*x^3+a)^(7/3)/(d*x^3+c)^2,x)
Output:
int(1/((a + b*x**3)**(1/3)*a**2*c**2 + 2*(a + b*x**3)**(1/3)*a**2*c*d*x**3 + (a + b*x**3)**(1/3)*a**2*d**2*x**6 + 2*(a + b*x**3)**(1/3)*a*b*c**2*x** 3 + 4*(a + b*x**3)**(1/3)*a*b*c*d*x**6 + 2*(a + b*x**3)**(1/3)*a*b*d**2*x* *9 + (a + b*x**3)**(1/3)*b**2*c**2*x**6 + 2*(a + b*x**3)**(1/3)*b**2*c*d*x **9 + (a + b*x**3)**(1/3)*b**2*d**2*x**12),x)