Integrand size = 22, antiderivative size = 67 \[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {x^2 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},2,\frac {3}{2},\frac {5}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 c \sqrt {c+d x^3}} \] Output:
1/2*x^2*(1+d*x^3/c)^(1/2)*AppellF1(2/3,2,3/2,5/3,-b*x^3/a,-d*x^3/c)/a^2/c/ (d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(216\) vs. \(2(67)=134\).
Time = 10.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.22 \[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {x^2 \left (-10 a \left (2 a^2 d^2+2 a b d^2 x^3+b^2 c \left (c+d x^3\right )\right )+5 \left (-b^2 c^2+6 a b c d+a^2 d^2\right ) \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+b d (b c+2 a d) x^3 \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{30 a^2 c (b c-a d)^2 \left (a+b x^3\right ) \sqrt {c+d x^3}} \] Input:
Integrate[x/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
-1/30*(x^2*(-10*a*(2*a^2*d^2 + 2*a*b*d^2*x^3 + b^2*c*(c + d*x^3)) + 5*(-(b ^2*c^2) + 6*a*b*c*d + a^2*d^2)*(a + b*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[2/ 3, 1/2, 1, 5/3, -((d*x^3)/c), -((b*x^3)/a)] + b*d*(b*c + 2*a*d)*x^3*(a + b *x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^ 3)/a)]))/(a^2*c*(b*c - a*d)^2*(a + b*x^3)*Sqrt[c + d*x^3])
Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1013, 1012}
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.
\(\displaystyle \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {x}{\left (b x^3+a\right )^2 \left (\frac {d x^3}{c}+1\right )^{3/2}}dx}{c \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^2 \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {2}{3},2,\frac {3}{2},\frac {5}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 c \sqrt {c+d x^3}}\) |
Input:
Int[x/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(x^2*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 2, 3/2, 5/3, -((b*x^3)/a), -((d*x^3 )/c)])/(2*a^2*c*Sqrt[c + d*x^3])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
Result contains higher order function than in optimal. Order 9 vs. order 6.
Time = 1.24 (sec) , antiderivative size = 986, normalized size of antiderivative = 14.72
method | result | size |
default | \(\text {Expression too large to display}\) | \(986\) |
elliptic | \(\text {Expression too large to display}\) | \(986\) |
Input:
int(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/3*d^2*x^2/c/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)+1/3*b^2/(a*d-b*c)^2/a*x^2*(d *x^3+c)^(1/2)/(b*x^3+a)-2/3*I*(-1/3*d^2/(a*d-b*c)^2/c-1/6*d*b/(a*d-b*c)^2/ a)*3^(1/2)/d*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c *d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2 /d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c* d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2) /(d*x^3+c)^(1/2)*((-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*E llipticE(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^( 1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d* (-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2))+1/d*(-c*d^2)^(1/3)* EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^ (1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d *(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)))+1/18*I/a/d^2*b*2^ (1/2)*sum((11*a*d-2*b*c)/(a*d-b*c)^3/_alpha*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1 /d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x -1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*( -1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3 ))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c* d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*Elliptic Pi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3...
Timed out. \[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x}{\left (a + b x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x/(b*x**3+a)**2/(d*x**3+c)**(3/2),x)
Output:
Integral(x/((a + b*x**3)**2*(c + d*x**3)**(3/2)), x)
\[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x}{{\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
integrate(x/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)), x)
\[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x}{{\left (b x^{3} + a\right )}^{2} {\left (d x^{3} + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(x/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)), x)
Timed out. \[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x}{{\left (b\,x^3+a\right )}^2\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \] Input:
int(x/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x)
Output:
int(x/((a + b*x^3)^2*(c + d*x^3)^(3/2)), x)
\[ \int \frac {x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x}{b^{2} d^{2} x^{12}+2 a b \,d^{2} x^{9}+2 b^{2} c d \,x^{9}+a^{2} d^{2} x^{6}+4 a b c d \,x^{6}+b^{2} c^{2} x^{6}+2 a^{2} c d \,x^{3}+2 a b \,c^{2} x^{3}+a^{2} c^{2}}d x \] Input:
int(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x)
Output:
int((sqrt(c + d*x**3)*x)/(a**2*c**2 + 2*a**2*c*d*x**3 + a**2*d**2*x**6 + 2 *a*b*c**2*x**3 + 4*a*b*c*d*x**6 + 2*a*b*d**2*x**9 + b**2*c**2*x**6 + 2*b** 2*c*d*x**9 + b**2*d**2*x**12),x)