Integrand size = 24, antiderivative size = 64 \[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {2}{3},2,\frac {1}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 x^2 \sqrt {c+d x^3}} \] Output:
-1/2*(1+d*x^3/c)^(1/2)*AppellF1(-2/3,2,1/2,1/3,-b*x^3/a,-d*x^3/c)/a^2/x^2/ (d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(411\) vs. \(2(64)=128\).
Time = 10.78 (sec) , antiderivative size = 411, normalized size of antiderivative = 6.42 \[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {b d (5 b c-3 a d) x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {a \left (32 a c \left (-10 b^2 c x^3 \left (3 c+d x^3\right )+3 a^2 d \left (2 c+3 d x^3\right )+3 a b \left (-2 c^2+7 c d x^3+2 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+24 x^3 \left (c+d x^3\right ) \left (-3 a^2 d+5 b^2 c x^3+3 a b \left (c-d x^3\right )\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}}{48 a^3 c (-b c+a d) x^2 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^3*(a + b*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
(b*d*(5*b*c - 3*a*d)*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -( (d*x^3)/c), -((b*x^3)/a)] + (a*(32*a*c*(-10*b^2*c*x^3*(3*c + d*x^3) + 3*a^ 2*d*(2*c + 3*d*x^3) + 3*a*b*(-2*c^2 + 7*c*d*x^3 + 2*d^2*x^6))*AppellF1[1/3 , 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 24*x^3*(c + d*x^3)*(-3*a^2*d + 5*b^2*c*x^3 + 3*a*b*(c - d*x^3))*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d* x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -((b *x^3)/a)])))/((a + b*x^3)*(-8*a*c*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 3*x^3*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -(( b*x^3)/a)] + a*d*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)]))) )/(48*a^3*c*(-(b*c) + a*d)*x^2*Sqrt[c + d*x^3])
Time = 0.34 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
| \(\displaystyle \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
| \(\Big \downarrow \) 1013 |
| \(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^3 \left (b x^3+a\right )^2 \sqrt {\frac {d x^3}{c}+1}}dx}{\sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {2}{3},2,\frac {1}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 x^2 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^3*(a + b*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
-1/2*(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 2, 1/2, 1/3, -((b*x^3)/a), -((d*x ^3)/c)])/(a^2*x^2*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 = 4.64 (sec) , antiderivative size = 809, normalized size of antiderivative = 12.64
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(809\) |
| default | \(\text {Expression too large to display}\) | \(1512\) |
| risch | \(\text {Expression too large to display}\) | \(1513\) |
Input:
int(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/2/c/a^2*(d*x^3+c)^(1/2)/x^2+1/3/(a*d-b*c)/a^2*b^2*x*(d*x^3+c)^(1/2)/(b* x^3+a)-2/3*I*(-1/4*d/c/a^2+1/6*d*b/(a*d-b*c)/a^2)*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)*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^2/d^2*b*2^(1/2)*sum ((13*a*d-10*b*c)/(a*d-b*c)^2/_alpha^2*(-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))*EllipticPi(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),1/2*b/d*(2*I*(-c*d^2)^(1/3)*_alpha^2*3^(1/2)*d-I *(-c*d^2)^(2/3)*_alpha*3^(1/2)+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d )/(a*d-b*c),(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)),_alpha=RootOf(_Z^3*b+a))
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\text {Timed out} \] Input:
integrate(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/x**3/(b*x**3+a)**2/(d*x**3+c)**(1/2),x)
Output:
Integral(1/(x**3*(a + b*x**3)**2*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)^2*sqrt(d*x^3 + c)*x^3), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^2*sqrt(d*x^3 + c)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^3\,{\left (b\,x^3+a\right )}^2\,\sqrt {d\,x^3+c}} \,d x \] Input:
int(1/(x^3*(a + b*x^3)^2*(c + d*x^3)^(1/2)),x)
Output:
int(1/(x^3*(a + b*x^3)^2*(c + d*x^3)^(1/2)), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-\left (\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \right ) a^{2} d \,x^{2}-10 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \right ) a b c \,x^{2}-\left (\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \right ) a b d \,x^{5}-10 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \right ) b^{2} c \,x^{5}-7 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \right ) a b d \,x^{2}-7 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \right ) b^{2} d \,x^{5}}{4 a c \,x^{2} \left (b \,x^{3}+a \right )} \] Input:
int(1/x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x)
Output:
( - 2*sqrt(c + d*x**3) - int(sqrt(c + d*x**3)/(a**2*c + a**2*d*x**3 + 2*a* b*c*x**3 + 2*a*b*d*x**6 + b**2*c*x**6 + b**2*d*x**9),x)*a**2*d*x**2 - 10*i nt(sqrt(c + d*x**3)/(a**2*c + a**2*d*x**3 + 2*a*b*c*x**3 + 2*a*b*d*x**6 + b**2*c*x**6 + b**2*d*x**9),x)*a*b*c*x**2 - int(sqrt(c + d*x**3)/(a**2*c + a**2*d*x**3 + 2*a*b*c*x**3 + 2*a*b*d*x**6 + b**2*c*x**6 + b**2*d*x**9),x)* a*b*d*x**5 - 10*int(sqrt(c + d*x**3)/(a**2*c + a**2*d*x**3 + 2*a*b*c*x**3 + 2*a*b*d*x**6 + b**2*c*x**6 + b**2*d*x**9),x)*b**2*c*x**5 - 7*int((sqrt(c + d*x**3)*x**3)/(a**2*c + a**2*d*x**3 + 2*a*b*c*x**3 + 2*a*b*d*x**6 + b** 2*c*x**6 + b**2*d*x**9),x)*a*b*d*x**2 - 7*int((sqrt(c + d*x**3)*x**3)/(a** 2*c + a**2*d*x**3 + 2*a*b*c*x**3 + 2*a*b*d*x**6 + b**2*c*x**6 + b**2*d*x** 9),x)*b**2*d*x**5)/(4*a*c*x**2*(a + b*x**3))