Integrand size = 24, antiderivative size = 64 \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {c+d x^3}} \] Output:
-1/2*(1+d*x^3/c)^(1/2)*AppellF1(-2/3,1,1/2,1/3,-b*x^3/a,-d*x^3/c)/a/x^2/(d *x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(339\) vs. \(2(64)=128\).
Time = 10.22 (sec) , antiderivative size = 339, normalized size of antiderivative = 5.30 \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-b d x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+\frac {8 a \left (-4 a c \left (2 a c+6 b c x^3+3 a d x^3+2 b d x^6\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}}{16 a^2 c x^2 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^3*(a + b*x^3)*Sqrt[c + d*x^3]),x]
Output:
(-(b*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)]) + (8*a*(-4*a*c*(2*a*c + 6*b*c*x^3 + 3*a*d*x^3 + 2*b*d*x^6)*Ap pellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 3*x^3*(a + b*x^3)*( 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)]))))/(16*a^2*c*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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^3 \left (b x^3+a\right ) \sqrt {\frac {d x^3}{c}+1}}dx}{\sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a x^2 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^3*(a + b*x^3)*Sqrt[c + d*x^3]),x]
Output:
-1/2*(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 1, 1/2, 1/3, -((b*x^3)/a), -((d*x ^3)/c)])/(a*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 = 2.02 (sec) , antiderivative size = 738, normalized size of antiderivative = 11.53
method | result | size |
default | \(\text {Expression too large to display}\) | \(738\) |
elliptic | \(\text {Expression too large to display}\) | \(739\) |
risch | \(\text {Expression too large to display}\) | \(740\) |
Input:
int(1/x^3/(b*x^3+a)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/a*(-1/2/c/x^2*(d*x^3+c)^(1/2)+1/6*I/c*3^(1/2)*(-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)))+1/3*I*b/a/d^2*2^(1/2)*sum(1/(a*d-b*c)/_ 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. \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\text {Timed out} \] Input:
integrate(1/x^3/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{3}\right ) \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/x**3/(b*x**3+a)/(d*x**3+c)**(1/2),x)
Output:
Integral(1/(x**3*(a + b*x**3)*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} \sqrt {d x^{3} + c} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a)*sqrt(d*x^3 + c)*x^3), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} \sqrt {d x^{3} + c} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^3+a)/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)*sqrt(d*x^3 + c)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^3\,\left (b\,x^3+a\right )\,\sqrt {d\,x^3+c}} \,d x \] Input:
int(1/(x^3*(a + b*x^3)*(c + d*x^3)^(1/2)),x)
Output:
int(1/(x^3*(a + b*x^3)*(c + d*x^3)^(1/2)), x)
\[ \int \frac {1}{x^3 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{b d \,x^{9}+a d \,x^{6}+b c \,x^{6}+a c \,x^{3}}d x \] Input:
int(1/x^3/(b*x^3+a)/(d*x^3+c)^(1/2),x)
Output:
int(sqrt(c + d*x**3)/(a*c*x**3 + a*d*x**6 + b*c*x**6 + b*d*x**9),x)