Integrand size = 21, antiderivative size = 59 \[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {x \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {1}{3},2,\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {c+d x^3}} \] Output:
x*(1+d*x^3/c)^(1/2)*AppellF1(1/3,2,1/2,4/3,-b*x^3/a,-d*x^3/c)/a^2/(d*x^3+c )^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(392\) vs. \(2(59)=118\).
Time = 10.21 (sec) , antiderivative size = 392, normalized size of antiderivative = 6.64 \[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {-8 a c x \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right ) \left (8 a \left (3 b c-3 a d+b d x^3\right )+b d x^3 \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )+3 b x^4 \left (8 a \left (c+d x^3\right )+d x^3 \left (a+b x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{24 a^2 (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3} \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )} \] Input:
Integrate[1/((a + b*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
(-8*a*c*x*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)]*(8*a*(3*b *c - 3*a*d + b*d*x^3) + b*d*x^3*(a + b*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[4 /3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)]) + 3*b*x^4*(8*a*(c + d*x^3) + d*x^3*(a + b*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3 )/c), -((b*x^3)/a)])*(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)]))/(2 4*a^2*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*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)])))
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {937, 936}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{\left (b x^3+a\right )^2 \sqrt {\frac {d x^3}{c}+1}}dx}{\sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {1}{3},2,\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 \sqrt {c+d x^3}}\) |
Input:
Int[1/((a + b*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
(x*Sqrt[1 + (d*x^3)/c]*AppellF1[1/3, 2, 1/2, 4/3, -((b*x^3)/a), -((d*x^3)/ c)])/(a^2*Sqrt[c + d*x^3])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
Result contains higher order function than in optimal. Order 9 vs. order 6.
Time = 1.14 (sec) , antiderivative size = 769, normalized size of antiderivative = 13.03
method | result | size |
default | \(\text {Expression too large to display}\) | \(769\) |
elliptic | \(\text {Expression too large to display}\) | \(769\) |
Input:
int(1/(b*x^3+a)^2/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3*b/(a*d-b*c)/a*x*(d*x^3+c)^(1/2)/(b*x^3+a)+1/9*I/(a*d-b*c)/a*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))+1/18*I/a/d^2*2^ (1/2)*sum((-7*a*d+4*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))*Ellipt icPi(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)*_alp ha-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. \[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/(b*x**3+a)**2/(d*x**3+c)**(1/2),x)
Output:
Integral(1/((a + b*x**3)**2*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c}} \,d x } \] Input:
integrate(1/(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)
\[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c}} \,d x } \] Input:
integrate(1/(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)
Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^2\,\sqrt {d\,x^3+c}} \,d x \] Input:
int(1/((a + b*x^3)^2*(c + d*x^3)^(1/2)),x)
Output:
int(1/((a + b*x^3)^2*(c + d*x^3)^(1/2)), x)
\[ \int \frac {1}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \] Input:
int(1/(b*x^3+a)^2/(d*x^3+c)^(1/2),x)
Output:
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)