Integrand size = 24, antiderivative size = 64 \[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {x^4 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},2,\frac {1}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 \sqrt {c+d x^3}} \] Output:
1/4*x^4*(1+d*x^3/c)^(1/2)*AppellF1(4/3,2,1/2,7/3,-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. \(238\) vs. \(2(64)=128\).
Time = 10.28 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.72 \[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {x \left (\frac {d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}+\frac {8 \left (c+d x^3+\frac {8 a c^2 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}\right )}{a+b x^3}\right )}{24 (-b c+a d) \sqrt {c+d x^3}} \] Input:
Integrate[x^3/((a + b*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
(x*((d*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -( (b*x^3)/a)])/a + (8*(c + d*x^3 + (8*a*c^2*AppellF1[1/3, 1/2, 1, 4/3, -((d* x^3)/c), -((b*x^3)/a)])/(-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)]))))/ (a + b*x^3)))/(24*(-(b*c) + a*d)*Sqrt[c + d*x^3])
Time = 0.35 (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.
\(\displaystyle \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {x^3}{\left (b x^3+a\right )^2 \sqrt {\frac {d x^3}{c}+1}}dx}{\sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^4 \sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (\frac {4}{3},2,\frac {1}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a^2 \sqrt {c+d x^3}}\) |
Input:
Int[x^3/((a + b*x^3)^2*Sqrt[c + d*x^3]),x]
Output:
(x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 2, 1/2, 7/3, -((b*x^3)/a), -((d*x^3 )/c)])/(4*a^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 = 2.04 (sec) , antiderivative size = 764, normalized size of antiderivative = 11.94
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(764\) |
default | \(\text {Expression too large to display}\) | \(1207\) |
Input:
int(x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3/(a*d-b*c)*x*(d*x^3+c)^(1/2)/(b*x^3+a)-1/9*I/(a*d-b*c)/b*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)*El lipticF(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/b/d^2*2^(1/2) *sum((a*d+2*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. \[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\text {Timed out} \] Input:
integrate(x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^{3}}{\left (a + b x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(x**3/(b*x**3+a)**2/(d*x**3+c)**(1/2),x)
Output:
Integral(x**3/((a + b*x**3)**2*sqrt(c + d*x**3)), x)
\[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c}} \,d x } \] Input:
integrate(x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="maxima")
Output:
integrate(x^3/((b*x^3 + a)^2*sqrt(d*x^3 + c)), x)
\[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c}} \,d x } \] Input:
integrate(x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
integrate(x^3/((b*x^3 + a)^2*sqrt(d*x^3 + c)), x)
Timed out. \[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^3}{{\left (b\,x^3+a\right )}^2\,\sqrt {d\,x^3+c}} \,d x \] Input:
int(x^3/((a + b*x^3)^2*(c + d*x^3)^(1/2)),x)
Output:
int(x^3/((a + b*x^3)^2*(c + d*x^3)^(1/2)), x)
\[ \int \frac {x^3}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \] Input:
int(x^3/(b*x^3+a)^2/(d*x^3+c)^(1/2),x)
Output:
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)