Integrand size = 24, antiderivative size = 64 \[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},1,-\frac {1}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {1+\frac {d x^3}{c}}} \] Output:
1/4*x^4*(d*x^3+c)^(1/2)*AppellF1(4/3,1,-1/2,7/3,-b*x^3/a,-d*x^3/c)/a/(1+d* x^3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(64)=128\).
Time = 7.29 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.77 \[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {x \left (\frac {(3 b c-5 a d) x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}+8 \left (c+d x^3+\frac {8 a^2 c^2 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}\right )\right )}{20 b \sqrt {c+d x^3}} \] Input:
Integrate[(x^3*Sqrt[c + d*x^3])/(a + b*x^3),x]
Output:
(x*(((3*b*c - 5*a*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^2*c^2*AppellF1[1/3, 1/2 , 1, 4/3, -((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)]))))))/(20*b*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.
\(\displaystyle \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {c+d x^3} \int \frac {x^3 \sqrt {\frac {d x^3}{c}+1}}{b x^3+a}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},1,-\frac {1}{2},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(x^3*Sqrt[c + d*x^3])/(a + b*x^3),x]
Output:
(x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 1, -1/2, 7/3, -((b*x^3)/a), -((d*x^3)/c )])/(4*a*Sqrt[1 + (d*x^3)/c])
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 = 2.26 (sec) , antiderivative size = 741, normalized size of antiderivative = 11.58
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(741\) |
risch | \(\text {Expression too large to display}\) | \(757\) |
default | \(\text {Expression too large to display}\) | \(1012\) |
Input:
int(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
2/5*x/b*(d*x^3+c)^(1/2)-2/3*I*(-(a*d-b*c)/b^2-2/5*c/b)*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)*Ellip ticF(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*a/b^2/d^2*2^(1/2) *sum(1/_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. \[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\text {Timed out} \] Input:
integrate(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\int \frac {x^{3} \sqrt {c + d x^{3}}}{a + b x^{3}}\, dx \] Input:
integrate(x**3*(d*x**3+c)**(1/2)/(b*x**3+a),x)
Output:
Integral(x**3*sqrt(c + d*x**3)/(a + b*x**3), x)
\[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{3}}{b x^{3} + a} \,d x } \] Input:
integrate(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="maxima")
Output:
integrate(sqrt(d*x^3 + c)*x^3/(b*x^3 + a), x)
\[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{3}}{b x^{3} + a} \,d x } \] Input:
integrate(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x, algorithm="giac")
Output:
integrate(sqrt(d*x^3 + c)*x^3/(b*x^3 + a), x)
Timed out. \[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\int \frac {x^3\,\sqrt {d\,x^3+c}}{b\,x^3+a} \,d x \] Input:
int((x^3*(c + d*x^3)^(1/2))/(a + b*x^3),x)
Output:
int((x^3*(c + d*x^3)^(1/2))/(a + b*x^3), x)
\[ \int \frac {x^3 \sqrt {c+d x^3}}{a+b x^3} \, dx=\frac {2 \sqrt {d \,x^{3}+c}\, x -2 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a c -5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a d +3 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b c}{5 b} \] Input:
int(x^3*(d*x^3+c)^(1/2)/(b*x^3+a),x)
Output:
(2*sqrt(c + d*x**3)*x - 2*int(sqrt(c + d*x**3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*c - 5*int((sqrt(c + d*x**3)*x**3)/(a*c + a*d*x**3 + b*c*x **3 + b*d*x**6),x)*a*d + 3*int((sqrt(c + d*x**3)*x**3)/(a*c + a*d*x**3 + b *c*x**3 + b*d*x**6),x)*b*c)/(5*b)