Integrand size = 21, antiderivative size = 60 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {c x \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {1}{3},1,-\frac {3}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a \sqrt {1+\frac {d x^3}{c}}} \] Output:
c*x*(d*x^3+c)^(1/2)*AppellF1(1/3,1,-3/2,4/3,-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. \(351\) vs. \(2(60)=120\).
Time = 10.35 (sec) , antiderivative size = 351, normalized size of antiderivative = 5.85 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {x \left (\frac {d (8 b c-5 a d) x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{a}+\frac {8 \left (-4 a c \left (2 a d^2 x^3+b \left (5 c^2+2 c d x^3+2 d^2 x^6\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 d x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}{\left (a+b x^3\right ) \left (-8 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+3 x^3 \left (2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}\right )}{20 b \sqrt {c+d x^3}} \] Input:
Integrate[(c + d*x^3)^(3/2)/(a + b*x^3),x]
Output:
(x*((d*(8*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*(-4*a*c*(2*a*d^2*x^3 + b*(5*c^2 + 2*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)] + 3*d*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)]))) ))/(20*b*Sqrt[c + d*x^3])
Time = 0.32 (sec) , antiderivative size = 60, 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.
\(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {c \sqrt {c+d x^3} \int \frac {\left (\frac {d x^3}{c}+1\right )^{3/2}}{b x^3+a}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {c x \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {1}{3},1,-\frac {3}{2},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(c + d*x^3)^(3/2)/(a + b*x^3),x]
Output:
(c*x*Sqrt[c + d*x^3]*AppellF1[1/3, 1, -3/2, 4/3, -((b*x^3)/a), -((d*x^3)/c )])/(a*Sqrt[1 + (d*x^3)/c])
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 = 2.15 (sec) , antiderivative size = 767, normalized size of antiderivative = 12.78
method | result | size |
risch | \(\text {Expression too large to display}\) | \(767\) |
default | \(\text {Expression too large to display}\) | \(776\) |
elliptic | \(\text {Expression too large to display}\) | \(776\) |
Input:
int((d*x^3+c)^(3/2)/(b*x^3+a),x,method=_RETURNVERBOSE)
Output:
2/5*d/b*x*(d*x^3+c)^(1/2)-1/5/b*(-2/3*I*(5*a*d-8*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)*Ellipt icF(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))+5/3*I*(a^2*d^2-2*a*b*c*d +b^2*c^2)/b/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))*Ell ipticPi(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. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\text {Timed out} \] Input:
integrate((d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\int \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{a + b x^{3}}\, dx \] Input:
integrate((d*x**3+c)**(3/2)/(b*x**3+a),x)
Output:
Integral((c + d*x**3)**(3/2)/(a + b*x**3), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{b x^{3} + a} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)/(b*x^3 + a), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{b x^{3} + a} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/(b*x^3+a),x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)/(b*x^3 + a), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{3/2}}{b\,x^3+a} \,d x \] Input:
int((c + d*x^3)^(3/2)/(a + b*x^3),x)
Output:
int((c + d*x^3)^(3/2)/(a + b*x^3), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 \sqrt {d \,x^{3}+c}\, d x -2 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a c d +5 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b \,c^{2}-5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,d^{2}+8 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b c d}{5 b} \] Input:
int((d*x^3+c)^(3/2)/(b*x^3+a),x)
Output:
(2*sqrt(c + d*x**3)*d*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*d + 5*int(sqrt(c + d*x**3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b*c**2 - 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**2 + 8*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*d)/(5*b)