Integrand size = 26, antiderivative size = 66 \[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},1,-\frac {1}{2},\frac {7}{3},-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{16 c \sqrt {1+\frac {d x^3}{c}}} \] Output:
1/16*x^4*(d*x^3+c)^(1/2)*AppellF1(4/3,-1/2,1,7/3,-d*x^3/c,-1/4*d*x^3/c)/c/ (1+d*x^3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(236\) vs. \(2(66)=132\).
Time = 6.37 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.58 \[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {x \left (-17 x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+32 \left (\frac {c}{d}+x^3+\frac {64 c^3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{d \left (4 c+d x^3\right ) \left (-16 c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+3 d x^3 \left (\operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+2 \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )\right )\right )}\right )\right )}{80 \sqrt {c+d x^3}} \] Input:
Integrate[(x^3*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
Output:
(x*(-17*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), - 1/4*(d*x^3)/c] + 32*(c/d + x^3 + (64*c^3*AppellF1[1/3, 1/2, 1, 4/3, -((d*x ^3)/c), -1/4*(d*x^3)/c])/(d*(4*c + d*x^3)*(-16*c*AppellF1[1/3, 1/2, 1, 4/3 , -((d*x^3)/c), -1/4*(d*x^3)/c] + 3*d*x^3*(AppellF1[4/3, 1/2, 2, 7/3, -((d *x^3)/c), -1/4*(d*x^3)/c] + 2*AppellF1[4/3, 3/2, 1, 7/3, -((d*x^3)/c), -1/ 4*(d*x^3)/c]))))))/(80*Sqrt[c + d*x^3])
Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.
\(\displaystyle \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {c+d x^3} \int \frac {x^3 \sqrt {\frac {d x^3}{c}+1}}{d x^3+4 c}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^4 \sqrt {c+d x^3} \operatorname {AppellF1}\left (\frac {4}{3},1,-\frac {1}{2},\frac {7}{3},-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{16 c \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[(x^3*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
Output:
(x^4*Sqrt[c + d*x^3]*AppellF1[4/3, 1, -1/2, 7/3, -1/4*(d*x^3)/c, -((d*x^3) /c)])/(16*c*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 = 1.69 (sec) , antiderivative size = 713, normalized size of antiderivative = 10.80
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(713\) |
risch | \(\text {Expression too large to display}\) | \(718\) |
default | \(\text {Expression too large to display}\) | \(1003\) |
Input:
int(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x,method=_RETURNVERBOSE)
Output:
2/5/d*x*(d*x^3+c)^(1/2)+34/15*I*c/d^2*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))-4/3*I*c/d^4*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),1/6/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)/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*d+4*c))
Leaf count of result is larger than twice the leaf count of optimal. 2387 vs. \(2 (52) = 104\).
Time = 1.77 (sec) , antiderivative size = 2387, normalized size of antiderivative = 36.17 \[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\text {Too large to display} \] Input:
integrate(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
Output:
-1/60*(10*(16/27)^(1/6)*d^2*(-c^5/d^8)^(1/6)*log((27*(16/27)^(5/6)*(d^9*x^ 8 - 7*c*d^8*x^5 - 8*c^2*d^7*x^2)*(-c^5/d^8)^(5/6) + 96*sqrt(1/3)*(c^2*d^6* x^7 - c^3*d^5*x^4 - 2*c^4*d^4*x)*sqrt(-c^5/d^8) + 4*(2*c^4*d^2*x^7 - 18*2^ (1/3)*c^3*d^4*x^5*(-c^5/d^8)^(1/3) - 32*c^5*d*x^4 - 16*c^6*x - 2^(2/3)*(5* c*d^7*x^6 - 20*c^2*d^6*x^3 - 16*c^3*d^5)*(-c^5/d^8)^(2/3))*sqrt(d*x^3 + c) - 2*(16/27)^(1/6)*(c^3*d^4*x^9 - 66*c^4*d^3*x^6 - 72*c^5*d^2*x^3 - 32*c^6 *d)*(-c^5/d^8)^(1/6))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - 10*(16/27)^(1/6)*d^2*(-c^5/d^8)^(1/6)*log(-(27*(16/27)^(5/6)*(d^9*x^8 - 7* c*d^8*x^5 - 8*c^2*d^7*x^2)*(-c^5/d^8)^(5/6) + 96*sqrt(1/3)*(c^2*d^6*x^7 - c^3*d^5*x^4 - 2*c^4*d^4*x)*sqrt(-c^5/d^8) - 4*(2*c^4*d^2*x^7 - 18*2^(1/3)* c^3*d^4*x^5*(-c^5/d^8)^(1/3) - 32*c^5*d*x^4 - 16*c^6*x - 2^(2/3)*(5*c*d^7* x^6 - 20*c^2*d^6*x^3 - 16*c^3*d^5)*(-c^5/d^8)^(2/3))*sqrt(d*x^3 + c) - 2*( 16/27)^(1/6)*(c^3*d^4*x^9 - 66*c^4*d^3*x^6 - 72*c^5*d^2*x^3 - 32*c^6*d)*(- c^5/d^8)^(1/6))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - 24*sqr t(d*x^3 + c)*d*x + 168*c*sqrt(d)*weierstrassPInverse(0, -4*c/d, x) - 5*(16 /27)^(1/6)*(sqrt(-3)*d^2 - d^2)*(-c^5/d^8)^(1/6)*log((27*(16/27)^(5/6)*(d^ 9*x^8 - 7*c*d^8*x^5 - 8*c^2*d^7*x^2 + sqrt(-3)*(d^9*x^8 - 7*c*d^8*x^5 - 8* c^2*d^7*x^2))*(-c^5/d^8)^(5/6) - 192*sqrt(1/3)*(c^2*d^6*x^7 - c^3*d^5*x^4 - 2*c^4*d^4*x)*sqrt(-c^5/d^8) + 4*(4*c^4*d^2*x^7 - 64*c^5*d*x^4 - 32*c^6*x + 2^(2/3)*(5*c*d^7*x^6 - 20*c^2*d^6*x^3 - 16*c^3*d^5 - sqrt(-3)*(5*c*d...
\[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int \frac {x^{3} \sqrt {c + d x^{3}}}{4 c + d x^{3}}\, dx \] Input:
integrate(x**3*(d*x**3+c)**(1/2)/(d*x**3+4*c),x)
Output:
Integral(x**3*sqrt(c + d*x**3)/(4*c + d*x**3), x)
\[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{3}}{d x^{3} + 4 \, c} \,d x } \] Input:
integrate(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
Output:
integrate(sqrt(d*x^3 + c)*x^3/(d*x^3 + 4*c), x)
\[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{3}}{d x^{3} + 4 \, c} \,d x } \] Input:
integrate(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
Output:
integrate(sqrt(d*x^3 + c)*x^3/(d*x^3 + 4*c), x)
Timed out. \[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int \frac {x^3\,\sqrt {d\,x^3+c}}{d\,x^3+4\,c} \,d x \] Input:
int((x^3*(c + d*x^3)^(1/2))/(4*c + d*x^3),x)
Output:
int((x^3*(c + d*x^3)^(1/2))/(4*c + d*x^3), x)
\[ \int \frac {x^3 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {2 \sqrt {d \,x^{3}+c}\, x -8 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) c^{2}-17 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{3}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) c d}{5 d} \] Input:
int(x^3*(d*x^3+c)^(1/2)/(d*x^3+4*c),x)
Output:
(2*sqrt(c + d*x**3)*x - 8*int(sqrt(c + d*x**3)/(4*c**2 + 5*c*d*x**3 + d**2 *x**6),x)*c**2 - 17*int((sqrt(c + d*x**3)*x**3)/(4*c**2 + 5*c*d*x**3 + d** 2*x**6),x)*c*d)/(5*d)