Integrand size = 26, antiderivative size = 66 \[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{8 c x^2 \sqrt {c+d x^3}} \] Output:
-1/8*(1+d*x^3/c)^(1/2)*AppellF1(-2/3,1/2,1,1/3,-d*x^3/c,-1/4*d*x^3/c)/c/x^ 2/(d*x^3+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(243\) vs. \(2(66)=132\).
Time = 11.20 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.68 \[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {-\frac {32 \left (c+d x^3\right )}{c^2}-\frac {d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{c^3}+\frac {2048 d x^3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{\left (4 c+d x^3\right ) \left (-16 c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+3 d x^3 \left (\operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+2 \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )\right )\right )}}{256 x^2 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^3*Sqrt[c + d*x^3]*(4*c + d*x^3)),x]
Output:
((-32*(c + d*x^3))/c^2 - (d^2*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1 , 7/3, -((d*x^3)/c), -1/4*(d*x^3)/c])/c^3 + (2048*d*x^3*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -1/4*(d*x^3)/c])/((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]))))/(256*x^2*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.
\(\displaystyle \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {\frac {d x^3}{c}+1} \int \frac {1}{x^3 \left (d x^3+4 c\right ) \sqrt {\frac {d x^3}{c}+1}}dx}{\sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt {\frac {d x^3}{c}+1} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{8 c x^2 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^3*Sqrt[c + d*x^3]*(4*c + d*x^3)),x]
Output:
-1/8*(Sqrt[1 + (d*x^3)/c]*AppellF1[-2/3, 1, 1/2, 1/3, -1/4*(d*x^3)/c, -((d *x^3)/c)])/(c*x^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 = 1.84 (sec) , antiderivative size = 716, normalized size of antiderivative = 10.85
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(716\) |
risch | \(\text {Expression too large to display}\) | \(720\) |
default | \(\text {Expression too large to display}\) | \(722\) |
Input:
int(1/x^3/(d*x^3+c)^(1/2)/(d*x^3+4*c),x,method=_RETURNVERBOSE)
Output:
-1/8*(d*x^3+c)^(1/2)/c^2/x^2+1/24*I/c^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))+1/36*I/c^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)*_al pha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alph a*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. 2381 vs. \(2 (52) = 104\).
Time = 1.84 (sec) , antiderivative size = 2381, normalized size of antiderivative = 36.08 \[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/x^3/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
Output:
1/288*(2*(1/108)^(1/6)*c^2*x^2*(-d^4/c^13)^(1/6)*log((d^6*x^9 - 66*c*d^5*x ^6 - 72*c^2*d^4*x^3 - 32*c^3*d^3 - 24*(1/4)^(2/3)*(c^9*d^3*x^8 - 7*c^10*d^ 2*x^5 - 8*c^11*d*x^2)*(-d^4/c^13)^(2/3) - 48*(1/4)^(1/3)*(c^5*d^4*x^7 - c^ 6*d^3*x^4 - 2*c^7*d^2*x)*(-d^4/c^13)^(1/3) + 6*(18*(1/108)^(1/6)*c^3*d^4*x ^5*(-d^4/c^13)^(1/6) + 36*(1/108)^(5/6)*(c^11*d^2*x^7 - 16*c^12*d*x^4 - 8* c^13*x)*(-d^4/c^13)^(5/6) + sqrt(1/3)*(5*c^7*d^3*x^6 - 20*c^8*d^2*x^3 - 16 *c^9*d)*sqrt(-d^4/c^13))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2 *d*x^3 + 64*c^3)) - 2*(1/108)^(1/6)*c^2*x^2*(-d^4/c^13)^(1/6)*log((d^6*x^9 - 66*c*d^5*x^6 - 72*c^2*d^4*x^3 - 32*c^3*d^3 - 24*(1/4)^(2/3)*(c^9*d^3*x^ 8 - 7*c^10*d^2*x^5 - 8*c^11*d*x^2)*(-d^4/c^13)^(2/3) - 48*(1/4)^(1/3)*(c^5 *d^4*x^7 - c^6*d^3*x^4 - 2*c^7*d^2*x)*(-d^4/c^13)^(1/3) - 6*(18*(1/108)^(1 /6)*c^3*d^4*x^5*(-d^4/c^13)^(1/6) + 36*(1/108)^(5/6)*(c^11*d^2*x^7 - 16*c^ 12*d*x^4 - 8*c^13*x)*(-d^4/c^13)^(5/6) + sqrt(1/3)*(5*c^7*d^3*x^6 - 20*c^8 *d^2*x^3 - 16*c^9*d)*sqrt(-d^4/c^13))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2 *x^6 + 48*c^2*d*x^3 + 64*c^3)) - 60*sqrt(d)*x^2*weierstrassPInverse(0, -4* c/d, x) + (1/108)^(1/6)*(sqrt(-3)*c^2*x^2 + c^2*x^2)*(-d^4/c^13)^(1/6)*log ((d^6*x^9 - 66*c*d^5*x^6 - 72*c^2*d^4*x^3 - 32*c^3*d^3 + 12*(1/4)^(2/3)*(c ^9*d^3*x^8 - 7*c^10*d^2*x^5 - 8*c^11*d*x^2 + sqrt(-3)*(c^9*d^3*x^8 - 7*c^1 0*d^2*x^5 - 8*c^11*d*x^2))*(-d^4/c^13)^(2/3) + 24*(1/4)^(1/3)*(c^5*d^4*x^7 - c^6*d^3*x^4 - 2*c^7*d^2*x - sqrt(-3)*(c^5*d^4*x^7 - c^6*d^3*x^4 - 2*...
\[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {1}{x^{3} \sqrt {c + d x^{3}} \cdot \left (4 c + d x^{3}\right )}\, dx \] Input:
integrate(1/x**3/(d*x**3+c)**(1/2)/(d*x**3+4*c),x)
Output:
Integral(1/(x**3*sqrt(c + d*x**3)*(4*c + d*x**3)), x)
\[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int { \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} x^{3}} \,d x } \] Input:
integrate(1/x^3/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
Output:
integrate(1/((d*x^3 + 4*c)*sqrt(d*x^3 + c)*x^3), x)
\[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int { \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} x^{3}} \,d x } \] Input:
integrate(1/x^3/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
Output:
integrate(1/((d*x^3 + 4*c)*sqrt(d*x^3 + c)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {1}{x^3\,\sqrt {d\,x^3+c}\,\left (d\,x^3+4\,c\right )} \,d x \] Input:
int(1/(x^3*(c + d*x^3)^(1/2)*(4*c + d*x^3)),x)
Output:
int(1/(x^3*(c + d*x^3)^(1/2)*(4*c + d*x^3)), x)
\[ \int \frac {1}{x^3 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d^{2} x^{9}+5 c d \,x^{6}+4 c^{2} x^{3}}d x \] Input:
int(1/x^3/(d*x^3+c)^(1/2)/(d*x^3+4*c),x)
Output:
int(sqrt(c + d*x**3)/(4*c**2*x**3 + 5*c*d*x**6 + d**2*x**9),x)