Integrand size = 26, antiderivative size = 66 \[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {2}{3},1,-\frac {1}{2},\frac {1}{3},-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{8 c x^2 \sqrt {1+\frac {d x^3}{c}}} \] Output:
-1/8*(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 /(1+d*x^3/c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(244\) vs. \(2(66)=132\).
Time = 11.16 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.70 \[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\frac {-32 c \left (c+d x^3\right )-d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+\frac {2048 c^3 d x^3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{\left (4 c+d x^3\right ) \left (16 c \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )-3 d x^3 \left (\operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},2,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+2 \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{2},1,\frac {7}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )\right )\right )}}{256 c^2 x^2 \sqrt {c+d x^3}} \] Input:
Integrate[Sqrt[c + d*x^3]/(x^3*(4*c + d*x^3)),x]
Output:
(-32*c*(c + d*x^3) - 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] + (2048*c^3*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*c^2*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.
\(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {\sqrt {c+d x^3} \int \frac {\sqrt {\frac {d x^3}{c}+1}}{x^3 \left (d x^3+4 c\right )}dx}{\sqrt {\frac {d x^3}{c}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt {c+d x^3} \operatorname {AppellF1}\left (-\frac {2}{3},1,-\frac {1}{2},\frac {1}{3},-\frac {d x^3}{4 c},-\frac {d x^3}{c}\right )}{8 c x^2 \sqrt {\frac {d x^3}{c}+1}}\) |
Input:
Int[Sqrt[c + d*x^3]/(x^3*(4*c + d*x^3)),x]
Output:
-1/8*(Sqrt[c + d*x^3]*AppellF1[-2/3, 1, -1/2, 1/3, -1/4*(d*x^3)/c, -((d*x^ 3)/c)])/(c*x^2*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 = 1.84 (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 | \(\text {Expression too large to display}\) | \(1002\) |
Input:
int((d*x^3+c)^(1/2)/x^3/(d*x^3+4*c),x,method=_RETURNVERBOSE)
Output:
-1/8/c/x^2*(d*x^3+c)^(1/2)+1/24*I/c*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))-1/12*I/d^2/c*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. 2361 vs. \(2 (52) = 104\).
Time = 1.15 (sec) , antiderivative size = 2361, normalized size of antiderivative = 35.77 \[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate((d*x^3+c)^(1/2)/x^3/(d*x^3+4*c),x, algorithm="fricas")
Output:
-1/96*(2*(1/108)^(1/6)*c*x^2*(-d^4/c^7)^(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^5*d^3*x^8 - 7*c^6*d^2*x^ 5 - 8*c^7*d*x^2)*(-d^4/c^7)^(2/3) - 48*(1/4)^(1/3)*(c^3*d^4*x^7 - c^4*d^3* x^4 - 2*c^5*d^2*x)*(-d^4/c^7)^(1/3) + 6*(18*(1/108)^(1/6)*c^2*d^4*x^5*(-d^ 4/c^7)^(1/6) + 36*(1/108)^(5/6)*(c^6*d^2*x^7 - 16*c^7*d*x^4 - 8*c^8*x)*(-d ^4/c^7)^(5/6) + sqrt(1/3)*(5*c^4*d^3*x^6 - 20*c^5*d^2*x^3 - 16*c^6*d)*sqrt (-d^4/c^7))*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*x^2*(-d^4/c^7)^(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^5*d^3*x^8 - 7*c^6*d^2*x^ 5 - 8*c^7*d*x^2)*(-d^4/c^7)^(2/3) - 48*(1/4)^(1/3)*(c^3*d^4*x^7 - c^4*d^3* x^4 - 2*c^5*d^2*x)*(-d^4/c^7)^(1/3) - 6*(18*(1/108)^(1/6)*c^2*d^4*x^5*(-d^ 4/c^7)^(1/6) + 36*(1/108)^(5/6)*(c^6*d^2*x^7 - 16*c^7*d*x^4 - 8*c^8*x)*(-d ^4/c^7)^(5/6) + sqrt(1/3)*(5*c^4*d^3*x^6 - 20*c^5*d^2*x^3 - 16*c^6*d)*sqrt (-d^4/c^7))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c ^3)) - 12*sqrt(d)*x^2*weierstrassPInverse(0, -4*c/d, x) + (1/108)^(1/6)*(s qrt(-3)*c*x^2 + c*x^2)*(-d^4/c^7)^(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^5*d^3*x^8 - 7*c^6*d^2*x^5 - 8* c^7*d*x^2 + sqrt(-3)*(c^5*d^3*x^8 - 7*c^6*d^2*x^5 - 8*c^7*d*x^2))*(-d^4/c^ 7)^(2/3) + 24*(1/4)^(1/3)*(c^3*d^4*x^7 - c^4*d^3*x^4 - 2*c^5*d^2*x - sqrt( -3)*(c^3*d^4*x^7 - c^4*d^3*x^4 - 2*c^5*d^2*x))*(-d^4/c^7)^(1/3) + 6*sqr...
\[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {c + d x^{3}}}{x^{3} \cdot \left (4 c + d x^{3}\right )}\, dx \] Input:
integrate((d*x**3+c)**(1/2)/x**3/(d*x**3+4*c),x)
Output:
Integral(sqrt(c + d*x**3)/(x**3*(4*c + d*x**3)), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} + 4 \, c\right )} x^{3}} \,d x } \] Input:
integrate((d*x^3+c)^(1/2)/x^3/(d*x^3+4*c),x, algorithm="maxima")
Output:
integrate(sqrt(d*x^3 + c)/((d*x^3 + 4*c)*x^3), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} + 4 \, c\right )} x^{3}} \,d x } \] Input:
integrate((d*x^3+c)^(1/2)/x^3/(d*x^3+4*c),x, algorithm="giac")
Output:
integrate(sqrt(d*x^3 + c)/((d*x^3 + 4*c)*x^3), x)
Timed out. \[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d\,x^3+c}}{x^3\,\left (d\,x^3+4\,c\right )} \,d x \] Input:
int((c + d*x^3)^(1/2)/(x^3*(4*c + d*x^3)),x)
Output:
int((c + d*x^3)^(1/2)/(x^3*(4*c + d*x^3)), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^3 \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d \,x^{6}+4 c \,x^{3}}d x \] Input:
int((d*x^3+c)^(1/2)/x^3/(d*x^3+4*c),x)
Output:
int(sqrt(c + d*x**3)/(4*c*x**3 + d*x**6),x)