Integrand size = 27, antiderivative size = 684 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=-\frac {13 \sqrt {c+d x^3}}{256 c x^4}-\frac {d \sqrt {c+d x^3}}{32 c^2 x}+\frac {d^{4/3} \sqrt {c+d x^3}}{32 c^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}-\frac {9 \sqrt {3} d^{4/3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{1024 c^{11/6}}+\frac {9 d^{4/3} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{1024 c^{11/6}}-\frac {9 d^{4/3} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{1024 c^{11/6}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{64 c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{16 \sqrt {2} \sqrt [4]{3} c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
-13/256*(d*x^3+c)^(1/2)/c/x^4-1/32*d*(d*x^3+c)^(1/2)/c^2/x+1/32*d^(4/3)*(d *x^3+c)^(1/2)/c^2/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)+3/8*(d*x^3+c)^(1/2)/x^4/ (-d*x^3+8*c)-9/1024*3^(1/2)*d^(4/3)*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3 )*x)/(d*x^3+c)^(1/2))/c^(11/6)+9/1024*d^(4/3)*arctanh(1/3*(c^(1/3)+d^(1/3) *x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(11/6)-9/1024*d^(4/3)*arctanh(1/3*(d*x^3+ c)^(1/2)/c^(1/2))/c^(11/6)-1/64*3^(1/4)*(1/2*6^(1/2)-1/2*2^(1/2))*d^(4/3)* (c^(1/3)+d^(1/3)*x)*((c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/((1+3^(1/2))* c^(1/3)+d^(1/3)*x)^2)^(1/2)*EllipticE(((1-3^(1/2))*c^(1/3)+d^(1/3)*x)/((1+ 3^(1/2))*c^(1/3)+d^(1/3)*x),I*3^(1/2)+2*I)/c^(5/3)/(c^(1/3)*(c^(1/3)+d^(1/ 3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1/2)/(d*x^3+c)^(1/2)+1/96*d^(4/3 )*(c^(1/3)+d^(1/3)*x)*((c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/((1+3^(1/2) )*c^(1/3)+d^(1/3)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*c^(1/3)+d^(1/3)*x)/(( 1+3^(1/2))*c^(1/3)+d^(1/3)*x),I*3^(1/2)+2*I)*2^(1/2)*3^(3/4)/c^(5/3)/(c^(1 /3)*(c^(1/3)+d^(1/3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1/2)/(d*x^3+c) ^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.29 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\sqrt {c+d x^3} \left (-\frac {1}{256 c x^4}-\frac {13 d}{512 c^2 x}-\frac {3 d^2 x^2}{512 c^2 \left (-8 c+d x^3\right )}\right )+\frac {145 d^2 x^2 \sqrt {\frac {c+d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{8192 c^2 \sqrt {c+d x^3}}-\frac {d^3 x^5 \sqrt {\frac {c+d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{2560 c^3 \sqrt {c+d x^3}} \] Input:
Integrate[(c + d*x^3)^(3/2)/(x^5*(8*c - d*x^3)^2),x]
Output:
Sqrt[c + d*x^3]*(-1/256*1/(c*x^4) - (13*d)/(512*c^2*x) - (3*d^2*x^2)/(512* c^2*(-8*c + d*x^3))) + (145*d^2*x^2*Sqrt[(c + d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)])/(8192*c^2*Sqrt[c + d*x^3]) - (d^3*x ^5*Sqrt[(c + d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), (d*x^3)/(8 *c)])/(2560*c^3*Sqrt[c + d*x^3])
Time = 1.96 (sec) , antiderivative size = 692, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {968, 27, 1053, 25, 27, 1053, 27, 1054, 2009}
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.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 968 |
| \(\displaystyle \frac {\int \frac {3 c d \left (17 d x^3+26 c\right )}{2 x^5 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{24 c d}+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{16} \int \frac {17 d x^3+26 c}{x^5 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {1}{16} \left (-\frac {\int -\frac {c d \left (65 d x^3+128 c\right )}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{32 c^2}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{16} \left (\frac {\int \frac {c d \left (65 d x^3+128 c\right )}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{32 c^2}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{16} \left (\frac {d \int \frac {65 d x^3+128 c}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{32 c}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {1}{16} \left (\frac {d \left (-\frac {\int -\frac {8 c d x \left (145 c-8 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{8 c^2}-\frac {16 \sqrt {c+d x^3}}{c x}\right )}{32 c}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{16} \left (\frac {d \left (\frac {d \int \frac {x \left (145 c-8 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{c}-\frac {16 \sqrt {c+d x^3}}{c x}\right )}{32 c}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {1}{16} \left (\frac {d \left (\frac {d \int \left (\frac {81 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {8 x}{\sqrt {d x^3+c}}\right )dx}{c}-\frac {16 \sqrt {c+d x^3}}{c x}\right )}{32 c}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{16} \left (\frac {d \left (\frac {d \left (\frac {16 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {8 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {9 \sqrt {3} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{2 d^{2/3}}+\frac {9 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{2 d^{2/3}}-\frac {9 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 d^{2/3}}+\frac {16 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{c}-\frac {16 \sqrt {c+d x^3}}{c x}\right )}{32 c}-\frac {13 \sqrt {c+d x^3}}{16 c x^4}\right )+\frac {3 \sqrt {c+d x^3}}{8 x^4 \left (8 c-d x^3\right )}\) |
Input:
Int[(c + d*x^3)^(3/2)/(x^5*(8*c - d*x^3)^2),x]
Output:
(3*Sqrt[c + d*x^3])/(8*x^4*(8*c - d*x^3)) + ((-13*Sqrt[c + d*x^3])/(16*c*x ^4) + (d*((-16*Sqrt[c + d*x^3])/(c*x) + (d*((16*Sqrt[c + d*x^3])/(d^(2/3)* ((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)) - (9*Sqrt[3]*c^(1/6)*ArcTan[(Sqrt[3]* c^(1/6)*(c^(1/3) + d^(1/3)*x))/Sqrt[c + d*x^3]])/(2*d^(2/3)) + (9*c^(1/6)* ArcTanh[(c^(1/3) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[c + d*x^3])])/(2*d^(2/3)) - (9*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2*d^(2/3)) - (8*3^(1/4 )*Sqrt[2 - Sqrt[3]]*c^(1/3)*(c^(1/3) + d^(1/3)*x)*Sqrt[(c^(2/3) - c^(1/3)* d^(1/3)*x + d^(2/3)*x^2)/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*EllipticE[ ArcSin[((1 - Sqrt[3])*c^(1/3) + d^(1/3)*x)/((1 + Sqrt[3])*c^(1/3) + d^(1/3 )*x)], -7 - 4*Sqrt[3]])/(d^(2/3)*Sqrt[(c^(1/3)*(c^(1/3) + d^(1/3)*x))/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*Sqrt[c + d*x^3]) + (16*Sqrt[2]*c^(1/3)* (c^(1/3) + d^(1/3)*x)*Sqrt[(c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2)/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*c^(1/3 ) + d^(1/3)*x)/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)], -7 - 4*Sqrt[3]])/(3^( 1/4)*d^(2/3)*Sqrt[(c^(1/3)*(c^(1/3) + d^(1/3)*x))/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*Sqrt[c + d*x^3])))/c))/(32*c))/16
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) *((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1)) Int [(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c *b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_ ))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b *x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( m + 1)) Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 ) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && LtQ[m, -1]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.90 (sec) , antiderivative size = 919, normalized size of antiderivative = 1.34
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(919\) |
| risch | \(\text {Expression too large to display}\) | \(1770\) |
| default | \(\text {Expression too large to display}\) | \(2691\) |
Input:
int((d*x^3+c)^(3/2)/x^5/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
Output:
3/512/c^2*x^2*d^2*(d*x^3+c)^(1/2)/(-d*x^3+8*c)-1/256*(d*x^3+c)^(1/2)/c/x^4 -13/512*d*(d*x^3+c)^(1/2)/c^2/x-1/96*I*d/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)*((-3/2/d*(-c*d^2) ^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*EllipticE(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/d*(-c*d^2)^(1/3)*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)))-3/512*I/d/c^2*2^(1/2)*sum(1/_alpha*(-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),-1/18/d*(2*I*(-c*d^2)^(1/...
Leaf count of result is larger than twice the leaf count of optimal. 2549 vs. \(2 (487) = 974\).
Time = 0.86 (sec) , antiderivative size = 2549, normalized size of antiderivative = 3.73 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((d*x^3+c)^(3/2)/x^5/(-d*x^3+8*c)^2,x, algorithm="fricas")
Output:
-1/4096*(128*(d^2*x^7 - 8*c*d*x^4)*sqrt(d)*weierstrassZeta(0, -4*c/d, weie rstrassPInverse(0, -4*c/d, x)) - 3*(c^2*d*x^7 - 8*c^3*x^4 + sqrt(-3)*(c^2* d*x^7 - 8*c^3*x^4))*(d^8/c^11)^(1/6)*log(6561*(d^9*x^9 + 318*c*d^8*x^6 + 1 200*c^2*d^7*x^3 + 640*c^3*d^6 - 9*(5*c^8*d^3*x^7 + 64*c^9*d^2*x^4 + 32*c^1 0*d*x + sqrt(-3)*(5*c^8*d^3*x^7 + 64*c^9*d^2*x^4 + 32*c^10*d*x))*(d^8/c^11 )^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c^11*x^2 - sqrt(-3)*(5*c ^10*d*x^5 + 32*c^11*x^2))*(d^8/c^11)^(5/6) - 2*(7*c^6*d^4*x^6 + 152*c^7*d^ 3*x^3 + 64*c^8*d^2)*sqrt(d^8/c^11) + (c^2*d^7*x^7 + 80*c^3*d^6*x^4 + 160*c ^4*d^5*x + sqrt(-3)*(c^2*d^7*x^7 + 80*c^3*d^6*x^4 + 160*c^4*d^5*x))*(d^8/c ^11)^(1/6)) - 9*(c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64*c^6*d^4*x^2 - sqrt(-3)* (c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64*c^6*d^4*x^2))*(d^8/c^11)^(1/3))/(d^3*x^ 9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) + 3*(c^2*d*x^7 - 8*c^3*x^4 + sqrt(-3)*(c^2*d*x^7 - 8*c^3*x^4))*(d^8/c^11)^(1/6)*log(6561*(d^9*x^9 + 318 *c*d^8*x^6 + 1200*c^2*d^7*x^3 + 640*c^3*d^6 - 9*(5*c^8*d^3*x^7 + 64*c^9*d^ 2*x^4 + 32*c^10*d*x + sqrt(-3)*(5*c^8*d^3*x^7 + 64*c^9*d^2*x^4 + 32*c^10*d *x))*(d^8/c^11)^(2/3) - 3*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c^11*x^2 - sqrt(-3)*(5*c^10*d*x^5 + 32*c^11*x^2))*(d^8/c^11)^(5/6) - 2*(7*c^6*d^4*x^ 6 + 152*c^7*d^3*x^3 + 64*c^8*d^2)*sqrt(d^8/c^11) + (c^2*d^7*x^7 + 80*c^3*d ^6*x^4 + 160*c^4*d^5*x + sqrt(-3)*(c^2*d^7*x^7 + 80*c^3*d^6*x^4 + 160*c^4* d^5*x))*(d^8/c^11)^(1/6)) - 9*(c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64*c^6*d^...
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\text {Timed out} \] Input:
integrate((d*x**3+c)**(3/2)/x**5/(-d*x**3+8*c)**2,x)
Output:
Timed out
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{5}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^5/(-d*x^3+8*c)^2,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)/((d*x^3 - 8*c)^2*x^5), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{5}} \,d x } \] Input:
integrate((d*x^3+c)^(3/2)/x^5/(-d*x^3+8*c)^2,x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)/((d*x^3 - 8*c)^2*x^5), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{3/2}}{x^5\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int((c + d*x^3)^(3/2)/(x^5*(8*c - d*x^3)^2),x)
Output:
int((c + d*x^3)^(3/2)/(x^5*(8*c - d*x^3)^2), x)
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^5 \left (8 c-d x^3\right )^2} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}+816 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{11}-15 c \,d^{2} x^{8}+48 c^{2} d \,x^{5}+64 c^{3} x^{2}}d x \right ) c^{2} d \,x^{4}-102 \left (\int \frac {\sqrt {d \,x^{3}+c}}{d^{3} x^{11}-15 c \,d^{2} x^{8}+48 c^{2} d \,x^{5}+64 c^{3} x^{2}}d x \right ) c \,d^{2} x^{7}+600 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{2} x^{4}-75 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) d^{3} x^{7}}{64 x^{4} \left (-d \,x^{3}+8 c \right )} \] Input:
int((d*x^3+c)^(3/2)/x^5/(-d*x^3+8*c)^2,x)
Output:
( - 2*sqrt(c + d*x**3) + 816*int(sqrt(c + d*x**3)/(64*c**3*x**2 + 48*c**2* d*x**5 - 15*c*d**2*x**8 + d**3*x**11),x)*c**2*d*x**4 - 102*int(sqrt(c + d* x**3)/(64*c**3*x**2 + 48*c**2*d*x**5 - 15*c*d**2*x**8 + d**3*x**11),x)*c*d **2*x**7 + 600*int((sqrt(c + d*x**3)*x)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d **2*x**6 + d**3*x**9),x)*c*d**2*x**4 - 75*int((sqrt(c + d*x**3)*x)/(64*c** 3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*d**3*x**7)/(64*x**4*(8 *c - d*x**3))