Integrand size = 27, antiderivative size = 665 \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=-\frac {\sqrt {c+d x^3}}{48 c^2 x}+\frac {\sqrt [3]{d} \sqrt {c+d x^3}}{48 c^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}-\frac {\sqrt [3]{d} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{48 \sqrt {3} c^{11/6}}+\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{144 c^{11/6}}-\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{144 c^{11/6}}-\frac {\sqrt {2-\sqrt {3}} \sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{32\ 3^{3/4} c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{24 \sqrt {2} \sqrt [4]{3} c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
-1/48*(d*x^3+c)^(1/2)/c^2/x+1/48*d^(1/3)*(d*x^3+c)^(1/2)/c^2/((1+3^(1/2))* c^(1/3)+d^(1/3)*x)+1/24*(d*x^3+c)^(1/2)/c/x/(-d*x^3+8*c)-1/144*d^(1/3)*arc tan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/(d*x^3+c)^(1/2))*3^(1/2)/c^(11/6)+ 1/144*d^(1/3)*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c ^(11/6)-1/144*d^(1/3)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(11/6)-1/96*3 ^(1/4)*(1/2*6^(1/2)-1/2*2^(1/2))*d^(1/3)*(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)*Ellip ticE(((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)+1/144*d^(1/3)*(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)*El lipticF(((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 = 11.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\frac {-80 c \left (6 c^2+5 c d x^3-d^2 x^6\right )+50 c d x^3 \left (8 c-d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+d^2 x^6 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{3840 c^3 \sqrt {c+d x^3} \left (8 c x-d x^4\right )} \] Input:
Integrate[Sqrt[c + d*x^3]/(x^2*(8*c - d*x^3)^2),x]
Output:
(-80*c*(6*c^2 + 5*c*d*x^3 - d^2*x^6) + 50*c*d*x^3*(8*c - d*x^3)*Sqrt[1 + ( d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] + d^2*x^ 6*(-8*c + d*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/ c), (d*x^3)/(8*c)])/(3840*c^3*Sqrt[c + d*x^3]*(8*c*x - d*x^4))
Time = 1.74 (sec) , antiderivative size = 669, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {969, 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.
| \(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 969 |
| \(\displaystyle \frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}-\frac {\int -\frac {5 d x^3+8 c}{2 x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{24 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {5 d x^3+8 c}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{48 c}+\frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {-\frac {\int -\frac {4 c d x \left (20 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{8 c^2}-\frac {\sqrt {c+d x^3}}{c x}}{48 c}+\frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {d \int \frac {x \left (20 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{2 c}-\frac {\sqrt {c+d x^3}}{c x}}{48 c}+\frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {\frac {d \int \left (\frac {12 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {x}{\sqrt {d x^3+c}}\right )dx}{2 c}-\frac {\sqrt {c+d x^3}}{c x}}{48 c}+\frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {d \left (\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {2 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {2 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {2 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{2 c}-\frac {\sqrt {c+d x^3}}{c x}}{48 c}+\frac {\sqrt {c+d x^3}}{24 c x \left (8 c-d x^3\right )}\) |
Input:
Int[Sqrt[c + d*x^3]/(x^2*(8*c - d*x^3)^2),x]
Output:
Sqrt[c + d*x^3]/(24*c*x*(8*c - d*x^3)) + (-(Sqrt[c + d*x^3]/(c*x)) + (d*(( 2*Sqrt[c + d*x^3])/(d^(2/3)*((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)) - (2*c^(1 /6)*ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + d^(1/3)*x))/Sqrt[c + d*x^3]])/(Sqrt [3]*d^(2/3)) + (2*c^(1/6)*ArcTanh[(c^(1/3) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[ c + d*x^3])])/(3*d^(2/3)) - (2*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c]) ])/(3*d^(2/3)) - (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 + S qrt[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] ) + (2*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[ArcSi n[((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])))/(2*c))/(48*c)
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[(-(e*x)^(m + 1))*(a + b*x^n)^(p + 1)*((c + d*x^n )^q/(a*e*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) Int[(e*x)^m*(a + b*x^n)^( p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + 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] && LtQ[0, 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 = 3.17 (sec) , antiderivative size = 898, normalized size of antiderivative = 1.35
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(898\) |
| risch | \(\text {Expression too large to display}\) | \(1758\) |
| default | \(\text {Expression too large to display}\) | \(2194\) |
Input:
int((d*x^3+c)^(1/2)/x^2/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
Output:
1/192*x^2/c^2*d*(d*x^3+c)^(1/2)/(-d*x^3+8*c)-1/64*(d*x^3+c)^(1/2)/c^2/x-1/ 144*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)*((-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)))-1/216*I/c^2/ d^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)^(1/3)*_alpha^2*3^(1/2)*d-I*(-c*d^2)^...
Leaf count of result is larger than twice the leaf count of optimal. 2390 vs. \(2 (472) = 944\).
Time = 0.46 (sec) , antiderivative size = 2390, normalized size of antiderivative = 3.59 \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((d*x^3+c)^(1/2)/x^2/(-d*x^3+8*c)^2,x, algorithm="fricas")
Output:
-1/1728*(36*(d*x^4 - 8*c*x)*sqrt(d)*weierstrassZeta(0, -4*c/d, weierstrass PInverse(0, -4*c/d, x)) - (c^2*d*x^4 - 8*c^3*x + sqrt(-3)*(c^2*d*x^4 - 8*c ^3*x))*(d^2/c^11)^(1/6)*log((d^4*x^9 + 318*c*d^3*x^6 + 1200*c^2*d^2*x^3 + 640*c^3*d - 9*(5*c^8*d^2*x^7 + 64*c^9*d*x^4 + 32*c^10*x + sqrt(-3)*(5*c^8* d^2*x^7 + 64*c^9*d*x^4 + 32*c^10*x))*(d^2/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 ^2/c^11)^(5/6) - 2*(7*c^6*d^2*x^6 + 152*c^7*d*x^3 + 64*c^8)*sqrt(d^2/c^11) + (c^2*d^3*x^7 + 80*c^3*d^2*x^4 + 160*c^4*d*x + sqrt(-3)*(c^2*d^3*x^7 + 8 0*c^3*d^2*x^4 + 160*c^4*d*x))*(d^2/c^11)^(1/6)) - 9*(c^4*d^3*x^8 + 38*c^5* d^2*x^5 + 64*c^6*d*x^2 - sqrt(-3)*(c^4*d^3*x^8 + 38*c^5*d^2*x^5 + 64*c^6*d *x^2))*(d^2/c^11)^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3 )) + (c^2*d*x^4 - 8*c^3*x + sqrt(-3)*(c^2*d*x^4 - 8*c^3*x))*(d^2/c^11)^(1/ 6)*log((d^4*x^9 + 318*c*d^3*x^6 + 1200*c^2*d^2*x^3 + 640*c^3*d - 9*(5*c^8* d^2*x^7 + 64*c^9*d*x^4 + 32*c^10*x + sqrt(-3)*(5*c^8*d^2*x^7 + 64*c^9*d*x^ 4 + 32*c^10*x))*(d^2/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^2/c^11)^(5/6) - 2*(7 *c^6*d^2*x^6 + 152*c^7*d*x^3 + 64*c^8)*sqrt(d^2/c^11) + (c^2*d^3*x^7 + 80* c^3*d^2*x^4 + 160*c^4*d*x + sqrt(-3)*(c^2*d^3*x^7 + 80*c^3*d^2*x^4 + 160*c ^4*d*x))*(d^2/c^11)^(1/6)) - 9*(c^4*d^3*x^8 + 38*c^5*d^2*x^5 + 64*c^6*d*x^ 2 - sqrt(-3)*(c^4*d^3*x^8 + 38*c^5*d^2*x^5 + 64*c^6*d*x^2))*(d^2/c^11)^...
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\int \frac {\sqrt {c + d x^{3}}}{x^{2} \left (- 8 c + d x^{3}\right )^{2}}\, dx \] Input:
integrate((d*x**3+c)**(1/2)/x**2/(-d*x**3+8*c)**2,x)
Output:
Integral(sqrt(c + d*x**3)/(x**2*(-8*c + d*x**3)**2), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{2}} \,d x } \] Input:
integrate((d*x^3+c)^(1/2)/x^2/(-d*x^3+8*c)^2,x, algorithm="maxima")
Output:
integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)^2*x^2), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{2}} \,d x } \] Input:
integrate((d*x^3+c)^(1/2)/x^2/(-d*x^3+8*c)^2,x, algorithm="giac")
Output:
integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)^2*x^2), x)
Timed out. \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\int \frac {\sqrt {d\,x^3+c}}{x^2\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int((c + d*x^3)^(1/2)/(x^2*(8*c - d*x^3)^2),x)
Output:
int((c + d*x^3)^(1/2)/(x^2*(8*c - d*x^3)^2), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (8 c-d x^3\right )^2} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d^{2} x^{8}-16 c d \,x^{5}+64 c^{2} x^{2}}d x \] Input:
int((d*x^3+c)^(1/2)/x^2/(-d*x^3+8*c)^2,x)
Output:
int(sqrt(c + d*x**3)/(64*c**2*x**2 - 16*c*d*x**5 + d**2*x**8),x)