Integrand size = 25, antiderivative size = 638 \[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {19 \sqrt {c+d x^3}}{8 d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {3 x^2 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}+\frac {9 \sqrt {3} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{16 d^{2/3}}-\frac {9 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{16 d^{2/3}}+\frac {9 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{16 d^{2/3}}-\frac {19 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{16 d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {19 \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{4 \sqrt {2} \sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
19/8*(d*x^3+c)^(1/2)/d^(2/3)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)+3*x^2*(d*x^3+ c)^(1/2)/(-8*d*x^3+64*c)+9/16*3^(1/2)*c^(1/6)*arctan(3^(1/2)*c^(1/6)*(c^(1 /3)+d^(1/3)*x)/(d*x^3+c)^(1/2))/d^(2/3)-9/16*c^(1/6)*arctanh(1/3*(c^(1/3)+ d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/d^(2/3)+9/16*c^(1/6)*arctanh(1/3*(d* x^3+c)^(1/2)/c^(1/2))/d^(2/3)-19/16*3^(1/4)*(1/2*6^(1/2)-1/2*2^(1/2))*c^(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)*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)/d^(2/3)/(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)+19/24*2 ^(1/2)*c^(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)*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)*3^(3/4)/d^(2/3)/ (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 = 10.15 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.22 \[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {x^2 \left (\frac {240 \left (c+d x^3\right )}{8 c-d x^3}-25 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-\frac {19 d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{c}\right )}{640 \sqrt {c+d x^3}} \] Input:
Integrate[(x*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]
Output:
(x^2*((240*(c + d*x^3))/(8*c - d*x^3) - 25*Sqrt[1 + (d*x^3)/c]*AppellF1[2/ 3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] - (19*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)])/c))/(640*Sqrt[ c + d*x^3])
Time = 1.63 (sec) , antiderivative size = 631, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {968, 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.
| \(\displaystyle \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 968 |
| \(\displaystyle \frac {\int -\frac {3 c d x \left (19 d x^3+10 c\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{24 c d}+\frac {3 x^2 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 x^2 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac {1}{16} \int \frac {x \left (19 d x^3+10 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {3 x^2 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac {1}{16} \int \left (\frac {162 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {19 x}{\sqrt {d x^3+c}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{16} \left (\frac {38 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {19 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {9 \sqrt {3} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{d^{2/3}}-\frac {9 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{d^{2/3}}+\frac {9 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^{2/3}}+\frac {38 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )+\frac {3 x^2 \sqrt {c+d x^3}}{8 \left (8 c-d x^3\right )}\) |
Input:
Int[(x*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]
Output:
(3*x^2*Sqrt[c + d*x^3])/(8*(8*c - d*x^3)) + ((38*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]])/d^(2/3) - (9*c^(1/6)*Arc Tanh[(c^(1/3) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[c + d*x^3])])/d^(2/3) + (9*c^ (1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^(2/3) - (19*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]) + (38*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]))/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_)*((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 = 1.13 (sec) , antiderivative size = 874, normalized size of antiderivative = 1.37
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(874\) |
| elliptic | \(\text {Expression too large to display}\) | \(874\) |
Input:
int(x*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
Output:
3/8*x^2*(d*x^3+c)^(1/2)/(-d*x^3+8*c)-19/24*I*3^(1/2)/d*(-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)))+3/8*I/d^3*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)*_a lpha^2*3^(1/2)*d-I*(-c*d^2)^(2/3)*_alpha*3^(1/2)+I*3^(1/2)*c*d-3*(-c*d^...
Leaf count of result is larger than twice the leaf count of optimal. 2335 vs. \(2 (449) = 898\).
Time = 1.06 (sec) , antiderivative size = 2335, normalized size of antiderivative = 3.66 \[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")
Output:
-1/64*(24*sqrt(d*x^3 + c)*d*x^2 + 152*(d*x^3 - 8*c)*sqrt(d)*weierstrassZet a(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) + 3*(d^2*x^3 - 8*c*d - sqr t(-3)*(d^2*x^3 - 8*c*d))*(c/d^4)^(1/6)*log(59049/4*((d^6*x^9 + 318*c*d^5*x ^6 + 1200*c^2*d^4*x^3 + 640*c^3*d^3 + sqrt(-3)*(d^6*x^9 + 318*c*d^5*x^6 + 1200*c^2*d^4*x^3 + 640*c^3*d^3))*(c/d^4)^(5/6) + 6*(2*c*d^2*x^7 + 160*c^2* d*x^4 + 320*c^3*x - 6*(5*c*d^4*x^5 + 32*c^2*d^3*x^2 - sqrt(-3)*(5*c*d^4*x^ 5 + 32*c^2*d^3*x^2))*(c/d^4)^(2/3) - (7*c*d^3*x^6 + 152*c^2*d^2*x^3 + 64*c ^3*d + sqrt(-3)*(7*c*d^3*x^6 + 152*c^2*d^2*x^3 + 64*c^3*d))*(c/d^4)^(1/3)) *sqrt(d*x^3 + c) - 36*(5*c*d^4*x^7 + 64*c^2*d^3*x^4 + 32*c^3*d^2*x)*sqrt(c /d^4) + 18*(c*d^3*x^8 + 38*c^2*d^2*x^5 + 64*c^3*d*x^2 - sqrt(-3)*(c*d^3*x^ 8 + 38*c^2*d^2*x^5 + 64*c^3*d*x^2))*(c/d^4)^(1/6))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 3*(d^2*x^3 - 8*c*d - sqrt(-3)*(d^2*x^3 - 8* c*d))*(c/d^4)^(1/6)*log(-59049/4*((d^6*x^9 + 318*c*d^5*x^6 + 1200*c^2*d^4* x^3 + 640*c^3*d^3 + sqrt(-3)*(d^6*x^9 + 318*c*d^5*x^6 + 1200*c^2*d^4*x^3 + 640*c^3*d^3))*(c/d^4)^(5/6) - 6*(2*c*d^2*x^7 + 160*c^2*d*x^4 + 320*c^3*x - 6*(5*c*d^4*x^5 + 32*c^2*d^3*x^2 - sqrt(-3)*(5*c*d^4*x^5 + 32*c^2*d^3*x^2 ))*(c/d^4)^(2/3) - (7*c*d^3*x^6 + 152*c^2*d^2*x^3 + 64*c^3*d + sqrt(-3)*(7 *c*d^3*x^6 + 152*c^2*d^2*x^3 + 64*c^3*d))*(c/d^4)^(1/3))*sqrt(d*x^3 + c) - 36*(5*c*d^4*x^7 + 64*c^2*d^3*x^4 + 32*c^3*d^2*x)*sqrt(c/d^4) + 18*(c*d^3* x^8 + 38*c^2*d^2*x^5 + 64*c^3*d*x^2 - sqrt(-3)*(c*d^3*x^8 + 38*c^2*d^2*...
\[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \] Input:
integrate(x*(d*x**3+c)**(3/2)/(-d*x**3+8*c)**2,x)
Output:
Integral(x*(c + d*x**3)**(3/2)/(-8*c + d*x**3)**2, x)
\[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x}{{\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)*x/(d*x^3 - 8*c)^2, x)
\[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x}{{\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)*x/(d*x^3 - 8*c)^2, x)
Timed out. \[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x\,{\left (d\,x^3+c\right )}^{3/2}}{{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int((x*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x)
Output:
int((x*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2, x)
\[ \int \frac {x \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2 \sqrt {d \,x^{3}+c}\, x^{2}+240 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{7}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{2}-30 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{7}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) d^{3} x^{3}-24 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{3}+3 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{2} d \,x^{3}}{-29 d \,x^{3}+232 c} \] Input:
int(x*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x)
Output:
(2*sqrt(c + d*x**3)*x**2 + 240*int((sqrt(c + d*x**3)*x**7)/(64*c**3 + 48*c **2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c*d**2 - 30*int((sqrt(c + d*x* *3)*x**7)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*d**3* x**3 - 24*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**3 + 3*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**2*d*x**3)/(29*(8*c - d*x**3))