Integrand size = 27, antiderivative size = 657 \[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {13 x^2 \sqrt {c+d x^3}}{21 d}+\frac {265 c \sqrt {c+d x^3}}{7 d^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {9 \sqrt {3} c^{7/6} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{d^{5/3}}-\frac {9 c^{7/6} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{d^{5/3}}+\frac {9 c^{7/6} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^{5/3}}-\frac {265 \sqrt [4]{3} \sqrt {2-\sqrt {3}} c^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{14 d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {265 \sqrt {2} c^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{7 \sqrt [4]{3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
13/21*x^2*(d*x^3+c)^(1/2)/d+265/7*c*(d*x^3+c)^(1/2)/d^(5/3)/((1+3^(1/2))*c ^(1/3)+d^(1/3)*x)+1/3*x^2*(d*x^3+c)^(3/2)/d/(-d*x^3+8*c)+9*3^(1/2)*c^(7/6) *arctan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/(d*x^3+c)^(1/2))/d^(5/3)-9*c^( 7/6)*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/d^(5/3)+9* c^(7/6)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^(5/3)-265/14*3^(1/4)*(1/2*6 ^(1/2)-1/2*2^(1/2))*c^(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)/d^( 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)+265/21*2^(1/2)*c^(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)*Ellipt icF(((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^(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 = 9.65 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.27 \[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=-\frac {16 x^2 \left (37 c^2+35 c d x^3-2 d^2 x^6\right )+74 c x^2 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+53 d x^5 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{112 d \left (-8 c+d x^3\right ) \sqrt {c+d x^3}} \] Input:
Integrate[(x^4*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]
Output:
-1/112*(16*x^2*(37*c^2 + 35*c*d*x^3 - 2*d^2*x^6) + 74*c*x^2*(-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 )] + 53*d*x^5*(-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)])/(d*(-8*c + d*x^3)*Sqrt[c + d*x^3])
Time = 1.78 (sec) , antiderivative size = 661, 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 = {967, 27, 1051, 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.
\(\displaystyle \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx\) |
\(\Big \downarrow \) 967 |
\(\displaystyle \frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {\int \frac {x \sqrt {d x^3+c} \left (13 d x^3+4 c\right )}{2 \left (8 c-d x^3\right )}dx}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {\int \frac {x \sqrt {d x^3+c} \left (13 d x^3+4 c\right )}{8 c-d x^3}dx}{6 d}\) |
\(\Big \downarrow \) 1051 |
\(\displaystyle \frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {-\frac {2 \int -\frac {3 c d x \left (265 d x^3+148 c\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{7 d}-\frac {26}{7} x^2 \sqrt {c+d x^3}}{6 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {\frac {3}{7} c \int \frac {x \left (265 d x^3+148 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx-\frac {26}{7} x^2 \sqrt {c+d x^3}}{6 d}\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle \frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {\frac {3}{7} c \int \left (\frac {2268 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {265 x}{\sqrt {d x^3+c}}\right )dx-\frac {26}{7} x^2 \sqrt {c+d x^3}}{6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^2 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {\frac {3}{7} c \left (-\frac {530 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {265 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {126 \sqrt {3} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{d^{2/3}}+\frac {126 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{d^{2/3}}-\frac {126 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^{2/3}}-\frac {530 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )-\frac {26}{7} x^2 \sqrt {c+d x^3}}{6 d}\) |
Input:
Int[(x^4*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]
Output:
(x^2*(c + d*x^3)^(3/2))/(3*d*(8*c - d*x^3)) - ((-26*x^2*Sqrt[c + d*x^3])/7 + (3*c*((-530*Sqrt[c + d*x^3])/(d^(2/3)*((1 + Sqrt[3])*c^(1/3) + d^(1/3)* x)) - (126*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) + (126*c^(1/6)*ArcTanh[(c^(1/3) + d^(1/3)*x)^2/( 3*c^(1/6)*Sqrt[c + d*x^3])])/d^(2/3) - (126*c^(1/6)*ArcTanh[Sqrt[c + d*x^3 ]/(3*Sqrt[c])])/d^(2/3) + (265*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]*Sq rt[c + d*x^3]) - (530*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])))/7) /(6*d)
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^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* ((c + d*x^n)^q/(b*n*(p + 1))), x] - Simp[e^n/(b*n*(p + 1)) Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*( q - 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] && IntBino mialQ[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[f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*g*(m + n*(p + q + 1) + 1))), x] + Simp[1/( b*(m + n*(p + q + 1) + 1)) Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)* Simp[c*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c , d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] && !(EqQ[q, 1] && Simpl erQ[e + f*x^n, c + d*x^n])
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 = 2.97 (sec) , antiderivative size = 897, normalized size of antiderivative = 1.37
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(897\) |
default | \(\text {Expression too large to display}\) | \(1749\) |
risch | \(\text {Expression too large to display}\) | \(1758\) |
Input:
int(x^4*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
Output:
3*c/d*x^2*(d*x^3+c)^(1/2)/(-d*x^3+8*c)+2/7*x^2*(d*x^3+c)^(1/2)/d-265/21*I* c/d^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)))+6*I*c/d^4*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)^(2/3)*_alp...
Leaf count of result is larger than twice the leaf count of optimal. 2568 vs. \(2 (470) = 940\).
Time = 3.53 (sec) , antiderivative size = 2568, normalized size of antiderivative = 3.91 \[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x^4*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")
Output:
-1/28*(1060*(c*d*x^3 - 8*c^2)*sqrt(d)*weierstrassZeta(0, -4*c/d, weierstra ssPInverse(0, -4*c/d, x)) + 21*(d^3*x^3 - 8*c*d^2 - sqrt(-3)*(d^3*x^3 - 8* c*d^2))*(c^7/d^10)^(1/6)*log(59049*((d^11*x^9 + 318*c*d^10*x^6 + 1200*c^2* d^9*x^3 + 640*c^3*d^8 + sqrt(-3)*(d^11*x^9 + 318*c*d^10*x^6 + 1200*c^2*d^9 *x^3 + 640*c^3*d^8))*(c^7/d^10)^(5/6) + 6*(2*c^6*d^2*x^7 + 160*c^7*d*x^4 + 320*c^8*x - 6*(5*c^2*d^8*x^5 + 32*c^3*d^7*x^2 - sqrt(-3)*(5*c^2*d^8*x^5 + 32*c^3*d^7*x^2))*(c^7/d^10)^(2/3) - (7*c^4*d^5*x^6 + 152*c^5*d^4*x^3 + 64 *c^6*d^3 + sqrt(-3)*(7*c^4*d^5*x^6 + 152*c^5*d^4*x^3 + 64*c^6*d^3))*(c^7/d ^10)^(1/3))*sqrt(d*x^3 + c) - 36*(5*c^3*d^7*x^7 + 64*c^4*d^6*x^4 + 32*c^5* d^5*x)*sqrt(c^7/d^10) + 18*(c^5*d^4*x^8 + 38*c^6*d^3*x^5 + 64*c^7*d^2*x^2 - sqrt(-3)*(c^5*d^4*x^8 + 38*c^6*d^3*x^5 + 64*c^7*d^2*x^2))*(c^7/d^10)^(1/ 6))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 21*(d^3*x^3 - 8* c*d^2 - sqrt(-3)*(d^3*x^3 - 8*c*d^2))*(c^7/d^10)^(1/6)*log(-59049*((d^11*x ^9 + 318*c*d^10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8 + sqrt(-3)*(d^11*x^9 + 318*c*d^10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8))*(c^7/d^10)^(5/6) - 6*( 2*c^6*d^2*x^7 + 160*c^7*d*x^4 + 320*c^8*x - 6*(5*c^2*d^8*x^5 + 32*c^3*d^7* x^2 - sqrt(-3)*(5*c^2*d^8*x^5 + 32*c^3*d^7*x^2))*(c^7/d^10)^(2/3) - (7*c^4 *d^5*x^6 + 152*c^5*d^4*x^3 + 64*c^6*d^3 + sqrt(-3)*(7*c^4*d^5*x^6 + 152*c^ 5*d^4*x^3 + 64*c^6*d^3))*(c^7/d^10)^(1/3))*sqrt(d*x^3 + c) - 36*(5*c^3*d^7 *x^7 + 64*c^4*d^6*x^4 + 32*c^5*d^5*x)*sqrt(c^7/d^10) + 18*(c^5*d^4*x^8 ...
\[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x^{4} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \] Input:
integrate(x**4*(d*x**3+c)**(3/2)/(-d*x**3+8*c)**2,x)
Output:
Integral(x**4*(c + d*x**3)**(3/2)/(-8*c + d*x**3)**2, x)
\[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{4}}{{\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x^4*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
Output:
integrate((d*x^3 + c)^(3/2)*x^4/(d*x^3 - 8*c)^2, x)
\[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{4}}{{\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x^4*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")
Output:
integrate((d*x^3 + c)^(3/2)*x^4/(d*x^3 - 8*c)^2, x)
Timed out. \[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x^4\,{\left (d\,x^3+c\right )}^{3/2}}{{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int((x^4*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x)
Output:
int((x^4*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2, x)
\[ \int \frac {x^4 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {6 \sqrt {d \,x^{3}+c}\, c \,x^{2}-4 \sqrt {d \,x^{3}+c}\, d \,x^{5}+1848 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{7}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{2} d^{2}-231 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{7}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{3} x^{3}-768 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{4}+96 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{3} d \,x^{3}}{14 d \left (-d \,x^{3}+8 c \right )} \] Input:
int(x^4*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x)
Output:
(6*sqrt(c + d*x**3)*c*x**2 - 4*sqrt(c + d*x**3)*d*x**5 + 1848*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**2*d**2 - 231*int((sqrt(c + d*x**3)*x**7)/(64*c**3 + 48*c**2*d*x**3 - 1 5*c*d**2*x**6 + d**3*x**9),x)*c*d**3*x**3 - 768*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**4 + 96*int((s qrt(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*d*x**3)/(14*d*(8*c - d*x**3))