Integrand size = 27, antiderivative size = 635 \[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 x^2}{27 c d \sqrt {c+d x^3}}+\frac {2 \sqrt {c+d x^3}}{27 c d^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {4 \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{27 \sqrt {3} c^{5/6} d^{5/3}}+\frac {4 \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{81 c^{5/6} d^{5/3}}-\frac {4 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 c^{5/6} d^{5/3}}-\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} c^{2/3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {2 \sqrt {2} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} c^{2/3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \]
4/81*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(5/6)/d^ (5/3)-4/81*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/6)/d^(5/3)-4/81*arcta n(c^(1/6)*(c^(1/3)+d^(1/3)*x)*3^(1/2)/(d*x^3+c)^(1/2))/c^(5/6)/d^(5/3)*3^( 1/2)-2/27*x^2/c/d/(d*x^3+c)^(1/2)+2/27*(d*x^3+c)^(1/2)/c/d^(5/3)/(d^(1/3)* x+c^(1/3)*(1+3^(1/2)))+2/81*(c^(1/3)+d^(1/3)*x)*EllipticF((d^(1/3)*x+c^(1/ 3)*(1-3^(1/2)))/(d^(1/3)*x+c^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*2^(1/2)*((c ^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/(d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^( 1/2)*3^(3/4)/c^(2/3)/d^(5/3)/(d*x^3+c)^(1/2)/(c^(1/3)*(c^(1/3)+d^(1/3)*x)/ (d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^(1/2)-1/27*(c^(1/3)+d^(1/3)*x)*Elliptic E((d^(1/3)*x+c^(1/3)*(1-3^(1/2)))/(d^(1/3)*x+c^(1/3)*(1+3^(1/2))),I*3^(1/2 )+2*I)*(1/2*6^(1/2)-1/2*2^(1/2))*((c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/ (d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(1/4)/c^(2/3)/d^(5/3)/(d*x^3+c) ^(1/2)/(c^(1/3)*(c^(1/3)+d^(1/3)*x)/(d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^(1/ 2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 8.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.20 \[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {x^2 \left (80 c-80 c \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}{1080 c^2 d \sqrt {c+d x^3}} \]
-1/1080*(x^2*(80*c - 80*c*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, - ((d*x^3)/c), (d*x^3)/(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)]))/(c^2*d*Sqrt[c + d*x^3])
Time = 1.02 (sec) , antiderivative size = 635, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {971, 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.
\(\displaystyle \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 971 |
\(\displaystyle \frac {2 \int \frac {x \left (32 c-d x^3\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c d}-\frac {2 x^2}{27 c d \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {x \left (32 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c d}-\frac {2 x^2}{27 c d \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle \frac {\int \left (\frac {24 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {x}{\sqrt {d x^3+c}}\right )dx}{27 c d}-\frac {2 x^2}{27 c d \sqrt {c+d x^3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {4 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {4 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {4 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{27 c d}-\frac {2 x^2}{27 c d \sqrt {c+d x^3}}\) |
(-2*x^2)/(27*c*d*Sqrt[c + d*x^3]) + ((2*Sqrt[c + d*x^3])/(d^(2/3)*((1 + Sq rt[3])*c^(1/3) + d^(1/3)*x)) - (4*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)) + (4*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)) - (4*c^(1/ 6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(3*d^(2/3)) - (3^(1/4)*Sqrt[2 - S qrt[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]) + (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[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]))/(27*c*d)
3.4.33.3.1 Defintions of rubi rules used
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 + 1)/(n*(b*c - a*d)*(p + 1))), x] - Simp[e^n/(n*(b*c - a*d) *(p + 1)) Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e , q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n, m - n + 1] && GtQ[m - n + 1, 0] && 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 = 4.36 (sec) , antiderivative size = 878, normalized size of antiderivative = 1.38
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(878\) |
default | \(\text {Expression too large to display}\) | \(1346\) |
-2/27/d*x^2/c/((x^3+c/d)*d)^(1/2)-2/81*I/d^2/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)*((-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)))-8/243*I/d^4/c*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 )*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.45 (sec) , antiderivative size = 2525, normalized size of antiderivative = 3.98 \[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \]
-1/243*(18*sqrt(d*x^3 + c)*d*x^2 + 18*(d*x^3 + c)*sqrt(d)*weierstrassZeta( 0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) - (c*d^3*x^3 + c^2*d^2 + sqr t(-3)*(c*d^3*x^3 + c^2*d^2))*(1/(c^5*d^10))^(1/6)*log((d^3*x^9 + 318*c*d^2 *x^6 + 1200*c^2*d*x^3 + 640*c^3 - 9*(5*c^4*d^9*x^7 + 64*c^5*d^8*x^4 + 32*c ^6*d^7*x + sqrt(-3)*(5*c^4*d^9*x^7 + 64*c^5*d^8*x^4 + 32*c^6*d^7*x))*(1/(c ^5*d^10))^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5*c^5*d^10*x^5 + 32*c^6*d^9*x^2 - sqrt(-3)*(5*c^5*d^10*x^5 + 32*c^6*d^9*x^2))*(1/(c^5*d^10))^(5/6) - 2*(7*c^ 3*d^7*x^6 + 152*c^4*d^6*x^3 + 64*c^5*d^5)*sqrt(1/(c^5*d^10)) + (c*d^4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x + sqrt(-3)*(c*d^4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x))*(1/(c^5*d^10))^(1/6)) - 9*(c^2*d^6*x^8 + 38*c^3*d^5*x^5 + 64*c^4*d^4*x^2 - sqrt(-3)*(c^2*d^6*x^8 + 38*c^3*d^5*x^5 + 64*c^4*d^4*x^2)) *(1/(c^5*d^10))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) + (c*d^3*x^3 + c^2*d^2 + sqrt(-3)*(c*d^3*x^3 + c^2*d^2))*(1/(c^5*d^10))^( 1/6)*log((d^3*x^9 + 318*c*d^2*x^6 + 1200*c^2*d*x^3 + 640*c^3 - 9*(5*c^4*d^ 9*x^7 + 64*c^5*d^8*x^4 + 32*c^6*d^7*x + sqrt(-3)*(5*c^4*d^9*x^7 + 64*c^5*d ^8*x^4 + 32*c^6*d^7*x))*(1/(c^5*d^10))^(2/3) - 3*sqrt(d*x^3 + c)*(6*(5*c^5 *d^10*x^5 + 32*c^6*d^9*x^2 - sqrt(-3)*(5*c^5*d^10*x^5 + 32*c^6*d^9*x^2))*( 1/(c^5*d^10))^(5/6) - 2*(7*c^3*d^7*x^6 + 152*c^4*d^6*x^3 + 64*c^5*d^5)*sqr t(1/(c^5*d^10)) + (c*d^4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x + sqrt(-3)*( c*d^4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x))*(1/(c^5*d^10))^(1/6)) - 9*...
Timed out. \[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {x^{4}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}} \,d x } \]
\[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {x^{4}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (d\,x^3+c\right )}^{3/2}\,\left (8\,c-d\,x^3\right )} \,d x \]