Integrand size = 26, antiderivative size = 697 \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{4 c x}+\frac {\sqrt [3]{d} \sqrt {c+d x^3}}{4 c \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {\sqrt [3]{d} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{4\ 2^{2/3} \sqrt {3} c^{5/6}}+\frac {\sqrt [3]{d} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{4\ 2^{2/3} \sqrt {3} c^{5/6}}-\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{4\ 2^{2/3} c^{5/6}}+\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{12\ 2^{2/3} c^{5/6}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{8 c^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt {2} \sqrt [4]{3} c^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \]
-1/8*d^(1/3)*arctanh(c^(1/6)*(c^(1/3)-2^(1/3)*d^(1/3)*x)/(d*x^3+c)^(1/2))* 2^(1/3)/c^(5/6)+1/24*d^(1/3)*arctanh((d*x^3+c)^(1/2)/c^(1/2))*2^(1/3)/c^(5 /6)-1/24*d^(1/3)*arctan(c^(1/6)*(c^(1/3)+2^(1/3)*d^(1/3)*x)*3^(1/2)/(d*x^3 +c)^(1/2))*2^(1/3)/c^(5/6)*3^(1/2)+1/24*d^(1/3)*arctan(1/3*(d*x^3+c)^(1/2) *3^(1/2)/c^(1/2))*2^(1/3)/c^(5/6)*3^(1/2)-1/4*(d*x^3+c)^(1/2)/c/x+1/4*d^(1 /3)*(d*x^3+c)^(1/2)/c/(d^(1/3)*x+c^(1/3)*(1+3^(1/2)))+1/12*d^(1/3)*(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)*((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)*2^(1/2)/(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/ 8*3^(1/4)*d^(1/3)*(c^(1/3)+d^(1/3)*x)*EllipticE((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)/c^(2/3)/(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 = 11.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.20 \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=\frac {-40 c \left (c+d x^3\right )+25 c d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{160 c^2 x \sqrt {c+d x^3}} \]
(-40*c*(c + d*x^3) + 25*c*d*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -1/4*(d*x^3)/c] + d^2*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[ 5/3, 1/2, 1, 8/3, -((d*x^3)/c), -1/4*(d*x^3)/c])/(160*c^2*x*Sqrt[c + d*x^3 ])
Time = 0.72 (sec) , antiderivative size = 690, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {975, 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.
\(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 975 |
\(\displaystyle \frac {\int \frac {d x \left (d x^3+10 c\right )}{2 \sqrt {d x^3+c} \left (d x^3+4 c\right )}dx}{4 c}-\frac {\sqrt {c+d x^3}}{4 c x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int \frac {x \left (d x^3+10 c\right )}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}dx}{8 c}-\frac {\sqrt {c+d x^3}}{4 c x}\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle \frac {d \int \left (\frac {x}{\sqrt {d x^3+c}}+\frac {6 c x}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}\right )dx}{8 c}-\frac {\sqrt {c+d x^3}}{4 c x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d \left (\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [3]{2} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {\sqrt [3]{2} \sqrt [6]{c} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^{2/3}}-\frac {\sqrt [3]{2} \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{d^{2/3}}+\frac {\sqrt [3]{2} \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{8 c}-\frac {\sqrt {c+d x^3}}{4 c x}\) |
-1/4*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^(1/3)*c^(1/6)*ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/ 3) + 2^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3]])/(Sqrt[3]*d^(2/3)) + (2^(1/3)*c^ (1/6)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(Sqrt[3]*d^(2/3)) - (2^(1 /3)*c^(1/6)*ArcTanh[(c^(1/6)*(c^(1/3) - 2^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3 ]])/d^(2/3) + (2^(1/3)*c^(1/6)*ArcTanh[Sqrt[c + d*x^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 + 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*Sqr t[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])))/(8*c)
3.3.66.3.1 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*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/ (a*e*(m + 1))), x] - Simp[1/(a*e^n*(m + 1)) Int[(e*x)^(m + n)*(a + b*x^n) ^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomi alQ[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 = 5.01 (sec) , antiderivative size = 868, normalized size of antiderivative = 1.25
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(868\) |
risch | \(\text {Expression too large to display}\) | \(872\) |
default | \(\text {Expression too large to display}\) | \(1306\) |
-1/4*(d*x^3+c)^(1/2)/c/x-1/12*I/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)))-1/12*I/d^2/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))*Ellipt icPi(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),1/6/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*(-c*d^2)^(2/3)*_al...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.53 (sec) , antiderivative size = 2253, normalized size of antiderivative = 3.23 \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=\text {Too large to display} \]
1/48*(2*(1/432)^(1/6)*c*x*(-d^2/c^5)^(1/6)*log((d^4*x^9 - 66*c*d^3*x^6 - 7 2*c^2*d^2*x^3 - 32*c^3*d + 48*(1/2)^(2/3)*(c^4*d^2*x^7 - c^5*d*x^4 - 2*c^6 *x)*(-d^2/c^5)^(2/3) + 12*(1/2)^(1/3)*(c^2*d^3*x^8 - 7*c^3*d^2*x^5 - 8*c^4 *d*x^2)*(-d^2/c^5)^(1/3) + 6*(1296*(1/432)^(5/6)*c^5*d*x^5*(-d^2/c^5)^(5/6 ) + sqrt(1/3)*(5*c^3*d^2*x^6 - 20*c^4*d*x^3 - 16*c^5)*sqrt(-d^2/c^5) + 2*( 1/432)^(1/6)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x)*(-d^2/c^5)^(1/6))*sq rt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - 2*(1/43 2)^(1/6)*c*x*(-d^2/c^5)^(1/6)*log((d^4*x^9 - 66*c*d^3*x^6 - 72*c^2*d^2*x^3 - 32*c^3*d + 48*(1/2)^(2/3)*(c^4*d^2*x^7 - c^5*d*x^4 - 2*c^6*x)*(-d^2/c^5 )^(2/3) + 12*(1/2)^(1/3)*(c^2*d^3*x^8 - 7*c^3*d^2*x^5 - 8*c^4*d*x^2)*(-d^2 /c^5)^(1/3) - 6*(1296*(1/432)^(5/6)*c^5*d*x^5*(-d^2/c^5)^(5/6) + sqrt(1/3) *(5*c^3*d^2*x^6 - 20*c^4*d*x^3 - 16*c^5)*sqrt(-d^2/c^5) + 2*(1/432)^(1/6)* (c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x)*(-d^2/c^5)^(1/6))*sqrt(d*x^3 + c) )/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - 12*sqrt(d)*x*weierst rassZeta(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) - (1/432)^(1/6)*(sq rt(-3)*c*x + c*x)*(-d^2/c^5)^(1/6)*log((d^4*x^9 - 66*c*d^3*x^6 - 72*c^2*d^ 2*x^3 - 32*c^3*d - 24*(1/2)^(2/3)*(c^4*d^2*x^7 - c^5*d*x^4 - 2*c^6*x + sqr t(-3)*(c^4*d^2*x^7 - c^5*d*x^4 - 2*c^6*x))*(-d^2/c^5)^(2/3) - 6*(1/2)^(1/3 )*(c^2*d^3*x^8 - 7*c^3*d^2*x^5 - 8*c^4*d*x^2 - sqrt(-3)*(c^2*d^3*x^8 - 7*c ^3*d^2*x^5 - 8*c^4*d*x^2))*(-d^2/c^5)^(1/3) + 6*sqrt(d*x^3 + c)*(648*(1...
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {c + d x^{3}}}{x^{2} \cdot \left (4 c + d x^{3}\right )}\, dx \]
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} + 4 \, c\right )} x^{2}} \,d x } \]
\[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} + 4 \, c\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {c+d x^3}}{x^2 \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d\,x^3+c}}{x^2\,\left (d\,x^3+4\,c\right )} \,d x \]