Integrand size = 26, antiderivative size = 697 \[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{4 c^2 x}+\frac {\sqrt [3]{d} \sqrt {c+d x^3}}{4 c^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {\sqrt [3]{d} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{12\ 2^{2/3} \sqrt {3} c^{11/6}}-\frac {\sqrt [3]{d} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{12\ 2^{2/3} \sqrt {3} c^{11/6}}+\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{12\ 2^{2/3} c^{11/6}}-\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{36\ 2^{2/3} c^{11/6}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{8 c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt {2} \sqrt [4]{3} c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
-1/4*(d*x^3+c)^(1/2)/c^2/x+1/4*d^(1/3)*(d*x^3+c)^(1/2)/c^2/((1+3^(1/2))*c^ (1/3)+d^(1/3)*x)+1/72*d^(1/3)*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+2^(1/3)*d^(1 /3)*x)/(d*x^3+c)^(1/2))*2^(1/3)*3^(1/2)/c^(11/6)-1/72*d^(1/3)*arctan(1/3*( d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))*2^(1/3)*3^(1/2)/c^(11/6)+1/24*d^(1/3)*arct anh(c^(1/6)*(c^(1/3)-2^(1/3)*d^(1/3)*x)/(d*x^3+c)^(1/2))*2^(1/3)/c^(11/6)- 1/72*d^(1/3)*arctanh((d*x^3+c)^(1/2)/c^(1/2))*2^(1/3)/c^(11/6)-1/8*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)*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)/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)+1/12*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)*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)*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 = 136, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {-40 c \left (c+d x^3\right )+5 c d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )+d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{160 c^3 x \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^2*Sqrt[c + d*x^3]*(4*c + d*x^3)),x]
Output:
(-40*c*(c + d*x^3) + 5*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^3*x*Sqrt[c + d*x^3] )
Time = 1.16 (sec) , antiderivative size = 697, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {980, 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.
| \(\displaystyle \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 980 |
| \(\displaystyle \frac {\int \frac {d x \left (d x^3+2 c\right )}{2 \sqrt {d x^3+c} \left (d x^3+4 c\right )}dx}{4 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \int \frac {x \left (d x^3+2 c\right )}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}dx}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {d \int \left (\frac {x}{\sqrt {d x^3+c}}-\frac {2 c x}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}\right )dx}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \left (\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [3]{2} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3 \sqrt {3} d^{2/3}}-\frac {\sqrt [3]{2} \sqrt [6]{c} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^{2/3}}+\frac {\sqrt [3]{2} \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {\sqrt [3]{2} \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{9 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x}\) |
Input:
Int[1/(x^2*Sqrt[c + d*x^3]*(4*c + d*x^3)),x]
Output:
-1/4*Sqrt[c + d*x^3]/(c^2*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]])/(3*Sqrt[3]*d^(2/3)) - (2^(1/3 )*c^(1/6)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(3*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]])/(3*d^(2/3)) - (2^(1/3)*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]) /(9*d^(2/3)) - (3^(1/4)*Sqrt[2 - Sqrt[3]]*c^(1/3)*(c^(1/3) + d^(1/3)*x)*Sq rt[(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 + Sqr t[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])))/(8*c^2)
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 + 1)/(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^n*(m + 1)) Int[(e*x)^(m + n)*( a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -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 = 1.95 (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 | \(\text {Expression too large to display}\) | \(874\) |
Input:
int(1/x^2/(d*x^3+c)^(1/2)/(d*x^3+4*c),x,method=_RETURNVERBOSE)
Output:
-1/4*(d*x^3+c)^(1/2)/c^2/x-1/12*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)))+1/36*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),1/6/d*(2*I*(-c*d^2)^(1/3)*_alpha^2 *3^(1/2)*d-I*(-c*d^2)^(2/3)*_alpha*3^(1/2)+I*3^(1/2)*c*d-3*(-c*d^2)^(2/...
Leaf count of result is larger than twice the leaf count of optimal. 2293 vs. \(2 (490) = 980\).
Time = 0.86 (sec) , antiderivative size = 2293, normalized size of antiderivative = 3.29 \[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/x^2/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
Output:
-1/144*(2*(1/432)^(1/6)*c^2*x*(-d^2/c^11)^(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^8*d^2*x^7 - c^9*d*x^4 - 2*c^10*x)*(-d^2/c^11)^(2/3) + 12*(1/2)^(1/3)*(c^4*d^3*x^8 - 7*c^5*d^2*x^5 - 8*c^6*d*x^2)*(-d^2/c^11)^(1/3) + 6*(1296*(1/432)^(5/6)*c^10*d*x^5*(-d^2/ c^11)^(5/6) + sqrt(1/3)*(5*c^6*d^2*x^6 - 20*c^7*d*x^3 - 16*c^8)*sqrt(-d^2/ c^11) + 2*(1/432)^(1/6)*(c^2*d^3*x^7 - 16*c^3*d^2*x^4 - 8*c^4*d*x)*(-d^2/c ^11)^(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)) - 2*(1/432)^(1/6)*c^2*x*(-d^2/c^11)^(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^8*d^2*x^7 - c^9*d*x^4 - 2 *c^10*x)*(-d^2/c^11)^(2/3) + 12*(1/2)^(1/3)*(c^4*d^3*x^8 - 7*c^5*d^2*x^5 - 8*c^6*d*x^2)*(-d^2/c^11)^(1/3) - 6*(1296*(1/432)^(5/6)*c^10*d*x^5*(-d^2/c ^11)^(5/6) + sqrt(1/3)*(5*c^6*d^2*x^6 - 20*c^7*d*x^3 - 16*c^8)*sqrt(-d^2/c ^11) + 2*(1/432)^(1/6)*(c^2*d^3*x^7 - 16*c^3*d^2*x^4 - 8*c^4*d*x)*(-d^2/c^ 11)^(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)) + 36*sqrt(d)*x*weierstrassZeta(0, -4*c/d, weierstrassPInverse(0, -4*c/ d, x)) - (1/432)^(1/6)*(sqrt(-3)*c^2*x + c^2*x)*(-d^2/c^11)^(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^8*d^2* x^7 - c^9*d*x^4 - 2*c^10*x + sqrt(-3)*(c^8*d^2*x^7 - c^9*d*x^4 - 2*c^10*x) )*(-d^2/c^11)^(2/3) - 6*(1/2)^(1/3)*(c^4*d^3*x^8 - 7*c^5*d^2*x^5 - 8*c^6*d *x^2 - sqrt(-3)*(c^4*d^3*x^8 - 7*c^5*d^2*x^5 - 8*c^6*d*x^2))*(-d^2/c^11...
\[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {1}{x^{2} \sqrt {c + d x^{3}} \cdot \left (4 c + d x^{3}\right )}\, dx \] Input:
integrate(1/x**2/(d*x**3+c)**(1/2)/(d*x**3+4*c),x)
Output:
Integral(1/(x**2*sqrt(c + d*x**3)*(4*c + d*x**3)), x)
\[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int { \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} x^{2}} \,d x } \] Input:
integrate(1/x^2/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
Output:
integrate(1/((d*x^3 + 4*c)*sqrt(d*x^3 + c)*x^2), x)
\[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int { \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} x^{2}} \,d x } \] Input:
integrate(1/x^2/(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
Output:
integrate(1/((d*x^3 + 4*c)*sqrt(d*x^3 + c)*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {1}{x^2\,\sqrt {d\,x^3+c}\,\left (d\,x^3+4\,c\right )} \,d x \] Input:
int(1/(x^2*(c + d*x^3)^(1/2)*(4*c + d*x^3)),x)
Output:
int(1/(x^2*(c + d*x^3)^(1/2)*(4*c + d*x^3)), x)
\[ \int \frac {1}{x^2 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{d^{2} x^{8}+5 c d \,x^{5}+4 c^{2} x^{2}}d x \] Input:
int(1/x^2/(d*x^3+c)^(1/2)/(d*x^3+4*c),x)
Output:
int(sqrt(c + d*x**3)/(4*c**2*x**2 + 5*c*d*x**5 + d**2*x**8),x)