Integrand size = 26, antiderivative size = 689 \[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {2 x^2 \sqrt {c+d x^3}}{7 d}-\frac {50 c \sqrt {c+d x^3}}{7 d^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {2 \sqrt [3]{2} c^{7/6} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{5/3}}+\frac {2 \sqrt [3]{2} c^{7/6} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^{5/3}}-\frac {2 \sqrt [3]{2} c^{7/6} \text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{d^{5/3}}+\frac {2 \sqrt [3]{2} c^{7/6} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 d^{5/3}}+\frac {25 \sqrt [4]{3} \sqrt {2-\sqrt {3}} c^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{7 d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {50 \sqrt {2} c^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{7 \sqrt [4]{3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
2/7*x^2*(d*x^3+c)^(1/2)/d-50/7*c*(d*x^3+c)^(1/2)/d^(5/3)/((1+3^(1/2))*c^(1 /3)+d^(1/3)*x)-2/3*2^(1/3)*c^(7/6)*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+2^(1/3) *d^(1/3)*x)/(d*x^3+c)^(1/2))*3^(1/2)/d^(5/3)+2/3*2^(1/3)*c^(7/6)*arctan(1/ 3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))*3^(1/2)/d^(5/3)-2*2^(1/3)*c^(7/6)*arcta nh(c^(1/6)*(c^(1/3)-2^(1/3)*d^(1/3)*x)/(d*x^3+c)^(1/2))/d^(5/3)+2/3*2^(1/3 )*c^(7/6)*arctanh((d*x^3+c)^(1/2)/c^(1/2))/d^(5/3)+25/7*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)-50/21*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)*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^(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 = 6.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.19 \[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {8 x^2 \left (c+d x^3\right )-8 c x^2 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )-5 d x^5 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{28 d \sqrt {c+d x^3}} \] Input:
Integrate[(x^4*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
Output:
(8*x^2*(c + d*x^3) - 8*c*x^2*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3 , -((d*x^3)/c), -1/4*(d*x^3)/c] - 5*d*x^5*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3 , 1/2, 1, 8/3, -((d*x^3)/c), -1/4*(d*x^3)/c])/(28*d*Sqrt[c + d*x^3])
Time = 1.23 (sec) , antiderivative size = 691, 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 = {978, 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.
| \(\displaystyle \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\) |
| \(\Big \downarrow \) 978 |
| \(\displaystyle \frac {2 x^2 \sqrt {c+d x^3}}{7 d}-\frac {2 \int \frac {c x \left (25 d x^3+16 c\right )}{2 \sqrt {d x^3+c} \left (d x^3+4 c\right )}dx}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x^2 \sqrt {c+d x^3}}{7 d}-\frac {c \int \frac {x \left (25 d x^3+16 c\right )}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}dx}{7 d}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {2 x^2 \sqrt {c+d x^3}}{7 d}-\frac {c \int \left (\frac {25 x}{\sqrt {d x^3+c}}-\frac {84 c x}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}\right )dx}{7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 x^2 \sqrt {c+d x^3}}{7 d}-\frac {c \left (\frac {50 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {25 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {14 \sqrt [3]{2} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}-\frac {14 \sqrt [3]{2} \sqrt [6]{c} \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} d^{2/3}}+\frac {14 \sqrt [3]{2} \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{d^{2/3}}-\frac {14 \sqrt [3]{2} \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 d^{2/3}}+\frac {50 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{7 d}\) |
Input:
Int[(x^4*Sqrt[c + d*x^3])/(4*c + d*x^3),x]
Output:
(2*x^2*Sqrt[c + d*x^3])/(7*d) - (c*((50*Sqrt[c + d*x^3])/(d^(2/3)*((1 + Sq rt[3])*c^(1/3) + d^(1/3)*x)) + (14*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)) - (14* 2^(1/3)*c^(1/6)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(Sqrt[3]*d^(2/3 )) + (14*2^(1/3)*c^(1/6)*ArcTanh[(c^(1/6)*(c^(1/3) - 2^(1/3)*d^(1/3)*x))/S qrt[c + d*x^3]])/d^(2/3) - (14*2^(1/3)*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/Sqr t[c]])/(3*d^(2/3)) - (25*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]) + (50*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]*Ellipti cF[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*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*(m + n*(p + q) + 1))), x] - Simp[e^n/(b*(m + n*(p + q) + 1)) Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c , d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && 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 = 1.88 (sec) , antiderivative size = 867, normalized size of antiderivative = 1.26
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(867\) |
| risch | \(\text {Expression too large to display}\) | \(872\) |
| default | \(\text {Expression too large to display}\) | \(1309\) |
Input:
int(x^4*(d*x^3+c)^(1/2)/(d*x^3+4*c),x,method=_RETURNVERBOSE)
Output:
2/7*x^2*(d*x^3+c)^(1/2)/d+50/21*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)))-4/3*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))*E llipticPi(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)^(2/3...
Leaf count of result is larger than twice the leaf count of optimal. 2442 vs. \(2 (488) = 976\).
Time = 3.01 (sec) , antiderivative size = 2442, normalized size of antiderivative = 3.54 \[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\text {Too large to display} \] Input:
integrate(x^4*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="fricas")
Output:
1/42*(12*sqrt(d*x^3 + c)*d*x^2 - 14*(4/27)^(1/6)*d^2*(-c^7/d^10)^(1/6)*log (32*(9*(4/27)^(5/6)*(d^11*x^9 - 66*c*d^10*x^6 - 72*c^2*d^9*x^3 - 32*c^3*d^ 8)*(-c^7/d^10)^(5/6) - 96*sqrt(1/3)*(c^3*d^7*x^7 - c^4*d^6*x^4 - 2*c^5*d^5 *x)*sqrt(-c^7/d^10) + 4*(9*4^(2/3)*c^2*d^8*x^5*(-c^7/d^10)^(2/3) + 2*c^6*d ^2*x^7 - 32*c^7*d*x^4 - 16*c^8*x + 4^(1/3)*(5*c^4*d^5*x^6 - 20*c^5*d^4*x^3 - 16*c^6*d^3)*(-c^7/d^10)^(1/3))*sqrt(d*x^3 + c) - 24*(4/27)^(1/6)*(c^5*d ^4*x^8 - 7*c^6*d^3*x^5 - 8*c^7*d^2*x^2)*(-c^7/d^10)^(1/6))/(d^3*x^9 + 12*c *d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 14*(4/27)^(1/6)*d^2*(-c^7/d^10)^(1/6) *log(-32*(9*(4/27)^(5/6)*(d^11*x^9 - 66*c*d^10*x^6 - 72*c^2*d^9*x^3 - 32*c ^3*d^8)*(-c^7/d^10)^(5/6) - 96*sqrt(1/3)*(c^3*d^7*x^7 - c^4*d^6*x^4 - 2*c^ 5*d^5*x)*sqrt(-c^7/d^10) - 4*(9*4^(2/3)*c^2*d^8*x^5*(-c^7/d^10)^(2/3) + 2* c^6*d^2*x^7 - 32*c^7*d*x^4 - 16*c^8*x + 4^(1/3)*(5*c^4*d^5*x^6 - 20*c^5*d^ 4*x^3 - 16*c^6*d^3)*(-c^7/d^10)^(1/3))*sqrt(d*x^3 + c) - 24*(4/27)^(1/6)*( c^5*d^4*x^8 - 7*c^6*d^3*x^5 - 8*c^7*d^2*x^2)*(-c^7/d^10)^(1/6))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 300*c*sqrt(d)*weierstrassZeta(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) + 7*(4/27)^(1/6)*(sqrt(-3)*d^2 - d^2)*(-c^7/d^10)^(1/6)*log(32*(9*(4/27)^(5/6)*(d^11*x^9 - 66*c*d^10*x^6 - 72*c^2*d^9*x^3 - 32*c^3*d^8 + sqrt(-3)*(d^11*x^9 - 66*c*d^10*x^6 - 72*c ^2*d^9*x^3 - 32*c^3*d^8))*(-c^7/d^10)^(5/6) + 192*sqrt(1/3)*(c^3*d^7*x^7 - c^4*d^6*x^4 - 2*c^5*d^5*x)*sqrt(-c^7/d^10) + 4*(4*c^6*d^2*x^7 - 64*c^7...
\[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int \frac {x^{4} \sqrt {c + d x^{3}}}{4 c + d x^{3}}\, dx \] Input:
integrate(x**4*(d*x**3+c)**(1/2)/(d*x**3+4*c),x)
Output:
Integral(x**4*sqrt(c + d*x**3)/(4*c + d*x**3), x)
\[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{4}}{d x^{3} + 4 \, c} \,d x } \] Input:
integrate(x^4*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")
Output:
integrate(sqrt(d*x^3 + c)*x^4/(d*x^3 + 4*c), x)
\[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{4}}{d x^{3} + 4 \, c} \,d x } \] Input:
integrate(x^4*(d*x^3+c)^(1/2)/(d*x^3+4*c),x, algorithm="giac")
Output:
integrate(sqrt(d*x^3 + c)*x^4/(d*x^3 + 4*c), x)
Timed out. \[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\int \frac {x^4\,\sqrt {d\,x^3+c}}{d\,x^3+4\,c} \,d x \] Input:
int((x^4*(c + d*x^3)^(1/2))/(4*c + d*x^3),x)
Output:
int((x^4*(c + d*x^3)^(1/2))/(4*c + d*x^3), x)
\[ \int \frac {x^4 \sqrt {c+d x^3}}{4 c+d x^3} \, dx=\frac {2 \sqrt {d \,x^{3}+c}\, x^{2}-25 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) c d -16 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \right ) c^{2}}{7 d} \] Input:
int(x^4*(d*x^3+c)^(1/2)/(d*x^3+4*c),x)
Output:
(2*sqrt(c + d*x**3)*x**2 - 25*int((sqrt(c + d*x**3)*x**4)/(4*c**2 + 5*c*d* x**3 + d**2*x**6),x)*c*d - 16*int((sqrt(c + d*x**3)*x)/(4*c**2 + 5*c*d*x** 3 + d**2*x**6),x)*c**2)/(7*d)