Integrand size = 40, antiderivative size = 251 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\frac {\left (c d^2+5 a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e}-\frac {\left (c^2 d^4-6 a c d^2 e^2-3 a^2 e^4\right ) \text {arctanh}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 \sqrt {c} \sqrt {d} e^{3/2}}-a^{3/2} \sqrt {d} e^{3/2} \text {arctanh}\left (\frac {2 a d e+\left (c d^2+a e^2\right ) x}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right ) \]
-1/8*(-3*a^2*e^4-6*a*c*d^2*e^2+c^2*d^4)*arctanh(1/2*(2*c*d*e*x+a*e^2+c*d^2 )/c^(1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/e^(3/2) /c^(1/2)/d^(1/2)-a^(3/2)*e^(3/2)*arctanh(1/2*(2*a*d*e+(a*e^2+c*d^2)*x)/a^( 1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))*d^(1/2)+1/4* (2*c*d*e*x+5*a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/e
Time = 0.53 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\frac {\sqrt {a e+c d x} \sqrt {d+e x} \left (\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x} \sqrt {d+e x} \left (5 a e^2+c d (d+2 e x)\right )+\left (-c^2 d^4+6 a c d^2 e^2+3 a^2 e^4\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )-8 a^{3/2} \sqrt {c} d e^3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sqrt {d+e x}}{\sqrt {d} \sqrt {a e+c d x}}\right )\right )}{4 \sqrt {c} \sqrt {d} e^{3/2} \sqrt {(a e+c d x) (d+e x)}} \]
(Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x ]*Sqrt[d + e*x]*(5*a*e^2 + c*d*(d + 2*e*x)) + (-(c^2*d^4) + 6*a*c*d^2*e^2 + 3*a^2*e^4)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e + c *d*x])] - 8*a^(3/2)*Sqrt[c]*d*e^3*ArcTanh[(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])/ (Sqrt[d]*Sqrt[a*e + c*d*x])]))/(4*Sqrt[c]*Sqrt[d]*e^(3/2)*Sqrt[(a*e + c*d* x)*(d + e*x)])
Time = 0.50 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1215, 1231, 27, 1269, 1092, 219, 1154, 219}
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.
| \(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx\) |
| \(\Big \downarrow \) 1215 |
| \(\displaystyle \int \frac {(a e+c d x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{x}dx\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {\left (5 a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e}-\frac {\int -\frac {c d \left (8 a^2 d e^3-\left (c^2 d^4-6 a c e^2 d^2-3 a^2 e^4\right ) x\right )}{2 x \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{4 c d e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {8 a^2 d e^3-\left (c^2 d^4-6 a c e^2 d^2-3 a^2 e^4\right ) x}{x \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{8 e}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (5 a e^2+c d^2+2 c d e x\right )}{4 e}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {8 a^2 d e^3 \int \frac {1}{x \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx-\left (-3 a^2 e^4-6 a c d^2 e^2+c^2 d^4\right ) \int \frac {1}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{8 e}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (5 a e^2+c d^2+2 c d e x\right )}{4 e}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {8 a^2 d e^3 \int \frac {1}{x \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx-2 \left (-3 a^2 e^4-6 a c d^2 e^2+c^2 d^4\right ) \int \frac {1}{4 c d e-\frac {\left (c d^2+2 c e x d+a e^2\right )^2}{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}d\frac {c d^2+2 c e x d+a e^2}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}}{8 e}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (5 a e^2+c d^2+2 c d e x\right )}{4 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {8 a^2 d e^3 \int \frac {1}{x \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx-\frac {\left (-3 a^2 e^4-6 a c d^2 e^2+c^2 d^4\right ) \text {arctanh}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{\sqrt {c} \sqrt {d} \sqrt {e}}}{8 e}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (5 a e^2+c d^2+2 c d e x\right )}{4 e}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {-16 a^2 d e^3 \int \frac {1}{4 a d e-\frac {\left (2 a d e+\left (c d^2+a e^2\right ) x\right )^2}{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}d\frac {2 a d e+\left (c d^2+a e^2\right ) x}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}-\frac {\left (-3 a^2 e^4-6 a c d^2 e^2+c^2 d^4\right ) \text {arctanh}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{\sqrt {c} \sqrt {d} \sqrt {e}}}{8 e}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (5 a e^2+c d^2+2 c d e x\right )}{4 e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-8 a^{3/2} \sqrt {d} e^{5/2} \text {arctanh}\left (\frac {x \left (a e^2+c d^2\right )+2 a d e}{2 \sqrt {a} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )-\frac {\left (-3 a^2 e^4-6 a c d^2 e^2+c^2 d^4\right ) \text {arctanh}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{\sqrt {c} \sqrt {d} \sqrt {e}}}{8 e}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (5 a e^2+c d^2+2 c d e x\right )}{4 e}\) |
((c*d^2 + 5*a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2] )/(4*e) + (-(((c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(Sqrt[c]*Sqrt[d]*Sqrt[e])) - 8*a^(3/2)*Sqrt[d]*e^(5/2)*ArcTa nh[(2*a*d*e + (c*d^2 + a*e^2)*x)/(2*Sqrt[a]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + ( c*d^2 + a*e^2)*x + c*d*e*x^2])])/(8*e)
3.5.50.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[(((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_))/( (d_) + (e_.)*(x_)), x_Symbol] :> Int[(a/d + c*(x/e))*(f + g*x)^n*(a + b*x + c*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ (c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c ^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x ] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || !R ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Integer Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(606\) vs. \(2(213)=426\).
Time = 0.64 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.42
| method | result | size |
| default | \(\frac {\frac {{\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{\frac {3}{2}}}{3}+\frac {\left (e^{2} a +c \,d^{2}\right ) \left (\frac {\left (2 c d e x +e^{2} a +c \,d^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}{4 c d e}+\frac {\left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \ln \left (\frac {\frac {1}{2} e^{2} a +\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{8 c d e \sqrt {c d e}}\right )}{2}+a d e \left (\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}+\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {1}{2} e^{2} a +\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 \sqrt {c d e}}-\frac {a d e \ln \left (\frac {2 a d e +\left (e^{2} a +c \,d^{2}\right ) x +2 \sqrt {a d e}\, \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}{x}\right )}{\sqrt {a d e}}\right )}{d}-\frac {\frac {\left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3}+\frac {\left (e^{2} a -c \,d^{2}\right ) \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+e^{2} a -c \,d^{2}\right ) \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{4 c d e}-\frac {\left (e^{2} a -c \,d^{2}\right )^{2} \ln \left (\frac {\frac {e^{2} a}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{8 c d e \sqrt {c d e}}\right )}{2}}{d}\) | \(607\) |
1/d*(1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*(a*e^2+c*d^2)*(1/4*(2 *c*d*e*x+a*e^2+c*d^2)/c/d/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+1/8*(4 *a*c*d^2*e^2-(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d* e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2))+a*d*e*((a *d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+1/2*(a*e^2+c*d^2)*ln((1/2*e^2*a+1/2* c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d *e)^(1/2)-a*d*e/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+2*(a*d*e)^(1/2)* (a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/x)))-1/d*(1/3*(c*d*e*(x+d/e)^2+(a *e^2-c*d^2)*(x+d/e))^(3/2)+1/2*(a*e^2-c*d^2)*(1/4*(2*c*d*e*(x+d/e)+e^2*a-c *d^2)/c/d/e*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(1/2)-1/8*(a*e^2-c*d^2 )^2/c/d/e*ln((1/2*e^2*a-1/2*c*d^2+c*d*e*(x+d/e))/(c*d*e)^(1/2)+(c*d*e*(x+d /e)^2+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)))
Time = 2.60 (sec) , antiderivative size = 1327, normalized size of antiderivative = 5.29 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\text {Too large to display} \]
[1/16*(8*sqrt(a*d*e)*a*c*d*e^3*log((8*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e ^2 + a^2*e^4)*x^2 - 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(a*d*e) + 8*(a*c*d^3*e + a^2*d*e^3)*x)/x^2) - (c ^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(c*d*e)*log(8*c^2*d^2*e^2*x^2 + c^ 2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^ 2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(c*d*e) + 8*(c^2*d^3*e + a*c*d*e^3)* x) + 4*(2*c^2*d^2*e^2*x + c^2*d^3*e + 5*a*c*d*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c*d*e^2), 1/8*(4*sqrt(a*d*e)*a*c*d*e^3*log((8*a^2*d ^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 - 4*sqrt(c*d*e*x^2 + a*d* e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^2 + a*e^2)*x)*sqrt(a*d*e) + 8*(a*c* d^3*e + a^2*d*e^3)*x)/x^2) + (c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(-c *d*e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d*e)/(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) + 2*(2*c^2*d^2*e^2*x + c^2*d^3*e + 5*a*c*d*e^3)*sqrt(c*d*e* x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c*d*e^2), 1/16*(16*sqrt(-a*d*e)*a*c*d*e ^3*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e + (c*d^ 2 + a*e^2)*x)*sqrt(-a*d*e)/(a*c*d^2*e^2*x^2 + a^2*d^2*e^2 + (a*c*d^3*e + a ^2*d*e^3)*x)) - (c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*sqrt(c*d*e)*log(8*c^ 2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*sqrt(c*d*e*x^2 + a*d *e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(c*d*e) + 8*(c^...
\[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}{x \left (d + e x\right )}\, dx \]
Exception generated. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Exception generated. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m operator + Error: Bad Argument Value
Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{x (d+e x)} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{x\,\left (d+e\,x\right )} \,d x \]