3.17.32 \(\int \frac {(d+e x)^3}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\)

Optimal. Leaf size=255 \[ \frac {5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{7/2} d^{7/2} \sqrt {e}}+\frac {5 \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 c^3 d^3}+\frac {5 (d+e x) \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{12 c^2 d^2}+\frac {(d+e x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d} \]

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Rubi [A]  time = 0.20, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {670, 640, 621, 206} \begin {gather*} \frac {5 \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 c^3 d^3}+\frac {5 (d+e x) \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{12 c^2 d^2}+\frac {5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\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 )}{16 c^{7/2} d^{7/2} \sqrt {e}}+\frac {(d+e x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(5*(c*d^2 - a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*c^3*d^3) + (5*(c*d^2 - a*e^2)*(d + e*x)*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(12*c^2*d^2) + ((d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e
*x^2])/(3*c*d) + (5*(c*d^2 - a*e^2)^3*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])])/(16*c^(7/2)*d^(7/2)*Sqrt[e])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac {\left (5 \left (d^2-\frac {a e^2}{c}\right )\right ) \int \frac {(d+e x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{6 d}\\ &=\frac {5 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac {(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac {\left (5 \left (d^2-\frac {a e^2}{c}\right )^2\right ) \int \frac {d+e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 d^2}\\ &=\frac {5 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 c^3 d^3}+\frac {5 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac {(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac {\left (5 \left (d^2-\frac {a e^2}{c}\right )^3\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 d^3}\\ &=\frac {5 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 c^3 d^3}+\frac {5 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac {(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac {\left (5 \left (d^2-\frac {a e^2}{c}\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d e-x^2} \, dx,x,\frac {c d^2+a e^2+2 c d e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 d^3}\\ &=\frac {5 \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 c^3 d^3}+\frac {5 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{12 c^2 d^2}+\frac {(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}+\frac {5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\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 )}{16 c^{7/2} d^{7/2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 214, normalized size = 0.84 \begin {gather*} \frac {\sqrt {(d+e x) (a e+c d x)} \left (\sqrt {c} \sqrt {d} \left (15 a^2 e^4-10 a c d e^2 (4 d+e x)+c^2 d^2 \left (33 d^2+26 d e x+8 e^2 x^2\right )\right )+\frac {15 \sqrt {c d} \left (c d^2-a e^2\right )^{5/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )}{\sqrt {e} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}\right )}{24 c^{7/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

(Sqrt[(a*e + c*d*x)*(d + e*x)]*(Sqrt[c]*Sqrt[d]*(15*a^2*e^4 - 10*a*c*d*e^2*(4*d + e*x) + c^2*d^2*(33*d^2 + 26*
d*e*x + 8*e^2*x^2)) + (15*Sqrt[c*d]*(c*d^2 - a*e^2)^(5/2)*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/
(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[e]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)])))/(24*c^(7
/2)*d^(7/2))

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IntegrateAlgebraic [B]  time = 10.37, size = 13174, normalized size = 51.66 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^3/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2],x]

[Out]

Result too large to show

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fricas [A]  time = 0.52, size = 534, normalized size = 2.09 \begin {gather*} \left [\frac {15 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {c d e} \log \left (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 + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 33 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 15 \, a^{2} c d e^{5} + 2 \, {\left (13 \, c^{3} d^{4} e^{2} - 5 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{96 \, c^{4} d^{4} e}, -\frac {15 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) - 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 33 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 15 \, a^{2} c d e^{5} + 2 \, {\left (13 \, c^{3} d^{4} e^{2} - 5 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{48 \, c^{4} d^{4} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(15*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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*(8*c^3*d^3*e^3*x^2 + 33*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 15*a^2*c*d*e^5 +
2*(13*c^3*d^4*e^2 - 5*a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e), -1/48*(15*(
c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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*(8*c^3*d^3*e^3*x^2 + 33*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 15*a^2*c*d*e^5 + 2*(13*c^3*d^4*e^2 - 5*a*c^2*d
^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e)]

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giac [A]  time = 0.98, size = 234, normalized size = 0.92 \begin {gather*} \frac {1}{24} \, \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, x {\left (\frac {4 \, x e^{2}}{c d} + \frac {{\left (13 \, c^{2} d^{3} e^{3} - 5 \, a c d e^{5}\right )} e^{\left (-2\right )}}{c^{3} d^{3}}\right )} + \frac {{\left (33 \, c^{2} d^{4} e^{2} - 40 \, a c d^{2} e^{4} + 15 \, a^{2} e^{6}\right )} e^{\left (-2\right )}}{c^{3} d^{3}}\right )} - \frac {5 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{16 \, c^{4} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)*(2*x*(4*x*e^2/(c*d) + (13*c^2*d^3*e^3 - 5*a*c*d*e^5)*e^(-2)/(
c^3*d^3)) + (33*c^2*d^4*e^2 - 40*a*c*d^2*e^4 + 15*a^2*e^6)*e^(-2)/(c^3*d^3)) - 5/16*(c^3*d^6 - 3*a*c^2*d^4*e^2
+ 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(c*d)*e^(-1/2)*log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - s
qrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^4*d^4)

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maple [B]  time = 0.07, size = 513, normalized size = 2.01 \begin {gather*} -\frac {5 a^{3} e^{6} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c^{3} d^{3}}+\frac {15 a^{2} e^{4} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c^{2} d}-\frac {15 a d \,e^{2} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c}+\frac {5 d^{3} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, e^{2} x^{2}}{3 c d}-\frac {5 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,e^{3} x}{12 c^{2} d^{2}}+\frac {13 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, e x}{12 c}+\frac {5 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{2} e^{4}}{8 c^{3} d^{3}}-\frac {5 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,e^{2}}{3 c^{2} d}+\frac {11 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, d}{8 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2),x)

[Out]

1/3*e^2*x^2/c/d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)-5/12*e^3/c^2/d^2*x*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(
1/2)*a+13/12*e/c*x*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+5/8*e^4/c^3/d^3*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(
1/2)*a^2-5/3*e^2/c^2/d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a+11/8/c*d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1
/2)-5/16*e^6/c^3/d^3*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(
c*d*e)^(1/2)*a^3+15/16*e^4/c^2/d*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)
*x)^(1/2))/(c*d*e)^(1/2)*a^2-15/16*e^2/c*d*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*
e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a+5/16*d^3*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+
(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2),x)

[Out]

int((d + e*x)^3/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

Integral((d + e*x)**3/sqrt((d + e*x)*(a*e + c*d*x)), x)

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