3.1955 \(\int \frac{(d+e x)^4}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=241 \[ \frac{5 e (d+e x) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c^2 d^2}+\frac{15 e \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c^3 d^3}+\frac{15 \sqrt{e} \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{7/2} d^{7/2}}-\frac{2 (d+e x)^3}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[Out]

(-2*(d + e*x)^3)/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (15*e*(c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2])/(4*c^3*d^3) + (5*e*(d + e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(2*c^2*d^2
) + (15*Sqrt[e]*(c*d^2 - a*e^2)^2*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])])/(8*c^(7/2)*d^(7/2))

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Rubi [A]  time = 0.161916, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {668, 670, 640, 621, 206} \[ \frac{5 e (d+e x) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c^2 d^2}+\frac{15 e \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c^3 d^3}+\frac{15 \sqrt{e} \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{7/2} d^{7/2}}-\frac{2 (d+e x)^3}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^3)/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (15*e*(c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2])/(4*c^3*d^3) + (5*e*(d + e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(2*c^2*d^2
) + (15*Sqrt[e]*(c*d^2 - a*e^2)^2*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])])/(8*c^(7/2)*d^(7/2))

Rule 668

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*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=-\frac{2 (d+e x)^3}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{(5 e) \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}{c d}\\ &=-\frac{2 (d+e x)^3}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 e (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac{\left (15 e \left (c d^2-a e^2\right )\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}{4 c^2 d^2}\\ &=-\frac{2 (d+e x)^3}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{15 e \left (c d^2-a e^2\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac{5 e (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac{\left (15 e \left (c d^2-a e^2\right )^2\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 c^3 d^3}\\ &=-\frac{2 (d+e x)^3}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{15 e \left (c d^2-a e^2\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac{5 e (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac{\left (15 e \left (c d^2-a e^2\right )^2\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 )}{4 c^3 d^3}\\ &=-\frac{2 (d+e x)^3}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{15 e \left (c d^2-a e^2\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac{5 e (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac{15 \sqrt{e} \left (c d^2-a e^2\right )^2 \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 )}{8 c^{7/2} d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0575939, size = 100, normalized size = 0.41 \[ -\frac{2 \left (c d^2-a e^2\right )^3 \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{c^4 d^4 \sqrt{(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*(c*d^2 - a*e^2)^3*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)]*Hypergeometric2F1[-5/2, -1/2, 1/2, (e*(a*e + c*d*x
))/(-(c*d^2) + a*e^2)])/(c^4*d^4*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [B]  time = 0.059, size = 1428, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

-49/16*d^2/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+15/8*e*d/c*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)
+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)-25/16*e^2*d^4/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+
(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a-11/8*e^6/c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*
x^2)^(1/2)*a^3+1/2*e^3*x^3/d/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-55/16*e^4/d^2/c^3/(a*d*e+(a*e^2+c*d^2)*
x+c*d*e*x^2)^(1/2)*a^2+15/8*e^9/d^3/c^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/
2)*x*a^4-5*e^7/d/c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a^3+9/4*e^5*d/
c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a^2-49/16*d^6*c/(-a^2*e^4+2*a*c*d
^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+73/16*e^2/c^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*
a+11/4*e^2/c*x^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+2*d^4*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c
*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+15/16*e^6/d^4/c^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^3
-15/8*e*d/c*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+21/8*e^4*d^2/c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(
a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^2+15/8*e^5/d^3/c^3*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*
e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)*a^2-49/8*e*d^5*c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c
*d^2)*x+c*d*e*x^2)^(1/2)*x+15/16*e^10/d^4/c^4/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^
2)^(1/2)*a^5-25/16*e^8/d^2/c^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^4+3*
e^3*d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a+15/4*e^3/d/c^2*x/(a*d*e+(
a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a-15/4*e^3/d/c^2*ln((1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+
c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)*a-15/8*e^5/d^3/c^3*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^2-5/4*
e^4/d^2/c^2*x^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.27712, size = 1216, normalized size = 5.05 \begin{align*} \left [\frac{15 \,{\left (a c^{2} d^{4} e - 2 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} +{\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x\right )} \sqrt{\frac{e}{c d}} \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} + 8 \,{\left (c^{2} d^{3} e + a c d e^{3}\right )} x + 4 \,{\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{\frac{e}{c d}}\right ) + 4 \,{\left (2 \, c^{2} d^{2} e^{2} x^{2} - 8 \, c^{2} d^{4} + 25 \, a c d^{2} e^{2} - 15 \, a^{2} e^{4} +{\left (9 \, c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{16 \,{\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}, -\frac{15 \,{\left (a c^{2} d^{4} e - 2 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} +{\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x\right )} \sqrt{-\frac{e}{c d}} \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{-\frac{e}{c d}}}{2 \,{\left (c d e^{2} x^{2} + a d e^{2} +{\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) - 2 \,{\left (2 \, c^{2} d^{2} e^{2} x^{2} - 8 \, c^{2} d^{4} + 25 \, a c d^{2} e^{2} - 15 \, a^{2} e^{4} +{\left (9 \, c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{8 \,{\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*(a*c^2*d^4*e - 2*a^2*c*d^2*e^3 + a^3*e^5 + (c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*x)*sqrt(e/(c*d)
)*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 + 8*(c^2*d^3*e + a*c*d*e^3)*x + 4*(2*c^2*d^2*e*x +
c^2*d^3 + a*c*d*e^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e/(c*d))) + 4*(2*c^2*d^2*e^2*x^2 - 8*c^
2*d^4 + 25*a*c*d^2*e^2 - 15*a^2*e^4 + (9*c^2*d^3*e - 5*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*
x))/(c^4*d^4*x + a*c^3*d^3*e), -1/8*(15*(a*c^2*d^4*e - 2*a^2*c*d^2*e^3 + a^3*e^5 + (c^3*d^5 - 2*a*c^2*d^3*e^2
+ a^2*c*d*e^4)*x)*sqrt(-e/(c*d))*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(-e/(c*d))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) - 2*(2*c^2*d^2*e^2*x^2 - 8*c^2*d^4 + 25*a*
c*d^2*e^2 - 15*a^2*e^4 + (9*c^2*d^3*e - 5*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*
x + a*c^3*d^3*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{4}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral((d + e*x)**4/((d + e*x)*(a*e + c*d*x))**(3/2), x)

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Giac [B]  time = 1.31402, size = 648, normalized size = 2.69 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (c^{4} d^{6} e^{5} - 2 \, a c^{3} d^{4} e^{7} + a^{2} c^{2} d^{2} e^{9}\right )} x}{c^{5} d^{7} e^{2} - 2 \, a c^{4} d^{5} e^{4} + a^{2} c^{3} d^{3} e^{6}} + \frac{11 \, c^{4} d^{7} e^{4} - 27 \, a c^{3} d^{5} e^{6} + 21 \, a^{2} c^{2} d^{3} e^{8} - 5 \, a^{3} c d e^{10}}{c^{5} d^{7} e^{2} - 2 \, a c^{4} d^{5} e^{4} + a^{2} c^{3} d^{3} e^{6}}\right )} x + \frac{c^{4} d^{8} e^{3} + 18 \, a c^{3} d^{6} e^{5} - 54 \, a^{2} c^{2} d^{4} e^{7} + 50 \, a^{3} c d^{2} e^{9} - 15 \, a^{4} e^{11}}{c^{5} d^{7} e^{2} - 2 \, a c^{4} d^{5} e^{4} + a^{2} c^{3} d^{3} e^{6}}\right )} x - \frac{8 \, c^{4} d^{9} e^{2} - 41 \, a c^{3} d^{7} e^{4} + 73 \, a^{2} c^{2} d^{5} e^{6} - 55 \, a^{3} c d^{3} e^{8} + 15 \, a^{4} d e^{10}}{c^{5} d^{7} e^{2} - 2 \, a c^{4} d^{5} e^{4} + a^{2} c^{3} d^{3} e^{6}}}{4 \, \sqrt{c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x}} - \frac{15 \,{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\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 )}{8 \, c^{4} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

1/4*(((2*(c^4*d^6*e^5 - 2*a*c^3*d^4*e^7 + a^2*c^2*d^2*e^9)*x/(c^5*d^7*e^2 - 2*a*c^4*d^5*e^4 + a^2*c^3*d^3*e^6)
+ (11*c^4*d^7*e^4 - 27*a*c^3*d^5*e^6 + 21*a^2*c^2*d^3*e^8 - 5*a^3*c*d*e^10)/(c^5*d^7*e^2 - 2*a*c^4*d^5*e^4 +
a^2*c^3*d^3*e^6))*x + (c^4*d^8*e^3 + 18*a*c^3*d^6*e^5 - 54*a^2*c^2*d^4*e^7 + 50*a^3*c*d^2*e^9 - 15*a^4*e^11)/(
c^5*d^7*e^2 - 2*a*c^4*d^5*e^4 + a^2*c^3*d^3*e^6))*x - (8*c^4*d^9*e^2 - 41*a*c^3*d^7*e^4 + 73*a^2*c^2*d^5*e^6 -
55*a^3*c*d^3*e^8 + 15*a^4*d*e^10)/(c^5*d^7*e^2 - 2*a*c^4*d^5*e^4 + a^2*c^3*d^3*e^6))/sqrt(c*d*x^2*e + a*d*e +
(c*d^2 + a*e^2)*x) - 15/8*(c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*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) - sqrt(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)