### 3.1966 $$\int \frac{(d+e x)^4}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx$$

Optimal. Leaf size=172 $-\frac{2 e (d+e x)}{c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{e^{3/2} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c^{5/2} d^{5/2}}-\frac{2 (d+e x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}$

[Out]

(-2*(d + e*x)^3)/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (2*e*(d + e*x))/(c^2*d^2*Sqrt[a*d*e +
(c*d^2 + a*e^2)*x + c*d*e*x^2]) + (e^(3/2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqr
t[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(c^(5/2)*d^(5/2))

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Rubi [A]  time = 0.0887596, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {668, 652, 621, 206} $-\frac{2 e (d+e x)}{c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{e^{3/2} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c^{5/2} d^{5/2}}-\frac{2 (d+e x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}$

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)^(5/2),x]

[Out]

(-2*(d + e*x)^3)/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (2*e*(d + e*x))/(c^2*d^2*Sqrt[a*d*e +
(c*d^2 + a*e^2)*x + c*d*e*x^2]) + (e^(3/2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqr
t[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(c^(5/2)*d^(5/2))

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 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{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] &&  !IntegerQ[p] && LtQ[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 )^{5/2}} \, dx &=-\frac{2 (d+e x)^3}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac{e \int \frac{(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{c d}\\ &=-\frac{2 (d+e x)^3}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac{2 e (d+e x)}{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{e^2 \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c^2 d^2}\\ &=-\frac{2 (d+e x)^3}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac{2 e (d+e x)}{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (2 e^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 )}{c^2 d^2}\\ &=-\frac{2 (d+e x)^3}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac{2 e (d+e x)}{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{e^{3/2} \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 )}{c^{5/2} d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0720668, size = 98, normalized size = 0.57 $-\frac{2 ((d+e x) (a e+c d x))^{3/2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{3 c d (a e+c d x)^3 \left (\frac{c d (d+e x)}{c d^2-a e^2}\right )^{3/2}}$

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)^(5/2),x]

[Out]

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

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Maple [B]  time = 0.061, size = 2538, normalized size = 14.8 \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)^(5/2),x)

[Out]

-1/48*e^6/d^4/c^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*a^3-31/8*e*d/c*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
3/2)+e^2*d^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-7/24*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)^(3/2)*a^3+5/16*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)^(3/2)*a-7/3*e^5*d^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d
^2)*x+c*d*e*x^2)^(1/2)*a^2-1/3*e^3*x^3/d/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-e^2/d^2/c^2*x/(a*d*e+(a*e^2
+c*d^2)*x+c*d*e*x^2)^(1/2)+1/2*e^3/d^3/c^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a+e^2/d^2/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)-1/24*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)^(3/2)*x*a^4+2/3*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)^(3/2)*x*a^3+16/3*e^4*d^4*c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a
*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a-5/4*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)^(3/2)*x*a^2-1/3*e^10/d^2/c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
1/2)*x*a^4+e^6/d^2/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^2-11/16*d^
2/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/48*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)^(3/2)+1/2*e*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)-25/16*e^2
/c^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*a-7/2*e^2/c*x^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*e/d/c
^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/6*e*d^7*c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^
2)*x+c*d*e*x^2)^(1/2)+7/16*e^4/d^2/c^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*a^2+3/2*e^3*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)*a-1/6*e^11/d^3/c^3/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^
2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^5-1/3*e^2*d^6*c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2
+c*d^2)*x+c*d*e*x^2)^(1/2)*x+5/2*e^3*d^5*c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2
)^(1/2)*a+1/8*e^5/d^3/c^3*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*a^2-7/24*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)^(3/2)*a^2-3/4*e^3/d/c^2*x/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*
a+5/2*e^9/d/c^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^4-7/3*e^7*d/c/(-a
^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^3+1/2*e^4/d^2/c^2*x^2/(a*d*e+(a*e^2+
c*d^2)*x+c*d*e*x^2)^(3/2)*a-1/24*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)^(3
/2)*x-1/48*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)^(3/2)*a^5+16/3*e^8/
c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a^3-10*e^6*d^2/(-a^2*e^4+2*a*c*
d^2*e^2-c^2*d^4)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a^2+2/3*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)^(3/2)*x*a+2*e^4/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+1/2*e^7/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)*a^
3+3/2*e^5/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)*a^2+5/16*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)^(3/2)*a^4

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.76715, size = 980, normalized size = 5.7 \begin{align*} \left [\frac{3 \,{\left (c^{2} d^{2} e x^{2} + 2 \, a c d e^{2} x + a^{2} e^{3}\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 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (4 \, c d e x + c d^{2} + 3 \, a e^{2}\right )}}{6 \,{\left (c^{4} d^{4} x^{2} + 2 \, a c^{3} d^{3} e x + a^{2} c^{2} d^{2} e^{2}\right )}}, -\frac{3 \,{\left (c^{2} d^{2} e x^{2} + 2 \, a c d e^{2} x + a^{2} e^{3}\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 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (4 \, c d e x + c d^{2} + 3 \, a e^{2}\right )}}{3 \,{\left (c^{4} d^{4} x^{2} + 2 \, a c^{3} d^{3} e x + a^{2} c^{2} d^{2} e^{2}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(c^2*d^2*e*x^2 + 2*a*c*d*e^2*x + a^2*e^3)*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*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(4*c*d*e*x + c*d^2 + 3*a*e^2)
)/(c^4*d^4*x^2 + 2*a*c^3*d^3*e*x + a^2*c^2*d^2*e^2), -1/3*(3*(c^2*d^2*e*x^2 + 2*a*c*d*e^2*x + a^2*e^3)*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*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(4*c*d*e*x + c*d^2
+ 3*a*e^2))/(c^4*d^4*x^2 + 2*a*c^3*d^3*e*x + a^2*c^2*d^2*e^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.49888, size = 837, normalized size = 4.87 \begin{align*} -\frac{2 \,{\left ({\left ({\left (\frac{4 \,{\left (c^{5} d^{9} e^{3} - 4 \, a c^{4} d^{7} e^{5} + 6 \, a^{2} c^{3} d^{5} e^{7} - 4 \, a^{3} c^{2} d^{3} e^{9} + a^{4} c d e^{11}\right )} x}{c^{6} d^{10} - 4 \, a c^{5} d^{8} e^{2} + 6 \, a^{2} c^{4} d^{6} e^{4} - 4 \, a^{3} c^{3} d^{4} e^{6} + a^{4} c^{2} d^{2} e^{8}} + \frac{3 \,{\left (3 \, c^{5} d^{10} e^{2} - 11 \, a c^{4} d^{8} e^{4} + 14 \, a^{2} c^{3} d^{6} e^{6} - 6 \, a^{3} c^{2} d^{4} e^{8} - a^{4} c d^{2} e^{10} + a^{5} e^{12}\right )}}{c^{6} d^{10} - 4 \, a c^{5} d^{8} e^{2} + 6 \, a^{2} c^{4} d^{6} e^{4} - 4 \, a^{3} c^{3} d^{4} e^{6} + a^{4} c^{2} d^{2} e^{8}}\right )} x + \frac{6 \,{\left (c^{5} d^{11} e - 3 \, a c^{4} d^{9} e^{3} + 2 \, a^{2} c^{3} d^{7} e^{5} + 2 \, a^{3} c^{2} d^{5} e^{7} - 3 \, a^{4} c d^{3} e^{9} + a^{5} d e^{11}\right )}}{c^{6} d^{10} - 4 \, a c^{5} d^{8} e^{2} + 6 \, a^{2} c^{4} d^{6} e^{4} - 4 \, a^{3} c^{3} d^{4} e^{6} + a^{4} c^{2} d^{2} e^{8}}\right )} x + \frac{c^{5} d^{12} - a c^{4} d^{10} e^{2} - 6 \, a^{2} c^{3} d^{8} e^{4} + 14 \, a^{3} c^{2} d^{6} e^{6} - 11 \, a^{4} c d^{4} e^{8} + 3 \, a^{5} d^{2} e^{10}}{c^{6} d^{10} - 4 \, a c^{5} d^{8} e^{2} + 6 \, a^{2} c^{4} d^{6} e^{4} - 4 \, a^{3} c^{3} d^{4} e^{6} + a^{4} c^{2} d^{2} e^{8}}\right )}}{3 \,{\left (c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}} - \frac{\sqrt{c d} e^{\frac{3}{2}} \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 )}{c^{3} d^{3}} \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)^(5/2),x, algorithm="giac")

[Out]

-2/3*(((4*(c^5*d^9*e^3 - 4*a*c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d*e^11)*x/(c^6*d^10 -
4*a*c^5*d^8*e^2 + 6*a^2*c^4*d^6*e^4 - 4*a^3*c^3*d^4*e^6 + a^4*c^2*d^2*e^8) + 3*(3*c^5*d^10*e^2 - 11*a*c^4*d^8
*e^4 + 14*a^2*c^3*d^6*e^6 - 6*a^3*c^2*d^4*e^8 - a^4*c*d^2*e^10 + a^5*e^12)/(c^6*d^10 - 4*a*c^5*d^8*e^2 + 6*a^2
*c^4*d^6*e^4 - 4*a^3*c^3*d^4*e^6 + a^4*c^2*d^2*e^8))*x + 6*(c^5*d^11*e - 3*a*c^4*d^9*e^3 + 2*a^2*c^3*d^7*e^5 +
2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*d*e^11)/(c^6*d^10 - 4*a*c^5*d^8*e^2 + 6*a^2*c^4*d^6*e^4 - 4*a^3*c^3
*d^4*e^6 + a^4*c^2*d^2*e^8))*x + (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 - 11*a^4*
c*d^4*e^8 + 3*a^5*d^2*e^10)/(c^6*d^10 - 4*a*c^5*d^8*e^2 + 6*a^2*c^4*d^6*e^4 - 4*a^3*c^3*d^4*e^6 + a^4*c^2*d^2*
e^8))/(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^(3/2) - sqrt(c*d)*e^(3/2)*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^3*d^3)