### 3.1909 $$\int (d+e x)^2 \sqrt{a d e+(c d^2+a e^2) x+c d e x^2} \, dx$$

Optimal. Leaf size=268 $\frac{5 \left (c d^2-a e^2\right )^2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c^3 d^3 e}+\frac{5 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{24 c^2 d^2}-\frac{5 \left (c d^2-a e^2\right )^4 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{128 c^{7/2} d^{7/2} e^{3/2}}+\frac{(d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 c d}$

[Out]

(5*(c*d^2 - a*e^2)^2*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(64*c^3*d^3*e) +
(5*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(24*c^2*d^2) + ((d + e*x)*(a*d*e + (c*d^2 +
a*e^2)*x + c*d*e*x^2)^(3/2))/(4*c*d) - (5*(c*d^2 - a*e^2)^4*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sq
rt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(128*c^(7/2)*d^(7/2)*e^(3/2))

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Rubi [A]  time = 0.18869, antiderivative size = 268, 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 = {670, 640, 612, 621, 206} $\frac{5 \left (c d^2-a e^2\right )^2 \left (a e^2+c d^2+2 c d e x\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 c^3 d^3 e}+\frac{5 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{24 c^2 d^2}-\frac{5 \left (c d^2-a e^2\right )^4 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{128 c^{7/2} d^{7/2} e^{3/2}}+\frac{(d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{4 c d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^2*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*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(64*c^3*d^3*e) +
(5*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(24*c^2*d^2) + ((d + e*x)*(a*d*e + (c*d^2 +
a*e^2)*x + c*d*e*x^2)^(3/2))/(4*c*d) - (5*(c*d^2 - a*e^2)^4*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sq
rt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(128*c^(7/2)*d^(7/2)*e^(3/2))

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 612

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

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 (d+e x)^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 c d}+\frac{\left (5 \left (d^2-\frac{a e^2}{c}\right )\right ) \int (d+e x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{8 d}\\ &=\frac{5 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 c^2 d^2}+\frac{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 c d}+\frac{\left (5 \left (d^2-\frac{a e^2}{c}\right )^2\right ) \int \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{16 d^2}\\ &=\frac{5 \left (c d^2-a e^2\right )^2 \left (c d^2+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}}{64 c^3 d^3 e}+\frac{5 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 c^2 d^2}+\frac{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 c d}-\frac{\left (5 \left (c d^2-a e^2\right )^4\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{128 c^3 d^3 e}\\ &=\frac{5 \left (c d^2-a e^2\right )^2 \left (c d^2+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}}{64 c^3 d^3 e}+\frac{5 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 c^2 d^2}+\frac{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 c d}-\frac{\left (5 \left (c d^2-a e^2\right )^4\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 )}{64 c^3 d^3 e}\\ &=\frac{5 \left (c d^2-a e^2\right )^2 \left (c d^2+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}}{64 c^3 d^3 e}+\frac{5 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 c^2 d^2}+\frac{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{4 c d}-\frac{5 \left (c d^2-a e^2\right )^4 \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 )}{128 c^{7/2} d^{7/2} e^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.950161, size = 316, normalized size = 1.18 $\frac{\sqrt{c d} \left (\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{c d} (d+e x) \left (a^2 c^2 d^2 e^3 \left (73 d^2-19 d e x-2 e^2 x^2\right )+5 a^3 c d e^5 (e x-11 d)+15 a^4 e^7+a c^3 d^3 e \left (191 d^2 e x+15 d^3+172 d e^2 x^2+56 e^3 x^3\right )+c^4 d^4 x \left (118 d^2 e x+15 d^3+136 d e^2 x^2+48 e^3 x^3\right )\right )-15 \left (c d^2-a e^2\right )^{9/2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )\right )}{192 c^{9/2} d^{9/2} e^{3/2} \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[c*d]*(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(15*a^4*e^7 + 5*a^3*c*d*e^5*(-11*d + e*x) + a^2*c^2*d^
2*e^3*(73*d^2 - 19*d*e*x - 2*e^2*x^2) + c^4*d^4*x*(15*d^3 + 118*d^2*e*x + 136*d*e^2*x^2 + 48*e^3*x^3) + a*c^3*
d^3*e*(15*d^3 + 191*d^2*e*x + 172*d*e^2*x^2 + 56*e^3*x^3)) - 15*(c*d^2 - a*e^2)^(9/2)*Sqrt[a*e + c*d*x]*Sqrt[(
c*d*(d + e*x))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*
e^2])]))/(192*c^(9/2)*d^(9/2)*e^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [B]  time = 0.05, size = 730, normalized size = 2.7 \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)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

[Out]

11/24/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-5/16*e^2/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a+5/64*e^
5/d^3/c^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^3-5/64*e^3/d/c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a
^2-5/64*e*d/c*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a-5/128/e*d^5*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)+5/32*e*a*d^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)+1/4*e*x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(
3/2)/d/c-5/24*e^2/d^2/c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*a+5/32*d^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)
^(1/2)*x+5/64/e*d^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5/32*e^4/d^2/c^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)
^(1/2)*x*a^2-5/128*e^7/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^4-15/64*e^3*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)*a^2+5/32*e^5*a^3/c^2/d*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)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84461, size = 1424, normalized size = 5.31 \begin{align*} \left [\frac{15 \,{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\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 (48 \, c^{4} d^{4} e^{4} x^{3} + 15 \, c^{4} d^{7} e + 73 \, a c^{3} d^{5} e^{3} - 55 \, a^{2} c^{2} d^{3} e^{5} + 15 \, a^{3} c d e^{7} + 8 \,{\left (17 \, c^{4} d^{5} e^{3} + a c^{3} d^{3} e^{5}\right )} x^{2} + 2 \,{\left (59 \, c^{4} d^{6} e^{2} + 18 \, a c^{3} d^{4} e^{4} - 5 \, a^{2} c^{2} d^{2} e^{6}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{768 \, c^{4} d^{4} e^{2}}, \frac{15 \,{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\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 (48 \, c^{4} d^{4} e^{4} x^{3} + 15 \, c^{4} d^{7} e + 73 \, a c^{3} d^{5} e^{3} - 55 \, a^{2} c^{2} d^{3} e^{5} + 15 \, a^{3} c d e^{7} + 8 \,{\left (17 \, c^{4} d^{5} e^{3} + a c^{3} d^{3} e^{5}\right )} x^{2} + 2 \,{\left (59 \, c^{4} d^{6} e^{2} + 18 \, a c^{3} d^{4} e^{4} - 5 \, a^{2} c^{2} d^{2} e^{6}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{384 \, c^{4} d^{4} e^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(15*(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*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*(48*c^4*d^4*e^4*x^3 + 15*c^4*d^7*e + 73*a*c^3*d^5*
e^3 - 55*a^2*c^2*d^3*e^5 + 15*a^3*c*d*e^7 + 8*(17*c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^2 + 2*(59*c^4*d^6*e^2 + 18*a*
c^3*d^4*e^4 - 5*a^2*c^2*d^2*e^6)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e^2), 1/384*(15*(c^4
*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*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*(48*c^4*d^4*e^4*x^3 + 15*c^4*d^7*e + 73*a*c^3*d^5*e^3 - 55*a^2*c^2*d^3*e^5 + 15*a^3*
c*d*e^7 + 8*(17*c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^2 + 2*(59*c^4*d^6*e^2 + 18*a*c^3*d^4*e^4 - 5*a^2*c^2*d^2*e^6)*x
)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e^2)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))*(d + e*x)**2, x)

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Giac [A]  time = 1.24141, size = 402, normalized size = 1.5 \begin{align*} \frac{1}{192} \, \sqrt{c d x^{2} e + c d^{2} x + a x e^{2} + a d e}{\left (2 \,{\left (4 \,{\left (6 \, x e^{2} + \frac{{\left (17 \, c^{3} d^{4} e^{4} + a c^{2} d^{2} e^{6}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} x + \frac{{\left (59 \, c^{3} d^{5} e^{3} + 18 \, a c^{2} d^{3} e^{5} - 5 \, a^{2} c d e^{7}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} x + \frac{{\left (15 \, c^{3} d^{6} e^{2} + 73 \, a c^{2} d^{4} e^{4} - 55 \, a^{2} c d^{2} e^{6} + 15 \, a^{3} e^{8}\right )} e^{\left (-3\right )}}{c^{3} d^{3}}\right )} + \frac{5 \,{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt{c d} e^{\left (-\frac{3}{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 + c d^{2} x + a x e^{2} + a d e}\right )} c d e - \sqrt{c d} a e^{\frac{5}{2}} \right |}\right )}{128 \, c^{4} d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*(4*(6*x*e^2 + (17*c^3*d^4*e^4 + a*c^2*d^2*e^6)*e^(-3)/(c^
3*d^3))*x + (59*c^3*d^5*e^3 + 18*a*c^2*d^3*e^5 - 5*a^2*c*d*e^7)*e^(-3)/(c^3*d^3))*x + (15*c^3*d^6*e^2 + 73*a*c
^2*d^4*e^4 - 55*a^2*c*d^2*e^6 + 15*a^3*e^8)*e^(-3)/(c^3*d^3)) + 5/128*(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d
^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*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 + c*d^2*x + a*x*e^2 + a*d*e))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^4*d^4)