### 3.2072 $$\int \frac{1}{(d+e x)^{7/2} (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx$$

Optimal. Leaf size=393 $-\frac{315 c^4 d^4 \sqrt{d+e x}}{64 \left (c d^2-a e^2\right )^5 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{105 c^3 d^3}{64 \sqrt{d+e x} \left (c d^2-a e^2\right )^4 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{21 c^2 d^2}{32 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{315 c^4 d^4 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{64 \left (c d^2-a e^2\right )^{11/2}}+\frac{3 c d}{8 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{4 (d+e x)^{7/2} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

[Out]

1/(4*(c*d^2 - a*e^2)*(d + e*x)^(7/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3*c*d)/(8*(c*d^2 - a*e^2)
^2*(d + e*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (21*c^2*d^2)/(32*(c*d^2 - a*e^2)^3*(d + e*x)
^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (105*c^3*d^3)/(64*(c*d^2 - a*e^2)^4*Sqrt[d + e*x]*Sqrt[a
*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (315*c^4*d^4*Sqrt[d + e*x])/(64*(c*d^2 - a*e^2)^5*Sqrt[a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2]) - (315*c^4*d^4*Sqrt[e]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/
(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/(64*(c*d^2 - a*e^2)^(11/2))

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Rubi [A]  time = 0.31957, antiderivative size = 393, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {672, 666, 660, 205} $-\frac{315 c^4 d^4 \sqrt{d+e x}}{64 \left (c d^2-a e^2\right )^5 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{105 c^3 d^3}{64 \sqrt{d+e x} \left (c d^2-a e^2\right )^4 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{21 c^2 d^2}{32 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{315 c^4 d^4 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{64 \left (c d^2-a e^2\right )^{11/2}}+\frac{3 c d}{8 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{4 (d+e x)^{7/2} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(7/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

1/(4*(c*d^2 - a*e^2)*(d + e*x)^(7/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (3*c*d)/(8*(c*d^2 - a*e^2)
^2*(d + e*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (21*c^2*d^2)/(32*(c*d^2 - a*e^2)^3*(d + e*x)
^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (105*c^3*d^3)/(64*(c*d^2 - a*e^2)^4*Sqrt[d + e*x]*Sqrt[a
*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (315*c^4*d^4*Sqrt[d + e*x])/(64*(c*d^2 - a*e^2)^5*Sqrt[a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2]) - (315*c^4*d^4*Sqrt[e]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/
(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/(64*(c*d^2 - a*e^2)^(11/2))

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(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] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 666

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

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*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]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^{7/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{(9 c d) \int \frac{1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )}\\ &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 c d}{8 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (21 c^2 d^2\right ) \int \frac{1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{16 \left (c d^2-a e^2\right )^2}\\ &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 c d}{8 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{21 c^2 d^2}{32 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (105 c^3 d^3\right ) \int \frac{1}{\sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{64 \left (c d^2-a e^2\right )^3}\\ &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 c d}{8 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{21 c^2 d^2}{32 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{105 c^3 d^3}{64 \left (c d^2-a e^2\right )^4 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (315 c^4 d^4\right ) \int \frac{\sqrt{d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{128 \left (c d^2-a e^2\right )^4}\\ &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 c d}{8 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{21 c^2 d^2}{32 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{105 c^3 d^3}{64 \left (c d^2-a e^2\right )^4 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{315 c^4 d^4 \sqrt{d+e x}}{64 \left (c d^2-a e^2\right )^5 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (315 c^4 d^4 e\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{128 \left (c d^2-a e^2\right )^5}\\ &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 c d}{8 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{21 c^2 d^2}{32 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{105 c^3 d^3}{64 \left (c d^2-a e^2\right )^4 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{315 c^4 d^4 \sqrt{d+e x}}{64 \left (c d^2-a e^2\right )^5 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (315 c^4 d^4 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}}\right )}{64 \left (c d^2-a e^2\right )^5}\\ &=\frac{1}{4 \left (c d^2-a e^2\right ) (d+e x)^{7/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{3 c d}{8 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{21 c^2 d^2}{32 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{105 c^3 d^3}{64 \left (c d^2-a e^2\right )^4 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{315 c^4 d^4 \sqrt{d+e x}}{64 \left (c d^2-a e^2\right )^5 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{315 c^4 d^4 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{c d^2-a e^2} \sqrt{d+e x}}\right )}{64 \left (c d^2-a e^2\right )^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0299, size = 81, normalized size = 0.21 $-\frac{2 c^4 d^4 \sqrt{d+e x} \, _2F_1\left (-\frac{1}{2},5;\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{\left (c d^2-a e^2\right )^5 \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(7/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.224, size = 767, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

-1/64*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(315*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x
+a*e)^(1/2)*x^4*c^4*d^4*e^5+1260*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*x^3*c^
4*d^5*e^4+1890*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*x^2*c^4*d^6*e^3-315*((a*
e^2-c*d^2)*e)^(1/2)*x^4*c^4*d^4*e^4+1260*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2
)*x*c^4*d^7*e^2-105*((a*e^2-c*d^2)*e)^(1/2)*x^3*a*c^3*d^3*e^5-1155*((a*e^2-c*d^2)*e)^(1/2)*x^3*c^4*d^5*e^3+315
*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*c^4*d^8*e+42*((a*e^2-c*d^2)*e)^(1/2)*x
^2*a^2*c^2*d^2*e^6-399*((a*e^2-c*d^2)*e)^(1/2)*x^2*a*c^3*d^4*e^4-1533*((a*e^2-c*d^2)*e)^(1/2)*x^2*c^4*d^6*e^2-
24*((a*e^2-c*d^2)*e)^(1/2)*x*a^3*c*d*e^7+156*((a*e^2-c*d^2)*e)^(1/2)*x*a^2*c^2*d^3*e^5-555*((a*e^2-c*d^2)*e)^(
1/2)*x*a*c^3*d^5*e^3-837*((a*e^2-c*d^2)*e)^(1/2)*x*c^4*d^7*e+16*((a*e^2-c*d^2)*e)^(1/2)*a^4*e^8-88*((a*e^2-c*d
^2)*e)^(1/2)*a^3*c*d^2*e^6+210*((a*e^2-c*d^2)*e)^(1/2)*a^2*c^2*d^4*e^4-325*((a*e^2-c*d^2)*e)^(1/2)*a*c^3*d^6*e
^2-128*((a*e^2-c*d^2)*e)^(1/2)*c^4*d^8)/(e*x+d)^(9/2)/(c*d*x+a*e)/(a*e^2-c*d^2)^5/((a*e^2-c*d^2)*e)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}}{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)*(e*x + d)^(7/2)), x)

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Fricas [B]  time = 2.67082, size = 4335, normalized size = 11.03 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/128*(315*(c^5*d^5*e^5*x^6 + a*c^4*d^9*e + (5*c^5*d^6*e^4 + a*c^4*d^4*e^6)*x^5 + 5*(2*c^5*d^7*e^3 + a*c^4*d^
5*e^5)*x^4 + 10*(c^5*d^8*e^2 + a*c^4*d^6*e^4)*x^3 + 5*(c^5*d^9*e + 2*a*c^4*d^7*e^3)*x^2 + (c^5*d^10 + 5*a*c^4*
d^8*e^2)*x)*sqrt(-e/(c*d^2 - a*e^2))*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 - 2*sqrt(c*d*e*x^2 + a*
d*e + (c*d^2 + a*e^2)*x)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-e/(c*d^2 - a*e^2)))/(e^2*x^2 + 2*d*e*x + d^2)) -
2*(315*c^4*d^4*e^4*x^4 + 128*c^4*d^8 + 325*a*c^3*d^6*e^2 - 210*a^2*c^2*d^4*e^4 + 88*a^3*c*d^2*e^6 - 16*a^4*e^8
+ 105*(11*c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^3 + 21*(73*c^4*d^6*e^2 + 19*a*c^3*d^4*e^4 - 2*a^2*c^2*d^2*e^6)*x^2 +
3*(279*c^4*d^7*e + 185*a*c^3*d^5*e^3 - 52*a^2*c^2*d^3*e^5 + 8*a^3*c*d*e^7)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2
+ a*e^2)*x)*sqrt(e*x + d))/(a*c^5*d^15*e - 5*a^2*c^4*d^13*e^3 + 10*a^3*c^3*d^11*e^5 - 10*a^4*c^2*d^9*e^7 + 5*
a^5*c*d^7*e^9 - a^6*d^5*e^11 + (c^6*d^11*e^5 - 5*a*c^5*d^9*e^7 + 10*a^2*c^4*d^7*e^9 - 10*a^3*c^3*d^5*e^11 + 5*
a^4*c^2*d^3*e^13 - a^5*c*d*e^15)*x^6 + (5*c^6*d^12*e^4 - 24*a*c^5*d^10*e^6 + 45*a^2*c^4*d^8*e^8 - 40*a^3*c^3*d
^6*e^10 + 15*a^4*c^2*d^4*e^12 - a^6*e^16)*x^5 + 5*(2*c^6*d^13*e^3 - 9*a*c^5*d^11*e^5 + 15*a^2*c^4*d^9*e^7 - 10
*a^3*c^3*d^7*e^9 + 3*a^5*c*d^3*e^13 - a^6*d*e^15)*x^4 + 10*(c^6*d^14*e^2 - 4*a*c^5*d^12*e^4 + 5*a^2*c^4*d^10*e
^6 - 5*a^4*c^2*d^6*e^10 + 4*a^5*c*d^4*e^12 - a^6*d^2*e^14)*x^3 + 5*(c^6*d^15*e - 3*a*c^5*d^13*e^3 + 10*a^3*c^3
*d^9*e^7 - 15*a^4*c^2*d^7*e^9 + 9*a^5*c*d^5*e^11 - 2*a^6*d^3*e^13)*x^2 + (c^6*d^16 - 15*a^2*c^4*d^12*e^4 + 40*
a^3*c^3*d^10*e^6 - 45*a^4*c^2*d^8*e^8 + 24*a^5*c*d^6*e^10 - 5*a^6*d^4*e^12)*x), -1/64*(315*(c^5*d^5*e^5*x^6 +
a*c^4*d^9*e + (5*c^5*d^6*e^4 + a*c^4*d^4*e^6)*x^5 + 5*(2*c^5*d^7*e^3 + a*c^4*d^5*e^5)*x^4 + 10*(c^5*d^8*e^2 +
a*c^4*d^6*e^4)*x^3 + 5*(c^5*d^9*e + 2*a*c^4*d^7*e^3)*x^2 + (c^5*d^10 + 5*a*c^4*d^8*e^2)*x)*sqrt(e/(c*d^2 - a*e
^2))*arctan(-sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(e/(c*d^2 - a*e^2))
/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) + (315*c^4*d^4*e^4*x^4 + 128*c^4*d^8 + 325*a*c^3*d^6*e^2 - 210
*a^2*c^2*d^4*e^4 + 88*a^3*c*d^2*e^6 - 16*a^4*e^8 + 105*(11*c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^3 + 21*(73*c^4*d^6*e
^2 + 19*a*c^3*d^4*e^4 - 2*a^2*c^2*d^2*e^6)*x^2 + 3*(279*c^4*d^7*e + 185*a*c^3*d^5*e^3 - 52*a^2*c^2*d^3*e^5 + 8
*a^3*c*d*e^7)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a*c^5*d^15*e - 5*a^2*c^4*d^13*e^3
+ 10*a^3*c^3*d^11*e^5 - 10*a^4*c^2*d^9*e^7 + 5*a^5*c*d^7*e^9 - a^6*d^5*e^11 + (c^6*d^11*e^5 - 5*a*c^5*d^9*e^7
+ 10*a^2*c^4*d^7*e^9 - 10*a^3*c^3*d^5*e^11 + 5*a^4*c^2*d^3*e^13 - a^5*c*d*e^15)*x^6 + (5*c^6*d^12*e^4 - 24*a*
c^5*d^10*e^6 + 45*a^2*c^4*d^8*e^8 - 40*a^3*c^3*d^6*e^10 + 15*a^4*c^2*d^4*e^12 - a^6*e^16)*x^5 + 5*(2*c^6*d^13*
e^3 - 9*a*c^5*d^11*e^5 + 15*a^2*c^4*d^9*e^7 - 10*a^3*c^3*d^7*e^9 + 3*a^5*c*d^3*e^13 - a^6*d*e^15)*x^4 + 10*(c^
6*d^14*e^2 - 4*a*c^5*d^12*e^4 + 5*a^2*c^4*d^10*e^6 - 5*a^4*c^2*d^6*e^10 + 4*a^5*c*d^4*e^12 - a^6*d^2*e^14)*x^3
+ 5*(c^6*d^15*e - 3*a*c^5*d^13*e^3 + 10*a^3*c^3*d^9*e^7 - 15*a^4*c^2*d^7*e^9 + 9*a^5*c*d^5*e^11 - 2*a^6*d^3*e
^13)*x^2 + (c^6*d^16 - 15*a^2*c^4*d^12*e^4 + 40*a^3*c^3*d^10*e^6 - 45*a^4*c^2*d^8*e^8 + 24*a^5*c*d^6*e^10 - 5*
a^6*d^4*e^12)*x)]

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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(1/(e*x+d)**(7/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 2\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

[undef, undef, undef, undef, undef, undef, undef, 2]