### 3.2071 $$\int \frac{1}{(d+e x)^{5/2} (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx$$

Optimal. Leaf size=331 $-\frac{35 c^3 d^3 \sqrt{d+e x}}{8 \left (c d^2-a e^2\right )^4 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{35 c^2 d^2}{24 \sqrt{d+e x} \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{35 c^3 d^3 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 \left (c d^2-a e^2\right )^{9/2}}+\frac{7 c d}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{3 (d+e x)^{5/2} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

[Out]

1/(3*(c*d^2 - a*e^2)*(d + e*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (7*c*d)/(12*(c*d^2 - a*e^2
)^2*(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (35*c^2*d^2)/(24*(c*d^2 - a*e^2)^3*Sqrt[d +
e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (35*c^3*d^3*Sqrt[d + e*x])/(8*(c*d^2 - a*e^2)^4*Sqrt[a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (35*c^3*d^3*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])])/(8*(c*d^2 - a*e^2)^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.246489, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {672, 666, 660, 205} $-\frac{35 c^3 d^3 \sqrt{d+e x}}{8 \left (c d^2-a e^2\right )^4 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{35 c^2 d^2}{24 \sqrt{d+e x} \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{35 c^3 d^3 \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 \left (c d^2-a e^2\right )^{9/2}}+\frac{7 c d}{12 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{3 (d+e x)^{5/2} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

1/(3*(c*d^2 - a*e^2)*(d + e*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (7*c*d)/(12*(c*d^2 - a*e^2
)^2*(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (35*c^2*d^2)/(24*(c*d^2 - a*e^2)^3*Sqrt[d +
e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (35*c^3*d^3*Sqrt[d + e*x])/(8*(c*d^2 - a*e^2)^4*Sqrt[a*d*
e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (35*c^3*d^3*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])])/(8*(c*d^2 - a*e^2)^(9/2))

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)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{1}{3 \left (c d^2-a e^2\right ) (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{(7 c d) \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}{6 \left (c d^2-a e^2\right )}\\ &=\frac{1}{3 \left (c d^2-a e^2\right ) (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{7 c d}{12 \left (c d^2-a e^2\right )^2 (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 (35 c^2 d^2\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}{24 \left (c d^2-a e^2\right )^2}\\ &=\frac{1}{3 \left (c d^2-a e^2\right ) (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{7 c d}{12 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{35 c^2 d^2}{24 \left (c d^2-a e^2\right )^3 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (35 c^3 d^3\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}{16 \left (c d^2-a e^2\right )^3}\\ &=\frac{1}{3 \left (c d^2-a e^2\right ) (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{7 c d}{12 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{35 c^2 d^2}{24 \left (c d^2-a e^2\right )^3 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{35 c^3 d^3 \sqrt{d+e x}}{8 \left (c d^2-a e^2\right )^4 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (35 c^3 d^3 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}{16 \left (c d^2-a e^2\right )^4}\\ &=\frac{1}{3 \left (c d^2-a e^2\right ) (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{7 c d}{12 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{35 c^2 d^2}{24 \left (c d^2-a e^2\right )^3 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{35 c^3 d^3 \sqrt{d+e x}}{8 \left (c d^2-a e^2\right )^4 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (35 c^3 d^3 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 )}{8 \left (c d^2-a e^2\right )^4}\\ &=\frac{1}{3 \left (c d^2-a e^2\right ) (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{7 c d}{12 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{35 c^2 d^2}{24 \left (c d^2-a e^2\right )^3 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{35 c^3 d^3 \sqrt{d+e x}}{8 \left (c d^2-a e^2\right )^4 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{35 c^3 d^3 \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 )}{8 \left (c d^2-a e^2\right )^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0259689, size = 81, normalized size = 0.24 $-\frac{2 c^3 d^3 \sqrt{d+e x} \, _2F_1\left (-\frac{1}{2},4;\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{\left (c d^2-a e^2\right )^4 \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.251, size = 559, normalized size = 1.7 \begin{align*}{\frac{1}{ \left ( 24\,cdx+24\,ae \right ) \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 105\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}{x}^{3}{c}^{3}{d}^{3}{e}^{4}+315\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}{x}^{2}{c}^{3}{d}^{4}{e}^{3}+315\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}x{c}^{3}{d}^{5}{e}^{2}-105\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{3}{c}^{3}{d}^{3}{e}^{3}+105\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}{c}^{3}{d}^{6}e-35\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{2}a{c}^{2}{d}^{2}{e}^{4}-280\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{2}{c}^{3}{d}^{4}{e}^{2}+14\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}x{a}^{2}cd{e}^{5}-98\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}xa{c}^{2}{d}^{3}{e}^{3}-231\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}x{c}^{3}{d}^{5}e-8\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{a}^{3}{e}^{6}+38\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{a}^{2}c{d}^{2}{e}^{4}-87\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}a{c}^{2}{d}^{4}{e}^{2}-48\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{c}^{3}{d}^{6} \right ) \left ( ex+d \right ) ^{-{\frac{7}{2}}}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

1/24*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(105*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^3*d^3*e^4+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^2*c^3*
d^4*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)*x*c^3*d^5*e^2-105*((a*e^2-c
*d^2)*e)^(1/2)*x^3*c^3*d^3*e^3+105*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^3*
d^6*e-35*((a*e^2-c*d^2)*e)^(1/2)*x^2*a*c^2*d^2*e^4-280*((a*e^2-c*d^2)*e)^(1/2)*x^2*c^3*d^4*e^2+14*((a*e^2-c*d^
2)*e)^(1/2)*x*a^2*c*d*e^5-98*((a*e^2-c*d^2)*e)^(1/2)*x*a*c^2*d^3*e^3-231*((a*e^2-c*d^2)*e)^(1/2)*x*c^3*d^5*e-8
*((a*e^2-c*d^2)*e)^(1/2)*a^3*e^6+38*((a*e^2-c*d^2)*e)^(1/2)*a^2*c*d^2*e^4-87*((a*e^2-c*d^2)*e)^(1/2)*a*c^2*d^4
*e^2-48*((a*e^2-c*d^2)*e)^(1/2)*c^3*d^6)/(e*x+d)^(7/2)/(c*d*x+a*e)/(a*e^2-c*d^2)^4/((a*e^2-c*d^2)*e)^(1/2)

________________________________________________________________________________________

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{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(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)^(5/2)), x)

________________________________________________________________________________________

Fricas [B]  time = 2.46908, size = 3243, normalized size = 9.8 \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)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(c^4*d^4*e^4*x^5 + a*c^3*d^7*e + (4*c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^4 + 2*(3*c^4*d^6*e^2 + 2*a*c^3*d
^4*e^4)*x^3 + 2*(2*c^4*d^7*e + 3*a*c^3*d^5*e^3)*x^2 + (c^4*d^8 + 4*a*c^3*d^6*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*(105*c^3*d^3*e^3*x^3 + 48*c^3*d^6 +
87*a*c^2*d^4*e^2 - 38*a^2*c*d^2*e^4 + 8*a^3*e^6 + 35*(8*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^2 + 7*(33*c^3*d^5*e +
14*a*c^2*d^3*e^3 - 2*a^2*c*d*e^5)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a*c^4*d^12*e
- 4*a^2*c^3*d^10*e^3 + 6*a^3*c^2*d^8*e^5 - 4*a^4*c*d^6*e^7 + a^5*d^4*e^9 + (c^5*d^9*e^4 - 4*a*c^4*d^7*e^6 + 6*
a^2*c^3*d^5*e^8 - 4*a^3*c^2*d^3*e^10 + a^4*c*d*e^12)*x^5 + (4*c^5*d^10*e^3 - 15*a*c^4*d^8*e^5 + 20*a^2*c^3*d^6
*e^7 - 10*a^3*c^2*d^4*e^9 + a^5*e^13)*x^4 + 2*(3*c^5*d^11*e^2 - 10*a*c^4*d^9*e^4 + 10*a^2*c^3*d^7*e^6 - 5*a^4*
c*d^3*e^10 + 2*a^5*d*e^12)*x^3 + 2*(2*c^5*d^12*e - 5*a*c^4*d^10*e^3 + 10*a^3*c^2*d^6*e^7 - 10*a^4*c*d^4*e^9 +
3*a^5*d^2*e^11)*x^2 + (c^5*d^13 - 10*a^2*c^3*d^9*e^4 + 20*a^3*c^2*d^7*e^6 - 15*a^4*c*d^5*e^8 + 4*a^5*d^3*e^10)
*x), -1/24*(105*(c^4*d^4*e^4*x^5 + a*c^3*d^7*e + (4*c^4*d^5*e^3 + a*c^3*d^3*e^5)*x^4 + 2*(3*c^4*d^6*e^2 + 2*a*
c^3*d^4*e^4)*x^3 + 2*(2*c^4*d^7*e + 3*a*c^3*d^5*e^3)*x^2 + (c^4*d^8 + 4*a*c^3*d^6*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)) + (105*c^3*d^3*e^3*x^3 + 48*c^3*d^6 + 87*a*c^2*d^4*e^2 - 38*a^2
*c*d^2*e^4 + 8*a^3*e^6 + 35*(8*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^2 + 7*(33*c^3*d^5*e + 14*a*c^2*d^3*e^3 - 2*a^2*c
*d*e^5)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a*c^4*d^12*e - 4*a^2*c^3*d^10*e^3 + 6*a
^3*c^2*d^8*e^5 - 4*a^4*c*d^6*e^7 + a^5*d^4*e^9 + (c^5*d^9*e^4 - 4*a*c^4*d^7*e^6 + 6*a^2*c^3*d^5*e^8 - 4*a^3*c^
2*d^3*e^10 + a^4*c*d*e^12)*x^5 + (4*c^5*d^10*e^3 - 15*a*c^4*d^8*e^5 + 20*a^2*c^3*d^6*e^7 - 10*a^3*c^2*d^4*e^9
+ a^5*e^13)*x^4 + 2*(3*c^5*d^11*e^2 - 10*a*c^4*d^9*e^4 + 10*a^2*c^3*d^7*e^6 - 5*a^4*c*d^3*e^10 + 2*a^5*d*e^12)
*x^3 + 2*(2*c^5*d^12*e - 5*a*c^4*d^10*e^3 + 10*a^3*c^2*d^6*e^7 - 10*a^4*c*d^4*e^9 + 3*a^5*d^2*e^11)*x^2 + (c^5
*d^13 - 10*a^2*c^3*d^9*e^4 + 20*a^3*c^2*d^7*e^6 - 15*a^4*c*d^5*e^8 + 4*a^5*d^3*e^10)*x)]

________________________________________________________________________________________

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)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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)^(5/2)/(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]