3.2077 $$\int \frac{1}{\sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx$$

Optimal. Leaf size=329 $\frac{35 c^2 d^2 e \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^4 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{35 c^2 d^2 e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}-\frac{35 c d e}{12 \sqrt{d+e x} \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{7 c d \sqrt{d+e x}}{6 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}+\frac{1}{2 \sqrt{d+e x} \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}$

[Out]

1/(2*(c*d^2 - a*e^2)*Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (7*c*d*Sqrt[d + e*x])/(6*(
c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (35*c*d*e)/(12*(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^2*d^2*e*Sqrt[d + e*x])/(4*(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^2*d^2*e^(3/2)*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])])/(4*(c*d^2 - a*e^2)^(9/2))

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Rubi [A]  time = 0.280659, antiderivative size = 329, 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} $\frac{35 c^2 d^2 e \sqrt{d+e x}}{4 \left (c d^2-a e^2\right )^4 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{35 c^2 d^2 e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{9/2}}-\frac{35 c d e}{12 \sqrt{d+e x} \left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{7 c d \sqrt{d+e x}}{6 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}+\frac{1}{2 \sqrt{d+e x} \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [C]  time = 0.0323592, size = 83, normalized size = 0.25 $-\frac{2 c^2 d^2 (d+e x)^{3/2} \, _2F_1\left (-\frac{3}{2},3;-\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{3 \left (c d^2-a e^2\right )^3 ((d+e x) (a e+c d x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.22, size = 668, normalized size = 2. \begin{align*} -{\frac{1}{12\, \left ( cdx+ae \right ) ^{2} \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}+105\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{2}a{c}^{2}{d}^{2}{e}^{5}\sqrt{cdx+ae}+210\,{\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}+210\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) xa{c}^{2}{d}^{3}{e}^{4}\sqrt{cdx+ae}+105\,{\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 ) a{c}^{2}{d}^{4}{e}^{3}\sqrt{cdx+ae}-140\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{2}a{c}^{2}{d}^{2}{e}^{4}-175\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{2}{c}^{3}{d}^{4}{e}^{2}-21\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}x{a}^{2}cd{e}^{5}-238\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}xa{c}^{2}{d}^{3}{e}^{3}-56\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}x{c}^{3}{d}^{5}e+6\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{a}^{3}{e}^{6}-39\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{a}^{2}c{d}^{2}{e}^{4}-80\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}a{c}^{2}{d}^{4}{e}^{2}+8\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{c}^{3}{d}^{6} \right ) \left ( ex+d \right ) ^{-{\frac{5}{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)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x)

[Out]

-1/12*(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+105*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^2*a*c^2*d^2*e^5*(c*d*x+
a*e)^(1/2)+210*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+210*arct
anh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x*a*c^2*d^3*e^4*(c*d*x+a*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*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))*a*c^2*d^4*e^3*(c*d*x+a*e)^(1/2)-140*((a*e^2-c*d^2)*e)^(1/
2)*x^2*a*c^2*d^2*e^4-175*((a*e^2-c*d^2)*e)^(1/2)*x^2*c^3*d^4*e^2-21*((a*e^2-c*d^2)*e)^(1/2)*x*a^2*c*d*e^5-238*
((a*e^2-c*d^2)*e)^(1/2)*x*a*c^2*d^3*e^3-56*((a*e^2-c*d^2)*e)^(1/2)*x*c^3*d^5*e+6*((a*e^2-c*d^2)*e)^(1/2)*a^3*e
^6-39*((a*e^2-c*d^2)*e)^(1/2)*a^2*c*d^2*e^4-80*((a*e^2-c*d^2)*e)^(1/2)*a*c^2*d^4*e^2+8*((a*e^2-c*d^2)*e)^(1/2)
*c^3*d^6)/(e*x+d)^(5/2)/(c*d*x+a*e)^2/(a*e^2-c*d^2)^4/((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{5}{2}} \sqrt{e x + d}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.079, size = 3603, normalized size = 10.95 \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)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")

[Out]

[1/24*(105*(c^4*d^4*e^4*x^5 + a^2*c^2*d^5*e^3 + (3*c^4*d^5*e^3 + 2*a*c^3*d^3*e^5)*x^4 + (3*c^4*d^6*e^2 + 6*a*c
^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^3 + (c^4*d^7*e + 6*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e^5)*x^2 + (2*a*c^3*d^6*e^2 +
3*a^2*c^2*d^4*e^4)*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 - 8*c^3*d^6 + 80*a*c^2*d^4*e^2 + 39*a^2*c*d^2*e^4 - 6*a^3*e^6 + 35*(5*c^3*d^
4*e^2 + 4*a*c^2*d^2*e^4)*x^2 + 7*(8*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 3*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^2*c^4*d^11*e^2 - 4*a^3*c^3*d^9*e^4 + 6*a^4*c^2*d^7*e^6 - 4*a^5*c*d^5*e^8
+ a^6*d^3*e^10 + (c^6*d^10*e^3 - 4*a*c^5*d^8*e^5 + 6*a^2*c^4*d^6*e^7 - 4*a^3*c^3*d^4*e^9 + a^4*c^2*d^2*e^11)*x
^5 + (3*c^6*d^11*e^2 - 10*a*c^5*d^9*e^4 + 10*a^2*c^4*d^7*e^6 - 5*a^4*c^2*d^3*e^10 + 2*a^5*c*d*e^12)*x^4 + (3*c
^6*d^12*e - 6*a*c^5*d^10*e^3 - 5*a^2*c^4*d^8*e^5 + 20*a^3*c^3*d^6*e^7 - 15*a^4*c^2*d^4*e^9 + 2*a^5*c*d^2*e^11
+ a^6*e^13)*x^3 + (c^6*d^13 + 2*a*c^5*d^11*e^2 - 15*a^2*c^4*d^9*e^4 + 20*a^3*c^3*d^7*e^6 - 5*a^4*c^2*d^5*e^8 -
6*a^5*c*d^3*e^10 + 3*a^6*d*e^12)*x^2 + (2*a*c^5*d^12*e - 5*a^2*c^4*d^10*e^3 + 10*a^4*c^2*d^6*e^7 - 10*a^5*c*d
^4*e^9 + 3*a^6*d^2*e^11)*x), 1/12*(105*(c^4*d^4*e^4*x^5 + a^2*c^2*d^5*e^3 + (3*c^4*d^5*e^3 + 2*a*c^3*d^3*e^5)*
x^4 + (3*c^4*d^6*e^2 + 6*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^3 + (c^4*d^7*e + 6*a*c^3*d^5*e^3 + 3*a^2*c^2*d^3*e
^5)*x^2 + (2*a*c^3*d^6*e^2 + 3*a^2*c^2*d^4*e^4)*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 - 8*c^3*d^6 + 80*a*c^2*d^4*e^2 + 39*a^2*c*d^2*e^4 - 6*a^3*e^6 + 35*(5*c^3*d^4*e
^2 + 4*a*c^2*d^2*e^4)*x^2 + 7*(8*c^3*d^5*e + 34*a*c^2*d^3*e^3 + 3*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^2*c^4*d^11*e^2 - 4*a^3*c^3*d^9*e^4 + 6*a^4*c^2*d^7*e^6 - 4*a^5*c*d^5*e^8 + a
^6*d^3*e^10 + (c^6*d^10*e^3 - 4*a*c^5*d^8*e^5 + 6*a^2*c^4*d^6*e^7 - 4*a^3*c^3*d^4*e^9 + a^4*c^2*d^2*e^11)*x^5
+ (3*c^6*d^11*e^2 - 10*a*c^5*d^9*e^4 + 10*a^2*c^4*d^7*e^6 - 5*a^4*c^2*d^3*e^10 + 2*a^5*c*d*e^12)*x^4 + (3*c^6*
d^12*e - 6*a*c^5*d^10*e^3 - 5*a^2*c^4*d^8*e^5 + 20*a^3*c^3*d^6*e^7 - 15*a^4*c^2*d^4*e^9 + 2*a^5*c*d^2*e^11 + a
^6*e^13)*x^3 + (c^6*d^13 + 2*a*c^5*d^11*e^2 - 15*a^2*c^4*d^9*e^4 + 20*a^3*c^3*d^7*e^6 - 5*a^4*c^2*d^5*e^8 - 6*
a^5*c*d^3*e^10 + 3*a^6*d*e^12)*x^2 + (2*a*c^5*d^12*e - 5*a^2*c^4*d^10*e^3 + 10*a^4*c^2*d^6*e^7 - 10*a^5*c*d^4*
e^9 + 3*a^6*d^2*e^11)*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)**(1/2)/(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x