### 3.2055 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{15/2}} \, dx$$

Optimal. Leaf size=301 $\frac{5 c^3 d^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 e^3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}-\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{32 e^3 (d+e x)^{5/2}}+\frac{5 c^4 d^4 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{64 e^{7/2} \left (c d^2-a e^2\right )^{3/2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}$

[Out]

(-5*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(32*e^3*(d + e*x)^(5/2)) + (5*c^3*d^3*Sqrt[a*d*e + (c
*d^2 + a*e^2)*x + c*d*e*x^2])/(64*e^3*(c*d^2 - a*e^2)*(d + e*x)^(3/2)) - (5*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c
*d*e*x^2)^(3/2))/(24*e^2*(d + e*x)^(9/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(4*e*(d + e*x)^(13/2
)) + (5*c^4*d^4*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*e^(7/2)*(c*d^2 - a*e^2)^(3/2))

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Rubi [A]  time = 0.234159, antiderivative size = 301, 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 = {662, 672, 660, 205} $\frac{5 c^3 d^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{64 e^3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}-\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{32 e^3 (d+e x)^{5/2}}+\frac{5 c^4 d^4 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{64 e^{7/2} \left (c d^2-a e^2\right )^{3/2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(15/2),x]

[Out]

(-5*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(32*e^3*(d + e*x)^(5/2)) + (5*c^3*d^3*Sqrt[a*d*e + (c
*d^2 + a*e^2)*x + c*d*e*x^2])/(64*e^3*(c*d^2 - a*e^2)*(d + e*x)^(3/2)) - (5*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c
*d*e*x^2)^(3/2))/(24*e^2*(d + e*x)^(9/2)) - (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(4*e*(d + e*x)^(13/2
)) + (5*c^4*d^4*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*e^(7/2)*(c*d^2 - a*e^2)^(3/2))

Rule 662

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

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 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{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{15/2}} \, dx &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}+\frac{(5 c d) \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{11/2}} \, dx}{8 e}\\ &=-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}+\frac{\left (5 c^2 d^2\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{7/2}} \, dx}{16 e^2}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{32 e^3 (d+e x)^{5/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}+\frac{\left (5 c^3 d^3\right ) \int \frac{1}{(d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{64 e^3}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{32 e^3 (d+e x)^{5/2}}+\frac{5 c^3 d^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 e^3 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}+\frac{\left (5 c^4 d^4\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 e^3 \left (c d^2-a e^2\right )}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{32 e^3 (d+e x)^{5/2}}+\frac{5 c^3 d^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 e^3 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}+\frac{\left (5 c^4 d^4\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 e^2 \left (c d^2-a e^2\right )}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{32 e^3 (d+e x)^{5/2}}+\frac{5 c^3 d^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{64 e^3 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{24 e^2 (d+e x)^{9/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{4 e (d+e x)^{13/2}}+\frac{5 c^4 d^4 \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 e^{7/2} \left (c d^2-a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0728909, size = 83, normalized size = 0.28 $\frac{2 c^4 d^4 ((d+e x) (a e+c d x))^{7/2} \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{7 (d+e x)^{7/2} \left (c d^2-a e^2\right )^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^(15/2),x]

[Out]

(2*c^4*d^4*((a*e + c*d*x)*(d + e*x))^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)
])/(7*(c*d^2 - a*e^2)^5*(d + e*x)^(7/2))

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Maple [B]  time = 0.292, size = 662, normalized size = 2.2 \begin{align*}{\frac{1}{192\,{e}^{3} \left ( a{e}^{2}-c{d}^{2} \right ) }\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{4}{c}^{4}{d}^{4}{e}^{4}+60\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{3}{c}^{4}{d}^{5}{e}^{3}+90\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{2}{c}^{4}{d}^{6}{e}^{2}+60\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) x{c}^{4}{d}^{7}e-15\,{x}^{3}{c}^{3}{d}^{3}{e}^{3}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{4}{d}^{8}-118\,{x}^{2}a{c}^{2}{d}^{2}{e}^{4}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+73\,{x}^{2}{c}^{3}{d}^{4}{e}^{2}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}-136\,x{a}^{2}cd{e}^{5}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+36\,xa{c}^{2}{d}^{3}{e}^{3}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+55\,x{c}^{3}{d}^{5}e\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}-48\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{a}^{3}{e}^{6}+8\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{a}^{2}c{d}^{2}{e}^{4}+10\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}a{c}^{2}{d}^{4}{e}^{2}+15\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{c}^{3}{d}^{6} \right ) \left ( ex+d \right ) ^{-{\frac{9}{2}}}{\frac{1}{\sqrt{cdx+ae}}}{\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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(15/2),x)

[Out]

1/192*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^4*c^4
*d^4*e^4+60*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^3*c^4*d^5*e^3+90*arctanh(e*(c*d*x+a*e)^(1/2
)/((a*e^2-c*d^2)*e)^(1/2))*x^2*c^4*d^6*e^2+60*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x*c^4*d^7*e
-15*x^3*c^3*d^3*e^3*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)
^(1/2))*c^4*d^8-118*x^2*a*c^2*d^2*e^4*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+73*x^2*c^3*d^4*e^2*(c*d*x+a*e)
^(1/2)*((a*e^2-c*d^2)*e)^(1/2)-136*x*a^2*c*d*e^5*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+36*x*a*c^2*d^3*e^3*
(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+55*x*c^3*d^5*e*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)-48*((a*e^2-
c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a^3*e^6+8*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a^2*c*d^2*e^4+10*((a*e^2
-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*a*c^2*d^4*e^2+15*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c^3*d^6)/(e*x+d)
^(9/2)/(c*d*x+a*e)^(1/2)/e^3/(a*e^2-c*d^2)/((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{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{\frac{15}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(15/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/2)/(e*x + d)^(15/2), x)

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Fricas [B]  time = 2.06199, size = 2495, normalized size = 8.29 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(15/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(c^4*d^4*e^5*x^5 + 5*c^4*d^5*e^4*x^4 + 10*c^4*d^6*e^3*x^3 + 10*c^4*d^7*e^2*x^2 + 5*c^4*d^8*e*x + c
^4*d^9)*sqrt(-c*d^2*e + a*e^3)*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)*sqrt(-c*d^2*e + a*e^3)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(15*c^4*d^8*e - 5*a*c^
3*d^6*e^3 - 2*a^2*c^2*d^4*e^5 - 56*a^3*c*d^2*e^7 + 48*a^4*e^9 - 15*(c^4*d^5*e^4 - a*c^3*d^3*e^6)*x^3 + (73*c^4
*d^6*e^3 - 191*a*c^3*d^4*e^5 + 118*a^2*c^2*d^2*e^7)*x^2 + (55*c^4*d^7*e^2 - 19*a*c^3*d^5*e^4 - 172*a^2*c^2*d^3
*e^6 + 136*a^3*c*d*e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c^2*d^9*e^4 - 2*a*c*d^7
*e^6 + a^2*d^5*e^8 + (c^2*d^4*e^9 - 2*a*c*d^2*e^11 + a^2*e^13)*x^5 + 5*(c^2*d^5*e^8 - 2*a*c*d^3*e^10 + a^2*d*e
^12)*x^4 + 10*(c^2*d^6*e^7 - 2*a*c*d^4*e^9 + a^2*d^2*e^11)*x^3 + 10*(c^2*d^7*e^6 - 2*a*c*d^5*e^8 + a^2*d^3*e^1
0)*x^2 + 5*(c^2*d^8*e^5 - 2*a*c*d^6*e^7 + a^2*d^4*e^9)*x), -1/192*(15*(c^4*d^4*e^5*x^5 + 5*c^4*d^5*e^4*x^4 + 1
0*c^4*d^6*e^3*x^3 + 10*c^4*d^7*e^2*x^2 + 5*c^4*d^8*e*x + c^4*d^9)*sqrt(c*d^2*e - a*e^3)*arctan(sqrt(c*d*e*x^2
+ a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d^2*e - a*e^3)*sqrt(e*x + d)/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)
) + (15*c^4*d^8*e - 5*a*c^3*d^6*e^3 - 2*a^2*c^2*d^4*e^5 - 56*a^3*c*d^2*e^7 + 48*a^4*e^9 - 15*(c^4*d^5*e^4 - a*
c^3*d^3*e^6)*x^3 + (73*c^4*d^6*e^3 - 191*a*c^3*d^4*e^5 + 118*a^2*c^2*d^2*e^7)*x^2 + (55*c^4*d^7*e^2 - 19*a*c^3
*d^5*e^4 - 172*a^2*c^2*d^3*e^6 + 136*a^3*c*d*e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)
)/(c^2*d^9*e^4 - 2*a*c*d^7*e^6 + a^2*d^5*e^8 + (c^2*d^4*e^9 - 2*a*c*d^2*e^11 + a^2*e^13)*x^5 + 5*(c^2*d^5*e^8
- 2*a*c*d^3*e^10 + a^2*d*e^12)*x^4 + 10*(c^2*d^6*e^7 - 2*a*c*d^4*e^9 + a^2*d^2*e^11)*x^3 + 10*(c^2*d^7*e^6 - 2
*a*c*d^5*e^8 + a^2*d^3*e^10)*x^2 + 5*(c^2*d^8*e^5 - 2*a*c*d^6*e^7 + a^2*d^4*e^9)*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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(15/2),x)

[Out]

Timed out

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Giac [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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(15/2),x, algorithm="giac")

[Out]

Timed out