### 3.2043 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^{9/2}} \, dx$$

Optimal. Leaf size=185 $\frac{3 c^2 d^2 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 e^{5/2} \sqrt{c d^2-a e^2}}-\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}$

[Out]

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

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Rubi [A]  time = 0.117634, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {662, 660, 205} $\frac{3 c^2 d^2 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 e^{5/2} \sqrt{c d^2-a e^2}}-\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e^2 (d+e x)^{3/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 e (d+e x)^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^(9/2),x]

[Out]

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

Mathematica [A]  time = 0.178477, size = 176, normalized size = 0.95 $\frac{3 c^2 d^2 (d+e x)^2 \sqrt{a e+c d x} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d^2-a e^2}}\right )-\sqrt{e} \sqrt{c d^2-a e^2} \left (2 a^2 e^3+a c d e (3 d+7 e x)+c^2 d^2 x (3 d+5 e x)\right )}{4 e^{5/2} (d+e x)^{3/2} \sqrt{c d^2-a e^2} \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^(9/2),x]

[Out]

(-(Sqrt[e]*Sqrt[c*d^2 - a*e^2]*(2*a^2*e^3 + c^2*d^2*x*(3*d + 5*e*x) + a*c*d*e*(3*d + 7*e*x))) + 3*c^2*d^2*Sqrt
[a*e + c*d*x]*(d + e*x)^2*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^2 - a*e^2]])/(4*e^(5/2)*Sqrt[c*d^2 - a*e
^2]*(d + e*x)^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [A]  time = 0.247, size = 281, normalized size = 1.5 \begin{align*} -{\frac{1}{4\,{e}^{2}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{2}{c}^{2}{d}^{2}{e}^{2}+6\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) x{c}^{2}{d}^{3}e+3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{2}{d}^{4}+5\,xcde\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}a{e}^{2}+3\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}c{d}^{2} \right ) \left ( ex+d \right ) ^{-{\frac{5}{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)^(3/2)/(e*x+d)^(9/2),x)

[Out]

-1/4*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^2*c^2*d
^2*e^2+6*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x*c^2*d^3*e+3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^
2-c*d^2)*e)^(1/2))*c^2*d^4+5*x*c*d*e*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+2*((a*e^2-c*d^2)*e)^(1/2)*(c*d*
x+a*e)^(1/2)*a*e^2+3*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c*d^2)/(e*x+d)^(5/2)/(c*d*x+a*e)^(1/2)/e^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{3}{2}}}{{\left (e x + d\right )}^{\frac{9}{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)^(3/2)/(e*x+d)^(9/2),x, algorithm="maxima")

[Out]

integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)/(e*x + d)^(9/2), x)

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Fricas [A]  time = 2.0764, size = 1323, normalized size = 7.15 \begin{align*} \left [-\frac{3 \,{\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt{-c d^{2} e + a e^{3}} \log \left (-\frac{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 +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{-c d^{2} e + a e^{3}} \sqrt{e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \,{\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} - 2 \, a^{2} e^{5} + 5 \,{\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{8 \,{\left (c d^{5} e^{3} - a d^{3} e^{5} +{\left (c d^{2} e^{6} - a e^{8}\right )} x^{3} + 3 \,{\left (c d^{3} e^{5} - a d e^{7}\right )} x^{2} + 3 \,{\left (c d^{4} e^{4} - a d^{2} e^{6}\right )} x\right )}}, -\frac{3 \,{\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt{c d^{2} e - a e^{3}} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{c d^{2} e - a e^{3}} \sqrt{e x + d}}{c d e^{2} x^{2} + a d e^{2} +{\left (c d^{2} e + a e^{3}\right )} x}\right ) +{\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} - 2 \, a^{2} e^{5} + 5 \,{\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{4 \,{\left (c d^{5} e^{3} - a d^{3} e^{5} +{\left (c d^{2} e^{6} - a e^{8}\right )} x^{3} + 3 \,{\left (c d^{3} e^{5} - a d e^{7}\right )} x^{2} + 3 \,{\left (c d^{4} e^{4} - a d^{2} e^{6}\right )} x\right )}}\right ] \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)^(3/2)/(e*x+d)^(9/2),x, algorithm="fricas")

[Out]

[-1/8*(3*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*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)*sqr
t(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(3*c^2*d^4*e - a*c*d^2*e^3 - 2*a^2*e^5 + 5*(c^2*d^3*e^2 - a*c*d*e^4
)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c*d^5*e^3 - a*d^3*e^5 + (c*d^2*e^6 - a*e^8)*x
^3 + 3*(c*d^3*e^5 - a*d*e^7)*x^2 + 3*(c*d^4*e^4 - a*d^2*e^6)*x), -1/4*(3*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2
+ 3*c^2*d^4*e*x + c^2*d^5)*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)) + (3*c^2*d^4*e - a*c*d^2*e^3 - 2*a^2*
e^5 + 5*(c^2*d^3*e^2 - a*c*d*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c*d^5*e^3 - a
*d^3*e^5 + (c*d^2*e^6 - a*e^8)*x^3 + 3*(c*d^3*e^5 - a*d*e^7)*x^2 + 3*(c*d^4*e^4 - a*d^2*e^6)*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)**(3/2)/(e*x+d)**(9/2),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)^(3/2)/(e*x+d)^(9/2),x, algorithm="giac")

[Out]

Timed out