3.1655 \(\int \frac{(d+e x)^{9/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=172 \[ -\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}+\frac{105 e^3 \sqrt{d+e x} (b d-a e)}{8 b^5}-\frac{105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{11/2}}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{35 e^3 (d+e x)^{3/2}}{8 b^4} \]

[Out]

(105*e^3*(b*d - a*e)*Sqrt[d + e*x])/(8*b^5) + (35*e^3*(d + e*x)^(3/2))/(8*b^4) - (21*e^2*(d + e*x)^(5/2))/(8*b
^3*(a + b*x)) - (3*e*(d + e*x)^(7/2))/(4*b^2*(a + b*x)^2) - (d + e*x)^(9/2)/(3*b*(a + b*x)^3) - (105*e^3*(b*d
- a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(11/2))

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Rubi [A]  time = 0.0973135, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {27, 47, 50, 63, 208} \[ -\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}+\frac{105 e^3 \sqrt{d+e x} (b d-a e)}{8 b^5}-\frac{105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{11/2}}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{35 e^3 (d+e x)^{3/2}}{8 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(105*e^3*(b*d - a*e)*Sqrt[d + e*x])/(8*b^5) + (35*e^3*(d + e*x)^(3/2))/(8*b^4) - (21*e^2*(d + e*x)^(5/2))/(8*b
^3*(a + b*x)) - (3*e*(d + e*x)^(7/2))/(4*b^2*(a + b*x)^2) - (d + e*x)^(9/2)/(3*b*(a + b*x)^3) - (105*e^3*(b*d
- a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(11/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^{9/2}}{(a+b x)^4} \, dx\\ &=-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{(3 e) \int \frac{(d+e x)^{7/2}}{(a+b x)^3} \, dx}{2 b}\\ &=-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{\left (21 e^2\right ) \int \frac{(d+e x)^{5/2}}{(a+b x)^2} \, dx}{8 b^2}\\ &=-\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{\left (105 e^3\right ) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{16 b^3}\\ &=\frac{35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{\left (105 e^3 (b d-a e)\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{16 b^4}\\ &=\frac{105 e^3 (b d-a e) \sqrt{d+e x}}{8 b^5}+\frac{35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{\left (105 e^3 (b d-a e)^2\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b^5}\\ &=\frac{105 e^3 (b d-a e) \sqrt{d+e x}}{8 b^5}+\frac{35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac{\left (105 e^2 (b d-a e)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{8 b^5}\\ &=\frac{105 e^3 (b d-a e) \sqrt{d+e x}}{8 b^5}+\frac{35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac{21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac{3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac{(d+e x)^{9/2}}{3 b (a+b x)^3}-\frac{105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0197256, size = 52, normalized size = 0.3 \[ \frac{2 e^3 (d+e x)^{11/2} \, _2F_1\left (4,\frac{11}{2};\frac{13}{2};-\frac{b (d+e x)}{a e-b d}\right )}{11 (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*(d + e*x)^(11/2)*Hypergeometric2F1[4, 11/2, 13/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(11*(-(b*d) + a*e)^
4)

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Maple [B]  time = 0.205, size = 525, normalized size = 3.1 \begin{align*}{\frac{2\,{e}^{3}}{3\,{b}^{4}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-8\,{\frac{{e}^{4}a\sqrt{ex+d}}{{b}^{5}}}+8\,{\frac{{e}^{3}\sqrt{ex+d}d}{{b}^{4}}}-{\frac{55\,{a}^{2}{e}^{5}}{8\,{b}^{3} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{55\,{e}^{4}ad}{4\,{b}^{2} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{55\,{e}^{3}{d}^{2}}{8\,b \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{35\,{a}^{3}{e}^{6}}{3\,{b}^{4} \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+35\,{\frac{{e}^{5} \left ( ex+d \right ) ^{3/2}{a}^{2}d}{{b}^{3} \left ( bxe+ae \right ) ^{3}}}-35\,{\frac{{e}^{4} \left ( ex+d \right ) ^{3/2}a{d}^{2}}{{b}^{2} \left ( bxe+ae \right ) ^{3}}}+{\frac{35\,{e}^{3}{d}^{3}}{3\,b \left ( bxe+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{41\,{e}^{7}{a}^{4}}{8\,{b}^{5} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{41\,{a}^{3}{e}^{6}d}{2\,{b}^{4} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{123\,{e}^{5}{d}^{2}{a}^{2}}{4\,{b}^{3} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{41\,{e}^{4}a{d}^{3}}{2\,{b}^{2} \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{41\,{e}^{3}{d}^{4}}{8\,b \left ( bxe+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{105\,{a}^{2}{e}^{5}}{8\,{b}^{5}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}-{\frac{105\,{e}^{4}ad}{4\,{b}^{4}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{105\,{e}^{3}{d}^{2}}{8\,{b}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

2/3*e^3*(e*x+d)^(3/2)/b^4-8*e^4/b^5*a*(e*x+d)^(1/2)+8*e^3/b^4*(e*x+d)^(1/2)*d-55/8*e^5/b^3/(b*e*x+a*e)^3*(e*x+
d)^(5/2)*a^2+55/4*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(5/2)*a*d-55/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*d^2-35/3*e^6/
b^4/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a^3+35*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a^2*d-35*e^4/b^2/(b*e*x+a*e)^3*(e*x
+d)^(3/2)*a*d^2+35/3*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(3/2)*d^3-41/8*e^7/b^5/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^4+41/2*e
^6/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^3*d-123/4*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d^2*a^2+41/2*e^4/b^2/(b*e*x
+a*e)^3*(e*x+d)^(1/2)*a*d^3-41/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d^4+105/8*e^5/b^5/((a*e-b*d)*b)^(1/2)*arcta
n(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2-105/4*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)
*b)^(1/2))*a*d+105/8*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*d^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.90122, size = 1528, normalized size = 8.88 \begin{align*} \left [-\frac{315 \,{\left (a^{3} b d e^{3} - a^{4} e^{4} +{\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \,{\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \,{\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e + 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) - 2 \,{\left (16 \, b^{4} e^{4} x^{4} - 8 \, b^{4} d^{4} - 18 \, a b^{3} d^{3} e - 63 \, a^{2} b^{2} d^{2} e^{2} + 420 \, a^{3} b d e^{3} - 315 \, a^{4} e^{4} + 16 \,{\left (13 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} - 3 \,{\left (55 \, b^{4} d^{2} e^{2} - 318 \, a b^{3} d e^{3} + 231 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \,{\left (25 \, b^{4} d^{3} e + 90 \, a b^{3} d^{2} e^{2} - 567 \, a^{2} b^{2} d e^{3} + 420 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{48 \,{\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, -\frac{315 \,{\left (a^{3} b d e^{3} - a^{4} e^{4} +{\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \,{\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \,{\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (16 \, b^{4} e^{4} x^{4} - 8 \, b^{4} d^{4} - 18 \, a b^{3} d^{3} e - 63 \, a^{2} b^{2} d^{2} e^{2} + 420 \, a^{3} b d e^{3} - 315 \, a^{4} e^{4} + 16 \,{\left (13 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} - 3 \,{\left (55 \, b^{4} d^{2} e^{2} - 318 \, a b^{3} d e^{3} + 231 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \,{\left (25 \, b^{4} d^{3} e + 90 \, a b^{3} d^{2} e^{2} - 567 \, a^{2} b^{2} d e^{3} + 420 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{24 \,{\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[-1/48*(315*(a^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^2*b^2*e^4)*x^2 + 3*(a^2*
b^2*d*e^3 - a^3*b*e^4)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e + 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b)
)/(b*x + a)) - 2*(16*b^4*e^4*x^4 - 8*b^4*d^4 - 18*a*b^3*d^3*e - 63*a^2*b^2*d^2*e^2 + 420*a^3*b*d*e^3 - 315*a^4
*e^4 + 16*(13*b^4*d*e^3 - 9*a*b^3*e^4)*x^3 - 3*(55*b^4*d^2*e^2 - 318*a*b^3*d*e^3 + 231*a^2*b^2*e^4)*x^2 - 2*(2
5*b^4*d^3*e + 90*a*b^3*d^2*e^2 - 567*a^2*b^2*d*e^3 + 420*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^8*x^3 + 3*a*b^7*x^2 +
3*a^2*b^6*x + a^3*b^5), -1/24*(315*(a^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^
2*b^2*e^4)*x^2 + 3*(a^2*b^2*d*e^3 - a^3*b*e^4)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*
e)/b)/(b*d - a*e)) - (16*b^4*e^4*x^4 - 8*b^4*d^4 - 18*a*b^3*d^3*e - 63*a^2*b^2*d^2*e^2 + 420*a^3*b*d*e^3 - 315
*a^4*e^4 + 16*(13*b^4*d*e^3 - 9*a*b^3*e^4)*x^3 - 3*(55*b^4*d^2*e^2 - 318*a*b^3*d*e^3 + 231*a^2*b^2*e^4)*x^2 -
2*(25*b^4*d^3*e + 90*a*b^3*d^2*e^2 - 567*a^2*b^2*d*e^3 + 420*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^8*x^3 + 3*a*b^7*x
^2 + 3*a^2*b^6*x + a^3*b^5)]

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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((e*x+d)**(9/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.24183, size = 486, normalized size = 2.83 \begin{align*} \frac{105 \,{\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{8 \, \sqrt{-b^{2} d + a b e} b^{5}} - \frac{165 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} d^{2} e^{3} - 280 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d^{3} e^{3} + 123 \, \sqrt{x e + d} b^{4} d^{4} e^{3} - 330 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{3} d e^{4} + 840 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} d^{2} e^{4} - 492 \, \sqrt{x e + d} a b^{3} d^{3} e^{4} + 165 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{2} b^{2} e^{5} - 840 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{2} d e^{5} + 738 \, \sqrt{x e + d} a^{2} b^{2} d^{2} e^{5} + 280 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} b e^{6} - 492 \, \sqrt{x e + d} a^{3} b d e^{6} + 123 \, \sqrt{x e + d} a^{4} e^{7}}{24 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{5}} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{8} e^{3} + 12 \, \sqrt{x e + d} b^{8} d e^{3} - 12 \, \sqrt{x e + d} a b^{7} e^{4}\right )}}{3 \, b^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

105/8*(b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)
*b^5) - 1/24*(165*(x*e + d)^(5/2)*b^4*d^2*e^3 - 280*(x*e + d)^(3/2)*b^4*d^3*e^3 + 123*sqrt(x*e + d)*b^4*d^4*e^
3 - 330*(x*e + d)^(5/2)*a*b^3*d*e^4 + 840*(x*e + d)^(3/2)*a*b^3*d^2*e^4 - 492*sqrt(x*e + d)*a*b^3*d^3*e^4 + 16
5*(x*e + d)^(5/2)*a^2*b^2*e^5 - 840*(x*e + d)^(3/2)*a^2*b^2*d*e^5 + 738*sqrt(x*e + d)*a^2*b^2*d^2*e^5 + 280*(x
*e + d)^(3/2)*a^3*b*e^6 - 492*sqrt(x*e + d)*a^3*b*d*e^6 + 123*sqrt(x*e + d)*a^4*e^7)/(((x*e + d)*b - b*d + a*e
)^3*b^5) + 2/3*((x*e + d)^(3/2)*b^8*e^3 + 12*sqrt(x*e + d)*b^8*d*e^3 - 12*sqrt(x*e + d)*a*b^7*e^4)/b^12