3.1725 \(\int \frac{(d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=270 \[ \frac{3 e^3 \sqrt{d+e x}}{64 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{3 e^4 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(3*e^3*Sqrt[d + e*x])/(64*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e*Sqrt[d + e*x])/(8*b^2*(a + b*x
)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^2*Sqrt[d + e*x])/(32*b^2*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^
2*x^2]) - (d + e*x)^(3/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*e^4*(a + b*x)*ArcTanh[(Sqrt[b]*
Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(5/2)*(b*d - a*e)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.139302, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {646, 47, 51, 63, 208} \[ \frac{3 e^3 \sqrt{d+e x}}{64 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{3 e^4 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*e^3*Sqrt[d + e*x])/(64*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e*Sqrt[d + e*x])/(8*b^2*(a + b*x
)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^2*Sqrt[d + e*x])/(32*b^2*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^
2*x^2]) - (d + e*x)^(3/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*e^4*(a + b*x)*ArcTanh[(Sqrt[b]*
Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(5/2)*(b*d - a*e)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 b^2 e \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{\left (a b+b^2 x\right )^4} \, dx}{8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^3 \sqrt{d+e x}} \, dx}{16 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (3 e^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^2 \sqrt{d+e x}} \, dx}{64 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{3 e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e^4 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{128 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{3 e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e^3 \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{64 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{3 e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e^4 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0343844, size = 67, normalized size = 0.25 \[ -\frac{2 e^4 (a+b x) (d+e x)^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{b (d+e x)}{b d-a e}\right )}{5 \sqrt{(a+b x)^2} (b d-a e)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.279, size = 477, normalized size = 1.8 \begin{align*}{\frac{bx+a}{64\, \left ( ae-bd \right ) ^{2}{b}^{2}} \left ( 3\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{4}{b}^{4}{e}^{4}+12\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{3}a{b}^{3}{e}^{4}+3\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{7/2}{b}^{3}+18\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}+11\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}a{b}^{2}e-11\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}{b}^{3}d+12\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{3}b{e}^{4}-11\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{a}^{2}b{e}^{2}+22\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}a{b}^{2}de-11\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{3}{d}^{2}+3\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{4}{e}^{4}-3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{3}{e}^{3}+9\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}bd{e}^{2}-9\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}a{b}^{2}{d}^{2}e+3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{3}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(b*x+a)*(3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^4*b^4*e^4+12*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b
)^(1/2))*x^3*a*b^3*e^4+3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*b^3+18*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*
x^2*a^2*b^2*e^4+11*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a*b^2*e-11*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^3*d+12*arc
tan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^3*b*e^4-11*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^2*b*e^2+22*((a*e-b
*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b^2*d*e-11*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^3*d^2+3*arctan(b*(e*x+d)^(1/2)/((a
*e-b*d)*b)^(1/2))*a^4*e^4-3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*e^3+9*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*
b*d*e^2-9*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^2*d^2*e+3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^3*d^3)/((a*e-b*d
)*b)^(1/2)/b^2/(a*e-b*d)^2/((b*x+a)^2)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{3}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(3/2)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)

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Fricas [B]  time = 1.85406, size = 2101, normalized size = 7.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

[1/128*(3*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*lo
g((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(16*b^5*d^4 - 40*a*b^4*d^3*e + 26
*a^2*b^3*d^2*e^2 + a^3*b^2*d*e^3 - 3*a^4*b*e^4 - 3*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (2*b^5*d^2*e^2 - 13*a*b^4*d*e
^3 + 11*a^2*b^3*e^4)*x^2 + (24*b^5*d^3*e - 68*a*b^4*d^2*e^2 + 55*a^2*b^3*d*e^3 - 11*a^3*b^2*e^4)*x)*sqrt(e*x +
d))/(a^4*b^6*d^3 - 3*a^5*b^5*d^2*e + 3*a^6*b^4*d*e^2 - a^7*b^3*e^3 + (b^10*d^3 - 3*a*b^9*d^2*e + 3*a^2*b^8*d*
e^2 - a^3*b^7*e^3)*x^4 + 4*(a*b^9*d^3 - 3*a^2*b^8*d^2*e + 3*a^3*b^7*d*e^2 - a^4*b^6*e^3)*x^3 + 6*(a^2*b^8*d^3
- 3*a^3*b^7*d^2*e + 3*a^4*b^6*d*e^2 - a^5*b^5*e^3)*x^2 + 4*(a^3*b^7*d^3 - 3*a^4*b^6*d^2*e + 3*a^5*b^5*d*e^2 -
a^6*b^4*e^3)*x), 1/64*(3*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b
^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (16*b^5*d^4 - 40*a*b^4*d^3*e + 26*a^2
*b^3*d^2*e^2 + a^3*b^2*d*e^3 - 3*a^4*b*e^4 - 3*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (2*b^5*d^2*e^2 - 13*a*b^4*d*e^3 +
11*a^2*b^3*e^4)*x^2 + (24*b^5*d^3*e - 68*a*b^4*d^2*e^2 + 55*a^2*b^3*d*e^3 - 11*a^3*b^2*e^4)*x)*sqrt(e*x + d))
/(a^4*b^6*d^3 - 3*a^5*b^5*d^2*e + 3*a^6*b^4*d*e^2 - a^7*b^3*e^3 + (b^10*d^3 - 3*a*b^9*d^2*e + 3*a^2*b^8*d*e^2
- a^3*b^7*e^3)*x^4 + 4*(a*b^9*d^3 - 3*a^2*b^8*d^2*e + 3*a^3*b^7*d*e^2 - a^4*b^6*e^3)*x^3 + 6*(a^2*b^8*d^3 - 3*
a^3*b^7*d^2*e + 3*a^4*b^6*d*e^2 - a^5*b^5*e^3)*x^2 + 4*(a^3*b^7*d^3 - 3*a^4*b^6*d^2*e + 3*a^5*b^5*d*e^2 - a^6*
b^4*e^3)*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((e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.24221, size = 568, normalized size = 2.1 \begin{align*} \frac{3 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{4}}{64 \,{\left (b^{4} d^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b^{3} d e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} b^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} + \frac{3 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{3} e^{4} - 11 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} d e^{4} - 11 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{4} + 3 \, \sqrt{x e + d} b^{3} d^{3} e^{4} + 11 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{2} e^{5} + 22 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{5} - 9 \, \sqrt{x e + d} a b^{2} d^{2} e^{5} - 11 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{6} + 9 \, \sqrt{x e + d} a^{2} b d e^{6} - 3 \, \sqrt{x e + d} a^{3} e^{7}}{64 \,{\left (b^{4} d^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b^{3} d e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} b^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

3/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^4*d^2*sgn((x*e + d)*b*e - b*d*e + a*e^2) - 2*a*b^3*d
*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^2*b^2*e^2*sgn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e))
+ 1/64*(3*(x*e + d)^(7/2)*b^3*e^4 - 11*(x*e + d)^(5/2)*b^3*d*e^4 - 11*(x*e + d)^(3/2)*b^3*d^2*e^4 + 3*sqrt(x*e
+ d)*b^3*d^3*e^4 + 11*(x*e + d)^(5/2)*a*b^2*e^5 + 22*(x*e + d)^(3/2)*a*b^2*d*e^5 - 9*sqrt(x*e + d)*a*b^2*d^2*
e^5 - 11*(x*e + d)^(3/2)*a^2*b*e^6 + 9*sqrt(x*e + d)*a^2*b*d*e^6 - 3*sqrt(x*e + d)*a^3*e^7)/((b^4*d^2*sgn((x*e
+ d)*b*e - b*d*e + a*e^2) - 2*a*b^3*d*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) + a^2*b^2*e^2*sgn((x*e + d)*b*e -
b*d*e + a*e^2))*((x*e + d)*b - b*d + a*e)^4)