3.1818 \(\int \frac{(A+B x) (d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=250 \[ \frac{5 e^2 \sqrt{d+e x} (-7 a B e+A b e+6 b B d)}{8 b^4 (b d-a e)}-\frac{5 e^2 (-7 a B e+A b e+6 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2} \sqrt{b d-a e}}-\frac{(d+e x)^{5/2} (-7 a B e+A b e+6 b B d)}{12 b^2 (a+b x)^2 (b d-a e)}-\frac{5 e (d+e x)^{3/2} (-7 a B e+A b e+6 b B d)}{24 b^3 (a+b x) (b d-a e)}-\frac{(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

(5*e^2*(6*b*B*d + A*b*e - 7*a*B*e)*Sqrt[d + e*x])/(8*b^4*(b*d - a*e)) - (5*e*(6*b*B*d + A*b*e - 7*a*B*e)*(d +
e*x)^(3/2))/(24*b^3*(b*d - a*e)*(a + b*x)) - ((6*b*B*d + A*b*e - 7*a*B*e)*(d + e*x)^(5/2))/(12*b^2*(b*d - a*e)
*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(7/2))/(3*b*(b*d - a*e)*(a + b*x)^3) - (5*e^2*(6*b*B*d + A*b*e - 7*a*B*
e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(9/2)*Sqrt[b*d - a*e])

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Rubi [A]  time = 0.201067, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {27, 78, 47, 50, 63, 208} \[ \frac{5 e^2 \sqrt{d+e x} (-7 a B e+A b e+6 b B d)}{8 b^4 (b d-a e)}-\frac{5 e^2 (-7 a B e+A b e+6 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2} \sqrt{b d-a e}}-\frac{(d+e x)^{5/2} (-7 a B e+A b e+6 b B d)}{12 b^2 (a+b x)^2 (b d-a e)}-\frac{5 e (d+e x)^{3/2} (-7 a B e+A b e+6 b B d)}{24 b^3 (a+b x) (b d-a e)}-\frac{(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*e^2*(6*b*B*d + A*b*e - 7*a*B*e)*Sqrt[d + e*x])/(8*b^4*(b*d - a*e)) - (5*e*(6*b*B*d + A*b*e - 7*a*B*e)*(d +
e*x)^(3/2))/(24*b^3*(b*d - a*e)*(a + b*x)) - ((6*b*B*d + A*b*e - 7*a*B*e)*(d + e*x)^(5/2))/(12*b^2*(b*d - a*e)
*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(7/2))/(3*b*(b*d - a*e)*(a + b*x)^3) - (5*e^2*(6*b*B*d + A*b*e - 7*a*B*
e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(9/2)*Sqrt[b*d - a*e])

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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{(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(A+B x) (d+e x)^{5/2}}{(a+b x)^4} \, dx\\ &=-\frac{(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(6 b B d+A b e-7 a B e) \int \frac{(d+e x)^{5/2}}{(a+b x)^3} \, dx}{6 b (b d-a e)}\\ &=-\frac{(6 b B d+A b e-7 a B e) (d+e x)^{5/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(5 e (6 b B d+A b e-7 a B e)) \int \frac{(d+e x)^{3/2}}{(a+b x)^2} \, dx}{24 b^2 (b d-a e)}\\ &=-\frac{5 e (6 b B d+A b e-7 a B e) (d+e x)^{3/2}}{24 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d+A b e-7 a B e) (d+e x)^{5/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}+\frac{\left (5 e^2 (6 b B d+A b e-7 a B e)\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{16 b^3 (b d-a e)}\\ &=\frac{5 e^2 (6 b B d+A b e-7 a B e) \sqrt{d+e x}}{8 b^4 (b d-a e)}-\frac{5 e (6 b B d+A b e-7 a B e) (d+e x)^{3/2}}{24 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d+A b e-7 a B e) (d+e x)^{5/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}+\frac{\left (5 e^2 (6 b B d+A b e-7 a B e)\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b^4}\\ &=\frac{5 e^2 (6 b B d+A b e-7 a B e) \sqrt{d+e x}}{8 b^4 (b d-a e)}-\frac{5 e (6 b B d+A b e-7 a B e) (d+e x)^{3/2}}{24 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d+A b e-7 a B e) (d+e x)^{5/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(5 e (6 b B d+A b e-7 a B e)) \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^4}\\ &=\frac{5 e^2 (6 b B d+A b e-7 a B e) \sqrt{d+e x}}{8 b^4 (b d-a e)}-\frac{5 e (6 b B d+A b e-7 a B e) (d+e x)^{3/2}}{24 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d+A b e-7 a B e) (d+e x)^{5/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}-\frac{5 e^2 (6 b B d+A b e-7 a B e) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{9/2} \sqrt{b d-a e}}\\ \end{align*}

Mathematica [C]  time = 0.092206, size = 99, normalized size = 0.4 \[ \frac{(d+e x)^{7/2} \left (\frac{7 (a B-A b)}{(a+b x)^3}-\frac{e^2 (-7 a B e+A b e+6 b B d) \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^3}\right )}{21 b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(7/2)*((7*(-(A*b) + a*B))/(a + b*x)^3 - (e^2*(6*b*B*d + A*b*e - 7*a*B*e)*Hypergeometric2F1[3, 7/2,
9/2, (b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^3))/(21*b*(b*d - a*e))

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Maple [B]  time = 0.021, size = 573, normalized size = 2.3 \begin{align*} 2\,{\frac{{e}^{2}B\sqrt{ex+d}}{{b}^{4}}}-{\frac{11\,{e}^{3}A}{8\,b \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{29\,{e}^{3}Ba}{8\,{b}^{2} \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{9\,{e}^{2}Bd}{4\,b \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{e}^{4}Aa}{3\,{b}^{2} \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{e}^{3}Ad}{3\,b \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{17\,B{a}^{2}{e}^{4}}{3\,{b}^{3} \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{29\,{e}^{3}Bad}{3\,{b}^{2} \left ( bex+ae \right ) ^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+4\,{\frac{{e}^{2}B \left ( ex+d \right ) ^{3/2}{d}^{2}}{b \left ( bex+ae \right ) ^{3}}}-{\frac{5\,A{a}^{2}{e}^{5}}{8\,{b}^{3} \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{5\,{e}^{4}Aad}{4\,{b}^{2} \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{5\,A{d}^{2}{e}^{3}}{8\,b \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{19\,B{e}^{5}{a}^{3}}{8\,{b}^{4} \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{13\,B{a}^{2}d{e}^{4}}{2\,{b}^{3} \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{47\,{e}^{3}Ba{d}^{2}}{8\,{b}^{2} \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}-{\frac{7\,B{d}^{3}{e}^{2}}{4\,b \left ( bex+ae \right ) ^{3}}\sqrt{ex+d}}+{\frac{5\,{e}^{3}A}{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}}}}-{\frac{35\,{e}^{3}Ba}{8\,{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{15\,{e}^{2}Bd}{4\,{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((B*x+A)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

2*e^2*B/b^4*(e*x+d)^(1/2)-11/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*A+29/8*e^3/b^2/(b*e*x+a*e)^3*(e*x+d)^(5/2)*B*
a-9/4*e^2/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*B*d-5/3*e^4/b^2/(b*e*x+a*e)^3*A*(e*x+d)^(3/2)*a+5/3*e^3/b/(b*e*x+a*e)^
3*A*(e*x+d)^(3/2)*d+17/3*e^4/b^3/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*a^2-29/3*e^3/b^2/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*
a*d+4*e^2/b/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*d^2-5/8*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*A*a^2+5/4*e^4/b^2/(b*e*x
+a*e)^3*(e*x+d)^(1/2)*A*a*d-5/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*A*d^2+19/8*e^5/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/
2)*B*a^3-13/2*e^4/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*a^2*d+47/8*e^3/b^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*a*d^2-7/4
*e^2/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*d^3+5/8*e^3/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^
(1/2))*A-35/8*e^3/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*B+15/4*e^2/b^3/((a*e-b
*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*B*d

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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((B*x+A)*(e*x+d)^(5/2)/(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 = 1.53373, size = 2188, normalized size = 8.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*(6*B*a^3*b*d*e^2 - (7*B*a^4 - A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (7*B*a*b^3 - A*b^4)*e^3)*x^3 + 3*(6*B*
a*b^3*d*e^2 - (7*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (7*B*a^3*b - A*a^2*b^2)*e^3)*x)*sqrt(b
^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(4*(B*a*b^4 + 2*A
*b^5)*d^3 + 2*(8*B*a^2*b^3 + A*a*b^4)*d^2*e - 5*(25*B*a^3*b^2 - A*a^2*b^3)*d*e^2 + 15*(7*B*a^4*b - A*a^3*b^2)*
e^3 - 48*(B*b^5*d*e^2 - B*a*b^4*e^3)*x^3 + 3*(18*B*b^5*d^2*e - (95*B*a*b^4 - 11*A*b^5)*d*e^2 + 11*(7*B*a^2*b^3
- A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (23*B*a*b^4 + 13*A*b^5)*d^2*e - (169*B*a^2*b^3 - 7*A*a*b^4)*d*e^2 + 20
*(7*B*a^3*b^2 - A*a^2*b^3)*e^3)*x)*sqrt(e*x + d))/(a^3*b^6*d - a^4*b^5*e + (b^9*d - a*b^8*e)*x^3 + 3*(a*b^8*d
- a^2*b^7*e)*x^2 + 3*(a^2*b^7*d - a^3*b^6*e)*x), 1/24*(15*(6*B*a^3*b*d*e^2 - (7*B*a^4 - A*a^3*b)*e^3 + (6*B*b^
4*d*e^2 - (7*B*a*b^3 - A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (7*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^
2*d*e^2 - (7*B*a^3*b - A*a^2*b^2)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*
x + b*d)) - (4*(B*a*b^4 + 2*A*b^5)*d^3 + 2*(8*B*a^2*b^3 + A*a*b^4)*d^2*e - 5*(25*B*a^3*b^2 - A*a^2*b^3)*d*e^2
+ 15*(7*B*a^4*b - A*a^3*b^2)*e^3 - 48*(B*b^5*d*e^2 - B*a*b^4*e^3)*x^3 + 3*(18*B*b^5*d^2*e - (95*B*a*b^4 - 11*A
*b^5)*d*e^2 + 11*(7*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (23*B*a*b^4 + 13*A*b^5)*d^2*e - (169*B*a^
2*b^3 - 7*A*a*b^4)*d*e^2 + 20*(7*B*a^3*b^2 - A*a^2*b^3)*e^3)*x)*sqrt(e*x + d))/(a^3*b^6*d - a^4*b^5*e + (b^9*d
- a*b^8*e)*x^3 + 3*(a*b^8*d - a^2*b^7*e)*x^2 + 3*(a^2*b^7*d - a^3*b^6*e)*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((B*x+A)*(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.21066, size = 500, normalized size = 2. \begin{align*} \frac{2 \, \sqrt{x e + d} B e^{2}}{b^{4}} + \frac{5 \,{\left (6 \, B b d e^{2} - 7 \, B a e^{3} + A b e^{3}\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^{4}} - \frac{54 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{3} d e^{2} - 96 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} d^{2} e^{2} + 42 \, \sqrt{x e + d} B b^{3} d^{3} e^{2} - 87 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{2} e^{3} + 33 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{3} e^{3} + 232 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{2} d e^{3} - 40 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} d e^{3} - 141 \, \sqrt{x e + d} B a b^{2} d^{2} e^{3} + 15 \, \sqrt{x e + d} A b^{3} d^{2} e^{3} - 136 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b e^{4} + 40 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{2} e^{4} + 156 \, \sqrt{x e + d} B a^{2} b d e^{4} - 30 \, \sqrt{x e + d} A a b^{2} d e^{4} - 57 \, \sqrt{x e + d} B a^{3} e^{5} + 15 \, \sqrt{x e + d} A a^{2} b e^{5}}{24 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x*e + d)*B*e^2/b^4 + 5/8*(6*B*b*d*e^2 - 7*B*a*e^3 + A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e
))/(sqrt(-b^2*d + a*b*e)*b^4) - 1/24*(54*(x*e + d)^(5/2)*B*b^3*d*e^2 - 96*(x*e + d)^(3/2)*B*b^3*d^2*e^2 + 42*s
qrt(x*e + d)*B*b^3*d^3*e^2 - 87*(x*e + d)^(5/2)*B*a*b^2*e^3 + 33*(x*e + d)^(5/2)*A*b^3*e^3 + 232*(x*e + d)^(3/
2)*B*a*b^2*d*e^3 - 40*(x*e + d)^(3/2)*A*b^3*d*e^3 - 141*sqrt(x*e + d)*B*a*b^2*d^2*e^3 + 15*sqrt(x*e + d)*A*b^3
*d^2*e^3 - 136*(x*e + d)^(3/2)*B*a^2*b*e^4 + 40*(x*e + d)^(3/2)*A*a*b^2*e^4 + 156*sqrt(x*e + d)*B*a^2*b*d*e^4
- 30*sqrt(x*e + d)*A*a*b^2*d*e^4 - 57*sqrt(x*e + d)*B*a^3*e^5 + 15*sqrt(x*e + d)*A*a^2*b*e^5)/(((x*e + d)*b -
b*d + a*e)^3*b^4)