3.1817 \(\int \frac{(A+B x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=284 \[ \frac{35 e^2 (d+e x)^{3/2} (-3 a B e+A b e+2 b B d)}{24 b^4 (b d-a e)}+\frac{35 e^2 \sqrt{d+e x} (-3 a B e+A b e+2 b B d)}{8 b^5}-\frac{35 e^2 \sqrt{b d-a e} (-3 a B e+A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{11/2}}-\frac{(d+e x)^{7/2} (-3 a B e+A b e+2 b B d)}{4 b^2 (a+b x)^2 (b d-a e)}-\frac{7 e (d+e x)^{5/2} (-3 a B e+A b e+2 b B d)}{8 b^3 (a+b x) (b d-a e)}-\frac{(d+e x)^{9/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

(35*e^2*(2*b*B*d + A*b*e - 3*a*B*e)*Sqrt[d + e*x])/(8*b^5) + (35*e^2*(2*b*B*d + A*b*e - 3*a*B*e)*(d + e*x)^(3/
2))/(24*b^4*(b*d - a*e)) - (7*e*(2*b*B*d + A*b*e - 3*a*B*e)*(d + e*x)^(5/2))/(8*b^3*(b*d - a*e)*(a + b*x)) - (
(2*b*B*d + A*b*e - 3*a*B*e)*(d + e*x)^(7/2))/(4*b^2*(b*d - a*e)*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(9/2))/(
3*b*(b*d - a*e)*(a + b*x)^3) - (35*e^2*Sqrt[b*d - a*e]*(2*b*B*d + A*b*e - 3*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e
*x])/Sqrt[b*d - a*e]])/(8*b^(11/2))

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Rubi [A]  time = 0.241614, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{35 e^2 (d+e x)^{3/2} (-3 a B e+A b e+2 b B d)}{24 b^4 (b d-a e)}+\frac{35 e^2 \sqrt{d+e x} (-3 a B e+A b e+2 b B d)}{8 b^5}-\frac{35 e^2 \sqrt{b d-a e} (-3 a B e+A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{11/2}}-\frac{(d+e x)^{7/2} (-3 a B e+A b e+2 b B d)}{4 b^2 (a+b x)^2 (b d-a e)}-\frac{7 e (d+e x)^{5/2} (-3 a B e+A b e+2 b B d)}{8 b^3 (a+b x) (b d-a e)}-\frac{(d+e x)^{9/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(35*e^2*(2*b*B*d + A*b*e - 3*a*B*e)*Sqrt[d + e*x])/(8*b^5) + (35*e^2*(2*b*B*d + A*b*e - 3*a*B*e)*(d + e*x)^(3/
2))/(24*b^4*(b*d - a*e)) - (7*e*(2*b*B*d + A*b*e - 3*a*B*e)*(d + e*x)^(5/2))/(8*b^3*(b*d - a*e)*(a + b*x)) - (
(2*b*B*d + A*b*e - 3*a*B*e)*(d + e*x)^(7/2))/(4*b^2*(b*d - a*e)*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(9/2))/(
3*b*(b*d - a*e)*(a + b*x)^3) - (35*e^2*Sqrt[b*d - a*e]*(2*b*B*d + A*b*e - 3*a*B*e)*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 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)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(A+B x) (d+e x)^{7/2}}{(a+b x)^4} \, dx\\ &=-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(2 b B d+A b e-3 a B e) \int \frac{(d+e x)^{7/2}}{(a+b x)^3} \, dx}{2 b (b d-a e)}\\ &=-\frac{(2 b B d+A b e-3 a B e) (d+e x)^{7/2}}{4 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(7 e (2 b B d+A b e-3 a B e)) \int \frac{(d+e x)^{5/2}}{(a+b x)^2} \, dx}{8 b^2 (b d-a e)}\\ &=-\frac{7 e (2 b B d+A b e-3 a B e) (d+e x)^{5/2}}{8 b^3 (b d-a e) (a+b x)}-\frac{(2 b B d+A b e-3 a B e) (d+e x)^{7/2}}{4 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}+\frac{\left (35 e^2 (2 b B d+A b e-3 a B e)\right ) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{16 b^3 (b d-a e)}\\ &=\frac{35 e^2 (2 b B d+A b e-3 a B e) (d+e x)^{3/2}}{24 b^4 (b d-a e)}-\frac{7 e (2 b B d+A b e-3 a B e) (d+e x)^{5/2}}{8 b^3 (b d-a e) (a+b x)}-\frac{(2 b B d+A b e-3 a B e) (d+e x)^{7/2}}{4 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}+\frac{\left (35 e^2 (2 b B d+A b e-3 a B e)\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{16 b^4}\\ &=\frac{35 e^2 (2 b B d+A b e-3 a B e) \sqrt{d+e x}}{8 b^5}+\frac{35 e^2 (2 b B d+A b e-3 a B e) (d+e x)^{3/2}}{24 b^4 (b d-a e)}-\frac{7 e (2 b B d+A b e-3 a B e) (d+e x)^{5/2}}{8 b^3 (b d-a e) (a+b x)}-\frac{(2 b B d+A b e-3 a B e) (d+e x)^{7/2}}{4 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}+\frac{\left (35 e^2 (b d-a e) (2 b B d+A b e-3 a B e)\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b^5}\\ &=\frac{35 e^2 (2 b B d+A b e-3 a B e) \sqrt{d+e x}}{8 b^5}+\frac{35 e^2 (2 b B d+A b e-3 a B e) (d+e x)^{3/2}}{24 b^4 (b d-a e)}-\frac{7 e (2 b B d+A b e-3 a B e) (d+e x)^{5/2}}{8 b^3 (b d-a e) (a+b x)}-\frac{(2 b B d+A b e-3 a B e) (d+e x)^{7/2}}{4 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(35 e (b d-a e) (2 b B d+A b e-3 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^5}\\ &=\frac{35 e^2 (2 b B d+A b e-3 a B e) \sqrt{d+e x}}{8 b^5}+\frac{35 e^2 (2 b B d+A b e-3 a B e) (d+e x)^{3/2}}{24 b^4 (b d-a e)}-\frac{7 e (2 b B d+A b e-3 a B e) (d+e x)^{5/2}}{8 b^3 (b d-a e) (a+b x)}-\frac{(2 b B d+A b e-3 a B e) (d+e x)^{7/2}}{4 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{9/2}}{3 b (b d-a e) (a+b x)^3}-\frac{35 e^2 \sqrt{b d-a e} (2 b B d+A b e-3 a B e) \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 = 0.0984059, size = 99, normalized size = 0.35 \[ \frac{(d+e x)^{9/2} \left (\frac{9 (a B-A b)}{(a+b x)^3}-\frac{3 e^2 (-3 a B e+A b e+2 b B d) \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^3}\right )}{27 b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

((d + e*x)^(9/2)*((9*(-(A*b) + a*B))/(a + b*x)^3 - (3*e^2*(2*b*B*d + A*b*e - 3*a*B*e)*Hypergeometric2F1[3, 9/2
, 11/2, (b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^3))/(27*b*(b*d - a*e))

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Maple [B]  time = 0.029, size = 905, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

-19/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*A*d^3-35/8*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*
d)*b)^(1/2))*A*a+29/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(5/2)*A*a-29/8*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*A*d-55/8*
e^4/b^3/(b*e*x+a*e)^3*(e*x+d)^(5/2)*B*a^2+17/3*e^5/b^3/(b*e*x+a*e)^3*A*(e*x+d)^(3/2)*a^2+105/8*e^4/b^5/((a*e-b
*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2*B+35/8*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*d-41/8*e^6/b^5/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*a^4+17/3*e^3/b/(b*e*x+a*e)^3*A*(e
*x+d)^(3/2)*d^2-35/3*e^5/b^4/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*a^3+19/8*e^6/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*A*a^3+
6*e^2/b/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*d^3-11/4*e^2/b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*d^4+35/4*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^2-13/4*e^2/b/(b*e*x+a*e)^3*(e*x+d)^(5/2)*B*d^2+2*e^3
/b^4*A*(e*x+d)^(1/2)+2/3*e^2/b^4*B*(e*x+d)^(3/2)+6*e^2/b^4*B*d*(e*x+d)^(1/2)-8*e^3/b^5*a*B*(e*x+d)^(1/2)-71/3*
e^3/b^2/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*a*d^2-57/8*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*A*d*a^2+81/8*e^3/b^2/(b*e
*x+a*e)^3*(e*x+d)^(5/2)*B*a*d-34/3*e^4/b^2/(b*e*x+a*e)^3*A*(e*x+d)^(3/2)*a*d-175/8*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))*B*d*a+88/3*e^4/b^3/(b*e*x+a*e)^3*B*(e*x+d)^(3/2)*a^2*d+57/8*e^4/b
^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*A*a*d^2+145/8*e^5/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*a^3*d-189/8*e^4/b^3/(b*e*x+
a*e)^3*(e*x+d)^(1/2)*B*d^2*a^2+107/8*e^3/b^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*B*d^3*a

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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)^(7/2)/(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 = 1.47759, size = 2125, normalized size = 7.48 \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)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[-1/48*(105*(2*B*a^3*b*d*e^2 - (3*B*a^4 - A*a^3*b)*e^3 + (2*B*b^4*d*e^2 - (3*B*a*b^3 - A*b^4)*e^3)*x^3 + 3*(2*
B*a*b^3*d*e^2 - (3*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 3*(2*B*a^2*b^2*d*e^2 - (3*B*a^3*b - A*a^2*b^2)*e^3)*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*b^4*e^3
*x^4 - 4*(B*a*b^3 + 2*A*b^4)*d^3 - 14*(2*B*a^2*b^2 + A*a*b^3)*d^2*e + 35*(9*B*a^3*b - A*a^2*b^2)*d*e^2 - 105*(
3*B*a^4 - A*a^3*b)*e^3 + 16*(10*B*b^4*d*e^2 - 3*(3*B*a*b^3 - A*b^4)*e^3)*x^3 - 3*(26*B*b^4*d^2*e - (241*B*a*b^
3 - 29*A*b^4)*d*e^2 + 77*(3*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 - 2*(6*B*b^4*d^3 + (41*B*a*b^3 + 19*A*b^4)*d^2*e - 7
*(61*B*a^2*b^2 - 7*A*a*b^3)*d*e^2 + 140*(3*B*a^3*b - A*a^2*b^2)*e^3)*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*(105*(2*B*a^3*b*d*e^2 - (3*B*a^4 - A*a^3*b)*e^3 + (2*B*b^4*d*e^2 - (3*B*a*b^3
- A*b^4)*e^3)*x^3 + 3*(2*B*a*b^3*d*e^2 - (3*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 3*(2*B*a^2*b^2*d*e^2 - (3*B*a^3*b
- A*a^2*b^2)*e^3)*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*b^
4*e^3*x^4 - 4*(B*a*b^3 + 2*A*b^4)*d^3 - 14*(2*B*a^2*b^2 + A*a*b^3)*d^2*e + 35*(9*B*a^3*b - A*a^2*b^2)*d*e^2 -
105*(3*B*a^4 - A*a^3*b)*e^3 + 16*(10*B*b^4*d*e^2 - 3*(3*B*a*b^3 - A*b^4)*e^3)*x^3 - 3*(26*B*b^4*d^2*e - (241*B
*a*b^3 - 29*A*b^4)*d*e^2 + 77*(3*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 - 2*(6*B*b^4*d^3 + (41*B*a*b^3 + 19*A*b^4)*d^2*
e - 7*(61*B*a^2*b^2 - 7*A*a*b^3)*d*e^2 + 140*(3*B*a^3*b - A*a^2*b^2)*e^3)*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((B*x+A)*(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.21409, size = 779, normalized size = 2.74 \begin{align*} \frac{35 \,{\left (2 \, B b^{2} d^{2} e^{2} - 5 \, B a b d e^{3} + A b^{2} d e^{3} + 3 \, B a^{2} e^{4} - A a b e^{4}\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{78 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{4} d^{2} e^{2} - 144 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{4} d^{3} e^{2} + 66 \, \sqrt{x e + d} B b^{4} d^{4} e^{2} - 243 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{3} d e^{3} + 87 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{4} d e^{3} + 568 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{3} d^{2} e^{3} - 136 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{4} d^{2} e^{3} - 321 \, \sqrt{x e + d} B a b^{3} d^{3} e^{3} + 57 \, \sqrt{x e + d} A b^{4} d^{3} e^{3} + 165 \,{\left (x e + d\right )}^{\frac{5}{2}} B a^{2} b^{2} e^{4} - 87 \,{\left (x e + d\right )}^{\frac{5}{2}} A a b^{3} e^{4} - 704 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b^{2} d e^{4} + 272 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{3} d e^{4} + 567 \, \sqrt{x e + d} B a^{2} b^{2} d^{2} e^{4} - 171 \, \sqrt{x e + d} A a b^{3} d^{2} e^{4} + 280 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{3} b e^{5} - 136 \,{\left (x e + d\right )}^{\frac{3}{2}} A a^{2} b^{2} e^{5} - 435 \, \sqrt{x e + d} B a^{3} b d e^{5} + 171 \, \sqrt{x e + d} A a^{2} b^{2} d e^{5} + 123 \, \sqrt{x e + d} B a^{4} e^{6} - 57 \, \sqrt{x e + d} A a^{3} b e^{6}}{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 b^{8} e^{2} + 9 \, \sqrt{x e + d} B b^{8} d e^{2} - 12 \, \sqrt{x e + d} B a b^{7} e^{3} + 3 \, \sqrt{x e + d} A b^{8} e^{3}\right )}}{3 \, b^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

35/8*(2*B*b^2*d^2*e^2 - 5*B*a*b*d*e^3 + A*b^2*d*e^3 + 3*B*a^2*e^4 - A*a*b*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^
2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^5) - 1/24*(78*(x*e + d)^(5/2)*B*b^4*d^2*e^2 - 144*(x*e + d)^(3/2)*B*b^4*
d^3*e^2 + 66*sqrt(x*e + d)*B*b^4*d^4*e^2 - 243*(x*e + d)^(5/2)*B*a*b^3*d*e^3 + 87*(x*e + d)^(5/2)*A*b^4*d*e^3
+ 568*(x*e + d)^(3/2)*B*a*b^3*d^2*e^3 - 136*(x*e + d)^(3/2)*A*b^4*d^2*e^3 - 321*sqrt(x*e + d)*B*a*b^3*d^3*e^3
+ 57*sqrt(x*e + d)*A*b^4*d^3*e^3 + 165*(x*e + d)^(5/2)*B*a^2*b^2*e^4 - 87*(x*e + d)^(5/2)*A*a*b^3*e^4 - 704*(x
*e + d)^(3/2)*B*a^2*b^2*d*e^4 + 272*(x*e + d)^(3/2)*A*a*b^3*d*e^4 + 567*sqrt(x*e + d)*B*a^2*b^2*d^2*e^4 - 171*
sqrt(x*e + d)*A*a*b^3*d^2*e^4 + 280*(x*e + d)^(3/2)*B*a^3*b*e^5 - 136*(x*e + d)^(3/2)*A*a^2*b^2*e^5 - 435*sqrt
(x*e + d)*B*a^3*b*d*e^5 + 171*sqrt(x*e + d)*A*a^2*b^2*d*e^5 + 123*sqrt(x*e + d)*B*a^4*e^6 - 57*sqrt(x*e + d)*A
*a^3*b*e^6)/(((x*e + d)*b - b*d + a*e)^3*b^5) + 2/3*((x*e + d)^(3/2)*B*b^8*e^2 + 9*sqrt(x*e + d)*B*b^8*d*e^2 -
12*sqrt(x*e + d)*B*a*b^7*e^3 + 3*sqrt(x*e + d)*A*b^8*e^3)/b^12