3.1877 \(\int \frac{(A+B x) (d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=359 \[ -\frac{5 e^2 \sqrt{d+e x} (-7 a B e-A b e+8 b B d)}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{5 e^3 (a+b x) (-7 a B e-A b e+8 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{(d+e x)^{5/2} (-7 a B e-A b e+8 b B d)}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{5 e (d+e x)^{3/2} (-7 a B e-A b e+8 b B d)}{96 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

(-5*e^2*(8*b*B*d - A*b*e - 7*a*B*e)*Sqrt[d + e*x])/(64*b^4*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(
8*b*B*d - A*b*e - 7*a*B*e)*(d + e*x)^(3/2))/(96*b^3*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((8
*b*B*d - A*b*e - 7*a*B*e)*(d + e*x)^(5/2))/(24*b^2*(b*d - a*e)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((
A*b - a*B)*(d + e*x)^(7/2))/(4*b*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e^3*(8*b*B*d - A*
b*e - 7*a*B*e)*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(9/2)*(b*d - a*e)^(3/2)*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.315328, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 78, 47, 63, 208} \[ -\frac{5 e^2 \sqrt{d+e x} (-7 a B e-A b e+8 b B d)}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{5 e^3 (a+b x) (-7 a B e-A b e+8 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{(d+e x)^{7/2} (A b-a B)}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{(d+e x)^{5/2} (-7 a B e-A b e+8 b B d)}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{5 e (d+e x)^{3/2} (-7 a B e-A b e+8 b B d)}{96 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e^2*(8*b*B*d - A*b*e - 7*a*B*e)*Sqrt[d + e*x])/(64*b^4*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e*(
8*b*B*d - A*b*e - 7*a*B*e)*(d + e*x)^(3/2))/(96*b^3*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((8
*b*B*d - A*b*e - 7*a*B*e)*(d + e*x)^(5/2))/(24*b^2*(b*d - a*e)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((
A*b - a*B)*(d + e*x)^(7/2))/(4*b*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*e^3*(8*b*B*d - A*
b*e - 7*a*B*e)*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(9/2)*(b*d - a*e)^(3/2)*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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 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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (d+e x)^{5/2}}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b^2 (8 b B d-A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{\left (a b+b^2 x\right )^4} \, dx}{8 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(8 b B d-A b e-7 a B e) (d+e x)^{5/2}}{24 b^2 (b d-a e) (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e (8 b B d-A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a b+b^2 x\right )^3} \, dx}{48 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e (8 b B d-A b e-7 a B e) (d+e x)^{3/2}}{96 b^3 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(8 b B d-A b e-7 a B e) (d+e x)^{5/2}}{24 b^2 (b d-a e) (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e^2 (8 b B d-A b e-7 a B e) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{\left (a b+b^2 x\right )^2} \, dx}{64 b^2 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^2 (8 b B d-A b e-7 a B e) \sqrt{d+e x}}{64 b^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (8 b B d-A b e-7 a B e) (d+e x)^{3/2}}{96 b^3 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(8 b B d-A b e-7 a B e) (d+e x)^{5/2}}{24 b^2 (b d-a e) (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e^3 (8 b B d-A b e-7 a B e) \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^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^2 (8 b B d-A b e-7 a B e) \sqrt{d+e x}}{64 b^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (8 b B d-A b e-7 a B e) (d+e x)^{3/2}}{96 b^3 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(8 b B d-A b e-7 a B e) (d+e x)^{5/2}}{24 b^2 (b d-a e) (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 e^2 (8 b B d-A b e-7 a B e) \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^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{5 e^2 (8 b B d-A b e-7 a B e) \sqrt{d+e x}}{64 b^4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (8 b B d-A b e-7 a B e) (d+e x)^{3/2}}{96 b^3 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(8 b B d-A b e-7 a B e) (d+e x)^{5/2}}{24 b^2 (b d-a e) (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^{7/2}}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e^3 (8 b B d-A b e-7 a B e) (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{9/2} (b d-a e)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.12469, size = 219, normalized size = 0.61 \[ \frac{-\frac{(a+b x) (-7 a B e-A b e+8 b B d) \left (b (d+e x) \sqrt{a e-b d} \left (15 a^2 e^2+10 a b e (d+4 e x)+b^2 \left (8 d^2+26 d e x+33 e^2 x^2\right )\right )-15 \sqrt{b} e^3 (a+b x)^3 \sqrt{d+e x} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )\right )}{3 \sqrt{a e-b d}}-16 b^4 (d+e x)^4 (A b-a B)}{64 b^5 (a+b x)^3 \sqrt{(a+b x)^2} \sqrt{d+e x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^4*(A*b - a*B)*(d + e*x)^4 - ((8*b*B*d - A*b*e - 7*a*B*e)*(a + b*x)*(b*Sqrt[-(b*d) + a*e]*(d + e*x)*(15*
a^2*e^2 + 10*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 26*d*e*x + 33*e^2*x^2)) - 15*Sqrt[b]*e^3*(a + b*x)^3*Sqrt[d + e*
x]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(3*Sqrt[-(b*d) + a*e]))/(64*b^5*(b*d - a*e)*(a + b*x)^
3*Sqrt[(a + b*x)^2]*Sqrt[d + e*x])

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Maple [B]  time = 0.02, size = 1273, normalized size = 3.6 \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)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/192*(-15*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*b*e^4-279*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*a*b^3*e-73*A*
((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a*b^3*e^2+73*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^4*d*e+105*B*arctan((e*x+d
)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^4*a*b^4*e^5-120*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^4*b^5*d*e^4+6
0*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^3*a*b^4*e^5+420*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2)
)*x^3*a^2*b^3*e^5+420*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a^4*b*e^5-55*A*((a*e-b*d)*b)^(1/2)*(e*x+
d)^(3/2)*a^2*b^2*e^3-55*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^4*d^2*e-385*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*
a^3*b*e^3-120*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^4*b*d*e^4-511*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2
)*a^2*b^2*e^2+60*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a^3*b^2*e^5+15*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^
(1/2)*b^4*d^3*e+90*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*a^2*b^3*e^5+630*B*arctan((e*x+d)^(1/2)*b/
((a*e-b*d)*b)^(1/2))*x^2*a^3*b^2*e^5+105*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^5*e^5+264*B*((a*e-b*d
)*b)^(1/2)*(e*x+d)^(7/2)*b^4*d-584*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^4*d^2+15*A*arctan((e*x+d)^(1/2)*b/((a
*e-b*d)*b)^(1/2))*x^4*b^5*e^5+15*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*b^4*e-105*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(
1/2)*a^4*e^4-120*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^4*d^4+15*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*
a^4*b*e^5+440*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^4*d^3+110*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b^3*d*e^2+
1210*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^2*b^2*d*e^2+435*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*b*d*e^3-720
*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*a^2*b^3*d*e^4+1095*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a*b^
3*d*e-480*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^3*a*b^4*d*e^4-675*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2
)*a^2*b^2*d^2*e^2+465*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^3*d^3*e-1265*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)
*a*b^3*d^2*e+45*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b^2*d*e^3-45*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^3
*d^2*e^2-480*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a^3*b^2*d*e^4)/e*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^4/
(a*e-b*d)/((b*x+a)^2)^(5/2)

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

[Out]

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

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Fricas [B]  time = 1.83546, size = 3206, normalized size = 8.93 \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)^(5/2),x, algorithm="fricas")

[Out]

[1/384*(15*(8*B*a^4*b*d*e^3 - (7*B*a^5 + A*a^4*b)*e^4 + (8*B*b^5*d*e^3 - (7*B*a*b^4 + A*b^5)*e^4)*x^4 + 4*(8*B
*a*b^4*d*e^3 - (7*B*a^2*b^3 + A*a*b^4)*e^4)*x^3 + 6*(8*B*a^2*b^3*d*e^3 - (7*B*a^3*b^2 + A*a^2*b^3)*e^4)*x^2 +
4*(8*B*a^3*b^2*d*e^3 - (7*B*a^4*b + A*a^3*b^2)*e^4)*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*(16*(B*a*b^5 + 3*A*b^6)*d^4 + 8*(B*a^2*b^4 - 7*A*a*b^5)*d^3*e + 2*
(13*B*a^3*b^3 - A*a^2*b^4)*d^2*e^2 - 5*(31*B*a^4*b^2 + A*a^3*b^3)*d*e^3 + 15*(7*B*a^5*b + A*a^4*b^2)*e^4 + 3*(
88*B*b^6*d^2*e^2 - (181*B*a*b^5 - 5*A*b^6)*d*e^3 + (93*B*a^2*b^4 - 5*A*a*b^5)*e^4)*x^3 + (208*B*b^6*d^3*e + 2*
(25*B*a*b^5 + 59*A*b^6)*d^2*e^2 - (769*B*a^2*b^4 + 191*A*a*b^5)*d*e^3 + 73*(7*B*a^3*b^3 + A*a^2*b^4)*e^4)*x^2
+ (64*B*b^6*d^4 + 8*(3*B*a*b^5 + 17*A*b^6)*d^3*e + 4*(25*B*a^2*b^4 - 43*A*a*b^5)*d^2*e^2 - (573*B*a^3*b^3 + 19
*A*a^2*b^4)*d*e^3 + 55*(7*B*a^4*b^2 + A*a^3*b^3)*e^4)*x)*sqrt(e*x + d))/(a^4*b^7*d^2 - 2*a^5*b^6*d*e + a^6*b^5
*e^2 + (b^11*d^2 - 2*a*b^10*d*e + a^2*b^9*e^2)*x^4 + 4*(a*b^10*d^2 - 2*a^2*b^9*d*e + a^3*b^8*e^2)*x^3 + 6*(a^2
*b^9*d^2 - 2*a^3*b^8*d*e + a^4*b^7*e^2)*x^2 + 4*(a^3*b^8*d^2 - 2*a^4*b^7*d*e + a^5*b^6*e^2)*x), 1/192*(15*(8*B
*a^4*b*d*e^3 - (7*B*a^5 + A*a^4*b)*e^4 + (8*B*b^5*d*e^3 - (7*B*a*b^4 + A*b^5)*e^4)*x^4 + 4*(8*B*a*b^4*d*e^3 -
(7*B*a^2*b^3 + A*a*b^4)*e^4)*x^3 + 6*(8*B*a^2*b^3*d*e^3 - (7*B*a^3*b^2 + A*a^2*b^3)*e^4)*x^2 + 4*(8*B*a^3*b^2*
d*e^3 - (7*B*a^4*b + A*a^3*b^2)*e^4)*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)) - (16*(B*a*b^5 + 3*A*b^6)*d^4 + 8*(B*a^2*b^4 - 7*A*a*b^5)*d^3*e + 2*(13*B*a^3*b^3 - A*a^2*b^4)*d^2*e^2
- 5*(31*B*a^4*b^2 + A*a^3*b^3)*d*e^3 + 15*(7*B*a^5*b + A*a^4*b^2)*e^4 + 3*(88*B*b^6*d^2*e^2 - (181*B*a*b^5 -
5*A*b^6)*d*e^3 + (93*B*a^2*b^4 - 5*A*a*b^5)*e^4)*x^3 + (208*B*b^6*d^3*e + 2*(25*B*a*b^5 + 59*A*b^6)*d^2*e^2 -
(769*B*a^2*b^4 + 191*A*a*b^5)*d*e^3 + 73*(7*B*a^3*b^3 + A*a^2*b^4)*e^4)*x^2 + (64*B*b^6*d^4 + 8*(3*B*a*b^5 + 1
7*A*b^6)*d^3*e + 4*(25*B*a^2*b^4 - 43*A*a*b^5)*d^2*e^2 - (573*B*a^3*b^3 + 19*A*a^2*b^4)*d*e^3 + 55*(7*B*a^4*b^
2 + A*a^3*b^3)*e^4)*x)*sqrt(e*x + d))/(a^4*b^7*d^2 - 2*a^5*b^6*d*e + a^6*b^5*e^2 + (b^11*d^2 - 2*a*b^10*d*e +
a^2*b^9*e^2)*x^4 + 4*(a*b^10*d^2 - 2*a^2*b^9*d*e + a^3*b^8*e^2)*x^3 + 6*(a^2*b^9*d^2 - 2*a^3*b^8*d*e + a^4*b^7
*e^2)*x^2 + 4*(a^3*b^8*d^2 - 2*a^4*b^7*d*e + a^5*b^6*e^2)*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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.44091, size = 873, normalized size = 2.43 \begin{align*} \frac{5 \,{\left (8 \, B b d e^{3} - 7 \, B a e^{4} - A b e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{64 \,{\left (b^{5} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{4} e \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{264 \,{\left (x e + d\right )}^{\frac{7}{2}} B b^{4} d e^{3} - 584 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{4} d^{2} e^{3} + 440 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{4} d^{3} e^{3} - 120 \, \sqrt{x e + d} B b^{4} d^{4} e^{3} - 279 \,{\left (x e + d\right )}^{\frac{7}{2}} B a b^{3} e^{4} + 15 \,{\left (x e + d\right )}^{\frac{7}{2}} A b^{4} e^{4} + 1095 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{3} d e^{4} + 73 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{4} d e^{4} - 1265 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{3} d^{2} e^{4} - 55 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{4} d^{2} e^{4} + 465 \, \sqrt{x e + d} B a b^{3} d^{3} e^{4} + 15 \, \sqrt{x e + d} A b^{4} d^{3} e^{4} - 511 \,{\left (x e + d\right )}^{\frac{5}{2}} B a^{2} b^{2} e^{5} - 73 \,{\left (x e + d\right )}^{\frac{5}{2}} A a b^{3} e^{5} + 1210 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b^{2} d e^{5} + 110 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{3} d e^{5} - 675 \, \sqrt{x e + d} B a^{2} b^{2} d^{2} e^{5} - 45 \, \sqrt{x e + d} A a b^{3} d^{2} e^{5} - 385 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{3} b e^{6} - 55 \,{\left (x e + d\right )}^{\frac{3}{2}} A a^{2} b^{2} e^{6} + 435 \, \sqrt{x e + d} B a^{3} b d e^{6} + 45 \, \sqrt{x e + d} A a^{2} b^{2} d e^{6} - 105 \, \sqrt{x e + d} B a^{4} e^{7} - 15 \, \sqrt{x e + d} A a^{3} b e^{7}}{192 \,{\left (b^{5} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{4} e \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((B*x+A)*(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

5/64*(8*B*b*d*e^3 - 7*B*a*e^4 - A*b*e^4)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d*sgn((x*e + d)*b*
e - b*d*e + a*e^2) - a*b^4*e*sgn((x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/192*(264*(x*e + d)^
(7/2)*B*b^4*d*e^3 - 584*(x*e + d)^(5/2)*B*b^4*d^2*e^3 + 440*(x*e + d)^(3/2)*B*b^4*d^3*e^3 - 120*sqrt(x*e + d)*
B*b^4*d^4*e^3 - 279*(x*e + d)^(7/2)*B*a*b^3*e^4 + 15*(x*e + d)^(7/2)*A*b^4*e^4 + 1095*(x*e + d)^(5/2)*B*a*b^3*
d*e^4 + 73*(x*e + d)^(5/2)*A*b^4*d*e^4 - 1265*(x*e + d)^(3/2)*B*a*b^3*d^2*e^4 - 55*(x*e + d)^(3/2)*A*b^4*d^2*e
^4 + 465*sqrt(x*e + d)*B*a*b^3*d^3*e^4 + 15*sqrt(x*e + d)*A*b^4*d^3*e^4 - 511*(x*e + d)^(5/2)*B*a^2*b^2*e^5 -
73*(x*e + d)^(5/2)*A*a*b^3*e^5 + 1210*(x*e + d)^(3/2)*B*a^2*b^2*d*e^5 + 110*(x*e + d)^(3/2)*A*a*b^3*d*e^5 - 67
5*sqrt(x*e + d)*B*a^2*b^2*d^2*e^5 - 45*sqrt(x*e + d)*A*a*b^3*d^2*e^5 - 385*(x*e + d)^(3/2)*B*a^3*b*e^6 - 55*(x
*e + d)^(3/2)*A*a^2*b^2*e^6 + 435*sqrt(x*e + d)*B*a^3*b*d*e^6 + 45*sqrt(x*e + d)*A*a^2*b^2*d*e^6 - 105*sqrt(x*
e + d)*B*a^4*e^7 - 15*sqrt(x*e + d)*A*a^3*b*e^7)/((b^5*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b^4*e*sgn((x*e
+ d)*b*e - b*d*e + a*e^2))*((x*e + d)*b - b*d + a*e)^4)