3.1782 \(\int \frac{A+B x}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=388 \[ -\frac{e^3 (a+b x) (B d-A e)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^5}-\frac{e^2 (a B e-4 A b e+3 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac{e^3 (a+b x) \log (a+b x) (a B e-5 A b e+4 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac{e^3 (a+b x) \log (d+e x) (a B e-5 A b e+4 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac{e (a B e-3 A b e+2 b B d)}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{a B e-2 A b e+b B d}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{A b-a B}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

[Out]

-((e^2*(3*b*B*d - 4*A*b*e + a*B*e))/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (A*b - a*B)/(4*(b*d - a*e
)^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*B*d - 2*A*b*e + a*B*e)/(3*(b*d - a*e)^3*(a + b*x)^2*Sqrt[a
^2 + 2*a*b*x + b^2*x^2]) + (e*(2*b*B*d - 3*A*b*e + a*B*e))/(2*(b*d - a*e)^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (e^3*(B*d - A*e)*(a + b*x))/((b*d - a*e)^5*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^3*(4*b*B*d -
5*A*b*e + a*B*e)*(a + b*x)*Log[a + b*x])/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^3*(4*b*B*d - 5*A*
b*e + a*B*e)*(a + b*x)*Log[d + e*x])/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.455119, antiderivative size = 388, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 77} \[ -\frac{e^3 (a+b x) (B d-A e)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^5}-\frac{e^2 (a B e-4 A b e+3 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac{e^3 (a+b x) \log (a+b x) (a B e-5 A b e+4 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac{e^3 (a+b x) \log (d+e x) (a B e-5 A b e+4 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac{e (a B e-3 A b e+2 b B d)}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac{a B e-2 A b e+b B d}{3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{A b-a B}{4 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e^2*(3*b*B*d - 4*A*b*e + a*B*e))/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (A*b - a*B)/(4*(b*d - a*e
)^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*B*d - 2*A*b*e + a*B*e)/(3*(b*d - a*e)^3*(a + b*x)^2*Sqrt[a
^2 + 2*a*b*x + b^2*x^2]) + (e*(2*b*B*d - 3*A*b*e + a*B*e))/(2*(b*d - a*e)^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (e^3*(B*d - A*e)*(a + b*x))/((b*d - a*e)^5*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^3*(4*b*B*d -
5*A*b*e + a*B*e)*(a + b*x)*Log[a + b*x])/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^3*(4*b*B*d - 5*A*
b*e + a*B*e)*(a + b*x)*Log[d + e*x])/((b*d - a*e)^6*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{A+B x}{(d+e x)^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}{\left (a b+b^2 x\right )^5 (d+e x)^2} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac{A b-a B}{b^4 (b d-a e)^2 (a+b x)^5}+\frac{b B d-2 A b e+a B e}{b^4 (b d-a e)^3 (a+b x)^4}+\frac{e (-2 b B d+3 A b e-a B e)}{b^4 (b d-a e)^4 (a+b x)^3}-\frac{e^2 (-3 b B d+4 A b e-a B e)}{b^4 (b d-a e)^5 (a+b x)^2}+\frac{e^3 (-4 b B d+5 A b e-a B e)}{b^4 (b d-a e)^6 (a+b x)}-\frac{e^4 (-B d+A e)}{b^5 (b d-a e)^5 (d+e x)^2}-\frac{e^4 (-4 b B d+5 A b e-a B e)}{b^5 (b d-a e)^6 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{e^2 (3 b B d-4 A b e+a B e)}{(b d-a e)^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A b-a B}{4 (b d-a e)^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b B d-2 A b e+a B e}{3 (b d-a e)^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (2 b B d-3 A b e+a B e)}{2 (b d-a e)^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^3 (B d-A e) (a+b x)}{(b d-a e)^5 (d+e x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e^3 (4 b B d-5 A b e+a B e) (a+b x) \log (a+b x)}{(b d-a e)^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^3 (4 b B d-5 A b e+a B e) (a+b x) \log (d+e x)}{(b d-a e)^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.269419, size = 250, normalized size = 0.64 \[ \frac{12 e^2 (a+b x)^2 (b d-a e) (-a B e+4 A b e-3 b B d)+\frac{12 e^3 (a+b x)^3 (b d-a e) (A e-B d)}{d+e x}-12 e^3 (a+b x)^3 \log (a+b x) (a B e-5 A b e+4 b B d)+12 e^3 (a+b x)^3 \log (d+e x) (a B e-5 A b e+4 b B d)+\frac{3 (a B-A b) (b d-a e)^4}{a+b x}-6 e (a+b x) (b d-a e)^2 (-a B e+3 A b e-2 b B d)-4 (b d-a e)^3 (a B e-2 A b e+b B d)}{12 \left ((a+b x)^2\right )^{3/2} (b d-a e)^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(b*d - a*e)^3*(b*B*d - 2*A*b*e + a*B*e) + (3*(-(A*b) + a*B)*(b*d - a*e)^4)/(a + b*x) - 6*e*(b*d - a*e)^2*(
-2*b*B*d + 3*A*b*e - a*B*e)*(a + b*x) + 12*e^2*(b*d - a*e)*(-3*b*B*d + 4*A*b*e - a*B*e)*(a + b*x)^2 + (12*e^3*
(b*d - a*e)*(-(B*d) + A*e)*(a + b*x)^3)/(d + e*x) - 12*e^3*(4*b*B*d - 5*A*b*e + a*B*e)*(a + b*x)^3*Log[a + b*x
] + 12*e^3*(4*b*B*d - 5*A*b*e + a*B*e)*(a + b*x)^3*Log[d + e*x])/(12*(b*d - a*e)^6*((a + b*x)^2)^(3/2))

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Maple [B]  time = 0.027, size = 1651, normalized size = 4.3 \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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/12*(48*B*ln(b*x+a)*a^4*b*d^2*e^3+48*B*ln(b*x+a)*x^2*a^4*b*e^5+60*A*ln(e*x+d)*x*a^4*b*e^5-60*A*ln(b*x+a)*x*a
^4*b*e^5+60*A*ln(e*x+d)*a^4*b*d*e^4-60*A*ln(b*x+a)*a^4*b*d*e^4-240*A*ln(b*x+a)*x^2*a^3*b^2*e^5-48*B*ln(e*x+d)*
x^2*a^4*b*e^5+60*A*ln(e*x+d)*x^4*b^5*d*e^4-240*A*ln(b*x+a)*x^4*a*b^4*e^5-60*A*ln(b*x+a)*x^4*b^5*d*e^4-120*A*x^
2*a*b^4*d^2*e^3-150*A*x^2*a^2*b^3*d*e^4+150*B*x^3*a*b^4*d^2*e^3-132*B*x^3*a^2*b^3*d*e^4-180*A*x^3*a*b^4*d*e^4-
36*B*x^4*a*b^4*d*e^4-48*B*ln(e*x+d)*a^4*b*d^2*e^3+144*B*x^2*a^2*b^3*d^2*e^3-178*B*x^2*a^3*b^2*d*e^4+94*B*x^2*a
*b^4*d^3*e^2+48*B*ln(b*x+a)*x^4*b^5*d^2*e^3+360*A*ln(e*x+d)*x^3*a^2*b^3*e^5-360*A*ln(b*x+a)*x^3*a^2*b^3*e^5-72
*B*ln(e*x+d)*x^3*a^3*b^2*e^5+72*B*ln(b*x+a)*x^3*a^3*b^2*e^5+240*A*ln(e*x+d)*x^2*a^3*b^2*e^5-12*B*ln(e*x+d)*x^5
*a*b^4*e^5-48*B*ln(e*x+d)*x^5*b^5*d*e^4+12*B*ln(b*x+a)*x^5*a*b^4*e^5+48*B*ln(b*x+a)*x^5*b^5*d*e^4+240*A*ln(e*x
+d)*x^4*a*b^4*e^5+20*B*a^3*b^2*d^2*e^3*x+136*B*a^2*b^3*d^3*e^2*x-31*B*a*b^4*d^4*e*x+20*A*a^3*b^2*d*e^4*x-180*A
*a^2*b^3*d^2*e^3*x+40*A*a*b^4*d^3*e^2*x-104*B*a^4*b*d*e^4*x-48*B*ln(e*x+d)*x^4*b^5*d^2*e^3+48*B*ln(b*x+a)*x^4*
a^2*b^3*e^5-48*B*ln(e*x+d)*x^4*a^2*b^3*e^5-37*B*a^5*d*e^4+B*a*b^4*d^5-25*B*a^5*e^5*x+4*B*b^5*d^5*x-5*A*b^5*d^4
*e*x+125*A*a^4*b*e^5*x-8*B*x^2*b^5*d^4*e+48*B*x^4*b^5*d^2*e^3+210*A*x^3*a^2*b^3*e^5-30*A*x^3*b^5*d^2*e^3-42*B*
x^3*a^3*b^2*e^5+24*B*x^3*b^5*d^3*e^2-12*B*x^4*a^2*b^3*e^5+60*A*x^4*a*b^4*e^5-60*A*x^4*b^5*d*e^4+60*A*ln(e*x+d)
*x^5*b^5*e^5-60*A*ln(b*x+a)*x^5*b^5*e^5-12*B*ln(e*x+d)*x*a^5*e^5+12*B*ln(b*x+a)*x*a^5*e^5-12*B*ln(e*x+d)*a^5*d
*e^4+12*B*ln(b*x+a)*a^5*d*e^4-52*B*x^2*a^4*b*e^5+260*A*x^2*a^3*b^2*e^5+10*A*x^2*b^5*d^3*e^2-8*B*a^2*b^3*d^4*e+
36*B*a^3*b^2*d^3*e^2+65*A*a^4*b*d*e^4+8*B*a^4*b*d^2*e^3-120*A*a^3*b^2*d^2*e^3+60*A*a^2*b^3*d^3*e^2-20*A*a*b^4*
d^4*e+12*A*a^5*e^5+3*A*b^5*d^5+360*A*ln(e*x+d)*x^2*a^2*b^3*d*e^4-360*A*ln(b*x+a)*x^2*a^2*b^3*d*e^4-264*B*ln(e*
x+d)*x^2*a^3*b^2*d*e^4-288*B*ln(e*x+d)*x^2*a^2*b^3*d^2*e^3+264*B*ln(b*x+a)*x^2*a^3*b^2*d*e^4+288*B*ln(b*x+a)*x
^2*a^2*b^3*d^2*e^3+240*A*ln(e*x+d)*x*a^3*b^2*d*e^4-240*A*ln(b*x+a)*x*a^3*b^2*d*e^4-96*B*ln(e*x+d)*x*a^4*b*d*e^
4-192*B*ln(e*x+d)*x*a^3*b^2*d^2*e^3+96*B*ln(b*x+a)*x*a^4*b*d*e^4+192*B*ln(b*x+a)*x*a^3*b^2*d^2*e^3-204*B*ln(e*
x+d)*x^4*a*b^4*d*e^4+204*B*ln(b*x+a)*x^4*a*b^4*d*e^4+240*A*ln(e*x+d)*x^3*a*b^4*d*e^4-240*A*ln(b*x+a)*x^3*a*b^4
*d*e^4-336*B*ln(e*x+d)*x^3*a^2*b^3*d*e^4-192*B*ln(e*x+d)*x^3*a*b^4*d^2*e^3+336*B*ln(b*x+a)*x^3*a^2*b^3*d*e^4+1
92*B*ln(b*x+a)*x^3*a*b^4*d^2*e^3)*(b*x+a)/(e*x+d)/(a*e-b*d)^6/((b*x+a)^2)^(5/2)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.69524, size = 3541, normalized size = 9.13 \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)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(12*A*a^5*e^5 + (B*a*b^4 + 3*A*b^5)*d^5 - 4*(2*B*a^2*b^3 + 5*A*a*b^4)*d^4*e + 12*(3*B*a^3*b^2 + 5*A*a^2*
b^3)*d^3*e^2 + 8*(B*a^4*b - 15*A*a^3*b^2)*d^2*e^3 - (37*B*a^5 - 65*A*a^4*b)*d*e^4 + 12*(4*B*b^5*d^2*e^3 - (3*B
*a*b^4 + 5*A*b^5)*d*e^4 - (B*a^2*b^3 - 5*A*a*b^4)*e^5)*x^4 + 6*(4*B*b^5*d^3*e^2 + 5*(5*B*a*b^4 - A*b^5)*d^2*e^
3 - 2*(11*B*a^2*b^3 + 15*A*a*b^4)*d*e^4 - 7*(B*a^3*b^2 - 5*A*a^2*b^3)*e^5)*x^3 - 2*(4*B*b^5*d^4*e - (47*B*a*b^
4 + 5*A*b^5)*d^3*e^2 - 12*(6*B*a^2*b^3 - 5*A*a*b^4)*d^2*e^3 + (89*B*a^3*b^2 + 75*A*a^2*b^3)*d*e^4 + 26*(B*a^4*
b - 5*A*a^3*b^2)*e^5)*x^2 + (4*B*b^5*d^5 - (31*B*a*b^4 + 5*A*b^5)*d^4*e + 8*(17*B*a^2*b^3 + 5*A*a*b^4)*d^3*e^2
+ 20*(B*a^3*b^2 - 9*A*a^2*b^3)*d^2*e^3 - 4*(26*B*a^4*b - 5*A*a^3*b^2)*d*e^4 - 25*(B*a^5 - 5*A*a^4*b)*e^5)*x +
12*(4*B*a^4*b*d^2*e^3 + (B*a^5 - 5*A*a^4*b)*d*e^4 + (4*B*b^5*d*e^4 + (B*a*b^4 - 5*A*b^5)*e^5)*x^5 + (4*B*b^5*
d^2*e^3 + (17*B*a*b^4 - 5*A*b^5)*d*e^4 + 4*(B*a^2*b^3 - 5*A*a*b^4)*e^5)*x^4 + 2*(8*B*a*b^4*d^2*e^3 + 2*(7*B*a^
2*b^3 - 5*A*a*b^4)*d*e^4 + 3*(B*a^3*b^2 - 5*A*a^2*b^3)*e^5)*x^3 + 2*(12*B*a^2*b^3*d^2*e^3 + (11*B*a^3*b^2 - 15
*A*a^2*b^3)*d*e^4 + 2*(B*a^4*b - 5*A*a^3*b^2)*e^5)*x^2 + (16*B*a^3*b^2*d^2*e^3 + 4*(2*B*a^4*b - 5*A*a^3*b^2)*d
*e^4 + (B*a^5 - 5*A*a^4*b)*e^5)*x)*log(b*x + a) - 12*(4*B*a^4*b*d^2*e^3 + (B*a^5 - 5*A*a^4*b)*d*e^4 + (4*B*b^5
*d*e^4 + (B*a*b^4 - 5*A*b^5)*e^5)*x^5 + (4*B*b^5*d^2*e^3 + (17*B*a*b^4 - 5*A*b^5)*d*e^4 + 4*(B*a^2*b^3 - 5*A*a
*b^4)*e^5)*x^4 + 2*(8*B*a*b^4*d^2*e^3 + 2*(7*B*a^2*b^3 - 5*A*a*b^4)*d*e^4 + 3*(B*a^3*b^2 - 5*A*a^2*b^3)*e^5)*x
^3 + 2*(12*B*a^2*b^3*d^2*e^3 + (11*B*a^3*b^2 - 15*A*a^2*b^3)*d*e^4 + 2*(B*a^4*b - 5*A*a^3*b^2)*e^5)*x^2 + (16*
B*a^3*b^2*d^2*e^3 + 4*(2*B*a^4*b - 5*A*a^3*b^2)*d*e^4 + (B*a^5 - 5*A*a^4*b)*e^5)*x)*log(e*x + d))/(a^4*b^6*d^7
- 6*a^5*b^5*d^6*e + 15*a^6*b^4*d^5*e^2 - 20*a^7*b^3*d^4*e^3 + 15*a^8*b^2*d^3*e^4 - 6*a^9*b*d^2*e^5 + a^10*d*e
^6 + (b^10*d^6*e - 6*a*b^9*d^5*e^2 + 15*a^2*b^8*d^4*e^3 - 20*a^3*b^7*d^3*e^4 + 15*a^4*b^6*d^2*e^5 - 6*a^5*b^5*
d*e^6 + a^6*b^4*e^7)*x^5 + (b^10*d^7 - 2*a*b^9*d^6*e - 9*a^2*b^8*d^5*e^2 + 40*a^3*b^7*d^4*e^3 - 65*a^4*b^6*d^3
*e^4 + 54*a^5*b^5*d^2*e^5 - 23*a^6*b^4*d*e^6 + 4*a^7*b^3*e^7)*x^4 + 2*(2*a*b^9*d^7 - 9*a^2*b^8*d^6*e + 12*a^3*
b^7*d^5*e^2 + 5*a^4*b^6*d^4*e^3 - 30*a^5*b^5*d^3*e^4 + 33*a^6*b^4*d^2*e^5 - 16*a^7*b^3*d*e^6 + 3*a^8*b^2*e^7)*
x^3 + 2*(3*a^2*b^8*d^7 - 16*a^3*b^7*d^6*e + 33*a^4*b^6*d^5*e^2 - 30*a^5*b^5*d^4*e^3 + 5*a^6*b^4*d^3*e^4 + 12*a
^7*b^3*d^2*e^5 - 9*a^8*b^2*d*e^6 + 2*a^9*b*e^7)*x^2 + (4*a^3*b^7*d^7 - 23*a^4*b^6*d^6*e + 54*a^5*b^5*d^5*e^2 -
65*a^6*b^4*d^4*e^3 + 40*a^7*b^3*d^3*e^4 - 9*a^8*b^2*d^2*e^5 - 2*a^9*b*d*e^6 + a^10*e^7)*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)**2/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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