3.1887 \(\int \frac{(A+B x) (d+e x)^m}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=126 \[ -\frac{e^2 (d+e x)^{m+1} (b (3 B d-A e (2-m))-a B e (m+1)) \, _2F_1\left (3,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{3 b (m+1) (b d-a e)^4}-\frac{(A b-a B) (d+e x)^{m+1}}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

-((A*b - a*B)*(d + e*x)^(1 + m))/(3*b*(b*d - a*e)*(a + b*x)^3) - (e^2*(b*(3*B*d - A*e*(2 - m)) - a*B*e*(1 + m)
)*(d + e*x)^(1 + m)*Hypergeometric2F1[3, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(3*b*(b*d - a*e)^4*(1 + m))

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Rubi [A]  time = 0.0725696, antiderivative size = 125, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {27, 78, 68} \[ -\frac{e^2 (d+e x)^{m+1} (-a B e (m+1)-A b e (2-m)+3 b B d) \, _2F_1\left (3,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{3 b (m+1) (b d-a e)^4}-\frac{(A b-a B) (d+e x)^{m+1}}{3 b (a+b x)^3 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b - a*B)*(d + e*x)^(1 + m))/(3*b*(b*d - a*e)*(a + b*x)^3) - (e^2*(3*b*B*d - A*b*e*(2 - m) - a*B*e*(1 + m)
)*(d + e*x)^(1 + m)*Hypergeometric2F1[3, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(3*b*(b*d - a*e)^4*(1 + m))

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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(A+B x) (d+e x)^m}{(a+b x)^4} \, dx\\ &=-\frac{(A b-a B) (d+e x)^{1+m}}{3 b (b d-a e) (a+b x)^3}+\frac{(3 b B d-A b e (2-m)-a B e (1+m)) \int \frac{(d+e x)^m}{(a+b x)^3} \, dx}{3 b (b d-a e)}\\ &=-\frac{(A b-a B) (d+e x)^{1+m}}{3 b (b d-a e) (a+b x)^3}-\frac{e^2 (3 b B d-A b e (2-m)-a B e (1+m)) (d+e x)^{1+m} \, _2F_1\left (3,1+m;2+m;\frac{b (d+e x)}{b d-a e}\right )}{3 b (b d-a e)^4 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0807949, size = 109, normalized size = 0.87 \[ \frac{(d+e x)^{m+1} \left (\frac{a B-A b}{(a+b x)^3}-\frac{e^2 (-a B e (m+1)+A b e (m-2)+3 b B d) \, _2F_1\left (3,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{(m+1) (b d-a e)^3}\right )}{3 b (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*((-(A*b) + a*B)/(a + b*x)^3 - (e^2*(3*b*B*d + A*b*e*(-2 + m) - a*B*e*(1 + m))*Hypergeometri
c2F1[3, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)^3*(1 + m))))/(3*b*(b*d - a*e))

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Maple [F]  time = 0.191, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( ex+d \right ) ^{m}}{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )}{\left (e x + d\right )}^{m}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x + A)*(e*x + d)^m/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )^{m}}{\left (a + b x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(d + e*x)**m/(a + b*x)**4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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