3.3186 \(\int \frac{(A+B x) (d+e x)^m}{(a+b x)^2} \, dx\)

Optimal. Leaf size=112 \[ \frac{(d+e x)^{m+1} (a B e (m+1)-b (A e m+B d)) \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{b (m+1) (b d-a e)^2}-\frac{(A b-a B) (d+e x)^{m+1}}{b (a+b x) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0568046, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {78, 68} \[ \frac{(d+e x)^{m+1} (a B e (m+1)-b (A e m+B d)) \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{b (m+1) (b d-a e)^2}-\frac{(A b-a B) (d+e x)^{m+1}}{b (a+b x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0484106, size = 98, normalized size = 0.88 \[ \frac{(d+e x)^{m+1} \left (\frac{(a B e (m+1)-b (A e m+B d)) \, _2F_1\left (1,m+1;m+2;\frac{b (d+e x)}{b d-a e}\right )}{m+1}+\frac{(a B-A b) (b d-a e)}{a+b x}\right )}{b (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(((-(A*b) + a*B)*(b*d - a*e))/(a + b*x) + ((a*B*e*(1 + m) - b*(B*d + A*e*m))*Hypergeometric
2F1[1, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(1 + m)))/(b*(b*d - a*e)^2)

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( ex+d \right ) ^{m}}{ \left ( bx+a \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*x+a)^2,x)

[Out]

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

[Out]

integrate((B*x + A)*(e*x + d)^m/(b*x + a)^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^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x + A)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2), 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 )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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