3.3200 \(\int \frac{(a+b x)^m}{(c+d x) (e+f x)^2} \, dx\)

Optimal. Leaf size=187 \[ \frac{d^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f)^2}+\frac{f (a+b x)^{m+1} (a d f-b (c f m+d e (1-m))) \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1}}{(e+f x) (b e-a f) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.175775, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {103, 156, 68} \[ \frac{d^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f)^2}+\frac{f (a+b x)^{m+1} (a d f-b c f m-b d e (1-m)) \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2 (d e-c f)^2}-\frac{f (a+b x)^{m+1}}{(e+f x) (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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

Mathematica [A]  time = 0.235035, size = 174, normalized size = 0.93 \[ \frac{(a+b x)^{m+1} \left (-\frac{d^2 (b e-a f) \, _2F_1\left (1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )}{(m+1) (b c-a d) (c f-d e)}+\frac{f (a d f-b c f m+b d e (m-1)) \, _2F_1\left (1,m+1;m+2;\frac{f (a+b x)}{a f-b e}\right )}{(m+1) (b e-a f) (d e-c f)}-\frac{f}{e+f x}\right )}{(b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.085, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{ \left ( dx+c \right ) \left ( fx+e \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m/(d*x+c)/(f*x+e)^2,x)

[Out]

int((b*x+a)^m/(d*x+c)/(f*x+e)^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^m/((d*x + c)*(f*x + e)^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 )}^{m}}{d f^{2} x^{3} + c e^{2} +{\left (2 \, d e f + c f^{2}\right )} x^{2} +{\left (d e^{2} + 2 \, c e f\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^m/(d*f^2*x^3 + c*e^2 + (2*d*e*f + c*f^2)*x^2 + (d*e^2 + 2*c*e*f)*x), 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)**m/(d*x+c)/(f*x+e)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)^m/((d*x + c)*(f*x + e)^2), x)