3.3199 \(\int \frac{(a+b x)^m}{(e+f x)^2} \, dx\)

Optimal. Leaf size=52 \[ \frac{b (a+b x)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

[Out]

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

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Rubi [A]  time = 0.0110129, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {68} \[ \frac{b (a+b x)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m/(e + f*x)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0170137, size = 52, normalized size = 1. \[ \frac{b (a+b x)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m/(e + f*x)^2,x]

[Out]

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

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{ \left ( fx+e \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**m/(e + f*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 )}^{m}}{{\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/(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((b*x + a)^m/(f*x + e)^2, x)