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

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Rubi [A]  time = 0.01, antiderivative size = 52, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 1.00 \[ \frac {b (a+b x)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

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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fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{2}}\,{d x} \]

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)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m}}{\left (f x +e \right )^{2}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m/(e + f*x)^2,x)

[Out]

int((a + b*x)^m/(e + f*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{m}}{\left (e + f x\right )^{2}}\, dx \]

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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