3.438 \(\int \frac{(d+e x)^m}{c d x+c e x^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac{(d+e x)^m \, _2F_1\left (1,m;m+1;\frac{e x}{d}+1\right )}{c d m} \]

[Out]

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

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Rubi [A]  time = 0.0151232, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {626, 12, 65} \[ -\frac{(d+e x)^m \, _2F_1\left (1,m;m+1;\frac{e x}{d}+1\right )}{c d m} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/(c*d*x + c*e*x^2),x]

[Out]

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

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

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

Mathematica [A]  time = 0.0084562, size = 32, normalized size = 1. \[ -\frac{(d+e x)^m \, _2F_1\left (1,m;m+1;\frac{e x}{d}+1\right )}{c d m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.547, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{ce{x}^{2}+cdx}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^m/(c*e*x^2 + c*d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(c*e*x**2+c*d*x),x)

[Out]

Integral((d + e*x)**m/(d*x + e*x**2), x)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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