3.1889 \(\int (1+2 x)^{-m} (2+3 x)^m \, dx\)

Optimal. Leaf size=47 \[ \frac{2^{-m-1} (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{1-m} \]

[Out]

(2^(-1 - m)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/(1 - m)

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Rubi [A]  time = 0.0144975, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {69} \[ \frac{2^{-m-1} (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{1-m} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^m/(1 + 2*x)^m,x]

[Out]

(2^(-1 - m)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/(1 - m)

Rule 69

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

Rubi steps

\begin{align*} \int (1+2 x)^{-m} (2+3 x)^m \, dx &=\frac{2^{-1-m} (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{1-m}\\ \end{align*}

Mathematica [A]  time = 0.012871, size = 47, normalized size = 1. \[ \frac{2^{-m-1} (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{1-m} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^m/(1 + 2*x)^m,x]

[Out]

(2^(-1 - m)*(1 + 2*x)^(1 - m)*Hypergeometric2F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/(1 - m)

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 2+3\,x \right ) ^{m}}{ \left ( 1+2\,x \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)^m/((1+2*x)^m),x)

[Out]

int((2+3*x)^m/((1+2*x)^m),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^m/((1+2*x)^m),x, algorithm="maxima")

[Out]

integrate((3*x + 2)^m/(2*x + 1)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^m/((1+2*x)^m),x, algorithm="fricas")

[Out]

integral((3*x + 2)^m/(2*x + 1)^m, x)

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Sympy [C]  time = 61.1545, size = 42, normalized size = 0.89 \begin{align*} \frac{3^{2 m} \left (x + \frac{2}{3}\right ) \left (x + \frac{2}{3}\right )^{m} e^{- i \pi m} \Gamma \left (m + 1\right ){{}_{2}F_{1}\left (\begin{matrix} m, m + 1 \\ m + 2 \end{matrix}\middle |{6 x + 4} \right )}}{\Gamma \left (m + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**m/((1+2*x)**m),x)

[Out]

3**(2*m)*(x + 2/3)*(x + 2/3)**m*exp(-I*pi*m)*gamma(m + 1)*hyper((m, m + 1), (m + 2,), 6*x + 4)/gamma(m + 2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^m/((1+2*x)^m),x, algorithm="giac")

[Out]

integrate((3*x + 2)^m/(2*x + 1)^m, x)