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)*hypergeom([-m, 1-m],[2-m],-3-6*x)/(1-m)

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Rubi [A]  time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 1.00 \[ \frac {2^{-m-1} (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{1-m} \]

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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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}}\,{d x} \]

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)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (2 x +1\right )^{-m} \left (3 x +2\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (3\,x+2\right )}^m}{{\left (2\,x+1\right )}^m} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 2)^m/(2*x + 1)^m,x)

[Out]

int((3*x + 2)^m/(2*x + 1)^m, x)

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sympy [C]  time = 25.03, size = 42, normalized size = 0.89 \[ \frac {3^{2 m} \left (x + \frac {2}{3}\right ) \left (x + \frac {2}{3}\right )^{m} e^{- i \pi m} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} m, m + 1 \\ m + 2 \end {matrix}\middle | {6 x + 4} \right )}}{\Gamma \left (m + 2\right )} \]

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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