\(\int (1+2 x)^{-m} (2+3 x)^m \, dx\) [1889]

Optimal result

Integrand size = 17, antiderivative size = 47 \[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\frac {2^{-1-m} (1+2 x)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (1+2 x))}{1-m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {71} \[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\frac {2^{-m-1} (2 x+1)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (2 x+1))}{1-m} \]

[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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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*} \text {integral}& = \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] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\frac {2^{-1-m} (1+2 x)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (1+2 x))}{1-m} \]

[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)

Maple [F]

\[\int \left (2+3 x \right )^{m} \left (1+2 x \right )^{-m}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\int { \frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}} \,d x } \]

[In]

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

[Out]

integral((3*x + 2)^m/(2*x + 1)^m, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 14.53 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\frac {3^{2 m} \left (x + \frac {2}{3}\right )^{m + 1} e^{- i \pi m} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} m, m + 1 \\ m + 2 \end {matrix}\middle | {6 x + 4} \right )}}{\Gamma \left (m + 2\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\int { \frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}} \,d x } \]

[In]

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

[Out]

integrate((3*x + 2)^m/(2*x + 1)^m, x)

Giac [F]

\[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\int { \frac {{\left (3 \, x + 2\right )}^{m}}{{\left (2 \, x + 1\right )}^{m}} \,d x } \]

[In]

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

[Out]

integrate((3*x + 2)^m/(2*x + 1)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (1+2 x)^{-m} (2+3 x)^m \, dx=\int \frac {{\left (3\,x+2\right )}^m}{{\left (2\,x+1\right )}^m} \,d x \]

[In]

int((3*x + 2)^m/(2*x + 1)^m,x)

[Out]

int((3*x + 2)^m/(2*x + 1)^m, x)