\(\int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx\) [3195]

Optimal result

Integrand size = 20, antiderivative size = 54 \[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=-\frac {5 (2+3 x)^{1+m}}{6 (1+m)}+\frac {11 (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )}{14 (1+m)} \]

[Out]

-5/6*(2+3*x)^(1+m)/(1+m)+11/14*(2+3*x)^(1+m)*hypergeom([1, 1+m],[2+m],4/7+6/7*x)/(1+m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {81, 70} \[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=\frac {11 (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2}{7} (3 x+2)\right )}{14 (m+1)}-\frac {5 (3 x+2)^{m+1}}{6 (m+1)} \]

[In]

Int[((2 + 3*x)^m*(3 + 5*x))/(1 - 2*x),x]

[Out]

(-5*(2 + 3*x)^(1 + m))/(6*(1 + m)) + (11*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7]
)/(14*(1 + m))

Rule 70

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

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rubi steps \begin{align*} \text {integral}& = -\frac {5 (2+3 x)^{1+m}}{6 (1+m)}+\frac {11}{2} \int \frac {(2+3 x)^m}{1-2 x} \, dx \\ & = -\frac {5 (2+3 x)^{1+m}}{6 (1+m)}+\frac {11 (2+3 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2}{7} (2+3 x)\right )}{14 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=\frac {(2+3 x)^{1+m} \left (-35+33 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )\right )}{42 (1+m)} \]

[In]

Integrate[((2 + 3*x)^m*(3 + 5*x))/(1 - 2*x),x]

[Out]

((2 + 3*x)^(1 + m)*(-35 + 33*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7]))/(42*(1 + m))

Maple [F]

\[\int \frac {\left (2+3 x \right )^{m} \left (3+5 x \right )}{1-2 x}d x\]

[In]

int((2+3*x)^m*(3+5*x)/(1-2*x),x)

[Out]

int((2+3*x)^m*(3+5*x)/(1-2*x),x)

Fricas [F]

\[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (5 \, x + 3\right )}}{2 \, x - 1} \,d x } \]

[In]

integrate((2+3*x)^m*(3+5*x)/(1-2*x),x, algorithm="fricas")

[Out]

integral(-(3*x + 2)^m*(5*x + 3)/(2*x - 1), x)

Sympy [F]

\[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=- \int \frac {3 \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {5 x \left (3 x + 2\right )^{m}}{2 x - 1}\, dx \]

[In]

integrate((2+3*x)**m*(3+5*x)/(1-2*x),x)

[Out]

-Integral(3*(3*x + 2)**m/(2*x - 1), x) - Integral(5*x*(3*x + 2)**m/(2*x - 1), x)

Maxima [F]

\[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (5 \, x + 3\right )}}{2 \, x - 1} \,d x } \]

[In]

integrate((2+3*x)^m*(3+5*x)/(1-2*x),x, algorithm="maxima")

[Out]

-integrate((3*x + 2)^m*(5*x + 3)/(2*x - 1), x)

Giac [F]

\[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (5 \, x + 3\right )}}{2 \, x - 1} \,d x } \]

[In]

integrate((2+3*x)^m*(3+5*x)/(1-2*x),x, algorithm="giac")

[Out]

integrate(-(3*x + 2)^m*(5*x + 3)/(2*x - 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(2+3 x)^m (3+5 x)}{1-2 x} \, dx=\int -\frac {{\left (3\,x+2\right )}^m\,\left (5\,x+3\right )}{2\,x-1} \,d x \]

[In]

int(-((3*x + 2)^m*(5*x + 3))/(2*x - 1),x)

[Out]

int(-((3*x + 2)^m*(5*x + 3))/(2*x - 1), x)