\(\int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx\) [3194]

Optimal result

Integrand size = 22, antiderivative size = 72 \[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=-\frac {155 (2+3 x)^{1+m}}{36 (1+m)}-\frac {25 (2+3 x)^{2+m}}{18 (2+m)}+\frac {121 (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )}{28 (1+m)} \]

[Out]

-155/36*(2+3*x)^(1+m)/(1+m)-25/18*(2+3*x)^(2+m)/(2+m)+121/28*(2+3*x)^(1+m)*hypergeom([1, 1+m],[2+m],4/7+6/7*x)
/(1+m)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {90, 70} \[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=\frac {121 (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2}{7} (3 x+2)\right )}{28 (m+1)}-\frac {155 (3 x+2)^{m+1}}{36 (m+1)}-\frac {25 (3 x+2)^{m+2}}{18 (m+2)} \]

[In]

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

[Out]

(-155*(2 + 3*x)^(1 + m))/(36*(1 + m)) - (25*(2 + 3*x)^(2 + m))/(18*(2 + m)) + (121*(2 + 3*x)^(1 + m)*Hypergeom
etric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7])/(28*(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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {155}{12} (2+3 x)^m+\frac {121 (2+3 x)^m}{4 (1-2 x)}-\frac {25}{6} (2+3 x)^{1+m}\right ) \, dx \\ & = -\frac {155 (2+3 x)^{1+m}}{36 (1+m)}-\frac {25 (2+3 x)^{2+m}}{18 (2+m)}+\frac {121}{4} \int \frac {(2+3 x)^m}{1-2 x} \, dx \\ & = -\frac {155 (2+3 x)^{1+m}}{36 (1+m)}-\frac {25 (2+3 x)^{2+m}}{18 (2+m)}+\frac {121 (2+3 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2}{7} (2+3 x)\right )}{28 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=\frac {(2+3 x)^{1+m} \left (-35 (82+30 x+m (51+30 x))+1089 (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )\right )}{252 (1+m) (2+m)} \]

[In]

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

[Out]

((2 + 3*x)^(1 + m)*(-35*(82 + 30*x + m*(51 + 30*x)) + 1089*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 +
3*x))/7]))/(252*(1 + m)*(2 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(25*x^2 + 30*x + 9)*(3*x + 2)^m/(2*x - 1), x)

Sympy [F]

\[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=- \int \frac {9 \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {30 x \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {25 x^{2} \left (3 x + 2\right )^{m}}{2 x - 1}\, dx \]

[In]

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

[Out]

-Integral(9*(3*x + 2)**m/(2*x - 1), x) - Integral(30*x*(3*x + 2)**m/(2*x - 1), x) - Integral(25*x**2*(3*x + 2)
**m/(2*x - 1), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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