\(\int \frac {(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx\) [3196]

Optimal result

Integrand size = 22, antiderivative size = 69 \[ \int \frac {(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx=\frac {2 (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )}{77 (1+m)}-\frac {5 (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,5 (2+3 x))}{11 (1+m)} \]

[Out]

2/77*(2+3*x)^(1+m)*hypergeom([1, 1+m],[2+m],4/7+6/7*x)/(1+m)-5/11*(2+3*x)^(1+m)*hypergeom([1, 1+m],[2+m],10+15
*x)/(1+m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, 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 = {88, 70} \[ \int \frac {(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx=\frac {2 (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2}{7} (3 x+2)\right )}{77 (m+1)}-\frac {5 (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,5 (3 x+2))}{11 (m+1)} \]

[In]

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

[Out]

(2*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7])/(77*(1 + m)) - (5*(2 + 3*x)^(1 + m)*
Hypergeometric2F1[1, 1 + m, 2 + m, 5*(2 + 3*x)])/(11*(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 88

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{11} \int \frac {(2+3 x)^m}{1-2 x} \, dx+\frac {5}{11} \int \frac {(2+3 x)^m}{3+5 x} \, dx \\ & = \frac {2 (2+3 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2}{7} (2+3 x)\right )}{77 (1+m)}-\frac {5 (2+3 x)^{1+m} \, _2F_1(1,1+m;2+m;5 (2+3 x))}{11 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \frac {(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx=-\frac {(2+3 x)^{1+m} \left (-2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )+35 \operatorname {Hypergeometric2F1}(1,1+m,2+m,5 (2+3 x))\right )}{77 (1+m)} \]

[In]

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

[Out]

-1/77*((2 + 3*x)^(1 + m)*(-2*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7] + 35*Hypergeometric2F1[1, 1 +
m, 2 + m, 5*(2 + 3*x)]))/(1 + m)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(3*x + 2)^m/(10*x^2 + x - 3), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.93 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.52 \[ \int \frac {(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx=- \frac {3^{m} m \left (x + \frac {2}{3}\right )^{m} \Phi \left (\frac {7}{6 \left (x + \frac {2}{3}\right )}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{11 \Gamma \left (1 - m\right )} + \frac {3^{m} m \left (x + \frac {2}{3}\right )^{m - 1} \Phi \left (\frac {1}{15 \left (x + \frac {2}{3}\right )}, 1, 1 - m\right ) \Gamma \left (1 - m\right )}{165 \Gamma \left (2 - m\right )} - \frac {3^{m} \left (x + \frac {2}{3}\right )^{m - 1} \Phi \left (\frac {1}{15 \left (x + \frac {2}{3}\right )}, 1, 1 - m\right ) \Gamma \left (1 - m\right )}{165 \Gamma \left (2 - m\right )} \]

[In]

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

[Out]

-3**m*m*(x + 2/3)**m*lerchphi(7/(6*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(11*gamma(1 - m)) + 3**m*m*(x +
2/3)**(m - 1)*lerchphi(1/(15*(x + 2/3)), 1, 1 - m)*gamma(1 - m)/(165*gamma(2 - m)) - 3**m*(x + 2/3)**(m - 1)*
lerchphi(1/(15*(x + 2/3)), 1, 1 - m)*gamma(1 - m)/(165*gamma(2 - m))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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