\(\int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^2} \, dx\) [3191]

Optimal result

Integrand size = 20, antiderivative size = 59 \[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^2} \, dx=-\frac {11 (2+3 x)^{1+m}}{5 (3+5 x)}+\frac {(2-33 m) (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,5 (2+3 x))}{5 (1+m)} \]

[Out]

-11/5*(2+3*x)^(1+m)/(3+5*x)+1/5*(2-33*m)*(2+3*x)^(1+m)*hypergeom([1, 1+m],[2+m],10+15*x)/(1+m)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, 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 = {79, 70} \[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^2} \, dx=\frac {(2-33 m) (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,5 (3 x+2))}{5 (m+1)}-\frac {11 (3 x+2)^{m+1}}{5 (5 x+3)} \]

[In]

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

[Out]

(-11*(2 + 3*x)^(1 + m))/(5*(3 + 5*x)) + ((2 - 33*m)*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 5*(2
+ 3*x)])/(5*(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 79

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

Rubi steps \begin{align*} \text {integral}& = -\frac {11 (2+3 x)^{1+m}}{5 (3+5 x)}-\frac {1}{5} (2-33 m) \int \frac {(2+3 x)^m}{3+5 x} \, dx \\ & = -\frac {11 (2+3 x)^{1+m}}{5 (3+5 x)}+\frac {(2-33 m) (2+3 x)^{1+m} \, _2F_1(1,1+m;2+m;5 (2+3 x))}{5 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^2} \, dx=-\frac {(2+3 x)^{1+m} (11 (1+m)+(-2+33 m) (3+5 x) \operatorname {Hypergeometric2F1}(1,1+m,2+m,5 (2+3 x)))}{5 (1+m) (3+5 x)} \]

[In]

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

[Out]

-1/5*((2 + 3*x)^(1 + m)*(11*(1 + m) + (-2 + 33*m)*(3 + 5*x)*Hypergeometric2F1[1, 1 + m, 2 + m, 5*(2 + 3*x)]))/
((1 + m)*(3 + 5*x))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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