3.3192 \(\int \frac{(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=59 \[ -\frac{3 (37-33 m) (3 x+2)^{m+1} \, _2F_1(2,m+1;m+2;5 (3 x+2))}{10 (m+1)}-\frac{11 (3 x+2)^{m+1}}{10 (5 x+3)^2} \]

[Out]

(-11*(2 + 3*x)^(1 + m))/(10*(3 + 5*x)^2) - (3*(37 - 33*m)*(2 + 3*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m,
5*(2 + 3*x)])/(10*(1 + m))

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Rubi [A]  time = 0.0219263, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {78, 68} \[ -\frac{3 (37-33 m) (3 x+2)^{m+1} \, _2F_1(2,m+1;m+2;5 (3 x+2))}{10 (m+1)}-\frac{11 (3 x+2)^{m+1}}{10 (5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*(2 + 3*x)^(1 + m))/(10*(3 + 5*x)^2) - (3*(37 - 33*m)*(2 + 3*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m,
5*(2 + 3*x)])/(10*(1 + m))

Rule 78

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] || LtQ
[p, n]))))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(1-2 x) (2+3 x)^m}{(3+5 x)^3} \, dx &=-\frac{11 (2+3 x)^{1+m}}{10 (3+5 x)^2}-\frac{1}{10} (37-33 m) \int \frac{(2+3 x)^m}{(3+5 x)^2} \, dx\\ &=-\frac{11 (2+3 x)^{1+m}}{10 (3+5 x)^2}-\frac{3 (37-33 m) (2+3 x)^{1+m} \, _2F_1(2,1+m;2+m;5 (2+3 x))}{10 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0185501, size = 50, normalized size = 0.85 \[ \frac{1}{10} (3 x+2)^{m+1} \left (\frac{3 (33 m-37) \, _2F_1(2,m+1;m+2;5 (3 x+2))}{m+1}-\frac{11}{(5 x+3)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 + 3*x)^(1 + m)*(-11/(3 + 5*x)^2 + (3*(-37 + 33*m)*Hypergeometric2F1[2, 1 + m, 2 + m, 5*(2 + 3*x)])/(1 + m)
))/10

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 1-2\,x \right ) \left ( 2+3\,x \right ) ^{m}}{ \left ( 3+5\,x \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (3 \, x + 2\right )}^{m}{\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (3 \, x + 2\right )}^{m}{\left (2 \, x - 1\right )}}{125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(3*x + 2)^m*(2*x - 1)/(125*x^3 + 225*x^2 + 135*x + 27), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{\left (3 x + 2\right )^{m}}{125 x^{3} + 225 x^{2} + 135 x + 27}\, dx - \int \frac{2 x \left (3 x + 2\right )^{m}}{125 x^{3} + 225 x^{2} + 135 x + 27}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-(3*x + 2)**m/(125*x**3 + 225*x**2 + 135*x + 27), x) - Integral(2*x*(3*x + 2)**m/(125*x**3 + 225*x**
2 + 135*x + 27), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (3 \, x + 2\right )}^{m}{\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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