3.3191 \(\int \frac {(1-2 x) (2+3 x)^m}{(3+5 x)^2} \, dx\)

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

[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)

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Rubi [A]  time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {78, 68} \[ \frac {(2-33 m) (3 x+2)^{m+1} \, _2F_1(1,m+1;m+2;5 (3 x+2))}{5 (m+1)}-\frac {11 (3 x+2)^{m+1}}{5 (5 x+3)} \]

Antiderivative was successfully verified.

[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 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]

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]))))

Rubi steps

\begin {align*} \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 {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*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.97 \[ -\frac {(3 x+2)^{m+1} ((33 m-2) (5 x+3) \, _2F_1(1,m+1;m+2;5 (3 x+2))+11 (m+1))}{5 (m+1) (5 x+3)} \]

Antiderivative was successfully verified.

[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))

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{25 \, x^{2} + 30 \, x + 9}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (-2 x +1\right ) \left (3 x +2\right )^{m}}{\left (5 x +3\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{{\left (5 \, x + 3\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int \frac {\left (2\,x-1\right )\,{\left (3\,x+2\right )}^m}{{\left (5\,x+3\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\left (3 x + 2\right )^{m}}{25 x^{2} + 30 x + 9}\right )\, dx - \int \frac {2 x \left (3 x + 2\right )^{m}}{25 x^{2} + 30 x + 9}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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