3.3193 \(\int \frac {(2+3 x)^m (3+5 x)^3}{1-2 x} \, dx\)

Optimal. Leaf size=90 \[ \frac {1331 (3 x+2)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {2}{7} (3 x+2)\right )}{56 (m+1)}-\frac {5135 (3 x+2)^{m+1}}{216 (m+1)}-\frac {725 (3 x+2)^{m+2}}{108 (m+2)}-\frac {125 (3 x+2)^{m+3}}{54 (m+3)} \]

[Out]

-5135/216*(2+3*x)^(1+m)/(1+m)-725/108*(2+3*x)^(2+m)/(2+m)-125/54*(2+3*x)^(3+m)/(3+m)+1331/56*(2+3*x)^(1+m)*hyp
ergeom([1, 1+m],[2+m],4/7+6/7*x)/(1+m)

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Rubi [A]  time = 0.03, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {88, 68} \[ \frac {1331 (3 x+2)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {2}{7} (3 x+2)\right )}{56 (m+1)}-\frac {5135 (3 x+2)^{m+1}}{216 (m+1)}-\frac {725 (3 x+2)^{m+2}}{108 (m+2)}-\frac {125 (3 x+2)^{m+3}}{54 (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5135*(2 + 3*x)^(1 + m))/(216*(1 + m)) - (725*(2 + 3*x)^(2 + m))/(108*(2 + m)) - (125*(2 + 3*x)^(3 + m))/(54*
(3 + m)) + (1331*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7])/(56*(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 88

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*} \int \frac {(2+3 x)^m (3+5 x)^3}{1-2 x} \, dx &=\int \left (-\frac {5135}{72} (2+3 x)^m+\frac {1331 (2+3 x)^m}{8 (1-2 x)}-\frac {725}{36} (2+3 x)^{1+m}-\frac {125}{18} (2+3 x)^{2+m}\right ) \, dx\\ &=-\frac {5135 (2+3 x)^{1+m}}{216 (1+m)}-\frac {725 (2+3 x)^{2+m}}{108 (2+m)}-\frac {125 (2+3 x)^{3+m}}{54 (3+m)}+\frac {1331}{8} \int \frac {(2+3 x)^m}{1-2 x} \, dx\\ &=-\frac {5135 (2+3 x)^{1+m}}{216 (1+m)}-\frac {725 (2+3 x)^{2+m}}{108 (2+m)}-\frac {125 (2+3 x)^{3+m}}{54 (3+m)}+\frac {1331 (2+3 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2}{7} (2+3 x)\right )}{56 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 71, normalized size = 0.79 \[ \frac {(3 x+2)^{m+1} \left (\frac {35937 \, _2F_1\left (1,m+1;m+2;\frac {2}{7} (3 x+2)\right )}{m+1}-\frac {3500 (3 x+2)^2}{m+3}-\frac {10150 (3 x+2)}{m+2}-\frac {35945}{m+1}\right )}{1512} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 + 3*x)^(1 + m)*(-35945/(1 + m) - (10150*(2 + 3*x))/(2 + m) - (3500*(2 + 3*x)^2)/(3 + m) + (35937*Hypergeom
etric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7])/(1 + m)))/1512

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )} {\left (3 \, x + 2\right )}^{m}}{2 \, x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x +3\right )^{3} \left (3 x +2\right )^{m}}{-2 x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {{\left (3\,x+2\right )}^m\,{\left (5\,x+3\right )}^3}{2\,x-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {27 \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {135 x \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {225 x^{2} \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {125 x^{3} \left (3 x + 2\right )^{m}}{2 x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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