3.3196 \(\int \frac{(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx\)

Optimal. Leaf size=69 \[ \frac{2 (3 x+2)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2}{7} (3 x+2)\right )}{77 (m+1)}-\frac{5 (3 x+2)^{m+1} \, _2F_1(1,m+1;m+2;5 (3 x+2))}{11 (m+1)} \]

[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))

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Rubi [A]  time = 0.0186439, antiderivative size = 69, normalized size of antiderivative = 1., 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 = {86, 68} \[ \frac{2 (3 x+2)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2}{7} (3 x+2)\right )}{77 (m+1)}-\frac{5 (3 x+2)^{m+1} \, _2F_1(1,m+1;m+2;5 (3 x+2))}{11 (m+1)} \]

Antiderivative was successfully verified.

[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 86

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]

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{(2+3 x)^m}{(1-2 x) (3+5 x)} \, dx &=\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]  time = 0.0386489, size = 55, normalized size = 0.8 \[ -\frac{(3 x+2)^{m+1} \left (35 \, _2F_1(1,m+1;m+2;5 (3 x+2))-2 \, _2F_1\left (1,m+1;m+2;\frac{2}{7} (3 x+2)\right )\right )}{77 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)]))/(77*(1 + m))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 2+3\,x \right ) ^{m}}{ \left ( 1-2\,x \right ) \left ( 3+5\,x \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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}}{10 \, x^{2} + x - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [C]  time = 1.29624, size = 112, normalized size = 1.62 \begin{align*} - \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} \Phi \left (\frac{1}{15 \left (x + \frac{2}{3}\right )}, 1, 1 - m\right ) \Gamma \left (1 - m\right )}{165 \left (x + \frac{2}{3}\right ) \Gamma \left (2 - m\right )} - \frac{3^{m} \left (x + \frac{2}{3}\right )^{m} \Phi \left (\frac{1}{15 \left (x + \frac{2}{3}\right )}, 1, 1 - m\right ) \Gamma \left (1 - m\right )}{165 \left (x + \frac{2}{3}\right ) \Gamma \left (2 - m\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*lerchphi(1/(15*(x + 2/3)), 1, 1 - m)*gamma(1 - m)/(165*(x + 2/3)*gamma(2 - m)) - 3**m*(x + 2/3)**m*le
rchphi(1/(15*(x + 2/3)), 1, 1 - m)*gamma(1 - m)/(165*(x + 2/3)*gamma(2 - m))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)