3.3197 \(\int \frac{(2+3 x)^m}{(1-2 x) (3+5 x)^2} \, dx\)

Optimal. Leaf size=94 \[ \frac{4 (3 x+2)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2}{7} (3 x+2)\right )}{847 (m+1)}-\frac{5 (33 m+2) (3 x+2)^{m+1} \, _2F_1(1,m+1;m+2;5 (3 x+2))}{121 (m+1)}-\frac{5 (3 x+2)^{m+1}}{11 (5 x+3)} \]

[Out]

(-5*(2 + 3*x)^(1 + m))/(11*(3 + 5*x)) + (4*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/
7])/(847*(1 + m)) - (5*(2 + 33*m)*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 5*(2 + 3*x)])/(121*(1 +
m))

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Rubi [A]  time = 0.0478595, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {103, 156, 68} \[ \frac{4 (3 x+2)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2}{7} (3 x+2)\right )}{847 (m+1)}-\frac{5 (33 m+2) (3 x+2)^{m+1} \, _2F_1(1,m+1;m+2;5 (3 x+2))}{121 (m+1)}-\frac{5 (3 x+2)^{m+1}}{11 (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(2 + 3*x)^(1 + m))/(11*(3 + 5*x)) + (4*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/
7])/(847*(1 + m)) - (5*(2 + 33*m)*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 5*(2 + 3*x)])/(121*(1 +
m))

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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)^2} \, dx &=-\frac{5 (2+3 x)^{1+m}}{11 (3+5 x)}-\frac{1}{11} \int \frac{(2+3 x)^m (-2-15 m+30 m x)}{(1-2 x) (3+5 x)} \, dx\\ &=-\frac{5 (2+3 x)^{1+m}}{11 (3+5 x)}+\frac{4}{121} \int \frac{(2+3 x)^m}{1-2 x} \, dx+\frac{1}{121} (5 (2+33 m)) \int \frac{(2+3 x)^m}{3+5 x} \, dx\\ &=-\frac{5 (2+3 x)^{1+m}}{11 (3+5 x)}+\frac{4 (2+3 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2}{7} (2+3 x)\right )}{847 (1+m)}-\frac{5 (2+33 m) (2+3 x)^{1+m} \, _2F_1(1,1+m;2+m;5 (2+3 x))}{121 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0286241, size = 82, normalized size = 0.87 \[ \frac{(3 x+2)^{m+1} \left (4 (5 x+3) \, _2F_1\left (1,m+1;m+2;\frac{2}{7} (3 x+2)\right )-35 (33 m+2) (5 x+3) \, _2F_1(1,m+1;m+2;5 (3 x+2))-385 (m+1)\right )}{847 (m+1) (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2 + 3*x)^(1 + m)*(-385*(1 + m) + 4*(3 + 5*x)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7] - 35*(2 + 3
3*m)*(3 + 5*x)*Hypergeometric2F1[1, 1 + m, 2 + m, 5*(2 + 3*x)]))/(847*(1 + m)*(3 + 5*x))

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Maple [F]  time = 0.055, 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 ) ^{2}}}\, 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)^2,x)

[Out]

int((2+3*x)^m/(1-2*x)/(3+5*x)^2,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 )}^{2}{\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)^2,x, algorithm="maxima")

[Out]

-integrate((3*x + 2)^m/((5*x + 3)^2*(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}}{50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9}, 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)^2,x, algorithm="fricas")

[Out]

integral(-(3*x + 2)^m/(50*x^3 + 35*x^2 - 12*x - 9), x)

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Sympy [C]  time = 2.38398, size = 413, normalized size = 4.39 \begin{align*} \frac{495 \cdot 45^{m} m^{2} \left (x + \frac{2}{3}\right ) \left (x + \frac{2}{3}\right )^{m} \Phi \left (\frac{1}{15 \left (x + \frac{2}{3}\right )}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - m\right )} - \frac{33 \cdot 45^{m} m^{2} \left (x + \frac{2}{3}\right )^{m} \Phi \left (\frac{1}{15 \left (x + \frac{2}{3}\right )}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - m\right )} + \frac{30 \cdot 45^{m} m \left (x + \frac{2}{3}\right ) \left (x + \frac{2}{3}\right )^{m} \Phi \left (\frac{1}{15 \left (x + \frac{2}{3}\right )}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - m\right )} - \frac{30 \cdot 45^{m} m \left (x + \frac{2}{3}\right ) \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 )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - m\right )} + \frac{495 \cdot 45^{m} m \left (x + \frac{2}{3}\right ) \left (x + \frac{2}{3}\right )^{m} \Gamma \left (- m\right )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - m\right )} - \frac{2 \cdot 45^{m} m \left (x + \frac{2}{3}\right )^{m} \Phi \left (\frac{1}{15 \left (x + \frac{2}{3}\right )}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - m\right )} + \frac{2 \cdot 45^{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 )}{1815 \cdot 15^{m} \left (x + \frac{2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \cdot 15^{m} \Gamma \left (1 - 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)**2,x)

[Out]

495*45**m*m**2*(x + 2/3)*(x + 2/3)**m*lerchphi(1/(15*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*15**m*(
x + 2/3)*gamma(1 - m) - 121*15**m*gamma(1 - m)) - 33*45**m*m**2*(x + 2/3)**m*lerchphi(1/(15*(x + 2/3)), 1, m*e
xp_polar(I*pi))*gamma(-m)/(1815*15**m*(x + 2/3)*gamma(1 - m) - 121*15**m*gamma(1 - m)) + 30*45**m*m*(x + 2/3)*
(x + 2/3)**m*lerchphi(1/(15*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*15**m*(x + 2/3)*gamma(1 - m) - 1
21*15**m*gamma(1 - m)) - 30*45**m*m*(x + 2/3)*(x + 2/3)**m*lerchphi(7/(6*(x + 2/3)), 1, m*exp_polar(I*pi))*gam
ma(-m)/(1815*15**m*(x + 2/3)*gamma(1 - m) - 121*15**m*gamma(1 - m)) + 495*45**m*m*(x + 2/3)*(x + 2/3)**m*gamma
(-m)/(1815*15**m*(x + 2/3)*gamma(1 - m) - 121*15**m*gamma(1 - m)) - 2*45**m*m*(x + 2/3)**m*lerchphi(1/(15*(x +
2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*15**m*(x + 2/3)*gamma(1 - m) - 121*15**m*gamma(1 - m)) + 2*45**m
*m*(x + 2/3)**m*lerchphi(7/(6*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*15**m*(x + 2/3)*gamma(1 - m) -
121*15**m*gamma(1 - 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 )}^{2}{\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)^2,x, algorithm="giac")

[Out]

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