3.32.97 \(\int \frac {(2+3 x)^m}{(1-2 x) (3+5 x)^2} \, dx\) [3197]

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

[Out]

-5/11*(2+3*x)^(1+m)/(3+5*x)+4/847*(2+3*x)^(1+m)*hypergeom([1, 1+m],[2+m],4/7+6/7*x)/(1+m)-5/121*(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 = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, 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 = {105, 162, 70} \begin {gather*} \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)} \end {gather*}

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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 105

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] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 162

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]

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*}

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Mathematica [A]
time = 0.08, size = 82, normalized size = 0.87 \begin {gather*} \frac {(2+3 x)^{1+m} \left (-385 (1+m)+4 (3+5 x) \, _2F_1\left (1,1+m;2+m;\frac {2}{7} (2+3 x)\right )-35 (2+33 m) (3+5 x) \, _2F_1(1,1+m;2+m;5 (2+3 x))\right )}{847 (1+m) (3+5 x)} \end {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (2+3 x \right )^{m}}{\left (1-2 x \right ) \left (3+5 x \right )^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 1.36, size = 366, normalized size = 3.89 \begin {gather*} \frac {495 \cdot 3^{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 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} - \frac {33 \cdot 3^{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 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} + \frac {30 \cdot 3^{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 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} - \frac {30 \cdot 3^{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 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} + \frac {495 \cdot 3^{m} m \left (x + \frac {2}{3}\right ) \left (x + \frac {2}{3}\right )^{m} \Gamma \left (- m\right )}{1815 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} - \frac {2 \cdot 3^{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 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} + \frac {2 \cdot 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 )}{1815 \left (x + \frac {2}{3}\right ) \Gamma \left (1 - m\right ) - 121 \Gamma \left (1 - m\right )} \end {gather*}

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*3**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*(x + 2/3
)*gamma(1 - m) - 121*gamma(1 - m)) - 33*3**m*m**2*(x + 2/3)**m*lerchphi(1/(15*(x + 2/3)), 1, m*exp_polar(I*pi)
)*gamma(-m)/(1815*(x + 2/3)*gamma(1 - m) - 121*gamma(1 - m)) + 30*3**m*m*(x + 2/3)*(x + 2/3)**m*lerchphi(1/(15
*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*(x + 2/3)*gamma(1 - m) - 121*gamma(1 - m)) - 30*3**m*m*(x +
2/3)*(x + 2/3)**m*lerchphi(7/(6*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*(x + 2/3)*gamma(1 - m) - 12
1*gamma(1 - m)) + 495*3**m*m*(x + 2/3)*(x + 2/3)**m*gamma(-m)/(1815*(x + 2/3)*gamma(1 - m) - 121*gamma(1 - m))
- 2*3**m*m*(x + 2/3)**m*lerchphi(1/(15*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815*(x + 2/3)*gamma(1 -
m) - 121*gamma(1 - m)) + 2*3**m*m*(x + 2/3)**m*lerchphi(7/(6*(x + 2/3)), 1, m*exp_polar(I*pi))*gamma(-m)/(1815
*(x + 2/3)*gamma(1 - m) - 121*gamma(1 - m))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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