\(\int \frac {(1+4 x)^m}{(2+3 x)^2 (1-5 x+3 x^2)} \, dx\) [936]

Optimal result

Integrand size = 27, antiderivative size = 199 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\frac {27 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {3 \left (117+47 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {3 \left (117-47 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {12 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{425 (1+m)} \]

[Out]

27/1445*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],-3/5-12/5*x)/(1+m)+12/425*(1+4*x)^(1+m)*hypergeom([2, 1+m],[2+m
],-3/5-12/5*x)/(1+m)+3/7514*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(117-47*13^(1/2)
)/(1+m)/(13+2*13^(1/2))+3/7514*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2)))*(117+47*13^(1
/2))/(1+m)/(13-2*13^(1/2))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {974, 70, 844} \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\frac {27 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{1445 (m+1)}+\frac {3 \left (117+47 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {3 \left (117-47 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {12 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{425 (m+1)} \]

[In]

Int[(1 + 4*x)^m/((2 + 3*x)^2*(1 - 5*x + 3*x^2)),x]

[Out]

(27*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(1445*(1 + m)) + (3*(117 + 47*Sqrt
[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(7514*(13 - 2*Sqr
t[13])*(1 + m)) + (3*(117 - 47*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(1
3 + 2*Sqrt[13])])/(7514*(13 + 2*Sqrt[13])*(1 + m)) + (12*(1 + 4*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m,
(-3*(1 + 4*x))/5])/(425*(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 844

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[m]

Rule 974

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 (1+4 x)^m}{17 (2+3 x)^2}+\frac {27 (1+4 x)^m}{289 (2+3 x)}+\frac {(46-27 x) (1+4 x)^m}{289 \left (1-5 x+3 x^2\right )}\right ) \, dx \\ & = \frac {1}{289} \int \frac {(46-27 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx+\frac {27}{289} \int \frac {(1+4 x)^m}{2+3 x} \, dx+\frac {3}{17} \int \frac {(1+4 x)^m}{(2+3 x)^2} \, dx \\ & = \frac {27 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {12 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{425 (1+m)}+\frac {1}{289} \int \left (\frac {\left (-27+\frac {141}{\sqrt {13}}\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-27-\frac {141}{\sqrt {13}}\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {27 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {12 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{425 (1+m)}-\frac {\left (3 \left (117-47 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx}{3757}-\frac {\left (3 \left (117+47 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{3757} \\ & = \frac {27 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {3 \left (117+47 \sqrt {13}\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {3 \left (117-47 \sqrt {13}\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {12 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{425 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.76 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\frac {(1+4 x)^{1+m} \left (10530 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )+25 \left (211+65 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+5275 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )-1625 \sqrt {13} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )+15912 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )\right )}{563550 (1+m)} \]

[In]

Integrate[(1 + 4*x)^m/((2 + 3*x)^2*(1 - 5*x + 3*x^2)),x]

[Out]

((1 + 4*x)^(1 + m)*(10530*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5] + 25*(211 + 65*Sqrt[13])*Hyperg
eometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])] + 5275*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x
)/(13 + 2*Sqrt[13])] - 1625*Sqrt[13]*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])] + 15912*
Hypergeometric2F1[2, 1 + m, 2 + m, (-3*(1 + 4*x))/5]))/(563550*(1 + m))

Maple [F]

\[\int \frac {\left (1+4 x \right )^{m}}{\left (2+3 x \right )^{2} \left (3 x^{2}-5 x +1\right )}d x\]

[In]

int((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1),x)

[Out]

int((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1),x)

Fricas [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}^{2}} \,d x } \]

[In]

integrate((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1),x, algorithm="fricas")

[Out]

integral((4*x + 1)^m/(27*x^4 - 9*x^3 - 39*x^2 - 8*x + 4), x)

Sympy [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x + 2\right )^{2} \cdot \left (3 x^{2} - 5 x + 1\right )}\, dx \]

[In]

integrate((1+4*x)**m/(2+3*x)**2/(3*x**2-5*x+1),x)

[Out]

Integral((4*x + 1)**m/((3*x + 2)**2*(3*x**2 - 5*x + 1)), x)

Maxima [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}^{2}} \,d x } \]

[In]

integrate((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1),x, algorithm="maxima")

[Out]

integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)*(3*x + 2)^2), x)

Giac [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}^{2}} \,d x } \]

[In]

integrate((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1),x, algorithm="giac")

[Out]

integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)*(3*x + 2)^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{{\left (3\,x+2\right )}^2\,\left (3\,x^2-5\,x+1\right )} \,d x \]

[In]

int((4*x + 1)^m/((3*x + 2)^2*(3*x^2 - 5*x + 1)),x)

[Out]

int((4*x + 1)^m/((3*x + 2)^2*(3*x^2 - 5*x + 1)), x)