\(\int \frac {(1+4 x)^m}{1-5 x+3 x^2} \, dx\) [934]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int \frac {(1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {3 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{\sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{\sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \]

[Out]

3/13*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2)))/(1+m)/(13-2*13^(1/2))*13^(1/2)-3/13*(1+
4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))/(1+m)*13^(1/2)/(13+2*13^(1/2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {725, 70} \[ \int \frac {(1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {3 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{\sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{\sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)} \]

[In]

Int[(1 + 4*x)^m/(1 - 5*x + 3*x^2),x]

[Out]

(3*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(Sqrt[13]*(13 - 2*Sq
rt[13])*(1 + m)) - (3*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(
Sqrt[13]*(13 + 2*Sqrt[13])*(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 725

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^
m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {6 (1+4 x)^m}{\sqrt {13} \left (5+\sqrt {13}-6 x\right )}-\frac {6 (1+4 x)^m}{\sqrt {13} \left (-5+\sqrt {13}+6 x\right )}\right ) \, dx \\ & = -\frac {6 \int \frac {(1+4 x)^m}{5+\sqrt {13}-6 x} \, dx}{\sqrt {13}}-\frac {6 \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{\sqrt {13}} \\ & = \frac {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{\sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{\sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int \frac {(1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {(1+4 x)^{1+m} \left (\left (13+2 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+\left (-13+2 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{39 \sqrt {13} (1+m)} \]

[In]

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

[Out]

((1 + 4*x)^(1 + m)*((13 + 2*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])] + (-13
+ 2*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])]))/(39*Sqrt[13]*(1 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((4*x + 1)^m/(3*x^2 - 5*x + 1),x)

[Out]

int((4*x + 1)^m/(3*x^2 - 5*x + 1), x)