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

Optimal. Leaf size=199 \[ \frac{27 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{1445 (m+1)}+\frac{3 \left (117+47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{7514 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{3 \left (117-47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{7514 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{12 (4 x+1)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{425 (m+1)} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.221183, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {960, 68, 830} \[ \frac{27 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{1445 (m+1)}+\frac{3 \left (117+47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{7514 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{3 \left (117-47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{7514 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{12 (4 x+1)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{425 (m+1)} \]

Antiderivative was successfully verified.

[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 960

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])

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]

Rule 830

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]

Rubi steps

\begin{align*} \int \frac{(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx &=\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]  time = 0.134042, size = 152, normalized size = 0.76 \[ \frac{(4 x+1)^{m+1} \left (10530 \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )+25 \left (211+65 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )-1625 \sqrt{13} \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )+5275 \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )+15912 \, _2F_1\left (2,m+1;m+2;-\frac{3}{5} (4 x+1)\right )\right )}{563550 (m+1)} \]

Antiderivative was successfully verified.

[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]  time = 1.34, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 4\,x+1 \right ) ^{m}}{ \left ( 2+3\,x \right ) ^{2} \left ( 3\,{x}^{2}-5\,x+1 \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}{\left (3 \, x + 2\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (4 \, x + 1\right )}^{m}}{27 \, x^{4} - 9 \, x^{3} - 39 \, x^{2} - 8 \, x + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (4 x + 1\right )^{m}}{\left (3 x + 2\right )^{2} \left (3 x^{2} - 5 x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}{\left (3 \, x + 2\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)