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

Optimal. Leaf size=340 \[ \frac{9 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{1445 (m+1)}-\frac{\left (\left (62+22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13+9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{7514 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (\left (62-22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13-9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{7514 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{(43-33 x) (4 x+1)^{m+1}}{663 \left (3 x^2-5 x+1\right )} \]

[Out]

((43 - 33*x)*(1 + 4*x)^(1 + m))/(663*(1 - 5*x + 3*x^2)) + (9*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 +
m, (-3*(1 + 4*x))/5])/(1445*(1 + m)) + (9*(13 + 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 +
m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(7514*(13 - 2*Sqrt[13])*(1 + m)) - ((81 + (62 + 22*Sqrt[13])*m)*(1 + 4*x
)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(221*Sqrt[13]*(13 - 2*Sqrt[13])
*(1 + m)) + (9*(13 - 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sq
rt[13])])/(7514*(13 + 2*Sqrt[13])*(1 + m)) + ((81 + (62 - 22*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[
1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(221*Sqrt[13]*(13 + 2*Sqrt[13])*(1 + m))

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Rubi [A]  time = 0.50447, antiderivative size = 340, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {960, 68, 822, 830} \[ \frac{9 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{1445 (m+1)}-\frac{\left (\left (62+22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13+9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{7514 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (\left (62-22 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (13-9 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{7514 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{(43-33 x) (4 x+1)^{m+1}}{663 \left (3 x^2-5 x+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((43 - 33*x)*(1 + 4*x)^(1 + m))/(663*(1 - 5*x + 3*x^2)) + (9*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 +
m, (-3*(1 + 4*x))/5])/(1445*(1 + m)) + (9*(13 + 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 +
m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(7514*(13 - 2*Sqrt[13])*(1 + m)) - ((81 + (62 + 22*Sqrt[13])*m)*(1 + 4*x
)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(221*Sqrt[13]*(13 - 2*Sqrt[13])
*(1 + m)) + (9*(13 - 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sq
rt[13])])/(7514*(13 + 2*Sqrt[13])*(1 + m)) + ((81 + (62 - 22*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[
1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(221*Sqrt[13]*(13 + 2*Sqrt[13])*(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 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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) \left (1-5 x+3 x^2\right )^2} \, dx &=\int \left (\frac{9 (1+4 x)^m}{289 (2+3 x)}+\frac{(7-3 x) (1+4 x)^m}{17 \left (1-5 x+3 x^2\right )^2}-\frac{3 (-7+3 x) (1+4 x)^m}{289 \left (1-5 x+3 x^2\right )}\right ) \, dx\\ &=-\left (\frac{3}{289} \int \frac{(-7+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx\right )+\frac{9}{289} \int \frac{(1+4 x)^m}{2+3 x} \, dx+\frac{1}{17} \int \frac{(7-3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx\\ &=\frac{(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac{9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{1445 (1+m)}-\frac{\int \frac{(1+4 x)^m (13 (81+172 m)-1716 m x)}{1-5 x+3 x^2} \, dx}{8619}-\frac{3}{289} \int \left (\frac{\left (3-\frac{27}{\sqrt{13}}\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (3+\frac{27}{\sqrt{13}}\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx\\ &=\frac{(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac{9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{1445 (1+m)}-\frac{\int \left (\frac{\left (-1716 m+6 \sqrt{13} (81+62 m)\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (-1716 m-6 \sqrt{13} (81+62 m)\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx}{8619}-\frac{\left (9 \left (13-9 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx}{3757}-\frac{\left (9 \left (13+9 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx}{3757}\\ &=\frac{(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac{9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac{9 \left (13+9 \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{9 \left (13-9 \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{\left (2 \left (81+\left (62-22 \sqrt{13}\right ) m\right )\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx}{221 \sqrt{13}}+\frac{\left (2 \left (81+\left (62+22 \sqrt{13}\right ) m\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx}{221 \sqrt{13}}\\ &=\frac{(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac{9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac{9 \left (13+9 \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{\left (81+\left (62+22 \sqrt{13}\right ) m\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{3 (1+4 x)}{13-2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13-2 \sqrt{13}\right ) (1+m)}+\frac{9 \left (13-9 \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{\left (81+\left (62-22 \sqrt{13}\right ) m\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{3 (1+4 x)}{13+2 \sqrt{13}}\right )}{221 \sqrt{13} \left (13+2 \sqrt{13}\right ) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.547949, size = 274, normalized size = 0.81 \[ \frac{(4 x+1)^{m+1} \left (\frac{9126 \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{m+1}+\frac{1755 \left (13+9 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )}{\left (13-2 \sqrt{13}\right ) (m+1)}+\frac{1755 \left (13-9 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )}{\left (13+2 \sqrt{13}\right ) (m+1)}+\frac{510 \sqrt{13} \left (\frac{\left (\left (62+22 \sqrt{13}\right ) m+81\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )}{2 \sqrt{13}-13}+\frac{\left (\left (62-22 \sqrt{13}\right ) m+81\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )}{13+2 \sqrt{13}}\right )}{m+1}+\frac{2210 (43-33 x)}{3 x^2-5 x+1}\right )}{1465230} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 4*x)^(1 + m)*((2210*(43 - 33*x))/(1 - 5*x + 3*x^2) + (9126*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4
*x))/5])/(1 + m) + (1755*(13 + 9*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/(
(13 - 2*Sqrt[13])*(1 + m)) + (1755*(13 - 9*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqr
t[13])])/((13 + 2*Sqrt[13])*(1 + m)) + (510*Sqrt[13]*(((81 + (62 + 22*Sqrt[13])*m)*Hypergeometric2F1[1, 1 + m,
2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/(-13 + 2*Sqrt[13]) + ((81 + (62 - 22*Sqrt[13])*m)*Hypergeometric2F1[1,
1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/(13 + 2*Sqrt[13])))/(1 + m)))/1465230

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Maple [F]  time = 1.341, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 4\,x+1 \right ) ^{m}}{ \left ( 2+3\,x \right ) \left ( 3\,{x}^{2}-5\,x+1 \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2}{\left (3 \, x + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (4 \, x + 1\right )}^{m}}{27 \, x^{5} - 72 \, x^{4} + 33 \, x^{3} + 32 \, x^{2} - 17 \, x + 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((4*x + 1)^m/(27*x^5 - 72*x^4 + 33*x^3 + 32*x^2 - 17*x + 2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2}{\left (3 \, x + 2\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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