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

Optimal. Leaf size=376 \[ \frac{162 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{24565 (m+1)}-\frac{\left (2 \left (211+65 \sqrt{13}\right ) m+423\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{3757 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (117+64 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{63869 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (\left (422-130 \sqrt{13}\right ) m+423\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{3757 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (117-64 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{63869 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{36 (4 x+1)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{7225 (m+1)}+\frac{(268-195 x) (4 x+1)^{m+1}}{11271 \left (3 x^2-5 x+1\right )} \]

[Out]

((268 - 195*x)*(1 + 4*x)^(1 + m))/(11271*(1 - 5*x + 3*x^2)) + (162*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 +
m, 2 + m, (-3*(1 + 4*x))/5])/(24565*(1 + m)) + (9*(117 + 64*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1
+ m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(63869*(13 - 2*Sqrt[13])*(1 + m)) - ((423 + 2*(211 + 65*Sqrt[13
])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(3757*Sqrt[13]*(1
3 - 2*Sqrt[13])*(1 + m)) + (9*(117 - 64*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 +
4*x))/(13 + 2*Sqrt[13])])/(63869*(13 + 2*Sqrt[13])*(1 + m)) + ((423 + (422 - 130*Sqrt[13])*m)*(1 + 4*x)^(1 +
m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(3757*Sqrt[13]*(13 + 2*Sqrt[13])*(1 +
m)) + (36*(1 + 4*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(7225*(1 + m))

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Rubi [A]  time = 0.493232, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {960, 68, 822, 830} \[ \frac{162 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{24565 (m+1)}-\frac{\left (2 \left (211+65 \sqrt{13}\right ) m+423\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{3757 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (117+64 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{63869 \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (\left (422-130 \sqrt{13}\right ) m+423\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{3757 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{9 \left (117-64 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{63869 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{36 (4 x+1)^{m+1} \, _2F_1\left (2,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{7225 (m+1)}+\frac{(268-195 x) (4 x+1)^{m+1}}{11271 \left (3 x^2-5 x+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((268 - 195*x)*(1 + 4*x)^(1 + m))/(11271*(1 - 5*x + 3*x^2)) + (162*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 +
m, 2 + m, (-3*(1 + 4*x))/5])/(24565*(1 + m)) + (9*(117 + 64*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1
+ m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(63869*(13 - 2*Sqrt[13])*(1 + m)) - ((423 + 2*(211 + 65*Sqrt[13
])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(3757*Sqrt[13]*(1
3 - 2*Sqrt[13])*(1 + m)) + (9*(117 - 64*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 +
4*x))/(13 + 2*Sqrt[13])])/(63869*(13 + 2*Sqrt[13])*(1 + m)) + ((423 + (422 - 130*Sqrt[13])*m)*(1 + 4*x)^(1 +
m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(3757*Sqrt[13]*(13 + 2*Sqrt[13])*(1 +
m)) + (36*(1 + 4*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(7225*(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)^2 \left (1-5 x+3 x^2\right )^2} \, dx &=\int \left (\frac{9 (1+4 x)^m}{289 (2+3 x)^2}+\frac{162 (1+4 x)^m}{4913 (2+3 x)}+\frac{(46-27 x) (1+4 x)^m}{289 \left (1-5 x+3 x^2\right )^2}-\frac{3 (1+4 x)^m (-109+54 x)}{4913 \left (1-5 x+3 x^2\right )}\right ) \, dx\\ &=-\frac{3 \int \frac{(1+4 x)^m (-109+54 x)}{1-5 x+3 x^2} \, dx}{4913}+\frac{1}{289} \int \frac{(46-27 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx+\frac{9}{289} \int \frac{(1+4 x)^m}{(2+3 x)^2} \, dx+\frac{162 \int \frac{(1+4 x)^m}{2+3 x} \, dx}{4913}\\ &=\frac{(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac{162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac{36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{7225 (1+m)}-\frac{\int \frac{(1+4 x)^m (13 (423+1072 m)-10140 m x)}{1-5 x+3 x^2} \, dx}{146523}-\frac{3 \int \left (\frac{\left (54-\frac{384}{\sqrt{13}}\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (54+\frac{384}{\sqrt{13}}\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx}{4913}\\ &=\frac{(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac{162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac{36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{7225 (1+m)}-\frac{\int \left (\frac{\left (-10140 m+6 \sqrt{13} (423+422 m)\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (-10140 m-6 \sqrt{13} (423+422 m)\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx}{146523}-\frac{\left (18 \left (117-64 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx}{63869}-\frac{\left (18 \left (117+64 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx}{63869}\\ &=\frac{(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac{162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac{9 \left (117+64 \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 )}{63869 \left (13-2 \sqrt{13}\right ) (1+m)}+\frac{9 \left (117-64 \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 )}{63869 \left (13+2 \sqrt{13}\right ) (1+m)}+\frac{36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{7225 (1+m)}-\frac{\left (2 \left (423+\left (422-130 \sqrt{13}\right ) m\right )\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx}{3757 \sqrt{13}}+\frac{\left (2 \left (1690 m+\sqrt{13} (423+422 m)\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx}{48841}\\ &=\frac{(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac{162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac{9 \left (117+64 \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 )}{63869 \left (13-2 \sqrt{13}\right ) (1+m)}-\frac{\left (1690 m+\sqrt{13} (423+422 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 )}{48841 \left (13-2 \sqrt{13}\right ) (1+m)}+\frac{9 \left (117-64 \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 )}{63869 \left (13+2 \sqrt{13}\right ) (1+m)}+\frac{\left (423+\left (422-130 \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 )}{3757 \sqrt{13} \left (13+2 \sqrt{13}\right ) (1+m)}+\frac{36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac{3}{5} (1+4 x)\right )}{7225 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.575377, size = 287, normalized size = 0.76 \[ \frac{(4 x+1)^{m+1} \left (\frac{1232010 \, _2F_1\left (1,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{m+1}+\frac{26325 \left (117+64 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )}{\left (13-2 \sqrt{13}\right ) (m+1)}+\frac{26325 \left (117-64 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )}{\left (13+2 \sqrt{13}\right ) (m+1)}-\frac{425 \left (\left (\left (2534+682 \sqrt{13}\right ) m+423 \left (2+\sqrt{13}\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )+\left (\left (2534-682 \sqrt{13}\right ) m-423 \left (\sqrt{13}-2\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )\right )}{m+1}+\frac{930852 \, _2F_1\left (2,m+1;m+2;-\frac{3}{5} (4 x+1)\right )}{m+1}+\frac{16575 (268-195 x)}{3 x^2-5 x+1}\right )}{186816825} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 4*x)^(1 + m)*((16575*(268 - 195*x))/(1 - 5*x + 3*x^2) + (1232010*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*
(1 + 4*x))/5])/(1 + m) + (26325*(117 + 64*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt
[13])])/((13 - 2*Sqrt[13])*(1 + m)) + (26325*(117 - 64*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)
/(13 + 2*Sqrt[13])])/((13 + 2*Sqrt[13])*(1 + m)) - (425*((423*(2 + Sqrt[13]) + (2534 + 682*Sqrt[13])*m)*Hyperg
eometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])] + (-423*(-2 + Sqrt[13]) + (2534 - 682*Sqrt[13])*m)*
Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])]))/(1 + m) + (930852*Hypergeometric2F1[2, 1 +
m, 2 + m, (-3*(1 + 4*x))/5])/(1 + m)))/186816825

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Maple [F]  time = 1.35, 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 ) ^{2}}}\, 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)^2,x)

[Out]

int((4*x+1)^m/(2+3*x)^2/(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 )}^{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)^2,x, algorithm="maxima")

[Out]

integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)^2*(3*x + 2)^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}}{81 \, x^{6} - 162 \, x^{5} - 45 \, x^{4} + 162 \, x^{3} + 13 \, x^{2} - 28 \, 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)^2,x, algorithm="fricas")

[Out]

integral((4*x + 1)^m/(81*x^6 - 162*x^5 - 45*x^4 + 162*x^3 + 13*x^2 - 28*x + 4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2,x)

[Out]

Timed out

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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 )}^{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)^2,x, algorithm="giac")

[Out]

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