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

Optimal. Leaf size=179 \[ -\frac {\left (2 \left (5+7 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (2 \left (5-7 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(20-21 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \]

[Out]

1/39*(20-21*x)*(1+4*x)^(1+m)/(3*x^2-5*x+1)+1/169*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/
2)))*(81+2*m*(5-7*13^(1/2)))/(1+m)*13^(1/2)/(13+2*13^(1/2))-1/169*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+
4*x)/(13-2*13^(1/2)))*(81+2*m*(5+7*13^(1/2)))/(1+m)/(13-2*13^(1/2))*13^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {822, 830, 68} \[ -\frac {\left (2 \left (5+7 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (2 \left (5-7 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(20-21 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((20 - 21*x)*(1 + 4*x)^(1 + m))/(39*(1 - 5*x + 3*x^2)) - ((81 + 2*(5 + 7*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hyperg
eometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(13*Sqrt[13]*(13 - 2*Sqrt[13])*(1 + m)) + ((81
+ 2*(5 - 7*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])
/(13*Sqrt[13]*(13 + 2*Sqrt[13])*(1 + m))

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 {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx &=\frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \frac {(1+4 x)^m (13 (81+80 m)-1092 m x)}{1-5 x+3 x^2} \, dx\\ &=\frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \left (\frac {\left (-1092 m+6 \sqrt {13} (81+10 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-1092 m-6 \sqrt {13} (81+10 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx\\ &=\frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}+\frac {1}{169} \left (2 \left (182 m+\sqrt {13} (81+10 m)\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx-\frac {1}{507} \left (-1092 m+6 \sqrt {13} (81+10 m)\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx\\ &=\frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (182 m+\sqrt {13} (81+10 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 )}{169 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (81+2 \left (5-7 \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 )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 149, normalized size = 0.83 \[ \frac {1}{507} (4 x+1)^{m+1} \left (\frac {3 \left (182 m+\sqrt {13} (10 m+81)\right ) \, _2F_1\left (1,m+1;m+2;\frac {12 x+3}{13-2 \sqrt {13}}\right )}{\left (2 \sqrt {13}-13\right ) (m+1)}-\frac {3 \left (182 m-\sqrt {13} (10 m+81)\right ) \, _2F_1\left (1,m+1;m+2;\frac {12 x+3}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (m+1)}+\frac {260-273 x}{3 x^2-5 x+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 4*x)^(1 + m)*((260 - 273*x)/(1 - 5*x + 3*x^2) + (3*(182*m + Sqrt[13]*(81 + 10*m))*Hypergeometric2F1[1, 1
+ m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/((-13 + 2*Sqrt[13])*(1 + m)) - (3*(182*m - Sqrt[13]*(81 + 10*m))*H
ypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sqrt[13])*(1 + m))))/507

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{9 \, x^{4} - 30 \, x^{3} + 31 \, x^{2} - 10 \, x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((4*x + 1)^m*(3*x + 2)/(9*x^4 - 30*x^3 + 31*x^2 - 10*x + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (3 x +2\right ) \left (4 x +1\right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (3\,x+2\right )\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (3 x + 2\right ) \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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