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

Optimal. Leaf size=164 \[ \frac {3 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {3}{5} (4 x+1)\right )}{85 (m+1)}+\frac {3 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{442 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {3 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{442 \left (13+2 \sqrt {13}\right ) (m+1)} \]

[Out]

3/85*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],-3/5-12/5*x)/(1+m)+3/442*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*
(1+4*x)/(13+2*13^(1/2)))*(13-9*13^(1/2))/(1+m)/(13+2*13^(1/2))+3/442*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*
(1+4*x)/(13-2*13^(1/2)))*(13+9*13^(1/2))/(1+m)/(13-2*13^(1/2))

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Rubi [A]  time = 0.20, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {960, 68, 830} \[ \frac {3 (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {3}{5} (4 x+1)\right )}{85 (m+1)}+\frac {3 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{442 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {3 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{442 \left (13+2 \sqrt {13}\right ) (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(85*(1 + m)) + (3*(13 + 9*Sqrt[13])
*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(442*(13 - 2*Sqrt[13])
*(1 + m)) + (3*(13 - 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sq
rt[13])])/(442*(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 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]

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])

Rubi steps

\begin {align*} \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx &=\int \left (\frac {3 (1+4 x)^m}{17 (2+3 x)}+\frac {(7-3 x) (1+4 x)^m}{17 \left (1-5 x+3 x^2\right )}\right ) \, dx\\ &=\frac {1}{17} \int \frac {(7-3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx+\frac {3}{17} \int \frac {(1+4 x)^m}{2+3 x} \, dx\\ &=\frac {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}+\frac {1}{17} \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 {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}-\frac {1}{221} \left (3 \left (13-9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx-\frac {1}{221} \left (3 \left (13+9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx\\ &=\frac {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}+\frac {3 \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 )}{442 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {3 \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 )}{442 \left (13+2 \sqrt {13}\right ) (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 110, normalized size = 0.67 \[ \frac {(4 x+1)^{m+1} \left (234 \, _2F_1\left (1,m+1;m+2;-\frac {3}{5} (4 x+1)\right )+5 \left (31+11 \sqrt {13}\right ) \, _2F_1\left (1,m+1;m+2;\frac {12 x+3}{13-2 \sqrt {13}}\right )+5 \left (31-11 \sqrt {13}\right ) \, _2F_1\left (1,m+1;m+2;\frac {12 x+3}{13+2 \sqrt {13}}\right )\right )}{6630 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 4*x)^(1 + m)*(234*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5] + 5*(31 + 11*Sqrt[13])*Hypergeome
tric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])] + 5*(31 - 11*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 +
m, (3 + 12*x)/(13 + 2*Sqrt[13])]))/(6630*(1 + m))

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (4 \, x + 1\right )}^{m}}{9 \, x^{3} - 9 \, x^{2} - 7 \, x + 2}, x\right ) \]

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),x, algorithm="fricas")

[Out]

integral((4*x + 1)^m/(9*x^3 - 9*x^2 - 7*x + 2), x)

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

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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (4 x +1\right )^{m}}{\left (3 x +2\right ) \left (3 x^{2}-5 x +1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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),x)

[Out]

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

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