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

Optimal. Leaf size=183 \[ -\frac{3 \left (5499-1631 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{26 \left (13-2 \sqrt{13}\right ) (m+1)}-\frac{3 \left (5499+1631 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{26 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{3687 (4 x+1)^{m+1}}{64 (m+1)}+\frac{207 (4 x+1)^{m+2}}{32 (m+2)}+\frac{27 (4 x+1)^{m+3}}{64 (m+3)} \]

[Out]

(3687*(1 + 4*x)^(1 + m))/(64*(1 + m)) + (207*(1 + 4*x)^(2 + m))/(32*(2 + m)) + (27*(1 + 4*x)^(3 + m))/(64*(3 +
m)) - (3*(5499 - 1631*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sq
rt[13])])/(26*(13 - 2*Sqrt[13])*(1 + m)) - (3*(5499 + 1631*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1
+ m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(26*(13 + 2*Sqrt[13])*(1 + m))

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Rubi [A]  time = 0.23311, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {1628, 68} \[ -\frac{3 \left (5499-1631 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{26 \left (13-2 \sqrt{13}\right ) (m+1)}-\frac{3 \left (5499+1631 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{26 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{3687 (4 x+1)^{m+1}}{64 (m+1)}+\frac{207 (4 x+1)^{m+2}}{32 (m+2)}+\frac{27 (4 x+1)^{m+3}}{64 (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3687*(1 + 4*x)^(1 + m))/(64*(1 + m)) + (207*(1 + 4*x)^(2 + m))/(32*(2 + m)) + (27*(1 + 4*x)^(3 + m))/(64*(3 +
m)) - (3*(5499 - 1631*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sq
rt[13])])/(26*(13 - 2*Sqrt[13])*(1 + m)) - (3*(5499 + 1631*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1
+ m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(26*(13 + 2*Sqrt[13])*(1 + m))

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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]

Rubi steps

\begin{align*} \int \frac{(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx &=\int \left (\frac{3687}{16} (1+4 x)^m+\frac{207}{8} (1+4 x)^{1+m}+\frac{27}{16} (1+4 x)^{2+m}+\frac{\left (1269+\frac{4893}{\sqrt{13}}\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (1269-\frac{4893}{\sqrt{13}}\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx\\ &=\frac{3687 (1+4 x)^{1+m}}{64 (1+m)}+\frac{207 (1+4 x)^{2+m}}{32 (2+m)}+\frac{27 (1+4 x)^{3+m}}{64 (3+m)}+\frac{1}{13} \left (3 \left (5499-1631 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx+\frac{1}{13} \left (3 \left (5499+1631 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx\\ &=\frac{3687 (1+4 x)^{1+m}}{64 (1+m)}+\frac{207 (1+4 x)^{2+m}}{32 (2+m)}+\frac{27 (1+4 x)^{3+m}}{64 (3+m)}-\frac{3 \left (5499-1631 \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 )}{26 \left (13-2 \sqrt{13}\right ) (1+m)}-\frac{3 \left (5499+1631 \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 )}{26 \left (13+2 \sqrt{13}\right ) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.284896, size = 151, normalized size = 0.83 \[ \frac{3}{832} (4 x+1)^{m+1} \left (-\frac{32 \left (1631 \sqrt{13}-5499\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )}{\left (2 \sqrt{13}-13\right ) (m+1)}-\frac{32 \left (5499+1631 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )}{\left (13+2 \sqrt{13}\right ) (m+1)}+\frac{117 (4 x+1)^2}{m+3}+\frac{1794 (4 x+1)}{m+2}+\frac{15977}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + 4*x)^(1 + m)*(15977/(1 + m) + (1794*(1 + 4*x))/(2 + m) + (117*(1 + 4*x)^2)/(3 + m) - (32*(-5499 + 1631
*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/((-13 + 2*Sqrt[13])*(1 + m)) - (3
2*(5499 + 1631*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sqrt[13])*
(1 + m))))/832

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Maple [F]  time = 1.43, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 4\,x+1 \right ) ^{m} \left ( 2+3\,x \right ) ^{4}}{3\,{x}^{2}-5\,x+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((2+3*x)^4*(4*x+1)^m/(3*x^2-5*x+1),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\right )}^{4}}{3 \, x^{2} - 5 \, x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}{\left (4 \, x + 1\right )}^{m}}{3 \, x^{2} - 5 \, x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*(4*x + 1)^m/(3*x^2 - 5*x + 1), 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((2+3*x)**4*(1+4*x)**m/(3*x**2-5*x+1),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\right )}^{4}}{3 \, x^{2} - 5 \, x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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