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

Optimal. Leaf size=147 \[ -\frac{3 \left (117-47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{26 \left (13-2 \sqrt{13}\right ) (m+1)}-\frac{3 \left (117+47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{26 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{3 (4 x+1)^{m+1}}{4 (m+1)} \]

[Out]

(3*(1 + 4*x)^(1 + m))/(4*(1 + m)) - (3*(117 - 47*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)) - (3*(117 + 47*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 = 0.152146, antiderivative size = 147, 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} \[ -\frac{3 \left (117-47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{26 \left (13-2 \sqrt{13}\right ) (m+1)}-\frac{3 \left (117+47 \sqrt{13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{26 \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{3 (4 x+1)^{m+1}}{4 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + 4*x)^(1 + m))/(4*(1 + m)) - (3*(117 - 47*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)) - (3*(117 + 47*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)^2 (1+4 x)^m}{1-5 x+3 x^2} \, dx &=\int \left (3 (1+4 x)^m+\frac{\left (27+\frac{141}{\sqrt{13}}\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (27-\frac{141}{\sqrt{13}}\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx\\ &=\frac{3 (1+4 x)^{1+m}}{4 (1+m)}+\frac{1}{13} \left (3 \left (117-47 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx+\frac{1}{13} \left (3 \left (117+47 \sqrt{13}\right )\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx\\ &=\frac{3 (1+4 x)^{1+m}}{4 (1+m)}-\frac{3 \left (117-47 \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 (117+47 \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 = 0.0903008, size = 91, normalized size = 0.62 \[ \frac{(4 x+1)^{m+1} \left (\left (58 \sqrt{13}-46\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )-2 \left (23+29 \sqrt{13}\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )+117\right )}{156 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 4*x)^(1 + m)*(117 + (-46 + 58*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])]
- 2*(23 + 29*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])]))/(156*(1 + m))

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Maple [F]  time = 1.404, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 4\,x+1 \right ) ^{m} \left ( 2+3\,x \right ) ^{2}}{3\,{x}^{2}-5\,x+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((4*x + 1)^m*(3*x + 2)^2/(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 (9 \, x^{2} + 12 \, x + 4\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)^2*(1+4*x)^m/(3*x^2-5*x+1),x, algorithm="fricas")

[Out]

integral((9*x^2 + 12*x + 4)*(4*x + 1)^m/(3*x^2 - 5*x + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((3*x + 2)**2*(4*x + 1)**m/(3*x**2 - 5*x + 1), 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\right )}^{2}}{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)^2*(1+4*x)^m/(3*x^2-5*x+1),x, algorithm="giac")

[Out]

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