3.10.32 \(\int \frac {(2+3 x)^2 (1+4 x)^m}{1-5 x+3 x^2} \, dx\) [932]

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

[Out]

3/4*(1+4*x)^(1+m)/(1+m)-3/26*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2)))*(117-47*13^(1/2
))/(1+m)/(13-2*13^(1/2))-3/26*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(117+47*13^(1/
2))/(1+m)/(13+2*13^(1/2))

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Rubi [A]
time = 0.10, antiderivative size = 147, normalized size of antiderivative = 1.00, 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 = {1642, 70} \begin {gather*} -\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)} \end {gather*}

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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 1642

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]

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*}

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Mathematica [A]
time = 0.11, size = 91, normalized size = 0.62 \begin {gather*} \frac {(1+4 x)^{1+m} \left (117+\left (-46+58 \sqrt {13}\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13-2 \sqrt {13}}\right )-2 \left (23+29 \sqrt {13}\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{156 (1+m)} \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (2+3 x \right )^{2} \left (1+4 x \right )^{m}}{3 x^{2}-5 x +1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (3 x + 2\right )^{2} \left (4 x + 1\right )^{m}}{3 x^{2} - 5 x + 1}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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