3.10.38 \(\int \frac {(2+3 x)^3 (1+4 x)^m}{(1-5 x+3 x^2)^2} \, dx\) [938]

Optimal. Leaf size=181 \[ \frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (1521+\sqrt {13} \left (1701-1168 m+568 \sqrt {13} m\right )\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{338 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (\sqrt {13} (1701-1168 m)-13 (117+568 m)\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{338 \left (13+2 \sqrt {13}\right ) (1+m)} \]

[Out]

1/39*(209-426*x)*(1+4*x)^(1+m)/(3*x^2-5*x+1)+1/338*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(
1/2)))*(-1521-7384*m+(1701-1168*m)*13^(1/2))/(1+m)/(13+2*13^(1/2))-1/338*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m
],3*(1+4*x)/(13-2*13^(1/2)))*(1521+13^(1/2)*(1701-1168*m+568*m*13^(1/2)))/(1+m)/(13-2*13^(1/2))

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Rubi [A]
time = 0.18, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1662, 844, 70} \begin {gather*} -\frac {\left (\sqrt {13} \left (568 \sqrt {13} m-1168 m+1701\right )+1521\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{338 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (\sqrt {13} (1701-1168 m)-13 (568 m+117)\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{338 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(209-426 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((209 - 426*x)*(1 + 4*x)^(1 + m))/(39*(1 - 5*x + 3*x^2)) - ((1521 + Sqrt[13]*(1701 - 1168*m + 568*Sqrt[13]*m))
*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(338*(13 - 2*Sqrt[13])
*(1 + m)) + ((Sqrt[13]*(1701 - 1168*m) - 13*(117 + 568*m))*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m
, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(338*(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 844

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 1662

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[Po
lynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 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)/((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)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Q + 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, m}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] &&  !
(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx &=\frac {(209-426 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 (1143+836 m)-39 (117+568 m) x)}{1-5 x+3 x^2} \, dx\\ &=\frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \left (\frac {\left (-39 (117+568 m)-3 \sqrt {13} (-1701+1168 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-39 (117+568 m)+3 \sqrt {13} (-1701+1168 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx\\ &=\frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{169} \left (\sqrt {13} (1701-1168 m)-13 (117+568 m)\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx+\frac {1}{169} \left (\sqrt {13} (1701-1168 m)+13 (117+568 m)\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx\\ &=\frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (\sqrt {13} (1701-1168 m)+13 (117+568 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 )}{338 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (\sqrt {13} (1701-1168 m)-13 (117+568 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 )}{338 \left (13+2 \sqrt {13}\right ) (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 252, normalized size = 1.39 \begin {gather*} \frac {(1+4 x)^{1+m} \left (\frac {5434-11076 x}{1-5 x+3 x^2}-\frac {351 \left (-13+27 \sqrt {13}\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {12 \left (\sqrt {13} (1215-292 m)+1846 m\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (13-2 \sqrt {13}\right ) (1+m)}-\frac {351 \left (13+27 \sqrt {13}\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}+\frac {12 \left (\sqrt {13} (1215-292 m)-1846 m\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}\right )}{1014} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 4*x)^(1 + m)*((5434 - 11076*x)/(1 - 5*x + 3*x^2) - (351*(-13 + 27*Sqrt[13])*Hypergeometric2F1[1, 1 + m,
2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/((-13 + 2*Sqrt[13])*(1 + m)) - (12*(Sqrt[13]*(1215 - 292*m) + 1846*m)*Hy
pergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/((13 - 2*Sqrt[13])*(1 + m)) - (351*(13 + 27*Sq
rt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sqrt[13])*(1 + m)) + (12*(S
qrt[13]*(1215 - 292*m) - 1846*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sq
rt[13])*(1 + m))))/1014

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (2+3 x \right )^{3} \left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((27*x^3 + 54*x^2 + 36*x + 8)*(4*x + 1)^m/(9*x^4 - 30*x^3 + 31*x^2 - 10*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 )^{3} \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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