3.31.60 \(\int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx\) [3060]

Optimal. Leaf size=188 \[ -\frac {1}{45} (88-m) (5-4 x)^2 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {2}{15} (5-4 x)^3 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{1-m} (2+3 x)^{1+m} \left (386850-25441 m+426 m^2-2 m^3-24 \left (4359-154 m+m^2\right ) x\right )}{1215}+\frac {2^{-1-m} \left (3528363-639760 m+29050 m^2-440 m^3+2 m^4\right ) (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{1215 (1-m)} \]

[Out]

-1/45*(88-m)*(5-4*x)^2*(1+2*x)^(1-m)*(2+3*x)^(1+m)-2/15*(5-4*x)^3*(1+2*x)^(1-m)*(2+3*x)^(1+m)-1/1215*(1+2*x)^(
1-m)*(2+3*x)^(1+m)*(386850-25441*m+426*m^2-2*m^3-24*(m^2-154*m+4359)*x)+1/1215*2^(-1-m)*(2*m^4-440*m^3+29050*m
^2-639760*m+3528363)*(1+2*x)^(1-m)*hypergeom([-m, 1-m],[2-m],-3-6*x)/(1-m)

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Rubi [A]
time = 0.15, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {102, 158, 152, 71} \begin {gather*} \frac {2^{-m-1} \left (2 m^4-440 m^3+29050 m^2-639760 m+3528363\right ) (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (2 x+1))}{1215 (1-m)}-\frac {(3 x+2)^{m+1} \left (-2 m^3-24 \left (m^2-154 m+4359\right ) x+426 m^2-25441 m+386850\right ) (2 x+1)^{1-m}}{1215}-\frac {2}{15} (5-4 x)^3 (3 x+2)^{m+1} (2 x+1)^{1-m}-\frac {1}{45} (88-m) (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{1-m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/45*((88 - m)*(5 - 4*x)^2*(1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m)) - (2*(5 - 4*x)^3*(1 + 2*x)^(1 - m)*(2 + 3*x)^
(1 + m))/15 - ((1 + 2*x)^(1 - m)*(2 + 3*x)^(1 + m)*(386850 - 25441*m + 426*m^2 - 2*m^3 - 24*(4359 - 154*m + m^
2)*x))/1215 + (2^(-1 - m)*(3528363 - 639760*m + 29050*m^2 - 440*m^3 + 2*m^4)*(1 + 2*x)^(1 - m)*Hypergeometric2
F1[1 - m, -m, 2 - m, -3*(1 + 2*x)])/(1215*(1 - m))

Rule 71

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

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 158

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int (5-4 x)^4 (1+2 x)^{-m} (2+3 x)^m \, dx &=-\frac {2}{15} (5-4 x)^3 (1+2 x)^{1-m} (2+3 x)^{1+m}+\frac {1}{30} \int (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^m (2 (397-10 m)-16 (88-m) x) \, dx\\ &=-\frac {1}{45} (88-m) (5-4 x)^2 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {2}{15} (5-4 x)^3 (1+2 x)^{1-m} (2+3 x)^{1+m}+\frac {1}{720} \int (5-4 x) (1+2 x)^{-m} (2+3 x)^m \left (16 \left (7627-609 m+5 m^2\right )-64 \left (4359-154 m+m^2\right ) x\right ) \, dx\\ &=-\frac {1}{45} (88-m) (5-4 x)^2 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {2}{15} (5-4 x)^3 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{1-m} (2+3 x)^{1+m} \left (386850-25441 m+426 m^2-2 m^3-24 \left (4359-154 m+m^2\right ) x\right )}{1215}+\frac {\left (3528363-639760 m+29050 m^2-440 m^3+2 m^4\right ) \int (1+2 x)^{-m} (2+3 x)^m \, dx}{1215}\\ &=-\frac {1}{45} (88-m) (5-4 x)^2 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {2}{15} (5-4 x)^3 (1+2 x)^{1-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{1-m} (2+3 x)^{1+m} \left (386850-25441 m+426 m^2-2 m^3-24 \left (4359-154 m+m^2\right ) x\right )}{1215}+\frac {2^{-1-m} \left (3528363-639760 m+29050 m^2-440 m^3+2 m^4\right ) (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{1215 (1-m)}\\ \end {align*}

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Mathematica [A]
time = 0.52, size = 159, normalized size = 0.85 \begin {gather*} \frac {2^{-1-m} (1+2 x)^{1-m} \left (2^{2+m} (-1+m) (2+3 x)^{1+m} \left (2 m^2 (-19+6 x)+3 m \left (1331-796 x+72 x^2\right )+2 \left (-54629+62064 x-19224 x^2+2592 x^3\right )\right )-4 \left (-123992+9065 m-175 m^2+m^3\right ) \, _2F_1(-1-m,1-m;2-m;-3-6 x)+23 \left (-164189+17822 m-490 m^2+4 m^3\right ) \, _2F_1(1-m,-m;2-m;-3-6 x)\right )}{1215 (-1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1 - m)*(1 + 2*x)^(1 - m)*(2^(2 + m)*(-1 + m)*(2 + 3*x)^(1 + m)*(2*m^2*(-19 + 6*x) + 3*m*(1331 - 796*x + 7
2*x^2) + 2*(-54629 + 62064*x - 19224*x^2 + 2592*x^3)) - 4*(-123992 + 9065*m - 175*m^2 + m^3)*Hypergeometric2F1
[-1 - m, 1 - m, 2 - m, -3 - 6*x] + 23*(-164189 + 17822*m - 490*m^2 + 4*m^3)*Hypergeometric2F1[1 - m, -m, 2 - m
, -3 - 6*x]))/(1215*(-1 + m))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (5-4 x \right )^{4} \left (2+3 x \right )^{m} \left (1+2 x \right )^{-m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((256*x^4 - 1280*x^3 + 2400*x^2 - 2000*x + 625)*(3*x + 2)^m/(2*x + 1)^m, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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