3.31.78 \(\int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx\) [3078]

Optimal. Leaf size=498 \[ -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (b (5 d e-c f (2-m))-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {f \left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f \left (d e (15+8 m)-c f \left (3-2 m-2 m^2\right )\right )+b^2 \left (11 d^2 e^2-c d e f (7-8 m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{6 (b e-a f)^3 (d e-c f)^3 (e+f x)}+\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 (b e-a f)^3 (d e-c f)^4 m} \]

[Out]

-1/3*f*(b*x+a)^(1+m)/(-a*f+b*e)/(-c*f+d*e)/((d*x+c)^m)/(f*x+e)^3-1/6*f*(b*(5*d*e-c*f*(2-m))-a*d*f*(3+m))*(b*x+
a)^(1+m)/(-a*f+b*e)^2/(-c*f+d*e)^2/((d*x+c)^m)/(f*x+e)^2-1/6*f*(a^2*d^2*f^2*(m^2+5*m+6)-a*b*d*f*(d*e*(15+8*m)-
c*f*(-2*m^2-2*m+3))+b^2*(11*d^2*e^2-c*d*e*f*(7-8*m)+c^2*f^2*(m^2-3*m+2)))*(b*x+a)^(1+m)/(-a*f+b*e)^3/(-c*f+d*e
)^3/((d*x+c)^m)/(f*x+e)+1/6*(3*a*b^2*d*f*(1+m)*(6*d^2*e^2+6*c*d*e*f*m-c^2*f^2*(1-m)*m)-3*a^2*b*d^2*f^2*(c*f*m+
3*d*e)*(m^2+3*m+2)+a^3*d^3*f^3*(m^3+6*m^2+11*m+6)-b^3*(6*d^3*e^3+18*c*d^2*e^2*f*m-9*c^2*d*e*f^2*(1-m)*m+c^3*f^
3*m*(m^2-3*m+2)))*(b*x+a)^m*hypergeom([1, -m],[1-m],(-a*f+b*e)*(d*x+c)/(-c*f+d*e)/(b*x+a))/(-a*f+b*e)^3/(-c*f+
d*e)^4/m/((d*x+c)^m)

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Rubi [A]
time = 0.49, antiderivative size = 497, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {105, 156, 12, 133} \begin {gather*} -\frac {f (a+b x)^{m+1} (c+d x)^{-m} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-a b d f \left (d e (8 m+15)-c f \left (-2 m^2-2 m+3\right )\right )+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-c d e f (7-8 m)+11 d^2 e^2\right )\right )}{6 (e+f x) (b e-a f)^3 (d e-c f)^3}+\frac {(a+b x)^m (c+d x)^{-m} \left (a^3 d^3 f^3 \left (m^3+6 m^2+11 m+6\right )-3 a^2 b d^2 f^2 \left (m^2+3 m+2\right ) (c f m+3 d e)+3 a b^2 d f (m+1) \left (-c^2 f^2 (1-m) m+6 c d e f m+6 d^2 e^2\right )-\left (b^3 \left (c^3 f^3 m \left (m^2-3 m+2\right )-9 c^2 d e f^2 (1-m) m+18 c d^2 e^2 f m+6 d^3 e^3\right )\right )\right ) \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{6 m (b e-a f)^3 (d e-c f)^4}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+3)-b c f (2-m)+5 b d e)}{6 (e+f x)^2 (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^m*(c + d*x)^(-1 - m))/(e + f*x)^4,x]

[Out]

-1/3*(f*(a + b*x)^(1 + m))/((b*e - a*f)*(d*e - c*f)*(c + d*x)^m*(e + f*x)^3) - (f*(5*b*d*e - b*c*f*(2 - m) - a
*d*f*(3 + m))*(a + b*x)^(1 + m))/(6*(b*e - a*f)^2*(d*e - c*f)^2*(c + d*x)^m*(e + f*x)^2) - (f*(a^2*d^2*f^2*(6
+ 5*m + m^2) - a*b*d*f*(d*e*(15 + 8*m) - c*f*(3 - 2*m - 2*m^2)) + b^2*(11*d^2*e^2 - c*d*e*f*(7 - 8*m) + c^2*f^
2*(2 - 3*m + m^2)))*(a + b*x)^(1 + m))/(6*(b*e - a*f)^3*(d*e - c*f)^3*(c + d*x)^m*(e + f*x)) + ((3*a*b^2*d*f*(
1 + m)*(6*d^2*e^2 + 6*c*d*e*f*m - c^2*f^2*(1 - m)*m) - 3*a^2*b*d^2*f^2*(3*d*e + c*f*m)*(2 + 3*m + m^2) + a^3*d
^3*f^3*(6 + 11*m + 6*m^2 + m^3) - b^3*(6*d^3*e^3 + 18*c*d^2*e^2*f*m - 9*c^2*d*e*f^2*(1 - m)*m + c^3*f^3*m*(2 -
3*m + m^2)))*(a + b*x)^m*Hypergeometric2F1[1, -m, 1 - m, ((b*e - a*f)*(c + d*x))/((d*e - c*f)*(a + b*x))])/(6
*(b*e - a*f)^3*(d*e - c*f)^4*m*(c + d*x)^m)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 105

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)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 133

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*c - a
*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rule 156

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

Rubi steps

\begin {align*} \int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^4} \, dx &=\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac {\int \frac {(a+b x)^m (c+d x)^{-m} (a d f (3+m)-b (d e+c f m)+2 b d f x)}{(e+f x)^4} \, dx}{(b c-a d) (d e-c f) m}\\ &=-\frac {f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}-\frac {\int \frac {(a+b x)^m (c+d x)^{-m} \left (-a b d f (3+2 m) (3 d e+c f m)+b^2 \left (3 d^2 e^2+6 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )+b d f (a d f (3+m)-b (3 d e+c f m)) x\right )}{(e+f x)^3} \, dx}{3 (b c-a d) (b e-a f) (d e-c f)^2 m}\\ &=-\frac {f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac {f \left (b^2 \left (6 d^2 e^2+7 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m (e+f x)^2}+\frac {\int \frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m}\\ &=-\frac {f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac {f \left (b^2 \left (6 d^2 e^2+7 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m (e+f x)^2}+\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) \int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m}\\ &=-\frac {f (a d f (3+m)-b (3 d e+c f m)) (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b c-a d) (b e-a f) (d e-c f)^2 m (e+f x)^3}+\frac {d (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (d e-c f) m (e+f x)^3}+\frac {f \left (b^2 \left (6 d^2 e^2+7 c d e f m-c^2 f^2 (2-m) m\right )+a^2 d^2 f^2 \left (6+5 m+m^2\right )-a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b c-a d) (b e-a f)^2 (d e-c f)^3 m (e+f x)^2}+\frac {\left (3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m-c^2 f^2 (1-m) m\right )-3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )+a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )-b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m-9 c^2 d e f^2 (1-m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (2,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (b e-a f)^4 (d e-c f)^3 m (1+m)}\\ \end {align*}

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Mathematica [A]
time = 1.31, size = 466, normalized size = 0.94 \begin {gather*} -\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (6 d (b e-a f)^4 (d e-c f)^2 (1+m)+2 f (-b e+a f)^3 (-d e+c f) (1+m) (-a d f (3+m)+b (3 d e+c f m)) (c+d x)-\frac {(e+f x) \left (f (b e-a f)^2 (1+m) \left (-b^2 \left (6 d^2 e^2+7 c d e f m+c^2 f^2 (-2+m) m\right )-a^2 d^2 f^2 \left (6+5 m+m^2\right )+a b d f (c f m (3+2 m)+d e (12+7 m))\right ) (c+d x)^2+(b c-a d) \left (-3 a b^2 d f (1+m) \left (6 d^2 e^2+6 c d e f m+c^2 f^2 (-1+m) m\right )+3 a^2 b d^2 f^2 (3 d e+c f m) \left (2+3 m+m^2\right )-a^3 d^3 f^3 \left (6+11 m+6 m^2+m^3\right )+b^3 \left (6 d^3 e^3+18 c d^2 e^2 f m+9 c^2 d e f^2 (-1+m) m+c^3 f^3 m \left (2-3 m+m^2\right )\right )\right ) (e+f x)^2 \, _2F_1\left (2,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{c+d x}\right )}{6 (b c-a d) (b e-a f)^4 (d e-c f)^2 (-d e+c f) m (1+m) (e+f x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(c + d*x)^(-1 - m))/(e + f*x)^4,x]

[Out]

-1/6*((a + b*x)^(1 + m)*(6*d*(b*e - a*f)^4*(d*e - c*f)^2*(1 + m) + 2*f*(-(b*e) + a*f)^3*(-(d*e) + c*f)*(1 + m)
*(-(a*d*f*(3 + m)) + b*(3*d*e + c*f*m))*(c + d*x) - ((e + f*x)*(f*(b*e - a*f)^2*(1 + m)*(-(b^2*(6*d^2*e^2 + 7*
c*d*e*f*m + c^2*f^2*(-2 + m)*m)) - a^2*d^2*f^2*(6 + 5*m + m^2) + a*b*d*f*(c*f*m*(3 + 2*m) + d*e*(12 + 7*m)))*(
c + d*x)^2 + (b*c - a*d)*(-3*a*b^2*d*f*(1 + m)*(6*d^2*e^2 + 6*c*d*e*f*m + c^2*f^2*(-1 + m)*m) + 3*a^2*b*d^2*f^
2*(3*d*e + c*f*m)*(2 + 3*m + m^2) - a^3*d^3*f^3*(6 + 11*m + 6*m^2 + m^3) + b^3*(6*d^3*e^3 + 18*c*d^2*e^2*f*m +
9*c^2*d*e*f^2*(-1 + m)*m + c^3*f^3*m*(2 - 3*m + m^2)))*(e + f*x)^2*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e -
c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/(c + d*x)))/((b*c - a*d)*(b*e - a*f)^4*(d*e - c*f)^2*(-(d*e) + c*f
)*m*(1 + m)*(c + d*x)^m*(e + f*x)^3)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-1-m}}{\left (f x +e \right )^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,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((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^4, 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((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m - 1)/(f^4*x^4 + 4*f^3*x^3*e + 6*f^2*x^2*e^2 + 4*f*x*e^3 + e^4), 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((b*x+a)**m*(d*x+c)**(-1-m)/(f*x+e)**4,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((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^4\,{\left (c+d\,x\right )}^{m+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m/((e + f*x)^4*(c + d*x)^(m + 1)),x)

[Out]

int((a + b*x)^m/((e + f*x)^4*(c + d*x)^(m + 1)), x)

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