3.31.68 \(\int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx\) [3068]

Optimal. Leaf size=309 \[ -\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (b (4 d e-c f (2-m))-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {(b c-a d) \left (2 a b d f (3 d e-c f (1-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (2,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (b e-a f)^4 (d e-c f)^2 (1+m)} \]

[Out]

-1/3*f*(b*x+a)^(1+m)*(d*x+c)^(1-m)/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)^3-1/6*f*(b*(4*d*e-c*f*(2-m))-a*d*f*(2+m))*(b*
x+a)^(1+m)*(d*x+c)^(1-m)/(-a*f+b*e)^2/(-c*f+d*e)^2/(f*x+e)^2-1/6*(-a*d+b*c)*(2*a*b*d*f*(3*d*e-c*f*(1-m))*(1+m)
-a^2*d^2*f^2*(m^2+3*m+2)-b^2*(6*d^2*e^2-6*c*d*e*f*(1-m)+c^2*f^2*(m^2-3*m+2)))*(b*x+a)^(1+m)*(d*x+c)^(-1-m)*hyp
ergeom([2, 1+m],[2+m],(-c*f+d*e)*(b*x+a)/(-a*f+b*e)/(d*x+c))/(-a*f+b*e)^4/(-c*f+d*e)^2/(1+m)

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 308, 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 = {105, 156, 12, 133} \begin {gather*} -\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (3 d e-c f (1-m))-\left (b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-6 c d e f (1-m)+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (m+1) (b e-a f)^4 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (m+2)-b c f (2-m)+4 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)^{1-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)^m*(e + f*x)^4),x]

[Out]

-1/3*(f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/((b*e - a*f)*(d*e - c*f)*(e + f*x)^3) - (f*(4*b*d*e - b*c*f*(2 -
m) - a*d*f*(2 + m))*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/(6*(b*e - a*f)^2*(d*e - c*f)^2*(e + f*x)^2) - ((b*c -
a*d)*(2*a*b*d*f*(3*d*e - c*f*(1 - m))*(1 + m) - a^2*d^2*f^2*(2 + 3*m + m^2) - b^2*(6*d^2*e^2 - 6*c*d*e*f*(1 -
m) + c^2*f^2*(2 - 3*m + m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e
- c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(6*(b*e - a*f)^4*(d*e - c*f)^2*(1 + 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)^{-m}}{(e+f x)^4} \, dx &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {\int \frac {(a+b x)^m (c+d x)^{-m} (-b (3 d e-c f (2-m))+a d f (2+m)+b d f x)}{(e+f x)^3} \, dx}{3 (b e-a f) (d e-c f)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (4 b d e-b c f (2-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}+\frac {\int \frac {\left (-2 a b d f (3 d e-c f (1-m)) (1+m)+a^2 d^2 f^2 \left (2+3 m+m^2\right )+b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^m (c+d x)^{-m}}{(e+f x)^2} \, dx}{6 (b e-a f)^2 (d e-c f)^2}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (4 b d e-b c f (2-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {\left (2 a b d f (3 d e-c f (1-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \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 e-a f)^2 (d e-c f)^2}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac {f (4 b d e-b c f (2-m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 (b e-a f)^2 (d e-c f)^2 (e+f x)^2}-\frac {(b c-a d) \left (2 a b d f (3 d e-c f (1-m)) (1+m)-a^2 d^2 f^2 \left (2+3 m+m^2\right )-b^2 \left (6 d^2 e^2-6 c d e f (1-m)+c^2 f^2 \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)^2 (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.45, size = 255, normalized size = 0.83 \begin {gather*} \frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (-\frac {2 f (-b e+a f) (-d e+c f) (c+d x)^2}{(e+f x)^3}-\frac {f (4 b d e+b c f (-2+m)-a d f (2+m)) (c+d x)^2}{(e+f x)^2}+\frac {(b c-a d) \left (-2 a b d f (3 d e+c f (-1+m)) (1+m)+a^2 d^2 f^2 \left (2+3 m+m^2\right )+b^2 \left (6 d^2 e^2+6 c d e f (-1+m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) \, _2F_1\left (2,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (1+m)}\right )}{6 (b e-a f)^2 (d e-c f)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*((-2*f*(-(b*e) + a*f)*(-(d*e) + c*f)*(c + d*x)^2)/(e + f*x)^3 - (f*(4*b*
d*e + b*c*f*(-2 + m) - a*d*f*(2 + m))*(c + d*x)^2)/(e + f*x)^2 + ((b*c - a*d)*(-2*a*b*d*f*(3*d*e + c*f*(-1 + m
))*(1 + m) + a^2*d^2*f^2*(2 + 3*m + m^2) + b^2*(6*d^2*e^2 + 6*c*d*e*f*(-1 + m) + c^2*f^2*(2 - 3*m + m^2)))*Hyp
ergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/((b*e - a*f)^2*(1 + m))))/(6
*(b*e - a*f)^2*(d*e - c*f)^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)^m)/(f*x+e)^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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)^m)/(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((b*x + a)^m/((f^4*x^4 + 4*f^3*x^3*e + 6*f^2*x^2*e^2 + 4*f*x*e^3 + e^4)*(d*x + c)^m), x)

________________________________________________________________________________________

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)**m)/(f*x+e)**4,x)

[Out]

Timed out

________________________________________________________________________________________

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)^m)/(f*x+e)^4,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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} \,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),x)

[Out]

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

________________________________________________________________________________________