3.3068 \(\int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^4} \, dx\)

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

[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)

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Rubi [A]  time = 0.32, 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 = {129, 151, 12, 131} \[ -\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (3 d e-c f (1-m))+b^2 \left (-\left (c^2 f^2 \left (m^2-3 m+2\right )-6 c d e f (1-m)+6 d^2 e^2\right )\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{6 (m+1) (b e-a f)^4 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (m+2)-b c f (2-m)+4 b d e)}{6 (e+f x)^2 (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m))/(3*(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 129

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, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rule 151

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] && LtQ[m, -1] && IntegerQ[m]

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*}

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Mathematica [A]  time = 0.53, size = 255, normalized size = 0.83 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m-1} \left (\frac {(b c-a d) \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (c f (m-1)+3 d e)+b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )+6 c d e f (m-1)+6 d^2 e^2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^2}-\frac {f (c+d x)^2 (-a d f (m+2)+b c f (m-2)+4 b d e)}{(e+f x)^2}-\frac {2 f (c+d x)^2 (a f-b e) (c f-d e)}{(e+f x)^3}\right )}{6 (b e-a f)^2 (d e-c f)^2} \]

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)

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fricas [F]  time = 1.27, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m}}{{\left (f^{4} x^{4} + 4 \, e f^{3} x^{3} + 6 \, e^{2} f^{2} x^{2} + 4 \, e^{3} f x + e^{4}\right )} {\left (d x + c\right )}^{m}}, x\right ) \]

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

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

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)

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maple [F]  time = 0.24, 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)

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

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)

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

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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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