3.31.77 \(\int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^3} \, dx\) [3077]

Optimal. Leaf size=283 \[ -\frac {f (a+b x)^{1+m} (c+d x)^{-m}}{2 (b e-a f) (d e-c f) (e+f x)^2}-\frac {f (b (3 d e-c f (1-m))-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{2 (b e-a f)^2 (d e-c f)^2 (e+f x)}+\frac {\left (2 a b d f (1+m) (2 d e+c f m)-b^2 \left (2 d^2 e^2+4 c d e f m-c^2 f^2 (1-m) m\right )-a^2 d^2 f^2 \left (2+3 m+m^2\right )\right ) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{2 (b e-a f)^2 (d e-c f)^3 m} \]

[Out]

-1/2*f*(b*x+a)^(1+m)/(-a*f+b*e)/(-c*f+d*e)/((d*x+c)^m)/(f*x+e)^2-1/2*f*(b*(3*d*e-c*f*(1-m))-a*d*f*(2+m))*(b*x+
a)^(1+m)/(-a*f+b*e)^2/(-c*f+d*e)^2/((d*x+c)^m)/(f*x+e)+1/2*(2*a*b*d*f*(1+m)*(c*f*m+2*d*e)-b^2*(2*d^2*e^2+4*c*d
*e*f*m-c^2*f^2*(1-m)*m)-a^2*d^2*f^2*(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)^2/(-c*f+d*e)^3/m/((d*x+c)^m)

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Rubi [A]
time = 0.17, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {(a+b x)^m (c+d x)^{-m} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f m+2 d e)-\left (b^2 \left (-c^2 f^2 (1-m) m+4 c d e f m+2 d^2 e^2\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 )}{2 m (b e-a f)^2 (d e-c f)^3}-\frac {f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+2)-b c f (1-m)+3 b d e)}{2 (e+f x) (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{-m}}{2 (e+f x)^2 (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)^3,x]

[Out]

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

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Mathematica [A]
time = 0.44, size = 260, normalized size = 0.92 \begin {gather*} \frac {(a+b x)^{1+m} (c+d x)^{-m} \left (-\frac {2 d}{(e+f x)^2}-\frac {f (-a d f (2+m)+b (2 d e+c f m)) (c+d x)}{(b e-a f) (d e-c f) (e+f x)^2}-\frac {(b c-a d) \left (-2 a b d f (1+m) (2 d e+c f m)+b^2 \left (2 d^2 e^2+4 c d e f m+c^2 f^2 (-1+m) m\right )+a^2 d^2 f^2 \left (2+3 m+m^2\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)^3 (-d e+c f) (1+m) (c+d x)}\right )}{2 (b c-a d) (-d e+c f) m} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(c + d*x)^(-1 - m))/(e + f*x)^3,x]

[Out]

((a + b*x)^(1 + m)*((-2*d)/(e + f*x)^2 - (f*(-(a*d*f*(2 + m)) + b*(2*d*e + c*f*m))*(c + d*x))/((b*e - a*f)*(d*
e - c*f)*(e + f*x)^2) - ((b*c - a*d)*(-2*a*b*d*f*(1 + m)*(2*d*e + c*f*m) + b^2*(2*d^2*e^2 + 4*c*d*e*f*m + c^2*
f^2*(-1 + m)*m) + a^2*d^2*f^2*(2 + 3*m + m^2))*Hypergeometric2F1[2, 1 + m, 2 + m, ((d*e - c*f)*(a + b*x))/((b*
e - a*f)*(c + d*x))])/((b*e - a*f)^3*(-(d*e) + c*f)*(1 + m)*(c + d*x))))/(2*(b*c - a*d)*(-(d*e) + c*f)*m*(c +
d*x)^m)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-1-m}}{\left (f x +e \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x+a)^m*(d*x+c)^(-1-m)/(f*x+e)^3,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)^3,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^3, 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)^3,x, algorithm="fricas")

[Out]

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

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 1)/(f*x + e)^3, 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 )}^3\,{\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)^3*(c + d*x)^(m + 1)),x)

[Out]

int((a + b*x)^m/((e + f*x)^3*(c + d*x)^(m + 1)), x)

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