3.3093 \(\int \frac {(a+b x)^m (c+d x)^{-3-m}}{e+f x} \, dx\)

Optimal. Leaf size=196 \[ -\frac {f^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (d e-c f)^3}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)+b (d e-c f (m+3)))}{(m+1) (m+2) (b c-a d)^2 (d e-c f)^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 208, normalized size of antiderivative = 1.06, 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 = {129, 155, 12, 131} \[ \frac {f^2 (a+b x)^{m+1} (c+d x)^{-m-1} \, _2F_1\left (1,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f) (d e-c f)^2}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d) (d e-c f)}+\frac {d (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b c f (m+3)+b d e)}{(m+1) (m+2) (b c-a d)^2 (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/((b*c - a*d)*(d*e - c*f)*(2 + m)) + (d*(b*d*e + a*d*f*(2 + m) - b*c*f
*(3 + m))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/((b*c - a*d)^2*(d*e - c*f)^2*(1 + m)*(2 + m)) + (f^2*(a + b*x)
^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[1, 1 + m, 2 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))
])/((b*e - a*f)*(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 155

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 + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 186, normalized size = 0.95 \[ -\frac {(a+b x)^{m+1} (c+d x)^{-m-2} \left (\frac {f^2 (m+2) (c+d x) (b c-a d) \, _2F_1\left (1,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f) (d e-c f)}-\frac {d (c+d x) (a d f (m+2)-b c f (m+3)+b d e)}{(m+1) (b c-a d) (c f-d e)}+d\right )}{(m+2) (b c-a d) (c f-d e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}}{f x + e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m -3}}{f x +e}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,x\right )}^m}{\left (e+f\,x\right )\,{\left (c+d\,x\right )}^{m+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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