3.3077 \(\int \frac {(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^3} \, dx\)

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

[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 = 0.29, antiderivative size = 300, 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, 151, 12, 131} \[ \frac {(a+b x)^{m+1} (c+d x)^{-m-1} \left (-a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f m+2 d e)+b^2 \left (-\left (-c^2 f^2 (1-m) m+4 c d e f m+2 d^2 e^2\right )\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 m (m+1) (b e-a f)^3 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (a d f (m+2)-b (c f m+2 d e))}{2 m (e+f x)^2 (b c-a d) (b e-a f) (d e-c f)^2}+\frac {d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x)^2 (b c-a d) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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) + (d*(a + b*x)^(1 + m))/((b*c - a*d)*(d*e - c*f)*m*(c + d*x)^m*(e + f*x)^2) + ((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)^(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))])/(2*(b*e - a*f)^3*(d*e - c*f)^2*m*(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)^{-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 = 0.53, size = 260, normalized size = 0.92 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (-\frac {(b c-a d) \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (c f m+2 d e)+b^2 \left (c^2 f^2 (m-1) m+4 c d e f m+2 d^2 e^2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (c+d x) (b e-a f)^3 (c f-d e)}-\frac {f (c+d x) (b (c f m+2 d e)-a d f (m+2))}{(e+f x)^2 (b e-a f) (d e-c f)}-\frac {2 d}{(e+f x)^2}\right )}{2 m (b c-a d) (c f-d e)} \]

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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fricas [F]  time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}}, x\right ) \]

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*e*f^2*x^2 + 3*e^2*f*x + e^3), x)

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

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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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m -1}}{\left (f x +e \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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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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)**(-1-m)/(f*x+e)**3,x)

[Out]

Timed out

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