3.31.67 \(\int \frac {(a+b x)^m (c+d x)^{-m}}{(e+f x)^3} \, dx\) [3067]

Optimal. Leaf size=174 \[ -\frac {f (a+b x)^{1+m} (c+d x)^{1-m}}{2 (b e-a f) (d e-c f) (e+f x)^2}+\frac {(b c-a d) (b (2 d e-c f (1-m))-a d f (1+m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (2,1+m;2+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{2 (b e-a f)^3 (d e-c f) (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 173, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {98, 133} \begin {gather*} \frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m-1} (-a d f (m+1)-b c f (1-m)+2 b d e) \, _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+1) (b e-a f)^3 (d e-c f)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-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)^m*(e + f*x)^3),x]

[Out]

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

Rule 98

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 146, normalized size = 0.84 \begin {gather*} \frac {(a+b x)^{1+m} (c+d x)^{-1-m} \left (-\frac {f (c+d x)^2}{(e+f x)^2}+\frac {(b c-a d) (2 b d e+b c f (-1+m)-a d f (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 )}{(b e-a f)^2 (1+m)}\right )}{2 (b e-a f) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

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

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