3.3049 \(\int (a+b x) (c+d x)^{-2+n} (e+f x)^{-n} \, dx\)

Optimal. Leaf size=117 \[ \frac {(b c-a d) (c+d x)^{n-1} (e+f x)^{1-n}}{d (1-n) (d e-c f)}+\frac {b (c+d x)^n (e+f x)^{-n} \left (\frac {d (e+f x)}{d e-c f}\right )^n \, _2F_1\left (n,n;n+1;-\frac {f (c+d x)}{d e-c f}\right )}{d^2 n} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {79, 70, 69} \[ \frac {(b c-a d) (c+d x)^{n-1} (e+f x)^{1-n}}{d (1-n) (d e-c f)}+\frac {b (c+d x)^n (e+f x)^{-n} \left (\frac {d (e+f x)}{d e-c f}\right )^n \, _2F_1\left (n,n;n+1;-\frac {f (c+d x)}{d e-c f}\right )}{d^2 n} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(c + d*x)^(-2 + n))/(e + f*x)^n,x]

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 115, normalized size = 0.98 \[ \frac {(c+d x)^{n-1} (e+f x)^{-n} \left (d^2 (e+f x) (b e-a f)-b (d e-c f)^2 \left (\frac {d (e+f x)}{d e-c f}\right )^n \, _2F_1\left (n-1,n-1;n;\frac {f (c+d x)}{c f-d e}\right )\right )}{d^2 f (n-1) (c f-d e)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(c + d*x)^(-2 + n))/(e + f*x)^n,x]

[Out]

((c + d*x)^(-1 + n)*(d^2*(b*e - a*f)*(e + f*x) - b*(d*e - c*f)^2*((d*(e + f*x))/(d*e - c*f))^n*Hypergeometric2
F1[-1 + n, -1 + n, n, (f*(c + d*x))/(-(d*e) + c*f)]))/(d^2*f*(-(d*e) + c*f)*(-1 + n)*(e + f*x)^n)

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fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )} {\left (d x + c\right )}^{n - 2}}{{\left (f x + e\right )}^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-2+n)/((f*x+e)^n),x, algorithm="fricas")

[Out]

integral((b*x + a)*(d*x + c)^(n - 2)/(f*x + e)^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\left (d x + c\right )}^{n - 2}}{{\left (f x + e\right )}^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-2+n)/((f*x+e)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)*(d*x + c)^(n - 2)/(f*x + e)^n, x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right ) \left (d x +c \right )^{n -2} \left (f x +e \right )^{-n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(d*x+c)^(n-2)/((f*x+e)^n),x)

[Out]

int((b*x+a)*(d*x+c)^(n-2)/((f*x+e)^n),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\left (d x + c\right )}^{n - 2}}{{\left (f x + e\right )}^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-2+n)/((f*x+e)^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(d*x + c)^(n - 2)/(f*x + e)^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{n-2}}{{\left (e+f\,x\right )}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(c + d*x)^(n - 2))/(e + f*x)^n,x)

[Out]

int(((a + b*x)*(c + d*x)^(n - 2))/(e + f*x)^n, 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)*(d*x+c)**(-2+n)/((f*x+e)**n),x)

[Out]

Timed out

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