3.3054 \(\int (a+b x)^{-n} (c+d x) (e+f x)^{-1+n} \, dx\)

Optimal. Leaf size=151 \[ \frac {(a+b x)^{-n} (e+f x)^{n+1} \left (-\frac {f (a+b x)}{b e-a f}\right )^n (b (c f-d e (1-n))-a d f n) \, _2F_1\left (n,n+1;n+2;\frac {b (e+f x)}{b e-a f}\right )}{f^2 n (n+1) (b e-a f)}+\frac {(a+b x)^{1-n} (d e-c f) (e+f x)^n}{f n (b e-a f)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 150, normalized size of antiderivative = 0.99, 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} \[ \frac {(a+b x)^{-n} (e+f x)^{n+1} \left (-\frac {f (a+b x)}{b e-a f}\right )^n (-a d f n+b c f-b d e (1-n)) \, _2F_1\left (n,n+1;n+2;\frac {b (e+f x)}{b e-a f}\right )}{f^2 n (n+1) (b e-a f)}+\frac {(a+b x)^{1-n} (d e-c f) (e+f x)^n}{f n (b e-a f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*e - c*f)*(a + b*x)^(1 - n)*(e + f*x)^n)/(f*(b*e - a*f)*n) + ((b*c*f - b*d*e*(1 - n) - a*d*f*n)*(-((f*(a +
b*x))/(b*e - a*f)))^n*(e + f*x)^(1 + n)*Hypergeometric2F1[n, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(f^2*(b
*e - a*f)*n*(1 + n)*(a + b*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)^{-n} (c+d x) (e+f x)^{-1+n} \, dx &=\frac {(d e-c f) (a+b x)^{1-n} (e+f x)^n}{f (b e-a f) n}-\frac {(b c f-d (b e (1-n)+a f n)) \int (a+b x)^{-n} (e+f x)^n \, dx}{f (-b e+a f) n}\\ &=\frac {(d e-c f) (a+b x)^{1-n} (e+f x)^n}{f (b e-a f) n}-\frac {\left ((b c f-d (b e (1-n)+a f n)) (a+b x)^{-n} \left (\frac {f (a+b x)}{-b e+a f}\right )^n\right ) \int (e+f x)^n \left (-\frac {a f}{b e-a f}-\frac {b f x}{b e-a f}\right )^{-n} \, dx}{f (-b e+a f) n}\\ &=\frac {(d e-c f) (a+b x)^{1-n} (e+f x)^n}{f (b e-a f) n}+\frac {(b c f-b d e (1-n)-a d f n) (a+b x)^{-n} \left (-\frac {f (a+b x)}{b e-a f}\right )^n (e+f x)^{1+n} \, _2F_1\left (n,1+n;2+n;\frac {b (e+f x)}{b e-a f}\right )}{f^2 (b e-a f) n (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 123, normalized size = 0.81 \[ \frac {(a+b x)^{-n} (e+f x)^n \left (\frac {(e+f x) \left (\frac {f (a+b x)}{a f-b e}\right )^n (a d f n-b (c f+d e (n-1))) \, _2F_1\left (n,n+1;n+2;\frac {b (e+f x)}{b e-a f}\right )}{n+1}+f (a+b x) (c f-d e)\right )}{f^2 n (a f-b e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d x + c\right )} {\left (f x + e\right )}^{n - 1}}{{\left (b x + a\right )}^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)*(f*x + e)^(n - 1)/(b*x + a)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*(f*x + e)^(n - 1)/(b*x + a)^n, x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right ) \left (b x +a \right )^{-n} \left (f x +e \right )^{n -1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*(f*x + e)^(n - 1)/(b*x + a)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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