3.3138 \(\int \frac{(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx\)

Optimal. Leaf size=154 \[ \frac{(a+b x)^{-n} (c+d x)^{n+1} \left (-\frac{d (a+b x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;\frac{b (c+d x)}{b c-a d}\right )}{4 b d^2 (n+1)}-\frac{(b c-a d) (a+b x)^{1-n} (c+d x)^{n-1} \, _2F_1\left (2,1-n;2-n;-\frac{d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)} \]

[Out]

-((b*c - a*d)*(a + b*x)^(1 - n)*(c + d*x)^(-1 + n)*Hypergeometric2F1[2, 1 - n, 2 - n, -((d*(a + b*x))/(b*(c +
d*x)))])/(4*b^3*d*(1 - n)) + ((-((d*(a + b*x))/(b*c - a*d)))^n*(c + d*x)^(1 + n)*Hypergeometric2F1[n, 1 + n, 2
+ n, (b*(c + d*x))/(b*c - a*d)])/(4*b*d^2*(1 + n)*(a + b*x)^n)

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Rubi [C]  time = 0.0781424, antiderivative size = 113, normalized size of antiderivative = 0.73, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {137, 136} \[ \frac{(a+b x)^{2-n} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} F_1\left (2-n;-n-1,2;3-n;-\frac{d (a+b x)}{b c-a d},-\frac{2 d (a+b x)}{b c-a d}\right )}{b^2 (2-n) (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x)^(2 - n)*(c + d*x)^n*AppellF1[2 - n, -1 - n, 2, 3 - n, -((d*(a + b*x))/(b*c - a*d)), (-2*d*(a + b*x)
)/(b*c - a*d)])/(b^2*(b*c - a*d)*(2 - n)*((b*(c + d*x))/(b*c - a*d))^n)

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx &=\frac{\left ((b c-a d) (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n}\right ) \int \frac{(a+b x)^{1-n} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^{1+n}}{(b c+a d+2 b d x)^2} \, dx}{b}\\ &=\frac{(a+b x)^{2-n} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} F_1\left (2-n;-1-n,2;3-n;-\frac{d (a+b x)}{b c-a d},-\frac{2 d (a+b x)}{b c-a d}\right )}{b^2 (b c-a d) (2-n)}\\ \end{align*}

Mathematica [C]  time = 0.707675, size = 234, normalized size = 1.52 \[ -\frac{(a+b x)^{-n} (c+d x)^n \left (2 d (a+b x) (a d+b (c+2 d x)) \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \, _2F_1\left (1-n,-n;2-n;\frac{d (a+b x)}{a d-b c}\right )-(n-1) (b c-a d)^2 \left (\frac{d (a+b x)}{a d+b (c+2 d x)}\right )^n \left (\frac{b (c+d x)}{a d+b (c+2 d x)}\right )^{-n} F_1\left (1;-n,n;2;\frac{a d-b c}{a d+b (c+2 d x)},\frac{b c-a d}{b c+a d+2 b d x}\right )\right )}{8 b^2 d^2 (n-1) (a d+b (c+2 d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((c + d*x)^n*(-(((b*c - a*d)^2*(-1 + n)*((d*(a + b*x))/(a*d + b*(c + 2*d*x)))^n*AppellF1[1, -n, n, 2, (-(b*c)
+ a*d)/(a*d + b*(c + 2*d*x)), (b*c - a*d)/(b*c + a*d + 2*b*d*x)])/((b*(c + d*x))/(a*d + b*(c + 2*d*x)))^n) +
(2*d*(a + b*x)*(a*d + b*(c + 2*d*x))*Hypergeometric2F1[1 - n, -n, 2 - n, (d*(a + b*x))/(-(b*c) + a*d)])/((b*(c
+ d*x))/(b*c - a*d))^n))/(8*b^2*d^2*(-1 + n)*(a + b*x)^n*(a*d + b*(c + 2*d*x)))

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Maple [F]  time = 0.074, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{1+n} \left ( bx+a \right ) ^{1-n}}{ \left ( 2\,bdx+ad+bc \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^2,x)

[Out]

int((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{-n + 1}{\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^2,x, algorithm="maxima")

[Out]

integrate((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(2*b*d*x + b*c + a*d)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{-n + 1}{\left (d x + c\right )}^{n + 1}}{4 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 2 \, a b c d + a^{2} d^{2} + 4 \,{\left (b^{2} c d + a b d^{2}\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^2,x, algorithm="fricas")

[Out]

integral((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(4*b^2*d^2*x^2 + b^2*c^2 + 2*a*b*c*d + a^2*d^2 + 4*(b^2*c*d + a*
b*d^2)*x), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1-n)*(d*x+c)**(1+n)/(2*b*d*x+a*d+b*c)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{-n + 1}{\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^2,x, algorithm="giac")

[Out]

integrate((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(2*b*d*x + b*c + a*d)^2, x)