\(\int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx\) [3138]

Optimal result

Integrand size = 35, antiderivative size = 154 \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=-\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{-1+n} \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {130, 72, 71, 133} \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\frac {(a+b x)^{-n} (c+d x)^{n+1} \left (-\frac {d (a+b x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,n+1,n+2,\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (n+1)}-\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{n-1} \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)} \]

[In]

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

[Out]

-1/4*((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)))])/(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)

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 130

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_))^2, x_Symbol] :> Dist[b*(d/f^2),
Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x] + Dist[(b*e - a*f)*((d*e - c*f)/f^2), Int[(a + b*x)^(m - 1)*(
(c + d*x)^(n - 1)/(e + f*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[m + n, 0] && EqQ[2*b*d*e
- f*(b*c + a*d), 0]

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*} \text {integral}& = \frac {\int (a+b x)^{-n} (c+d x)^n \, dx}{4 b d}-\frac {(b c-a d)^2 \int \frac {(a+b x)^{-n} (c+d x)^n}{(b c+a d+2 b d x)^2} \, dx}{4 b d} \\ & = -\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{-1+n} \, _2F_1\left (2,1-n;2-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)}+\frac {\left ((a+b x)^{-n} \left (\frac {d (a+b x)}{-b c+a d}\right )^n\right ) \int (c+d x)^n \left (-\frac {a d}{b c-a d}-\frac {b d x}{b c-a d}\right )^{-n} \, dx}{4 b d} \\ & = -\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{-1+n} \, _2F_1\left (2,1-n;2-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{1+n} \, _2F_1\left (n,1+n;2+n;\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (1+n)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 1.28 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.52 \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=-\frac {(a+b x)^{-n} (c+d x)^n \left (-(b c-a d)^2 (-1+n) \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^n \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^{-n} \operatorname {AppellF1}\left (1,-n,n,2,\frac {-b c+a d}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )+2 d (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (a d+b (c+2 d x)) \operatorname {Hypergeometric2F1}\left (1-n,-n,2-n,\frac {d (a+b x)}{-b c+a d}\right )\right )}{8 b^2 d^2 (-1+n) (a d+b (c+2 d x))} \]

[In]

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

[Out]

-1/8*((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))/(b^2*d^2*(-1 + n)*(a + b*x)^n*(a*d + b*(c + 2*d*x)))

Maple [F]

\[\int \frac {\left (b x +a \right )^{1-n} \left (d x +c \right )^{1+n}}{\left (2 b d x +a d +b c \right )^{2}}d x\]

[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)

Fricas [F]

\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]

[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)

Sympy [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^{1-n}\,{\left (c+d\,x\right )}^{n+1}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^2} \,d x \]

[In]

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

[Out]

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