3.32.39 \(\int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^3} \, dx\) [3139]
Optimal. Leaf size=230 \[ -\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^n}{8 b^2 d (b c+a d+2 b d x)^2}-\frac {(1+2 n) (a+b x)^{1-n} (c+d x)^n}{8 b^2 d (b c+a d+2 b d x)}-\frac {\left (1-2 n^2\right ) (a+b x)^{-n} (c+d x)^n \, _2F_1\left (1,n;1+n;-\frac {b (c+d x)}{d (a+b x)}\right )}{8 b^2 d^2 n}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^n \, _2F_1\left (n,n;1+n;\frac {b (c+d x)}{b c-a d}\right )}{8 b^2 d^2 n} \]
[Out]
-1/8*(-a*d+b*c)*(b*x+a)^(1-n)*(d*x+c)^n/b^2/d/(2*b*d*x+a*d+b*c)^2-1/8*(1+2*n)*(b*x+a)^(1-n)*(d*x+c)^n/b^2/d/(2
*b*d*x+a*d+b*c)-1/8*(-2*n^2+1)*(d*x+c)^n*hypergeom([1, n],[1+n],-b*(d*x+c)/d/(b*x+a))/b^2/d^2/n/((b*x+a)^n)+1/
8*(-d*(b*x+a)/(-a*d+b*c))^n*(d*x+c)^n*hypergeom([n, n],[1+n],b*(d*x+c)/(-a*d+b*c))/b^2/d^2/n/((b*x+a)^n)
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Rubi [A]
time = 0.33, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {132, 72, 71,
1627, 156, 12, 133} \begin {gather*} -\frac {\left (1-2 n^2\right ) (a+b x)^{-n} (c+d x)^n \, _2F_1\left (1,n;n+1;-\frac {b (c+d x)}{d (a+b x)}\right )}{8 b^2 d^2 n}+\frac {(a+b x)^{-n} (c+d x)^n \left (-\frac {d (a+b x)}{b c-a d}\right )^n \, _2F_1\left (n,n;n+1;\frac {b (c+d x)}{b c-a d}\right )}{8 b^2 d^2 n}-\frac {(2 n+1) (a+b x)^{1-n} (c+d x)^n}{8 b^2 d (a d+b c+2 b d x)}-\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^n}{8 b^2 d (a d+b c+2 b d x)^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((a + b*x)^(1 - n)*(c + d*x)^(1 + n))/(b*c + a*d + 2*b*d*x)^3,x]
[Out]
-1/8*((b*c - a*d)*(a + b*x)^(1 - n)*(c + d*x)^n)/(b^2*d*(b*c + a*d + 2*b*d*x)^2) - ((1 + 2*n)*(a + b*x)^(1 - n
)*(c + d*x)^n)/(8*b^2*d*(b*c + a*d + 2*b*d*x)) - ((1 - 2*n^2)*(c + d*x)^n*Hypergeometric2F1[1, n, 1 + n, -((b*
(c + d*x))/(d*(a + b*x)))])/(8*b^2*d^2*n*(a + b*x)^n) + ((-((d*(a + b*x))/(b*c - a*d)))^n*(c + d*x)^n*Hypergeo
metric2F1[n, n, 1 + n, (b*(c + d*x))/(b*c - a*d)])/(8*b^2*d^2*n*(a + b*x)^n)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 132
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[b*d^(m
+ n)*f^p, Int[(a + b*x)^(m - 1)/(c + d*x)^m, x], x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandTo
Sum[(a + b*x)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 0] || SumSimplerQ[m, -1] || !(GtQ[n, 0] || SumSimplerQ[n,
-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]
Rule 156
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Rule 1627
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[{
Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[b*R*(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[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*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f
*R*(m + 1) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x], x], x]] /; FreeQ[{a, b,
c, d, e, f, n, p}, x] && PolyQ[Px, x] && ILtQ[m, -1]
Rubi steps
\begin {align*} \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^3} \, 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)^3} \, 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,3;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 (2-n)}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 1.40, size = 239, normalized size = 1.04 \begin {gather*} \frac {(a+b x)^{-n} (c+d x)^n \left (\frac {(b c-a d)^3 \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 (2;-n,n;3;\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 )}{(a d+b (c+2 d x))^2}-\frac {4 b \left (\frac {d (a+b x)}{-b c+a d}\right )^n (c+d x) F_1\left (1+n;n,1;2+n;\frac {b (c+d x)}{b c-a d},\frac {2 b (c+d x)}{b c-a d}\right )}{1+n}\right )}{16 b^2 d^2 (b c-a d)} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x)^(1 - n)*(c + d*x)^(1 + n))/(b*c + a*d + 2*b*d*x)^3,x]
[Out]
((c + d*x)^n*(((b*c - a*d)^3*((d*(a + b*x))/(a*d + b*(c + 2*d*x)))^n*AppellF1[2, -n, n, 3, (-(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*(a*d + b*(c +
2*d*x))^2) - (4*b*((d*(a + b*x))/(-(b*c) + a*d))^n*(c + d*x)*AppellF1[1 + n, n, 1, 2 + n, (b*(c + d*x))/(b*c -
a*d), (2*b*(c + d*x))/(b*c - a*d)])/(1 + n)))/(16*b^2*d^2*(b*c - a*d)*(a + b*x)^n)
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{1-n} \left (d x +c \right )^{1+n}}{\left (2 b d x +a d +b c \right )^{3}}\, dx\]
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)^3,x)
[Out]
int((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^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)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^3,x, algorithm="maxima")
[Out]
integrate((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(2*b*d*x + b*c + a*d)^3, 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)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^3,x, algorithm="fricas")
[Out]
integral((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(8*b^3*d^3*x^3 + b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d
^3 + 12*(b^3*c*d^2 + a*b^2*d^3)*x^2 + 6*(b^3*c^2*d + 2*a*b^2*c*d^2 + a^2*b*d^3)*x), 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)**(1-n)*(d*x+c)**(1+n)/(2*b*d*x+a*d+b*c)**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)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^3,x, algorithm="giac")
[Out]
integrate((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(2*b*d*x + b*c + a*d)^3, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + b*x)^(1 - n)*(c + d*x)^(n + 1))/(a*d + b*c + 2*b*d*x)^3,x)
[Out]
int(((a + b*x)^(1 - n)*(c + d*x)^(n + 1))/(a*d + b*c + 2*b*d*x)^3, x)
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