3.3137 \(\int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{b c+a d+2 b d x} \, dx\)

Optimal. Leaf size=245 \[ \frac {(b c-a d)^2 (a+b x)^{1-n} (c+d x)^{n-1} \, _2F_1\left (1,n-1;n;-\frac {b (c+d x)}{d (a+b x)}\right )}{8 b^3 d (1-n)}+\frac {(3-2 n) (b c-a d) (a+b x)^{2-n} (c+d x)^{n-1}}{8 b^3 (1-n)}+\frac {d (a+b x)^{3-n} (c+d x)^{n-1}}{4 b^3}-\frac {\left (1-2 n^2\right ) (b c-a d)^2 (a+b x)^{-n} (c+d x)^n \left (-\frac {d (a+b x)}{b c-a d}\right )^n \, _2F_1\left (n-1,n;n+1;\frac {b (c+d x)}{b c-a d}\right )}{8 b^2 d^2 (1-n) n} \]

[Out]

1/8*(-a*d+b*c)*(3-2*n)*(b*x+a)^(2-n)*(d*x+c)^(-1+n)/b^3/(1-n)+1/4*d*(b*x+a)^(3-n)*(d*x+c)^(-1+n)/b^3+1/8*(-a*d
+b*c)^2*(b*x+a)^(1-n)*(d*x+c)^(-1+n)*hypergeom([1, -1+n],[n],-b*(d*x+c)/d/(b*x+a))/b^3/d/(1-n)-1/8*(-a*d+b*c)^
2*(-2*n^2+1)*(-d*(b*x+a)/(-a*d+b*c))^n*(d*x+c)^n*hypergeom([n, -1+n],[1+n],b*(d*x+c)/(-a*d+b*c))/b^2/d^2/(1-n)
/n/((b*x+a)^n)

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Rubi [A]  time = 0.27, antiderivative size = 319, normalized size of antiderivative = 1.30, number of steps used = 10, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {105, 70, 69, 131} \[ -\frac {(b c-a d)^2 (a+b x)^{-n} (c+d x)^n \, _2F_1\left (1,-n;1-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^2 d^2 n}+\frac {(b c-a d)^2 (a+b x)^{-n} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \, _2F_1\left (-n,-n;1-n;-\frac {d (a+b x)}{b c-a d}\right )}{8 b^2 d^2 n}-\frac {(b c-a d) (a+b x)^{-n} (c+d x)^{n+1} \left (-\frac {d (a+b x)}{b c-a d}\right )^n \, _2F_1\left (n,n+1;n+2;\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (n+1)}+\frac {(a+b x)^{-n} (c+d x)^{n+2} \left (-\frac {d (a+b x)}{b c-a d}\right )^n \, _2F_1\left (n,n+2;n+3;\frac {b (c+d x)}{b c-a d}\right )}{2 d^2 (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)^2*(c + d*x)^n*Hypergeometric2F1[1, -n, 1 - n, -((d*(a + b*x))/(b*(c + d*x)))])/(8*b^2*d^2*n*(a +
b*x)^n) + ((b*c - a*d)^2*(c + d*x)^n*Hypergeometric2F1[-n, -n, 1 - n, -((d*(a + b*x))/(b*c - a*d))])/(8*b^2*d
^2*n*(a + b*x)^n*((b*(c + d*x))/(b*c - a*d))^n) - ((b*c - a*d)*(-((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) + ((-((d*(a +
b*x))/(b*c - a*d)))^n*(c + d*x)^(2 + n)*Hypergeometric2F1[n, 2 + n, 3 + n, (b*(c + d*x))/(b*c - a*d)])/(2*d^2
*(2 + 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 105

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

Rule 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{b c+a d+2 b d x} \, dx &=\frac {\int (a+b x)^{-n} (c+d x)^{1+n} \, dx}{2 d}-\frac {(b c-a d) \int \frac {(a+b x)^{-n} (c+d x)^{1+n}}{b c+a d+2 b d x} \, dx}{2 d}\\ &=-\frac {(b c-a d) \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} \, dx}{4 b d}+\frac {\left ((a+b x)^{-n} \left (\frac {d (a+b x)}{-b c+a d}\right )^n\right ) \int (c+d x)^{1+n} \left (-\frac {a d}{b c-a d}-\frac {b d x}{b c-a d}\right )^{-n} \, dx}{2 d}\\ &=\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{2+n} \, _2F_1\left (n,2+n;3+n;\frac {b (c+d x)}{b c-a d}\right )}{2 d^2 (2+n)}-\frac {(b c-a d)^2 \int (a+b x)^{-1-n} (c+d x)^n \, dx}{8 b d^2}+\frac {(b c-a d)^3 \int \frac {(a+b x)^{-1-n} (c+d x)^n}{b c+a d+2 b d x} \, dx}{8 b d^2}-\frac {\left ((b c-a d) (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)^2 (a+b x)^{-n} (c+d x)^n \, _2F_1\left (1,-n;1-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^2 d^2 n}-\frac {(b c-a d) (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)}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{2+n} \, _2F_1\left (n,2+n;3+n;\frac {b (c+d x)}{b c-a d}\right )}{2 d^2 (2+n)}-\frac {\left ((b c-a d)^2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^{-1-n} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \, dx}{8 b d^2}\\ &=-\frac {(b c-a d)^2 (a+b x)^{-n} (c+d x)^n \, _2F_1\left (1,-n;1-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^2 d^2 n}+\frac {(b c-a d)^2 (a+b x)^{-n} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \, _2F_1\left (-n,-n;1-n;-\frac {d (a+b x)}{b c-a d}\right )}{8 b^2 d^2 n}-\frac {(b c-a d) (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)}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{2+n} \, _2F_1\left (n,2+n;3+n;\frac {b (c+d x)}{b c-a d}\right )}{2 d^2 (2+n)}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 257, normalized size = 1.05 \[ \frac {(a d-b c) (a+b x)^{-n} (c+d x)^n \left (\frac {\left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (4 d n (n+1) (a+b x) \, _2F_1\left (-n-1,1-n;2-n;\frac {d (a+b x)}{a d-b c}\right )-(n-1) \left ((n+1) (b c-a d) \, _2F_1\left (-n,-n;1-n;\frac {d (a+b x)}{a d-b c}\right )-2 b n (c+d x) \left (-\frac {b d (a+b x) (c+d x)}{(b c-a d)^2}\right )^n \, _2F_1\left (n,n+1;n+2;\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{n^2-1}+(b c-a d) \, _2F_1\left (1,-n;1-n;-\frac {d (a+b x)}{b (c+d x)}\right )\right )}{8 b^2 d^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-(b*c) + a*d)*(c + d*x)^n*((b*c - a*d)*Hypergeometric2F1[1, -n, 1 - n, -((d*(a + b*x))/(b*(c + d*x)))] + (4*
d*n*(1 + n)*(a + b*x)*Hypergeometric2F1[-1 - n, 1 - n, 2 - n, (d*(a + b*x))/(-(b*c) + a*d)] - (-1 + n)*((b*c -
a*d)*(1 + n)*Hypergeometric2F1[-n, -n, 1 - n, (d*(a + b*x))/(-(b*c) + a*d)] - 2*b*n*(c + d*x)*(-((b*d*(a + b*
x)*(c + d*x))/(b*c - a*d)^2))^n*Hypergeometric2F1[n, 1 + n, 2 + n, (b*(c + d*x))/(b*c - a*d)]))/((-1 + n^2)*((
b*(c + d*x))/(b*c - a*d))^n)))/(8*b^2*d^2*n*(a + b*x)^n)

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{2 \, b d x + b c + a d}, x\right ) \]

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),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{2 \, b d x + b c + a d}\,{d x} \]

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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{-n +1} \left (d x +c \right )^{n +1}}{2 b d x +a d +b c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{2 \, b d x + b c + a d}\,{d x} \]

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),x, algorithm="maxima")

[Out]

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

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

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),x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

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),x)

[Out]

Exception raised: HeuristicGCDFailed

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