3.10.99 \(\int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx\) [999]

Optimal. Leaf size=50 \[ -\frac {(b x)^{-1-2 m} \, _2F_1\left (\frac {1}{2} (-1-2 m),-m;\frac {1}{2} (1-2 m);a^2 x^2\right )}{b (1+2 m)} \]

[Out]

-(b*x)^(-1-2*m)*hypergeom([-m, -1/2-m],[1/2-m],a^2*x^2)/b/(1+2*m)

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Rubi [A]
time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {126, 371} \begin {gather*} -\frac {(b x)^{-2 m-1} \, _2F_1\left (\frac {1}{2} (-2 m-1),-m;\frac {1}{2} (1-2 m);a^2 x^2\right )}{b (2 m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x)^(-2 - 2*m)*(1 - a*x)^m*(1 + a*x)^m,x]

[Out]

-(((b*x)^(-1 - 2*m)*Hypergeometric2F1[(-1 - 2*m)/2, -m, (1 - 2*m)/2, a^2*x^2])/(b*(1 + 2*m)))

Rule 126

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin {align*} \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx &=\int (b x)^{-2-2 m} \left (1-a^2 x^2\right )^m \, dx\\ &=-\frac {(b x)^{-1-2 m} \, _2F_1\left (\frac {1}{2} (-1-2 m),-m;\frac {1}{2} (1-2 m);a^2 x^2\right )}{b (1+2 m)}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 44, normalized size = 0.88 \begin {gather*} -\frac {(b x)^{-1-2 m} \, _2F_1\left (-\frac {1}{2}-m,-m;\frac {1}{2}-m;a^2 x^2\right )}{b+2 b m} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x)^(-2 - 2*m)*(1 - a*x)^m*(1 + a*x)^m,x]

[Out]

-(((b*x)^(-1 - 2*m)*Hypergeometric2F1[-1/2 - m, -m, 1/2 - m, a^2*x^2])/(b + 2*b*m))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (b x \right )^{-2-2 m} \left (-a x +1\right )^{m} \left (a x +1\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x)

[Out]

int((b*x)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,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)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x, algorithm="maxima")

[Out]

integrate((a*x + 1)^m*(-a*x + 1)^m*(b*x)^(-2*m - 2), 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)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x, algorithm="fricas")

[Out]

integral((a*x + 1)^m*(-a*x + 1)^m*(b*x)^(-2*m - 2), 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)**(-2-2*m)*(-a*x+1)**m*(a*x+1)**m,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)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x, algorithm="giac")

[Out]

integrate((a*x + 1)^m*(-a*x + 1)^m*(b*x)^(-2*m - 2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (1-a\,x\right )}^m\,{\left (a\,x+1\right )}^m}{{\left (b\,x\right )}^{2\,m+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - a*x)^m*(a*x + 1)^m)/(b*x)^(2*m + 2),x)

[Out]

int(((1 - a*x)^m*(a*x + 1)^m)/(b*x)^(2*m + 2), x)

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