3.313 \(\int \frac{(g x)^m (1-a^2 x^2)^p}{1+a x} \, dx\)

Optimal. Leaf size=89 \[ \frac{(g x)^{m+1} \, _2F_1\left (\frac{m+1}{2},1-p;\frac{m+3}{2};a^2 x^2\right )}{g (m+1)}-\frac{a (g x)^{m+2} \, _2F_1\left (\frac{m+2}{2},1-p;\frac{m+4}{2};a^2 x^2\right )}{g^2 (m+2)} \]

[Out]

((g*x)^(1 + m)*Hypergeometric2F1[(1 + m)/2, 1 - p, (3 + m)/2, a^2*x^2])/(g*(1 + m)) - (a*(g*x)^(2 + m)*Hyperge
ometric2F1[(2 + m)/2, 1 - p, (4 + m)/2, a^2*x^2])/(g^2*(2 + m))

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Rubi [A]  time = 0.0555549, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {890, 82, 125, 364} \[ \frac{(g x)^{m+1} \, _2F_1\left (\frac{m+1}{2},1-p;\frac{m+3}{2};a^2 x^2\right )}{g (m+1)}-\frac{a (g x)^{m+2} \, _2F_1\left (\frac{m+2}{2},1-p;\frac{m+4}{2};a^2 x^2\right )}{g^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[((g*x)^m*(1 - a^2*x^2)^p)/(1 + a*x),x]

[Out]

((g*x)^(1 + m)*Hypergeometric2F1[(1 + m)/2, 1 - p, (3 + m)/2, a^2*x^2])/(g*(1 + m)) - (a*(g*x)^(2 + m)*Hyperge
ometric2F1[(2 + m)/2, 1 - p, (4 + m)/2, a^2*x^2])/(g^2*(2 + m))

Rule 890

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(d + e*x)^
(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c*
d^2 + a*e^2, 0] &&  !IntegerQ[p] && GtQ[a, 0] && GtQ[d, 0] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rule 82

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

Rule 125

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[m - n, 0] && GtQ[a, 0] && GtQ
[c, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(g x)^m \left (1-a^2 x^2\right )^p}{1+a x} \, dx &=\int (g x)^m (1-a x)^p (1+a x)^{-1+p} \, dx\\ &=-\frac{a \int (g x)^{1+m} (1-a x)^{-1+p} (1+a x)^{-1+p} \, dx}{g}+\int (g x)^m (1-a x)^{-1+p} (1+a x)^{-1+p} \, dx\\ &=-\frac{a \int (g x)^{1+m} \left (1-a^2 x^2\right )^{-1+p} \, dx}{g}+\int (g x)^m \left (1-a^2 x^2\right )^{-1+p} \, dx\\ &=\frac{(g x)^{1+m} \, _2F_1\left (\frac{1+m}{2},1-p;\frac{3+m}{2};a^2 x^2\right )}{g (1+m)}-\frac{a (g x)^{2+m} \, _2F_1\left (\frac{2+m}{2},1-p;\frac{4+m}{2};a^2 x^2\right )}{g^2 (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0456823, size = 77, normalized size = 0.87 \[ x (g x)^m \left (\frac{\, _2F_1\left (\frac{m+1}{2},1-p;\frac{m+3}{2};a^2 x^2\right )}{m+1}-\frac{a x \, _2F_1\left (\frac{m}{2}+1,1-p;\frac{m}{2}+2;a^2 x^2\right )}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((g*x)^m*(1 - a^2*x^2)^p)/(1 + a*x),x]

[Out]

x*(g*x)^m*(-((a*x*Hypergeometric2F1[1 + m/2, 1 - p, 2 + m/2, a^2*x^2])/(2 + m)) + Hypergeometric2F1[(1 + m)/2,
1 - p, (3 + m)/2, a^2*x^2]/(1 + m))

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Maple [F]  time = 0.674, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{p}}{ax+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(-a^2*x^2+1)^p/(a*x+1),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^p*(g*x)^m/(a*x + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-a^2*x^2 + 1)^p*(g*x)^m/(a*x + 1), x)

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Sympy [C]  time = 10.2144, size = 308, normalized size = 3.46 \begin{align*} \frac{0^{p} g^{m} m x^{m} \Phi \left (\frac{1}{a^{2} x^{2}}, 1, \frac{m e^{i \pi }}{2}\right ) \Gamma \left (- \frac{m}{2}\right )}{4 a \Gamma \left (1 - \frac{m}{2}\right )} - \frac{0^{p} g^{m} m x^{m} \Phi \left (\frac{1}{a^{2} x^{2}}, 1, \frac{1}{2} - \frac{m}{2}\right ) \Gamma \left (\frac{1}{2} - \frac{m}{2}\right )}{4 a^{2} x \Gamma \left (\frac{3}{2} - \frac{m}{2}\right )} + \frac{0^{p} g^{m} x^{m} \Phi \left (\frac{1}{a^{2} x^{2}}, 1, \frac{1}{2} - \frac{m}{2}\right ) \Gamma \left (\frac{1}{2} - \frac{m}{2}\right )}{4 a^{2} x \Gamma \left (\frac{3}{2} - \frac{m}{2}\right )} - \frac{a^{2 p} g^{m} p x^{m} x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- \frac{m}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, - \frac{m}{2} - p \\ - \frac{m}{2} - p + 1 \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 a \Gamma \left (p + 1\right ) \Gamma \left (- \frac{m}{2} - p + 1\right )} + \frac{a^{2 p} g^{m} p x^{m} x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (- \frac{m}{2} - p + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, - \frac{m}{2} - p + \frac{1}{2} \\ - \frac{m}{2} - p + \frac{3}{2} \end{matrix}\middle |{\frac{1}{a^{2} x^{2}}} \right )}}{2 a^{2} x \Gamma \left (p + 1\right ) \Gamma \left (- \frac{m}{2} - p + \frac{3}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)**m*(-a**2*x**2+1)**p/(a*x+1),x)

[Out]

0**p*g**m*m*x**m*lerchphi(1/(a**2*x**2), 1, m*exp_polar(I*pi)/2)*gamma(-m/2)/(4*a*gamma(1 - m/2)) - 0**p*g**m*
m*x**m*lerchphi(1/(a**2*x**2), 1, 1/2 - m/2)*gamma(1/2 - m/2)/(4*a**2*x*gamma(3/2 - m/2)) + 0**p*g**m*x**m*ler
chphi(1/(a**2*x**2), 1, 1/2 - m/2)*gamma(1/2 - m/2)/(4*a**2*x*gamma(3/2 - m/2)) - a**(2*p)*g**m*p*x**m*x**(2*p
)*exp(I*pi*p)*gamma(p)*gamma(-m/2 - p)*hyper((1 - p, -m/2 - p), (-m/2 - p + 1,), 1/(a**2*x**2))/(2*a*gamma(p +
1)*gamma(-m/2 - p + 1)) + a**(2*p)*g**m*p*x**m*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(-m/2 - p + 1/2)*hyper((1 -
p, -m/2 - p + 1/2), (-m/2 - p + 3/2,), 1/(a**2*x**2))/(2*a**2*x*gamma(p + 1)*gamma(-m/2 - p + 3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(-a^2*x^2+1)^p/(a*x+1),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^p*(g*x)^m/(a*x + 1), x)