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

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Rubi [A]  time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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])

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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-a^{2} x^{2} + 1\right )}^{p} \left (g x\right )^{m}}{a x + 1}, x\right ) \]

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

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)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x \right )^{m} \left (-a^{2} x^{2}+1\right )^{p}}{a x +1}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{p} \left (g x\right )^{m}}{a x + 1}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (g\,x\right )}^m\,{\left (1-a^2\,x^2\right )}^p}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.81, size = 308, normalized size = 3.46 \[ \frac {0^{p} g^{m} m x^{m} \Phi \left (\frac {1}{a^{2} x^{2}}, 1, \frac {m e^{i \pi }}{2}\right ) \Gamma \left (- \frac {m}{2}\right )}{4 a \Gamma \left (1 - \frac {m}{2}\right )} - \frac {0^{p} g^{m} m x^{m} \Phi \left (\frac {1}{a^{2} x^{2}}, 1, \frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {1}{2} - \frac {m}{2}\right )}{4 a^{2} x \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )} + \frac {0^{p} g^{m} x^{m} \Phi \left (\frac {1}{a^{2} x^{2}}, 1, \frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {1}{2} - \frac {m}{2}\right )}{4 a^{2} x \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )} - \frac {a^{2 p} g^{m} p x^{m} x^{2 p} e^{i \pi p} \Gamma \relax (p) \Gamma \left (- \frac {m}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, - \frac {m}{2} - p \\ - \frac {m}{2} - p + 1 \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 a \Gamma \left (p + 1\right ) \Gamma \left (- \frac {m}{2} - p + 1\right )} + \frac {a^{2 p} g^{m} p x^{m} x^{2 p} e^{i \pi p} \Gamma \relax (p) \Gamma \left (- \frac {m}{2} - p + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} 1 - p, - \frac {m}{2} - p + \frac {1}{2} \\ - \frac {m}{2} - p + \frac {3}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 a^{2} x \Gamma \left (p + 1\right ) \Gamma \left (- \frac {m}{2} - p + \frac {3}{2}\right )} \]

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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