3.4.86 \(\int \frac {x^m}{(1-\frac {\sqrt {a} x}{\sqrt {-b}})^2 (1+\frac {\sqrt {a} x}{\sqrt {-b}})^2} \, dx\) [386]

Optimal. Leaf size=36 \[ \frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {a x^2}{b}\right )}{1+m} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {74, 371} \begin {gather*} \frac {x^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {a x^2}{b}\right )}{m+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 74

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

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

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Mathematica [A]
time = 0.01, size = 38, normalized size = 1.06 \begin {gather*} \frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{2};1+\frac {1+m}{2};-\frac {a x^2}{b}\right )}{1+m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(b^2*x^m/(a^2*x^4 + 2*a*b*x^2 + b^2), x)

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Sympy [C] Result contains complex when optimal does not.
time = 3.62, size = 541, normalized size = 15.03 \begin {gather*} \frac {a b^{2} m^{2} x^{m} \Phi \left (\frac {b e^{i \pi }}{a x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} - \frac {4 a b^{2} m x^{m} \Phi \left (\frac {b e^{i \pi }}{a x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} + \frac {2 a b^{2} m x^{m} \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} + \frac {3 a b^{2} x^{m} \Phi \left (\frac {b e^{i \pi }}{a x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} - \frac {6 a b^{2} x^{m} \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} + \frac {b^{3} m^{2} x^{m} \Phi \left (\frac {b e^{i \pi }}{a x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x^{3} \cdot \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} - \frac {4 b^{3} m x^{m} \Phi \left (\frac {b e^{i \pi }}{a x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x^{3} \cdot \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} + \frac {3 b^{3} x^{m} \Phi \left (\frac {b e^{i \pi }}{a x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {3}{2} - \frac {m}{2}\right )}{x^{3} \cdot \left (8 a^{3} x^{2} \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) + 8 a^{2} b \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b**2*m**2*x**m*lerchphi(b*exp_polar(I*pi)/(a*x**2), 1, 3/2 - m/2)*gamma(3/2 - m/2)/(x*(8*a**3*x**2*gamma(5/2
- m/2) + 8*a**2*b*gamma(5/2 - m/2))) - 4*a*b**2*m*x**m*lerchphi(b*exp_polar(I*pi)/(a*x**2), 1, 3/2 - m/2)*gam
ma(3/2 - m/2)/(x*(8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*gamma(5/2 - m/2))) + 2*a*b**2*m*x**m*gamma(3/2 - m/2
)/(x*(8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*gamma(5/2 - m/2))) + 3*a*b**2*x**m*lerchphi(b*exp_polar(I*pi)/(a
*x**2), 1, 3/2 - m/2)*gamma(3/2 - m/2)/(x*(8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*gamma(5/2 - m/2))) - 6*a*b*
*2*x**m*gamma(3/2 - m/2)/(x*(8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*gamma(5/2 - m/2))) + b**3*m**2*x**m*lerch
phi(b*exp_polar(I*pi)/(a*x**2), 1, 3/2 - m/2)*gamma(3/2 - m/2)/(x**3*(8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*
gamma(5/2 - m/2))) - 4*b**3*m*x**m*lerchphi(b*exp_polar(I*pi)/(a*x**2), 1, 3/2 - m/2)*gamma(3/2 - m/2)/(x**3*(
8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*gamma(5/2 - m/2))) + 3*b**3*x**m*lerchphi(b*exp_polar(I*pi)/(a*x**2),
1, 3/2 - m/2)*gamma(3/2 - m/2)/(x**3*(8*a**3*x**2*gamma(5/2 - m/2) + 8*a**2*b*gamma(5/2 - m/2)))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{1,[1]%%%},[2]%%%}+%%%{%%{[%%%{%%{[-2,0]:[1,0,%%%{1,[
1]%%%}]%%},

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^m}{{\left (\frac {\sqrt {a}\,x}{\sqrt {-b}}-1\right )}^2\,{\left (\frac {\sqrt {a}\,x}{\sqrt {-b}}+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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