∫ √ ____ −5+x √ ___ 3+x_ (−1+x)(−25+x2) dx

Optimal. Leaf size=54 \[ \frac {1}{6} \tan ^{-1}\left (\frac {1}{4} \sqrt {x-5} \sqrt {x+3}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt {5} \sqrt {x+3}}{\sqrt {x-5}}\right )}{3 \sqrt {5}} \]

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Rubi [A] time = 0.10, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1586, 178, 92, 203, 93, 206} \[ \frac {1}{6} \tan ^{-1}\left (\frac {1}{4} \sqrt {x-5} \sqrt {x+3}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt {5} \sqrt {x+3}}{\sqrt {x-5}}\right )}{3 \sqrt {5}} \]

Int[(Sqrt[-5 + x]*Sqrt[3 + x])/((-1 + x)*(-25 + x^2)),x]

ArcTan[(Sqrt[-5 + x]*Sqrt[3 + x])/4]/6 + ArcTanh[(Sqrt[5]*Sqrt[3 + x])/Sqrt[-5 + x]]/(3*Sqrt[5])

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I

nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin

ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^

(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbo

l] :> Dist[(b*e - a*f)/(b*c - a*d), Int[((e + f*x)^(p - 1)*(g + h*x)^q)/(a + b*x), x], x] - Dist[(d*e - c*f)/(

b*c - a*d), Int[((e + f*x)^(p - 1)*(g + h*x)^q)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, q}, x] &&

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;

FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]

/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[

q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

\begin {align*} \int \frac {\sqrt {-5+x} \sqrt {3+x}}{(-1+x) \left (-25+x^2\right )} \, dx &=\int \frac {\sqrt {3+x}}{\sqrt {-5+x} (-1+x) (5+x)} \, dx\\ &=\frac {1}{3} \int \frac {1}{\sqrt {-5+x} \sqrt {3+x} (5+x)} \, dx+\frac {2}{3} \int \frac {1}{\sqrt {-5+x} (-1+x) \sqrt {3+x}} \, dx\\ &=\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{2-10 x^2} \, dx,x,\frac {\sqrt {3+x}}{\sqrt {-5+x}}\right )+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{16+x^2} \, dx,x,\sqrt {-5+x} \sqrt {3+x}\right )\\ &=\frac {1}{6} \tan ^{-1}\left (\frac {1}{4} \sqrt {-5+x} \sqrt {3+x}\right )+\frac {\tanh ^{-1}\left (\frac {\sqrt {5} \sqrt {3+x}}{\sqrt {-5+x}}\right )}{3 \sqrt {5}}\\ \end {align*}

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Mathematica [A] time = 0.15, size = 47, normalized size = 0.87 \[ \frac {1}{15} \left (5 \tan ^{-1}\left (\sqrt {\frac {x-5}{x+3}}\right )+\sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {\frac {x-5}{x+3}}}{\sqrt {5}}\right )\right ) \]

Integrate[(Sqrt[-5 + x]*Sqrt[3 + x])/((-1 + x)*(-25 + x^2)),x]

(5*ArcTan[Sqrt[(-5 + x)/(3 + x)]] + Sqrt[5]*ArcTanh[Sqrt[(-5 + x)/(3 + x)]/Sqrt[5]])/15

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IntegrateAlgebraic [B] time = 0.52, size = 121, normalized size = 2.24 \[ \frac {1}{6} \tan ^{-1}\left (\frac {1}{4} \sqrt {x-5} \sqrt {x+3}\right )-\frac {\tanh ^{-1}\left (\frac {-\frac {11 x}{\sqrt {5}}+\frac {\sqrt {x-5} \left (3 \sqrt {x+3}+2\right )}{\sqrt {5}}+6 \sqrt {5} \sqrt {x+3}-7 \sqrt {5}}{\sqrt {x-5} \left (2 \sqrt {x+3}-6\right )-2 \sqrt {x+3}+6}\right )}{3 \sqrt {5}} \]

IntegrateAlgebraic[(Sqrt[-5 + x]*Sqrt[3 + x])/((-1 + x)*(-25 + x^2)),x]

ArcTan[(Sqrt[-5 + x]*Sqrt[3 + x])/4]/6 - ArcTanh[(-7*Sqrt[5] - (11*x)/Sqrt[5] + 6*Sqrt[5]*Sqrt[3 + x] + (Sqrt[

-5 + x]*(2 + 3*Sqrt[3 + x]))/Sqrt[5])/(6 - 2*Sqrt[3 + x] + Sqrt[-5 + x]*(-6 + 2*Sqrt[3 + x]))]/(3*Sqrt[5])

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fricas [A] time = 0.90, size = 65, normalized size = 1.20 \[ \frac {1}{30} \, \sqrt {5} \log \left (\frac {\sqrt {x + 3} \sqrt {x - 5} {\left (3 \, \sqrt {5} + 5\right )} + \sqrt {5} {\left (3 \, x + 5\right )} + 9 \, x + 15}{x + 5}\right ) + \frac {1}{3} \, \arctan \left (\frac {1}{4} \, \sqrt {x + 3} \sqrt {x - 5} - \frac {1}{4} \, x + \frac {1}{4}\right ) \]

integrate((-5+x)^(1/2)*(3+x)^(1/2)/(-1+x)/(x^2-25),x, algorithm="fricas")

1/30*sqrt(5)*log((sqrt(x + 3)*sqrt(x - 5)*(3*sqrt(5) + 5) + sqrt(5)*(3*x + 5) + 9*x + 15)/(x + 5)) + 1/3*arcta

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giac [B] time = 0.96, size = 74, normalized size = 1.37 \[ -\frac {1}{30} \, \sqrt {5} \log \left (\frac {{\left (\sqrt {x + 3} - \sqrt {x - 5}\right )}^{2} - 4 \, \sqrt {5} + 12}{{\left (\sqrt {x + 3} - \sqrt {x - 5}\right )}^{2} + 4 \, \sqrt {5} + 12}\right ) - \frac {1}{3} \, \arctan \left (\frac {1}{8} \, {\left (\sqrt {x + 3} - \sqrt {x - 5}\right )}^{2}\right ) \]

integrate((-5+x)^(1/2)*(3+x)^(1/2)/(-1+x)/(x^2-25),x, algorithm="giac")

-1/30*sqrt(5)*log(((sqrt(x + 3) - sqrt(x - 5))^2 - 4*sqrt(5) + 12)/((sqrt(x + 3) - sqrt(x - 5))^2 + 4*sqrt(5)

+ 12)) - 1/3*arctan(1/8*(sqrt(x + 3) - sqrt(x - 5))^2)

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method	result	size

default	\(\frac {\sqrt {-5+x}\, \sqrt {3+x}\, \left (\sqrt {5}\, \arctanh \left (\frac {\left (5+3 x \right ) \sqrt {5}}{5 \sqrt {x^{2}-2 x -15}}\right )-5 \arctan \left (\frac {4}{\sqrt {x^{2}-2 x -15}}\right )\right )}{30 \sqrt {x^{2}-2 x -15}}\)	\(64\)

int((-5+x)^(1/2)*(3+x)^(1/2)/(-1+x)/(x^2-25),x,method=_RETURNVERBOSE)

1/30*(-5+x)^(1/2)*(3+x)^(1/2)*(5^(1/2)*arctanh(1/5*(5+3*x)*5^(1/2)/(x^2-2*x-15)^(1/2))-5*arctan(4/(x^2-2*x-15)

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x + 3} \sqrt {x - 5}}{{\left (x^{2} - 25\right )} {\left (x - 1\right )}}\,{d x} \]

integrate((-5+x)^(1/2)*(3+x)^(1/2)/(-1+x)/(x^2-25),x, algorithm="maxima")

integrate(sqrt(x + 3)*sqrt(x - 5)/((x^2 - 25)*(x - 1)), x)

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mupad [B] time = 0.61, size = 95, normalized size = 1.76 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {x+3}\,\sqrt {x-5}-2\,\sqrt {2}\,\sqrt {x-5}}{x-2\,\sqrt {2}\,\sqrt {x+3}+3}\right )}{3}-\frac {\sqrt {5}\,\mathrm {atanh}\left (-\frac {\sqrt {5}\,\sqrt {x+3}\,\sqrt {x-5}-2\,\sqrt {2}\,\sqrt {5}\,\sqrt {x-5}}{5\,x-10\,\sqrt {2}\,\sqrt {x+3}+15}\right )}{15} \]

int(((x + 3)^(1/2)*(x - 5)^(1/2))/((x^2 - 25)*(x - 1)),x)

atan(((x + 3)^(1/2)*(x - 5)^(1/2) - 2*2^(1/2)*(x - 5)^(1/2))/(x - 2*2^(1/2)*(x + 3)^(1/2) + 3))/3 - (5^(1/2)*a

tanh(-(5^(1/2)*(x + 3)^(1/2)*(x - 5)^(1/2) - 2*2^(1/2)*5^(1/2)*(x - 5)^(1/2))/(5*x - 10*2^(1/2)*(x + 3)^(1/2)

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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x + 3}}{\sqrt {x - 5} \left (x - 1\right ) \left (x + 5\right )}\, dx \]

integrate((-5+x)**(1/2)*(3+x)**(1/2)/(-1+x)/(x**2-25),x)

Integral(sqrt(x + 3)/(sqrt(x - 5)*(x - 1)*(x + 5)), x)

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3.220 \(\int \frac {\sqrt {-5+x} \sqrt {3+x}}{(-1+x) (-25+x^2)} \, dx\)