3.825 \(\int x \sec (5-x^2) \, dx\)

Optimal. Leaf size=13 \[ -\frac {1}{2} \tanh ^{-1}\left (\sin \left (5-x^2\right )\right ) \]

[Out]

1/2*arctanh(sin(x^2-5))

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4204, 3770} \[ -\frac {1}{2} \tanh ^{-1}\left (\sin \left (5-x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sec[5 - x^2],x]

[Out]

-ArcTanh[Sin[5 - x^2]]/2

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int x \sec \left (5-x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \sec (5-x) \, dx,x,x^2\right )\\ &=-\frac {1}{2} \tanh ^{-1}\left (\sin \left (5-x^2\right )\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 13, normalized size = 1.00 \[ -\frac {1}{2} \tanh ^{-1}\left (\sin \left (5-x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sec[5 - x^2],x]

[Out]

-1/2*ArcTanh[Sin[5 - x^2]]

________________________________________________________________________________________

fricas [B]  time = 0.69, size = 25, normalized size = 1.92 \[ \frac {1}{4} \, \log \left (\sin \left (x^{2} - 5\right ) + 1\right ) - \frac {1}{4} \, \log \left (-\sin \left (x^{2} - 5\right ) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(x^2-5),x, algorithm="fricas")

[Out]

1/4*log(sin(x^2 - 5) + 1) - 1/4*log(-sin(x^2 - 5) + 1)

________________________________________________________________________________________

giac [B]  time = 0.16, size = 41, normalized size = 3.15 \[ \frac {1}{8} \, \log \left ({\left | \frac {1}{\sin \left (x^{2} - 5\right )} + \sin \left (x^{2} - 5\right ) + 2 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | \frac {1}{\sin \left (x^{2} - 5\right )} + \sin \left (x^{2} - 5\right ) - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(x^2-5),x, algorithm="giac")

[Out]

1/8*log(abs(1/sin(x^2 - 5) + sin(x^2 - 5) + 2)) - 1/8*log(abs(1/sin(x^2 - 5) + sin(x^2 - 5) - 2))

________________________________________________________________________________________

maple [A]  time = 0.00, size = 17, normalized size = 1.31 \[ \frac {\ln \left (\sec \left (x^{2}-5\right )+\tan \left (x^{2}-5\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sec(x^2-5),x)

[Out]

1/2*ln(sec(x^2-5)+tan(x^2-5))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 16, normalized size = 1.23 \[ \frac {1}{2} \, \log \left (\sec \left (x^{2} - 5\right ) + \tan \left (x^{2} - 5\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(x^2-5),x, algorithm="maxima")

[Out]

1/2*log(sec(x^2 - 5) + tan(x^2 - 5))

________________________________________________________________________________________

mupad [B]  time = 3.53, size = 15, normalized size = 1.15 \[ -\mathrm {atan}\left ({\mathrm {e}}^{-5{}\mathrm {i}}\,{\mathrm {e}}^{x^2\,1{}\mathrm {i}}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/cos(x^2 - 5),x)

[Out]

-atan(exp(-5i)*exp(x^2*1i))*1i

________________________________________________________________________________________

sympy [A]  time = 1.03, size = 15, normalized size = 1.15 \[ \frac {\log {\left (\tan {\left (x^{2} - 5 \right )} + \sec {\left (x^{2} - 5 \right )} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(x**2-5),x)

[Out]

log(tan(x**2 - 5) + sec(x**2 - 5))/2

________________________________________________________________________________________