3.770 \(\int x \sec (x^2) \tan (x^2) \, dx\)

Optimal. Leaf size=8 \[ \frac {\sec \left (x^2\right )}{2} \]

[Out]

1/2*sec(x^2)

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Rubi [A]  time = 0.06, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6715, 2606, 8} \[ \frac {\sec \left (x^2\right )}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sec[x^2]*Tan[x^2],x]

[Out]

Sec[x^2]/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 8, normalized size = 1.00 \[ \frac {\sec \left (x^2\right )}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sec[x^2]*Tan[x^2],x]

[Out]

Sec[x^2]/2

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fricas [A]  time = 0.77, size = 8, normalized size = 1.00 \[ \frac {1}{2 \, \cos \left (x^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/cos(x^2)

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giac [A]  time = 0.12, size = 8, normalized size = 1.00 \[ \frac {1}{2 \, \cos \left (x^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/cos(x^2)

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maple [A]  time = 0.03, size = 7, normalized size = 0.88 \[ \frac {\sec \left (x^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sec(x^2)*tan(x^2),x)

[Out]

1/2*sec(x^2)

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maxima [B]  time = 0.33, size = 56, normalized size = 7.00 \[ \frac {\cos \left (2 \, x^{2}\right ) \cos \left (x^{2}\right ) + \sin \left (2 \, x^{2}\right ) \sin \left (x^{2}\right ) + \cos \left (x^{2}\right )}{\cos \left (2 \, x^{2}\right )^{2} + \sin \left (2 \, x^{2}\right )^{2} + 2 \, \cos \left (2 \, x^{2}\right ) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(2*x^2)*cos(x^2) + sin(2*x^2)*sin(x^2) + cos(x^2))/(cos(2*x^2)^2 + sin(2*x^2)^2 + 2*cos(2*x^2) + 1)

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mupad [B]  time = 0.07, size = 8, normalized size = 1.00 \[ \frac {1}{2\,\cos \left (x^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*tan(x^2))/cos(x^2),x)

[Out]

1/(2*cos(x^2))

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sympy [A]  time = 0.28, size = 5, normalized size = 0.62 \[ \frac {\sec {\left (x^{2} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sec(x**2)*tan(x**2),x)

[Out]

sec(x**2)/2

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