3.772 \(\int x \tan (1+x^2) \, dx\)

Optimal. Leaf size=11 \[ -\frac {1}{2} \log \left (\cos \left (x^2+1\right )\right ) \]

[Out]

-1/2*ln(cos(x^2+1))

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3747, 3475} \[ -\frac {1}{2} \log \left (\cos \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Tan[1 + x^2],x]

[Out]

-Log[Cos[1 + x^2]]/2

Rule 3475

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

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[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 \tan \left (1+x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \tan (1+x) \, dx,x,x^2\right )\\ &=-\frac {1}{2} \log \left (\cos \left (1+x^2\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 11, normalized size = 1.00 \[ -\frac {1}{2} \log \left (\cos \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Tan[1 + x^2],x]

[Out]

-1/2*Log[Cos[1 + x^2]]

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fricas [A]  time = 0.79, size = 15, normalized size = 1.36 \[ -\frac {1}{4} \, \log \left (\frac {1}{\tan \left (x^{2} + 1\right )^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(x^2+1),x, algorithm="fricas")

[Out]

-1/4*log(1/(tan(x^2 + 1)^2 + 1))

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giac [A]  time = 0.17, size = 10, normalized size = 0.91 \[ -\frac {1}{2} \, \log \left ({\left | \cos \left (x^{2} + 1\right ) \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(x^2+1),x, algorithm="giac")

[Out]

-1/2*log(abs(cos(x^2 + 1)))

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maple [A]  time = 0.00, size = 10, normalized size = 0.91 \[ -\frac {\ln \left (\cos \left (x^{2}+1\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tan(x^2+1),x)

[Out]

-1/2*ln(cos(x^2+1))

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maxima [A]  time = 0.33, size = 9, normalized size = 0.82 \[ \frac {1}{2} \, \log \left (\sec \left (x^{2} + 1\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(x^2+1),x, algorithm="maxima")

[Out]

1/2*log(sec(x^2 + 1))

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mupad [B]  time = 0.28, size = 13, normalized size = 1.18 \[ \frac {\ln \left ({\mathrm {tan}\left (x^2+1\right )}^2+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*tan(x^2 + 1),x)

[Out]

log(tan(x^2 + 1)^2 + 1)/4

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sympy [A]  time = 0.12, size = 12, normalized size = 1.09 \[ \frac {\log {\left (\tan ^{2}{\left (x^{2} + 1 \right )} + 1 \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*tan(x**2+1),x)

[Out]

log(tan(x**2 + 1)**2 + 1)/4

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