3.29 \(\int t \sec ^2(t) \, dt\)

Optimal. Leaf size=8 \[ t \tan (t)+\log (\cos (t)) \]

[Out]

Log[Cos[t]] + t*Tan[t]

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Rubi [A]  time = 0.0180556, antiderivative size = 8, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4184, 3475} \[ t \tan (t)+\log (\cos (t)) \]

Antiderivative was successfully verified.

[In]

Int[t*Sec[t]^2,t]

[Out]

Log[Cos[t]] + t*Tan[t]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

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

Rubi steps

\begin{align*} \int t \sec ^2(t) \, dt &=t \tan (t)-\int \tan (t) \, dt\\ &=\log (\cos (t))+t \tan (t)\\ \end{align*}

Mathematica [A]  time = 0.004618, size = 8, normalized size = 1. \[ t \tan (t)+\log (\cos (t)) \]

Antiderivative was successfully verified.

[In]

Integrate[t*Sec[t]^2,t]

[Out]

Log[Cos[t]] + t*Tan[t]

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Maple [A]  time = 0.004, size = 9, normalized size = 1.1 \begin{align*} \ln \left ( \cos \left ( t \right ) \right ) +t\tan \left ( t \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(t*sec(t)^2,t)

[Out]

ln(cos(t))+t*tan(t)

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Maxima [B]  time = 1.42639, size = 100, normalized size = 12.5 \begin{align*} \frac{{\left (\cos \left (2 \, t\right )^{2} + \sin \left (2 \, t\right )^{2} + 2 \, \cos \left (2 \, t\right ) + 1\right )} \log \left (\cos \left (2 \, t\right )^{2} + \sin \left (2 \, t\right )^{2} + 2 \, \cos \left (2 \, t\right ) + 1\right ) + 4 \, t \sin \left (2 \, t\right )}{2 \,{\left (\cos \left (2 \, t\right )^{2} + \sin \left (2 \, t\right )^{2} + 2 \, \cos \left (2 \, t\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((cos(2*t)^2 + sin(2*t)^2 + 2*cos(2*t) + 1)*log(cos(2*t)^2 + sin(2*t)^2 + 2*cos(2*t) + 1) + 4*t*sin(2*t))/
(cos(2*t)^2 + sin(2*t)^2 + 2*cos(2*t) + 1)

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Fricas [B]  time = 2.08662, size = 55, normalized size = 6.88 \begin{align*} \frac{\cos \left (t\right ) \log \left (-\cos \left (t\right )\right ) + t \sin \left (t\right )}{\cos \left (t\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(t)*log(-cos(t)) + t*sin(t))/cos(t)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int t \sec ^{2}{\left (t \right )}\, dt \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(t*sec(t)**2,t)

[Out]

Integral(t*sec(t)**2, t)

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Giac [B]  time = 1.10245, size = 139, normalized size = 17.38 \begin{align*} \frac{\log \left (\frac{4 \,{\left (\tan \left (\frac{1}{2} \, t\right )^{4} - 2 \, \tan \left (\frac{1}{2} \, t\right )^{2} + 1\right )}}{\tan \left (\frac{1}{2} \, t\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, t\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, t\right )^{2} - 4 \, t \tan \left (\frac{1}{2} \, t\right ) - \log \left (\frac{4 \,{\left (\tan \left (\frac{1}{2} \, t\right )^{4} - 2 \, \tan \left (\frac{1}{2} \, t\right )^{2} + 1\right )}}{\tan \left (\frac{1}{2} \, t\right )^{4} + 2 \, \tan \left (\frac{1}{2} \, t\right )^{2} + 1}\right )}{2 \,{\left (\tan \left (\frac{1}{2} \, t\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(4*(tan(1/2*t)^4 - 2*tan(1/2*t)^2 + 1)/(tan(1/2*t)^4 + 2*tan(1/2*t)^2 + 1))*tan(1/2*t)^2 - 4*t*tan(1/2
*t) - log(4*(tan(1/2*t)^4 - 2*tan(1/2*t)^2 + 1)/(tan(1/2*t)^4 + 2*tan(1/2*t)^2 + 1)))/(tan(1/2*t)^2 - 1)