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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 8, normalized size of antiderivative = 1.00, 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 3475

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

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]

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*}

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Mathematica [A]  time = 0.01, size = 8, normalized size = 1.00 \[ t \tan (t)+\log (\cos (t)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Cos[t]] + t*Tan[t]

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fricas [B]  time = 0.44, size = 18, normalized size = 2.25 \[ \frac {\cos \relax (t) \log \left (-\cos \relax (t)\right ) + t \sin \relax (t)}{\cos \relax (t)} \]

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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giac [B]  time = 0.93, size = 103, normalized size = 12.88 \[ \frac {\log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, t\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, t\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, t\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, t\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, t\right )^{2} - 4 \, t \tan \left (\frac {1}{2} \, t\right ) - \log \left (\frac {4 \, {\left (\tan \left (\frac {1}{2} \, t\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, t\right )^{2} + 1\right )}}{\tan \left (\frac {1}{2} \, t\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, t\right )^{2} + 1}\right )}{2 \, {\left (\tan \left (\frac {1}{2} \, t\right )^{2} - 1\right )}} \]

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)

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maple [A]  time = 0.02, size = 9, normalized size = 1.12 \[ t \tan \relax (t )+\ln \left (\cos \relax (t )\right ) \]

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 = 0.97, size = 74, normalized size = 9.25 \[ \frac {{\left (\cos \left (2 \, t\right )^{2} + \sin \left (2 \, t\right )^{2} + 2 \, \cos \left (2 \, t\right ) + 1\right )} \log \left (\cos \left (2 \, t\right )^{2} + \sin \left (2 \, t\right )^{2} + 2 \, \cos \left (2 \, t\right ) + 1\right ) + 4 \, t \sin \left (2 \, t\right )}{2 \, {\left (\cos \left (2 \, t\right )^{2} + \sin \left (2 \, t\right )^{2} + 2 \, \cos \left (2 \, t\right ) + 1\right )}} \]

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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mupad [B]  time = 0.02, size = 8, normalized size = 1.00 \[ \ln \left (\cos \relax (t)\right )+t\,\mathrm {tan}\relax (t) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(t/cos(t)^2,t)

[Out]

log(cos(t)) + t*tan(t)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int t \sec ^{2}{\relax (t )}\, dt \]

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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