3.275 \(\int (\sec (x)+\tan (x))^3 \, dx\)

Optimal. Leaf size=18 \[ \frac {2}{1-\sin (x)}+\log (1-\sin (x)) \]

[Out]

ln(1-sin(x))+2/(1-sin(x))

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Rubi [A]  time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4391, 2667, 43} \[ \frac {2}{1-\sin (x)}+\log (1-\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x] + Tan[x])^3,x]

[Out]

Log[1 - Sin[x]] + 2/(1 - Sin[x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int (\sec (x)+\tan (x))^3 \, dx &=\int \sec ^3(x) (1+\sin (x))^3 \, dx\\ &=\operatorname {Subst}\left (\int \frac {1+x}{(1-x)^2} \, dx,x,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {2}{(-1+x)^2}+\frac {1}{-1+x}\right ) \, dx,x,\sin (x)\right )\\ &=\log (1-\sin (x))+\frac {2}{1-\sin (x)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.72 \[ \frac {\tan ^2(x)}{2}+\frac {3 \sec ^2(x)}{2}-\tanh ^{-1}(\sin (x))+\log (\cos (x))+2 \tan (x) \sec (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x] + Tan[x])^3,x]

[Out]

-ArcTanh[Sin[x]] + Log[Cos[x]] + (3*Sec[x]^2)/2 + 2*Sec[x]*Tan[x] + Tan[x]^2/2

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fricas [A]  time = 0.91, size = 21, normalized size = 1.17 \[ \frac {{\left (\sin \relax (x) - 1\right )} \log \left (-\sin \relax (x) + 1\right ) - 2}{\sin \relax (x) - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))^3,x, algorithm="fricas")

[Out]

((sin(x) - 1)*log(-sin(x) + 1) - 2)/(sin(x) - 1)

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giac [B]  time = 0.17, size = 48, normalized size = 2.67 \[ -\frac {3 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 10 \, \tan \left (\frac {1}{2} \, x\right ) + 3}{{\left (\tan \left (\frac {1}{2} \, x\right ) - 1\right )}^{2}} - \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + 2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))^3,x, algorithm="giac")

[Out]

-(3*tan(1/2*x)^2 - 10*tan(1/2*x) + 3)/(tan(1/2*x) - 1)^2 - log(tan(1/2*x)^2 + 1) + 2*log(abs(tan(1/2*x) - 1))

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maple [B]  time = 0.10, size = 45, normalized size = 2.50 \[ \frac {\sec \relax (x ) \tan \relax (x )}{2}-\ln \left (\sec \relax (x )+\tan \relax (x )\right )+\frac {3}{2 \cos \relax (x )^{2}}+\frac {3 \left (\sin ^{3}\relax (x )\right )}{2 \cos \relax (x )^{2}}+\frac {3 \sin \relax (x )}{2}+\frac {\left (\tan ^{2}\relax (x )\right )}{2}+\ln \left (\cos \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sec(x)+tan(x))^3,x)

[Out]

1/2*sec(x)*tan(x)-ln(sec(x)+tan(x))+3/2/cos(x)^2+3/2*sin(x)^3/cos(x)^2+3/2*sin(x)+1/2*tan(x)^2+ln(cos(x))

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maxima [B]  time = 0.33, size = 52, normalized size = 2.89 \[ \frac {3}{2} \, \tan \relax (x)^{2} - \frac {2 \, \sin \relax (x)}{\sin \relax (x)^{2} - 1} - \frac {1}{2 \, {\left (\sin \relax (x)^{2} - 1\right )}} + \frac {1}{2} \, \log \left (\sin \relax (x)^{2} - 1\right ) - \frac {1}{2} \, \log \left (\sin \relax (x) + 1\right ) + \frac {1}{2} \, \log \left (\sin \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))^3,x, algorithm="maxima")

[Out]

3/2*tan(x)^2 - 2*sin(x)/(sin(x)^2 - 1) - 1/2/(sin(x)^2 - 1) + 1/2*log(sin(x)^2 - 1) - 1/2*log(sin(x) + 1) + 1/
2*log(sin(x) - 1)

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mupad [B]  time = 2.40, size = 43, normalized size = 2.39 \[ 2\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )-\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )+\frac {4\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\mathrm {tan}\left (\frac {x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {x}{2}\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x) + 1/cos(x))^3,x)

[Out]

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

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sympy [B]  time = 4.81, size = 44, normalized size = 2.44 \[ \frac {\log {\left (\sin {\relax (x )} - 1 \right )}}{2} - \frac {\log {\left (\sin {\relax (x )} + 1 \right )}}{2} - \frac {\log {\left (\sec ^{2}{\relax (x )} \right )}}{2} + 2 \sec ^{2}{\relax (x )} - \frac {4 \sin {\relax (x )}}{2 \sin ^{2}{\relax (x )} - 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))**3,x)

[Out]

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

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