3.274 \(\int (\sec (x)+\tan (x))^4 \, dx\)

Optimal. Leaf size=30 \[ x+\frac {2 \cos ^3(x)}{3 (1-\sin (x))^3}-\frac {2 \cos (x)}{1-\sin (x)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4391, 2670, 2680, 8} \[ x+\frac {2 \cos ^3(x)}{3 (1-\sin (x))^3}-\frac {2 \cos (x)}{1-\sin (x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x] + Tan[x])^4,x]

[Out]

x + (2*Cos[x]^3)/(3*(1 - Sin[x])^3) - (2*Cos[x])/(1 - Sin[x])

Rule 8

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

Rule 2670

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

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))^4 \, dx &=\int \sec ^4(x) (1+\sin (x))^4 \, dx\\ &=\int \frac {\cos ^4(x)}{(1-\sin (x))^4} \, dx\\ &=\frac {2 \cos ^3(x)}{3 (1-\sin (x))^3}-\int \frac {\cos ^2(x)}{(1-\sin (x))^2} \, dx\\ &=\frac {2 \cos ^3(x)}{3 (1-\sin (x))^3}-\frac {2 \cos (x)}{1-\sin (x)}+\int 1 \, dx\\ &=x+\frac {2 \cos ^3(x)}{3 (1-\sin (x))^3}-\frac {2 \cos (x)}{1-\sin (x)}\\ \end {align*}

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Mathematica [B]  time = 0.13, size = 64, normalized size = 2.13 \[ -\frac {-3 (3 x+8) \cos \left (\frac {x}{2}\right )+(3 x+16) \cos \left (\frac {3 x}{2}\right )+6 \sin \left (\frac {x}{2}\right ) (2 x+x \cos (x)+4)}{6 \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x] + Tan[x])^4,x]

[Out]

-1/6*(-3*(8 + 3*x)*Cos[x/2] + (16 + 3*x)*Cos[(3*x)/2] + 6*(4 + 2*x + x*Cos[x])*Sin[x/2])/(Cos[x/2] - Sin[x/2])
^3

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fricas [B]  time = 0.91, size = 61, normalized size = 2.03 \[ \frac {{\left (3 \, x + 8\right )} \cos \relax (x)^{2} - {\left (3 \, x - 4\right )} \cos \relax (x) + {\left ({\left (3 \, x - 8\right )} \cos \relax (x) + 6 \, x - 4\right )} \sin \relax (x) - 6 \, x - 4}{3 \, {\left (\cos \relax (x)^{2} + {\left (\cos \relax (x) + 2\right )} \sin \relax (x) - \cos \relax (x) - 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))^4,x, algorithm="fricas")

[Out]

1/3*((3*x + 8)*cos(x)^2 - (3*x - 4)*cos(x) + ((3*x - 8)*cos(x) + 6*x - 4)*sin(x) - 6*x - 4)/(cos(x)^2 + (cos(x
) + 2)*sin(x) - cos(x) - 2)

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giac [A]  time = 0.15, size = 20, normalized size = 0.67 \[ x - \frac {8 \, {\left (3 \, \tan \left (\frac {1}{2} \, x\right ) - 1\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right ) - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))^4,x, algorithm="giac")

[Out]

x - 8/3*(3*tan(1/2*x) - 1)/(tan(1/2*x) - 1)^3

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maple [B]  time = 0.10, size = 71, normalized size = 2.37 \[ -\left (-\frac {2}{3}-\frac {\left (\sec ^{2}\relax (x )\right )}{3}\right ) \tan \relax (x )+\frac {4}{3 \cos \relax (x )^{3}}+\frac {2 \left (\sin ^{3}\relax (x )\right )}{\cos \relax (x )^{3}}+\frac {4 \left (\sin ^{4}\relax (x )\right )}{3 \cos \relax (x )^{3}}-\frac {4 \left (\sin ^{4}\relax (x )\right )}{3 \cos \relax (x )}-\frac {4 \left (2+\sin ^{2}\relax (x )\right ) \cos \relax (x )}{3}+\frac {\left (\tan ^{3}\relax (x )\right )}{3}-\tan \relax (x )+x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sec(x)+tan(x))^4,x)

[Out]

-(-2/3-1/3*sec(x)^2)*tan(x)+4/3/cos(x)^3+2*sin(x)^3/cos(x)^3+4/3*sin(x)^4/cos(x)^3-4/3*sin(x)^4/cos(x)-4/3*(2+
sin(x)^2)*cos(x)+1/3*tan(x)^3-tan(x)+x

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maxima [A]  time = 0.44, size = 28, normalized size = 0.93 \[ \frac {8}{3} \, \tan \relax (x)^{3} + x - \frac {4 \, {\left (3 \, \cos \relax (x)^{2} - 1\right )}}{3 \, \cos \relax (x)^{3}} + \frac {4}{3 \, \cos \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))^4,x, algorithm="maxima")

[Out]

8/3*tan(x)^3 + x - 4/3*(3*cos(x)^2 - 1)/cos(x)^3 + 4/3/cos(x)^3

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mupad [B]  time = 2.37, size = 20, normalized size = 0.67 \[ x-\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {8}{3}}{{\left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x) + 1/cos(x))^4,x)

[Out]

x - (8*tan(x/2) - 8/3)/(tan(x/2) - 1)^3

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sympy [A]  time = 4.10, size = 44, normalized size = 1.47 \[ x + \frac {\sin ^{3}{\relax (x )}}{3 \cos ^{3}{\relax (x )}} - \frac {\sin {\relax (x )}}{\cos {\relax (x )}} + \frac {7 \tan ^{3}{\relax (x )}}{3} + \tan {\relax (x )} + \frac {8 \sec ^{3}{\relax (x )}}{3} - 4 \sec {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((sec(x)+tan(x))**4,x)

[Out]

x + sin(x)**3/(3*cos(x)**3) - sin(x)/cos(x) + 7*tan(x)**3/3 + tan(x) + 8*sec(x)**3/3 - 4*sec(x)

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