3.342 \(\int (\sin (x)+\tan (x))^3 \, dx\)

Optimal. Leaf size=38 \[ \frac{\cos ^3(x)}{3}+\frac{3 \cos ^2(x)}{2}+2 \cos (x)+\frac{\sec ^2(x)}{2}+3 \sec (x)-2 \log (\cos (x)) \]

[Out]

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

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Rubi [A]  time = 0.049694, antiderivative size = 38, normalized size of antiderivative = 1., 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 = {4397, 2707, 75} \[ \frac{\cos ^3(x)}{3}+\frac{3 \cos ^2(x)}{2}+2 \cos (x)+\frac{\sec ^2(x)}{2}+3 \sec (x)-2 \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[(Sin[x] + Tan[x])^3,x]

[Out]

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

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(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]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int (\sin (x)+\tan (x))^3 \, dx &=\int (1+\cos (x))^3 \tan ^3(x) \, dx\\ &=-\operatorname{Subst}\left (\int \frac{(1-x) (1+x)^4}{x^3} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-2+\frac{1}{x^3}+\frac{3}{x^2}+\frac{2}{x}-3 x-x^2\right ) \, dx,x,\cos (x)\right )\\ &=2 \cos (x)+\frac{3 \cos ^2(x)}{2}+\frac{\cos ^3(x)}{3}-2 \log (\cos (x))+3 \sec (x)+\frac{\sec ^2(x)}{2}\\ \end{align*}

Mathematica [A]  time = 0.041343, size = 40, normalized size = 1.05 \[ \frac{9 \cos (x)}{4}+\frac{3}{4} \cos (2 x)+\frac{1}{12} \cos (3 x)+\frac{\sec ^2(x)}{2}+3 \sec (x)-2 \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[x] + Tan[x])^3,x]

[Out]

(9*Cos[x])/4 + (3*Cos[2*x])/4 + Cos[3*x]/12 - 2*Log[Cos[x]] + 3*Sec[x] + Sec[x]^2/2

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Maple [A]  time = 0.02, size = 39, normalized size = 1. \begin{align*}{\frac{ \left ( 16+8\, \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \cos \left ( x \right ) }{3}}-{\frac{3\, \left ( \sin \left ( x \right ) \right ) ^{2}}{2}}-2\,\ln \left ( \cos \left ( x \right ) \right ) +3\,{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{\cos \left ( x \right ) }}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.984777, size = 57, normalized size = 1.5 \begin{align*} \frac{1}{3} \, \cos \left (x\right )^{3} - \frac{3}{2} \, \sin \left (x\right )^{2} - \frac{1}{2 \,{\left (\sin \left (x\right )^{2} - 1\right )}} + \frac{3}{\cos \left (x\right )} + 2 \, \cos \left (x\right ) - \log \left (\sin \left (x\right )^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.23008, size = 151, normalized size = 3.97 \begin{align*} \frac{4 \, \cos \left (x\right )^{5} + 18 \, \cos \left (x\right )^{4} + 24 \, \cos \left (x\right )^{3} - 24 \, \cos \left (x\right )^{2} \log \left (-\cos \left (x\right )\right ) - 9 \, \cos \left (x\right )^{2} + 36 \, \cos \left (x\right ) + 6}{12 \, \cos \left (x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*cos(x)^5 + 18*cos(x)^4 + 24*cos(x)^3 - 24*cos(x)^2*log(-cos(x)) - 9*cos(x)^2 + 36*cos(x) + 6)/cos(x)^2

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Sympy [A]  time = 6.66531, size = 46, normalized size = 1.21 \begin{align*} - 3 \log{\left (\cos{\left (x \right )} \right )} - \frac{\log{\left (\sec ^{2}{\left (x \right )} \right )}}{2} + \frac{\cos ^{3}{\left (x \right )}}{3} + \frac{3 \cos ^{2}{\left (x \right )}}{2} + 2 \cos{\left (x \right )} + \frac{\sec ^{2}{\left (x \right )}}{2} + \frac{3}{\cos{\left (x \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(cos(x)) - log(sec(x)**2)/2 + cos(x)**3/3 + 3*cos(x)**2/2 + 2*cos(x) + sec(x)**2/2 + 3/cos(x)

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Giac [B]  time = 1.65285, size = 234, normalized size = 6.16 \begin{align*} \frac{\tan \left (\frac{1}{2} \, x\right )^{4} \tan \left (x\right )^{4} - 2 \, \log \left (\frac{4}{\tan \left (x\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, x\right )^{4} \tan \left (x\right )^{2} - 10 \, \tan \left (\frac{1}{2} \, x\right )^{4} \tan \left (x\right )^{2} - 2 \, \log \left (\frac{4}{\tan \left (x\right )^{2} + 1}\right ) \tan \left (\frac{1}{2} \, x\right )^{4} - 8 \, \tan \left (\frac{1}{2} \, x\right )^{4} - 3 \, \tan \left (\frac{1}{2} \, x\right )^{2} \tan \left (x\right )^{2} - \tan \left (x\right )^{4} + 2 \, \log \left (\frac{4}{\tan \left (x\right )^{2} + 1}\right ) \tan \left (x\right )^{2} - 3 \, \tan \left (\frac{1}{2} \, x\right )^{2} - 11 \, \tan \left (x\right )^{2} + 2 \, \log \left (\frac{4}{\tan \left (x\right )^{2} + 1}\right ) - 13}{2 \,{\left (\tan \left (\frac{1}{2} \, x\right )^{4} \tan \left (x\right )^{2} + \tan \left (\frac{1}{2} \, x\right )^{4} - \tan \left (x\right )^{2} - 1\right )}} + \frac{1}{12} \, \cos \left (3 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(tan(1/2*x)^4*tan(x)^4 - 2*log(4/(tan(x)^2 + 1))*tan(1/2*x)^4*tan(x)^2 - 10*tan(1/2*x)^4*tan(x)^2 - 2*log(
4/(tan(x)^2 + 1))*tan(1/2*x)^4 - 8*tan(1/2*x)^4 - 3*tan(1/2*x)^2*tan(x)^2 - tan(x)^4 + 2*log(4/(tan(x)^2 + 1))
*tan(x)^2 - 3*tan(1/2*x)^2 - 11*tan(x)^2 + 2*log(4/(tan(x)^2 + 1)) - 13)/(tan(1/2*x)^4*tan(x)^2 + tan(1/2*x)^4
- tan(x)^2 - 1) + 1/12*cos(3*x)