3.15 \(\int \cos (c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=34 \[ \frac{a^2 \sin (c+d x)}{d}+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+2 a^2 x \]

[Out]

2*a^2*x + (a^2*ArcTanh[Sin[c + d*x]])/d + (a^2*Sin[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.0508976, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3788, 8, 4045, 3770} \[ \frac{a^2 \sin (c+d x)}{d}+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^2,x]

[Out]

2*a^2*x + (a^2*ArcTanh[Sin[c + d*x]])/d + (a^2*Sin[c + d*x])/d

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 8

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 3770

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

Rubi steps

\begin{align*} \int \cos (c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int 1 \, dx+\int \cos (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=2 a^2 x+\frac{a^2 \sin (c+d x)}{d}+a^2 \int \sec (c+d x) \, dx\\ &=2 a^2 x+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.0132871, size = 47, normalized size = 1.38 \[ \frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \sin (c) \cos (d x)}{d}+\frac{a^2 \cos (c) \sin (d x)}{d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Sec[c + d*x])^2,x]

[Out]

2*a^2*x + (a^2*ArcTanh[Sin[c + d*x]])/d + (a^2*Cos[d*x]*Sin[c])/d + (a^2*Cos[c]*Sin[d*x])/d

________________________________________________________________________________________

Maple [A]  time = 0.055, size = 51, normalized size = 1.5 \begin{align*} 2\,{a}^{2}x+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*sec(d*x+c))^2,x)

[Out]

2*a^2*x+1/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+a^2*sin(d*x+c)/d+2/d*a^2*c

________________________________________________________________________________________

Maxima [A]  time = 1.15053, size = 70, normalized size = 2.06 \begin{align*} \frac{4 \,{\left (d x + c\right )} a^{2} + a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*(4*(d*x + c)*a^2 + a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*a^2*sin(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.70983, size = 131, normalized size = 3.85 \begin{align*} \frac{4 \, a^{2} d x + a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(4*a^2*d*x + a^2*log(sin(d*x + c) + 1) - a^2*log(-sin(d*x + c) + 1) + 2*a^2*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \cos{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(2*cos(c + d*x)*sec(c + d*x), x) + Integral(cos(c + d*x)*sec(c + d*x)**2, x) + Integral(cos(c +
d*x), x))

________________________________________________________________________________________

Giac [B]  time = 1.3963, size = 107, normalized size = 3.15 \begin{align*} \frac{2 \,{\left (d x + c\right )} a^{2} + a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

(2*(d*x + c)*a^2 + a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*a^2*tan
(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1))/d