3.25 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=59 \[ \frac{3 a^3 \sin (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{7 a^3 x}{2} \]

[Out]

(7*a^3*x)/2 + (a^3*ArcTanh[Sin[c + d*x]])/d + (3*a^3*Sin[c + d*x])/d + (a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.066875, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3791, 2637, 2635, 8, 3770} \[ \frac{3 a^3 \sin (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{7 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3,x]

[Out]

(7*a^3*x)/2 + (a^3*ArcTanh[Sin[c + d*x]])/d + (3*a^3*Sin[c + d*x])/d + (a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

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 ^2(c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (3 a^3+3 a^3 \cos (c+d x)+a^3 \cos ^2(c+d x)+a^3 \sec (c+d x)\right ) \, dx\\ &=3 a^3 x+a^3 \int \cos ^2(c+d x) \, dx+a^3 \int \sec (c+d x) \, dx+\left (3 a^3\right ) \int \cos (c+d x) \, dx\\ &=3 a^3 x+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a^3 \sin (c+d x)}{d}+\frac{a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{1}{2} a^3 \int 1 \, dx\\ &=\frac{7 a^3 x}{2}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a^3 \sin (c+d x)}{d}+\frac{a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0700981, size = 81, normalized size = 1.37 \[ \frac{a^3 \left (12 \sin (c+d x)+\sin (2 (c+d x))-4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+14 d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3,x]

[Out]

(a^3*(14*d*x - 4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 12*Si
n[c + d*x] + Sin[2*(c + d*x)]))/(4*d)

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Maple [A]  time = 0.06, size = 72, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}x}{2}}+{\frac{7\,{a}^{3}c}{2\,d}}+3\,{\frac{{a}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^3,x)

[Out]

1/2*a^3*cos(d*x+c)*sin(d*x+c)/d+7/2*a^3*x+7/2/d*a^3*c+3*a^3*sin(d*x+c)/d+1/d*a^3*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.10061, size = 100, normalized size = 1.69 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} + 12 \,{\left (d x + c\right )} a^{3} + 2 \, a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 + 12*(d*x + c)*a^3 + 2*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c)
- 1)) + 12*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.75841, size = 159, normalized size = 2.69 \begin{align*} \frac{7 \, a^{3} d x + a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (a^{3} \cos \left (d x + c\right ) + 6 \, a^{3}\right )} \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)^2*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(7*a^3*d*x + a^3*log(sin(d*x + c) + 1) - a^3*log(-sin(d*x + c) + 1) + (a^3*cos(d*x + c) + 6*a^3)*sin(d*x +
c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.39002, size = 135, normalized size = 2.29 \begin{align*} \frac{7 \,{\left (d x + c\right )} a^{3} + 2 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(7*(d*x + c)*a^3 + 2*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2
*(5*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d