3.35 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=73 \[ \frac{4 a^4 \sin (c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{4 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^4 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{13 a^4 x}{2} \]

[Out]

(13*a^4*x)/2 + (4*a^4*ArcTanh[Sin[c + d*x]])/d + (4*a^4*Sin[c + d*x])/d + (a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d
) + (a^4*Tan[c + d*x])/d

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Rubi [A]  time = 0.078554, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3791, 2637, 2635, 8, 3770, 3767} \[ \frac{4 a^4 \sin (c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{4 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^4 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{13 a^4 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4,x]

[Out]

(13*a^4*x)/2 + (4*a^4*ArcTanh[Sin[c + d*x]])/d + (4*a^4*Sin[c + d*x])/d + (a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d
) + (a^4*Tan[c + d*x])/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]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \, dx &=\int \left (6 a^4+4 a^4 \cos (c+d x)+a^4 \cos ^2(c+d x)+4 a^4 \sec (c+d x)+a^4 \sec ^2(c+d x)\right ) \, dx\\ &=6 a^4 x+a^4 \int \cos ^2(c+d x) \, dx+a^4 \int \sec ^2(c+d x) \, dx+\left (4 a^4\right ) \int \cos (c+d x) \, dx+\left (4 a^4\right ) \int \sec (c+d x) \, dx\\ &=6 a^4 x+\frac{4 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a^4 \sin (c+d x)}{d}+\frac{a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{1}{2} a^4 \int 1 \, dx-\frac{a^4 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac{13 a^4 x}{2}+\frac{4 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{4 a^4 \sin (c+d x)}{d}+\frac{a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^4 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [B]  time = 1.77213, size = 241, normalized size = 3.3 \[ \frac{1}{64} a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \left (\frac{16 \sin (c) \cos (d x)}{d}+\frac{\sin (2 c) \cos (2 d x)}{d}+\frac{16 \cos (c) \sin (d x)}{d}+\frac{\cos (2 c) \sin (2 d x)}{d}+\frac{4 \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{16 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{16 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+26 x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4,x]

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(26*x - (16*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (16*Log
[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (16*Cos[d*x]*Sin[c])/d + (Cos[2*d*x]*Sin[2*c])/d + (16*Cos[c]*Sin[d
*x])/d + (Cos[2*c]*Sin[2*d*x])/d + (4*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)
/2])) + (4*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/64

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Maple [A]  time = 0.061, size = 86, normalized size = 1.2 \begin{align*}{\frac{{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}x}{2}}+{\frac{13\,{a}^{4}c}{2\,d}}+4\,{\frac{{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{4}\tan \left ( dx+c \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))^4,x)

[Out]

1/2*a^4*cos(d*x+c)*sin(d*x+c)/d+13/2*a^4*x+13/2/d*a^4*c+4*a^4*sin(d*x+c)/d+4/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+a
^4*tan(d*x+c)/d

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Maxima [A]  time = 1.06882, size = 115, normalized size = 1.58 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 24 \,{\left (d x + c\right )} a^{4} + 8 \, a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a^{4} \sin \left (d x + c\right ) + 4 \, a^{4} \tan \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))^4,x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 + 24*(d*x + c)*a^4 + 8*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c)
- 1)) + 16*a^4*sin(d*x + c) + 4*a^4*tan(d*x + c))/d

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Fricas [A]  time = 1.7898, size = 270, normalized size = 3.7 \begin{align*} \frac{13 \, a^{4} d x \cos \left (d x + c\right ) + 4 \, a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} \cos \left (d x + c\right ) + 2 \, a^{4}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \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))^4,x, algorithm="fricas")

[Out]

1/2*(13*a^4*d*x*cos(d*x + c) + 4*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 4*a^4*cos(d*x + c)*log(-sin(d*x + c)
+ 1) + (a^4*cos(d*x + c)^2 + 8*a^4*cos(d*x + c) + 2*a^4)*sin(d*x + c))/(d*cos(d*x + c))

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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))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.41339, size = 174, normalized size = 2.38 \begin{align*} \frac{13 \,{\left (d x + c\right )} a^{4} + 8 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{4 \, a^{4} \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} + \frac{2 \,{\left (7 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{4} \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))^4,x, algorithm="giac")

[Out]

1/2*(13*(d*x + c)*a^4 + 8*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
4*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(7*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*a^4*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d