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

Optimal. Leaf size=73 \[ -\frac{a^4 \sin ^3(c+d x)}{3 d}+\frac{7 a^4 \sin (c+d x)}{d}+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a^4 \sin (c+d x) \cos (c+d x)}{d}+6 a^4 x \]

[Out]

6*a^4*x + (a^4*ArcTanh[Sin[c + d*x]])/d + (7*a^4*Sin[c + d*x])/d + (2*a^4*Cos[c + d*x]*Sin[c + d*x])/d - (a^4*
Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.080568, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2757, 2637, 2635, 8, 2633, 3770} \[ -\frac{a^4 \sin ^3(c+d x)}{3 d}+\frac{7 a^4 \sin (c+d x)}{d}+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 a^4 \sin (c+d x) \cos (c+d x)}{d}+6 a^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

6*a^4*x + (a^4*ArcTanh[Sin[c + d*x]])/d + (7*a^4*Sin[c + d*x])/d + (2*a^4*Cos[c + d*x]*Sin[c + d*x])/d - (a^4*
Sin[c + d*x]^3)/(3*d)

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, 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 2633

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

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 (a+a \cos (c+d x))^4 \sec (c+d x) \, dx &=\int \left (4 a^4+6 a^4 \cos (c+d x)+4 a^4 \cos ^2(c+d x)+a^4 \cos ^3(c+d x)+a^4 \sec (c+d x)\right ) \, dx\\ &=4 a^4 x+a^4 \int \cos ^3(c+d x) \, dx+a^4 \int \sec (c+d x) \, dx+\left (4 a^4\right ) \int \cos ^2(c+d x) \, dx+\left (6 a^4\right ) \int \cos (c+d x) \, dx\\ &=4 a^4 x+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{6 a^4 \sin (c+d x)}{d}+\frac{2 a^4 \cos (c+d x) \sin (c+d x)}{d}+\left (2 a^4\right ) \int 1 \, dx-\frac{a^4 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=6 a^4 x+\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{7 a^4 \sin (c+d x)}{d}+\frac{2 a^4 \cos (c+d x) \sin (c+d x)}{d}-\frac{a^4 \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.104197, size = 91, normalized size = 1.25 \[ \frac{a^4 \left (81 \sin (c+d x)+12 \sin (2 (c+d x))+\sin (3 (c+d x))-12 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+72 d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(72*d*x - 12*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 81*
Sin[c + d*x] + 12*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.066, size = 94, normalized size = 1.3 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{4}}{3\,d}}+{\frac{20\,{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+6\,{a}^{4}x+6\,{\frac{{a}^{4}c}{d}}+{\frac{{a}^{4}\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((a+cos(d*x+c)*a)^4*sec(d*x+c),x)

[Out]

1/3/d*sin(d*x+c)*cos(d*x+c)^2*a^4+20/3*a^4*sin(d*x+c)/d+2*a^4*cos(d*x+c)*sin(d*x+c)/d+6*a^4*x+6/d*a^4*c+1/d*a^
4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.14451, size = 120, normalized size = 1.64 \begin{align*} -\frac{{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 12 \,{\left (d x + c\right )} a^{4} - 3 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 18 \, a^{4} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c),x, algorithm="maxima")

[Out]

-1/3*((sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 - 12*(d*x + c)*a^4 - 3*a^
4*log(sec(d*x + c) + tan(d*x + c)) - 18*a^4*sin(d*x + c))/d

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Fricas [A]  time = 1.70575, size = 201, normalized size = 2.75 \begin{align*} \frac{36 \, a^{4} d x + 3 \, a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{4} \cos \left (d x + c\right ) + 20 \, a^{4}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c),x, algorithm="fricas")

[Out]

1/6*(36*a^4*d*x + 3*a^4*log(sin(d*x + c) + 1) - 3*a^4*log(-sin(d*x + c) + 1) + 2*(a^4*cos(d*x + c)^2 + 6*a^4*c
os(d*x + c) + 20*a^4)*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((a+a*cos(d*x+c))**4*sec(d*x+c),x)

[Out]

Timed out

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Giac [A]  time = 1.4796, size = 157, normalized size = 2.15 \begin{align*} \frac{18 \,{\left (d x + c\right )} a^{4} + 3 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 38 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, 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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c),x, algorithm="giac")

[Out]

1/3*(18*(d*x + c)*a^4 + 3*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) +
2*(15*a^4*tan(1/2*d*x + 1/2*c)^5 + 38*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 + 1)^3)/d