\(\int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx\) [36]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=6 a^4 x+\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {7 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \cos (c+d x) \sin (c+d x)}{d}-\frac {a^4 \sin ^3(c+d x)}{3 d} \]

[Out]

6*a^4*x+a^4*arctanh(sin(d*x+c))/d+7*a^4*sin(d*x+c)/d+2*a^4*cos(d*x+c)*sin(d*x+c)/d-1/3*a^4*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2836, 2717, 2715, 8, 2713, 3855} \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {a^4 \sin ^3(c+d x)}{3 d}+\frac {7 a^4 \sin (c+d x)}{d}+\frac {2 a^4 \sin (c+d x) \cos (c+d x)}{d}+6 a^4 x \]

[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 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2717

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

Rule 2836

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \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 \text {arctanh}(\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 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d} \\ & = 6 a^4 x+\frac {a^4 \text {arctanh}(\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] (verified)

Time = 0.75 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {a^4 \left (72 d x-12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+81 \sin (c+d x)+12 \sin (2 (c+d x))+\sin (3 (c+d x))\right )}{12 d} \]

[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)

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(\frac {a^{4} \left (72 d x +\sin \left (3 d x +3 c \right )+12 \sin \left (2 d x +2 c \right )+81 \sin \left (d x +c \right )+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )\right )}{12 d}\)	\(70\)
derivativedivides	\(\frac {\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} \sin \left (d x +c \right )+4 a^{4} \left (d x +c \right )+a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(93\)
default	\(\frac {\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 a^{4} \sin \left (d x +c \right )+4 a^{4} \left (d x +c \right )+a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(93\)
parts	\(\frac {a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {4 a^{4} \left (d x +c \right )}{d}+\frac {6 a^{4} \sin \left (d x +c \right )}{d}+\frac {4 a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(104\)
risch	\(6 a^{4} x -\frac {27 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {27 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{4} \sin \left (3 d x +3 c \right )}{12 d}+\frac {a^{4} \sin \left (2 d x +2 c \right )}{d}\)	\(118\)
norman	\(\frac {6 a^{4} x +\frac {18 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {130 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {106 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {10 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+24 a^{4} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+36 a^{4} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 a^{4} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a^{4} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(206\)

[In]

int((a+cos(d*x+c)*a)^4*sec(d*x+c),x,method=_RETURNVERBOSE)

[Out]

1/12*a^4*(72*d*x+sin(3*d*x+3*c)+12*sin(2*d*x+2*c)+81*sin(d*x+c)+12*ln(tan(1/2*d*x+1/2*c)+1)-12*ln(tan(1/2*d*x+
1/2*c)-1))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10 \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {36 \, a^{4} d x + 3 \, a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{4} \cos \left (d x + c\right ) + 20 \, a^{4}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[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

Sympy [F]

\[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=a^{4} \left (\int 4 \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sec {\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*cos(d*x+c))**4*sec(d*x+c),x)

[Out]

a**4*(Integral(4*cos(c + d*x)*sec(c + d*x), x) + Integral(6*cos(c + d*x)**2*sec(c + d*x), x) + Integral(4*cos(
c + d*x)**3*sec(c + d*x), x) + Integral(cos(c + d*x)**4*sec(c + d*x), x) + Integral(sec(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=-\frac {{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 12 \, {\left (d x + c\right )} a^{4} - 3 \, a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 18 \, a^{4} \sin \left (d x + c\right )}{3 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.59 \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=\frac {18 \, {\left (d x + c\right )} a^{4} + 3 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 38 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 14.00 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^4 \sec (c+d x) \, dx=6\,a^4\,x+\frac {20\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \]

[In]

int((a + a*cos(c + d*x))^4/cos(c + d*x),x)

[Out]

6*a^4*x + (20*a^4*sin(c + d*x))/(3*d) + (2*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (a^4*cos(c +
d*x)^2*sin(c + d*x))/(3*d) + (2*a^4*cos(c + d*x)*sin(c + d*x))/d