\(\int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx\) [11]

Optimal result

Integrand size = 19, antiderivative size = 85 \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x)}{d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^3(c+d x)}{3 d} \]

[Out]

3/8*a*arctanh(sin(d*x+c))/d+a*tan(d*x+c)/d+3/8*a*sec(d*x+c)*tan(d*x+c)/d+1/4*a*sec(d*x+c)^3*tan(d*x+c)/d+1/3*a
*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2827, 3853, 3855, 3852} \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

[In]

Int[(a + a*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + (a*Tan[c + d*x])/d + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Sec[c + d*
x]^3*Tan[c + d*x])/(4*d) + (a*Tan[c + d*x]^3)/(3*d)

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3852

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]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*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}& = a \int \sec ^4(c+d x) \, dx+a \int \sec ^5(c+d x) \, dx \\ & = \frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} (3 a) \int \sec ^3(c+d x) \, dx-\frac {a \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {a \tan (c+d x)}{d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^3(c+d x)}{3 d}+\frac {1}{8} (3 a) \int \sec (c+d x) \, dx \\ & = \frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x)}{d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a \left (9 \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (9 \sec (c+d x)+6 \sec ^3(c+d x)+8 \left (3+\tan ^2(c+d x)\right )\right )\right )}{24 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(a*(9*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(9*Sec[c + d*x] + 6*Sec[c + d*x]^3 + 8*(3 + Tan[c + d*x]^2))))/(24*
d)

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(73\)
default	\(\frac {-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(73\)
parts	\(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(75\)
risch	\(-\frac {i a \left (9 \,{\mathrm e}^{7 i \left (d x +c \right )}+33 \,{\mathrm e}^{5 i \left (d x +c \right )}-48 \,{\mathrm e}^{4 i \left (d x +c \right )}-33 \,{\mathrm e}^{3 i \left (d x +c \right )}-64 \,{\mathrm e}^{2 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}-16\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\)	\(128\)
parallelrisch	\(\frac {a \left (9 \left (-\cos \left (4 d x +4 c \right )-4 \cos \left (2 d x +2 c \right )-3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \sin \left (4 d x +4 c \right )+66 \sin \left (d x +c \right )+64 \sin \left (2 d x +2 c \right )+18 \sin \left (3 d x +3 c \right )\right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(147\)
norman	\(\frac {\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {3 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {10 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {3 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(153\)

[In]

int((a+cos(d*x+c)*a)*sec(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)
+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {9 \, a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, a \cos \left (d x + c\right )^{3} + 9 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + 6 \, a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+a*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(9*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*a*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*a*cos(d*x +
c)^3 + 9*a*cos(d*x + c)^2 + 8*a*cos(d*x + c) + 6*a)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=a \left (\int \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*cos(d*x+c))*sec(d*x+c)**5,x)

[Out]

a*(Integral(cos(c + d*x)*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a - 3 \, a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

integrate((a+a*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a - 3*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*
sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.29 \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {9 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 49 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 39 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate((a+a*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(9*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(9*a*tan(1/2*d*x + 1
/2*c)^7 - 49*a*tan(1/2*d*x + 1/2*c)^5 + 31*a*tan(1/2*d*x + 1/2*c)^3 - 39*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 17.61 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.53 \[ \int (a+a \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {-\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {49\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}-\frac {31\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {13\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]

[In]

int((a + a*cos(c + d*x))/cos(c + d*x)^5,x)

[Out]

((13*a*tan(c/2 + (d*x)/2))/4 - (31*a*tan(c/2 + (d*x)/2)^3)/12 + (49*a*tan(c/2 + (d*x)/2)^5)/12 - (3*a*tan(c/2
+ (d*x)/2)^7)/4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x
)/2)^8 + 1)) + (3*a*atanh(tan(c/2 + (d*x)/2)))/(4*d)