\(\int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx\) [3]

Optimal result

Integrand size = 19, antiderivative size = 87 \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {a \cos (c+d x)}{d}+\frac {a \cos ^2(c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos ^4(c+d x)}{4 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \log (\cos (c+d x))}{d} \]

[Out]

-a*cos(d*x+c)/d+a*cos(d*x+c)^2/d+2/3*a*cos(d*x+c)^3/d-1/4*a*cos(d*x+c)^4/d-1/5*a*cos(d*x+c)^5/d-a*ln(cos(d*x+c
))/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3957, 2915, 12, 90} \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \cos ^4(c+d x)}{4 d}+\frac {2 a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^2(c+d x)}{d}-\frac {a \cos (c+d x)}{d}-\frac {a \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

-((a*Cos[c + d*x])/d) + (a*Cos[c + d*x]^2)/d + (2*a*Cos[c + d*x]^3)/(3*d) - (a*Cos[c + d*x]^4)/(4*d) - (a*Cos[
c + d*x]^5)/(5*d) - (a*Log[Cos[c + d*x]])/d

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\int (-a-a \cos (c+d x)) \sin ^4(c+d x) \tan (c+d x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {a (-a-x)^2 (-a+x)^3}{x} \, dx,x,-a \cos (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(-a-x)^2 (-a+x)^3}{x} \, dx,x,-a \cos (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (a^4-\frac {a^5}{x}+2 a^3 x-2 a^2 x^2-a x^3+x^4\right ) \, dx,x,-a \cos (c+d x)\right )}{a^4 d} \\ & = -\frac {a \cos (c+d x)}{d}+\frac {a \cos ^2(c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos ^4(c+d x)}{4 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \log (\cos (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {5 a \cos (c+d x)}{8 d}+\frac {5 a \cos (3 (c+d x))}{48 d}-\frac {a \cos (5 (c+d x))}{80 d}-\frac {a \left (-\cos ^2(c+d x)+\frac {1}{4} \cos ^4(c+d x)+\log (\cos (c+d x))\right )}{d} \]

[In]

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

[Out]

(-5*a*Cos[c + d*x])/(8*d) + (5*a*Cos[3*(c + d*x)])/(48*d) - (a*Cos[5*(c + d*x)])/(80*d) - (a*(-Cos[c + d*x]^2
+ Cos[c + d*x]^4/4 + Log[Cos[c + d*x]]))/d

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}}{d}\)	\(67\)
default	\(\frac {a \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}}{d}\)	\(67\)
parts	\(-\frac {a \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5 d}+\frac {a \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(69\)
parallelrisch	\(-\frac {a \left (-480 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+480 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+480 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+421+6 \cos \left (5 d x +5 c \right )+15 \cos \left (4 d x +4 c \right )-50 \cos \left (3 d x +3 c \right )-180 \cos \left (2 d x +2 c \right )+300 \cos \left (d x +c \right )\right )}{480 d}\)	\(103\)
risch	\(i a x +\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {2 i a c}{d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {5 a \cos \left (d x +c \right )}{8 d}-\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \cos \left (3 d x +3 c \right )}{48 d}\)	\(120\)
norman	\(\frac {-\frac {16 a}{15 d}-\frac {22 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {62 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {10 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {a \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(148\)

[In]

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

[Out]

1/d*(a*(-1/4*sin(d*x+c)^4-1/2*sin(d*x+c)^2-ln(cos(d*x+c)))-1/5*a*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c
))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{5} + 15 \, a \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 60 \, a \cos \left (d x + c\right )^{2} + 60 \, a \cos \left (d x + c\right ) + 60 \, a \log \left (-\cos \left (d x + c\right )\right )}{60 \, d} \]

[In]

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

[Out]

-1/60*(12*a*cos(d*x + c)^5 + 15*a*cos(d*x + c)^4 - 40*a*cos(d*x + c)^3 - 60*a*cos(d*x + c)^2 + 60*a*cos(d*x +
c) + 60*a*log(-cos(d*x + c)))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79 \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{5} + 15 \, a \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 60 \, a \cos \left (d x + c\right )^{2} + 60 \, a \cos \left (d x + c\right ) + 60 \, a \log \left (\cos \left (d x + c\right )\right )}{60 \, d} \]

[In]

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

[Out]

-1/60*(12*a*cos(d*x + c)^5 + 15*a*cos(d*x + c)^4 - 40*a*cos(d*x + c)^3 - 60*a*cos(d*x + c)^2 + 60*a*cos(d*x +
c) + 60*a*log(cos(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (81) = 162\).

Time = 0.31 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.31 \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=\frac {60 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {201 \, a - \frac {1125 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2610 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1970 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (201*a - 1125*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2610*a*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2 - 1970*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*a*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^
4 - 137*a*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {a\,\cos \left (c+d\,x\right )-a\,{\cos \left (c+d\,x\right )}^2-\frac {2\,a\,{\cos \left (c+d\,x\right )}^3}{3}+\frac {a\,{\cos \left (c+d\,x\right )}^4}{4}+\frac {a\,{\cos \left (c+d\,x\right )}^5}{5}+a\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

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

[Out]

-(a*cos(c + d*x) - a*cos(c + d*x)^2 - (2*a*cos(c + d*x)^3)/3 + (a*cos(c + d*x)^4)/4 + (a*cos(c + d*x)^5)/5 + a
*log(cos(c + d*x)))/d