\(\int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx\) [161]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (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, 2916, 12, 780} \[ \int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {a \cos ^5(c+d x)}{5 d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {a \cos (c+d x)}{d}-\frac {b \cos ^4(c+d x)}{4 d}+\frac {b \cos ^2(c+d x)}{d}-\frac {b \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

-((a*Cos[c + d*x])/d) + (b*Cos[c + d*x]^2)/d + (2*a*Cos[c + d*x]^3)/(3*d) - (b*Cos[c + d*x]^4)/(4*d) - (a*Cos[
c + d*x]^5)/(5*d) - (b*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 780

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[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 (-b-a \cos (c+d x)) \sin ^4(c+d x) \tan (c+d x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {a (-b+x) \left (a^2-x^2\right )^2}{x} \, dx,x,-a \cos (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(-b+x) \left (a^2-x^2\right )^2}{x} \, dx,x,-a \cos (c+d x)\right )}{a^4 d} \\ & = \frac {\text {Subst}\left (\int \left (a^4-\frac {a^4 b}{x}+2 a^2 b x-2 a^2 x^2-b x^3+x^4\right ) \, dx,x,-a \cos (c+d x)\right )}{a^4 d} \\ & = -\frac {a \cos (c+d x)}{d}+\frac {b \cos ^2(c+d x)}{d}+\frac {2 a \cos ^3(c+d x)}{3 d}-\frac {b \cos ^4(c+d x)}{4 d}-\frac {a \cos ^5(c+d x)}{5 d}-\frac {b \log (\cos (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {5 a \cos (c+d x)}{8 d}+\frac {5 a \cos (3 (c+d x))}{48 d}-\frac {a \cos (5 (c+d x))}{80 d}-\frac {b \left (-\cos ^2(c+d x)+\frac {1}{4} \cos ^4(c+d x)+\log (\cos (c+d x))\right )}{d} \]

[In]

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

Maple [A] (verified)

Time = 1.58 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}+b \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(67\)
default	\(\frac {-\frac {a \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}+b \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(67\)
parts	\(-\frac {a \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5 d}+\frac {b \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(69\)
parallelrisch	\(\frac {-300 a \cos \left (d x +c \right )-6 a \cos \left (5 d x +5 c \right )+50 a \cos \left (3 d x +3 c \right )-15 b \cos \left (4 d x +4 c \right )+180 \cos \left (2 d x +2 c \right ) b +480 b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-480 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-480 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-256 a -165 b}{480 d}\)	\(115\)
risch	\(i b x +\frac {3 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {2 i b c}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {5 a \cos \left (d x +c \right )}{8 d}-\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \cos \left (3 d x +3 c \right )}{48 d}\)	\(120\)
norman	\(\frac {-\frac {16 a}{15 d}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {10 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {\left (16 a +6 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {2 \left (16 a +15 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(160\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{5} + 15 \, b \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 60 \, b \cos \left (d x + c\right )^{2} + 60 \, a \cos \left (d x + c\right ) + 60 \, b \log \left (-\cos \left (d x + c\right )\right )}{60 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.79 \[ \int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {12 \, a \cos \left (d x + c\right )^{5} + 15 \, b \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 60 \, b \cos \left (d x + c\right )^{2} + 60 \, a \cos \left (d x + c\right ) + 60 \, b \log \left (\cos \left (d x + c\right )\right )}{60 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (81) = 162\).

Time = 0.33 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.85 \[ \int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx=\frac {60 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, b \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {64 \, a + 137 \, b - \frac {320 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {805 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {640 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1970 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1970 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {805 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(60*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*b*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (64*a + 137*b - 320*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 805*b*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) + 640*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1970*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)
^2 - 1970*b*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*b*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 137*
b*(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 = 13.94 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int (a+b \sec (c+d x)) \sin ^5(c+d x) \, dx=-\frac {a\,\cos \left (c+d\,x\right )-\frac {2\,a\,{\cos \left (c+d\,x\right )}^3}{3}+\frac {a\,{\cos \left (c+d\,x\right )}^5}{5}-b\,{\cos \left (c+d\,x\right )}^2+\frac {b\,{\cos \left (c+d\,x\right )}^4}{4}+b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

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

[Out]

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