\(\int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx\) [40]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 134 \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=\frac {5 a^3 \cos (c+d x)}{d}+\frac {5 a^3 \cos ^2(c+d x)}{2 d}-\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {3 a^3 \cos ^4(c+d x)}{4 d}-\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2915, 12, 90} \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=-\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {3 a^3 \cos ^4(c+d x)}{4 d}-\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {5 a^3 \cos ^2(c+d x)}{2 d}+\frac {5 a^3 \cos (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d}-\frac {a^3 \log (\cos (c+d x))}{d} \]

[In]

Int[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^5,x]

[Out]

(5*a^3*Cos[c + d*x])/d + (5*a^3*Cos[c + d*x]^2)/(2*d) - (a^3*Cos[c + d*x]^3)/(3*d) - (3*a^3*Cos[c + d*x]^4)/(4
*d) - (a^3*Cos[c + d*x]^5)/(5*d) - (a^3*Log[Cos[c + d*x]])/d + (3*a^3*Sec[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(
2*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))^3 \sin ^2(c+d x) \tan ^3(c+d x) \, dx \\ & = \frac {\text {Subst}\left (\int \frac {a^3 (-a-x)^2 (-a+x)^5}{x^3} \, dx,x,-a \cos (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(-a-x)^2 (-a+x)^5}{x^3} \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (-5 a^4-\frac {a^7}{x^3}+\frac {3 a^6}{x^2}-\frac {a^5}{x}+5 a^3 x+a^2 x^2-3 a x^3+x^4\right ) \, dx,x,-a \cos (c+d x)\right )}{a^2 d} \\ & = \frac {5 a^3 \cos (c+d x)}{d}+\frac {5 a^3 \cos ^2(c+d x)}{2 d}-\frac {a^3 \cos ^3(c+d x)}{3 d}-\frac {3 a^3 \cos ^4(c+d x)}{4 d}-\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=-\frac {a^3 (-120-12350 \cos (c+d x)-2074 \cos (3 (c+d x))-330 \cos (4 (c+d x))+82 \cos (5 (c+d x))+45 \cos (6 (c+d x))+6 \cos (7 (c+d x))+960 \log (\cos (c+d x))+15 \cos (2 (c+d x)) (31+64 \log (\cos (c+d x)))) \sec ^2(c+d x)}{1920 d} \]

[In]

Integrate[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^5,x]

[Out]

-1/1920*(a^3*(-120 - 12350*Cos[c + d*x] - 2074*Cos[3*(c + d*x)] - 330*Cos[4*(c + d*x)] + 82*Cos[5*(c + d*x)] +
 45*Cos[6*(c + d*x)] + 6*Cos[7*(c + d*x)] + 960*Log[Cos[c + d*x]] + 15*Cos[2*(c + d*x)]*(31 + 64*Log[Cos[c + d
*x]]))*Sec[c + d*x]^2)/d

Maple [A] (verified)

Time = 2.11 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.28

method result size
derivativedivides \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}+2 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{3} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}}{d}\) \(172\)
default \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}+2 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{3} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}}{d}\) \(172\)
parallelrisch \(\frac {a^{3} \left (960 \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-960 \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-960 \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-45 \cos \left (6 d x +6 c \right )-6 \cos \left (7 d x +7 c \right )+12350 \cos \left (d x +c \right )+6733 \cos \left (2 d x +2 c \right )+2074 \cos \left (3 d x +3 c \right )+330 \cos \left (4 d x +4 c \right )-82 \cos \left (5 d x +5 c \right )+7318\right )}{960 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) \(173\)
parts \(-\frac {a^{3} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5 d}+\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}\) \(180\)
norman \(\frac {\frac {8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {224 a^{3}}{15 d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}-\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}+\frac {134 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{15 d}+\frac {214 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{5 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}+\frac {a^{3} \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) \(217\)
risch \(i a^{3} x -\frac {7 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{96 d}+\frac {7 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {37 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {37 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {7 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}-\frac {7 a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{96 d}+\frac {2 i a^{3} c}{d}+\frac {2 a^{3} \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {a^{3} \cos \left (5 d x +5 c \right )}{80 d}-\frac {3 a^{3} \cos \left (4 d x +4 c \right )}{32 d}\) \(228\)

[In]

int((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(1/2*sin(d*x+c)^6/cos(d*x+c)^2+1/2*sin(d*x+c)^4+sin(d*x+c)^2+2*ln(cos(d*x+c)))+3*a^3*(sin(d*x+c)^6/co
s(d*x+c)+(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+3*a^3*(-1/4*sin(d*x+c)^4-1/2*sin(d*x+c)^2-ln(cos(d*x+
c)))-1/5*a^3*(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 = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.97 \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=-\frac {96 \, a^{3} \cos \left (d x + c\right )^{7} + 360 \, a^{3} \cos \left (d x + c\right )^{6} + 160 \, a^{3} \cos \left (d x + c\right )^{5} - 1200 \, a^{3} \cos \left (d x + c\right )^{4} - 2400 \, a^{3} \cos \left (d x + c\right )^{3} + 480 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + 465 \, a^{3} \cos \left (d x + c\right )^{2} - 1440 \, a^{3} \cos \left (d x + c\right ) - 240 \, a^{3}}{480 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/480*(96*a^3*cos(d*x + c)^7 + 360*a^3*cos(d*x + c)^6 + 160*a^3*cos(d*x + c)^5 - 1200*a^3*cos(d*x + c)^4 - 24
00*a^3*cos(d*x + c)^3 + 480*a^3*cos(d*x + c)^2*log(-cos(d*x + c)) + 465*a^3*cos(d*x + c)^2 - 1440*a^3*cos(d*x
+ c) - 240*a^3)/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**3*sin(d*x+c)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79 \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=-\frac {12 \, a^{3} \cos \left (d x + c\right )^{5} + 45 \, a^{3} \cos \left (d x + c\right )^{4} + 20 \, a^{3} \cos \left (d x + c\right )^{3} - 150 \, a^{3} \cos \left (d x + c\right )^{2} - 300 \, a^{3} \cos \left (d x + c\right ) + 60 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac {30 \, {\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )}}{\cos \left (d x + c\right )^{2}}}{60 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/60*(12*a^3*cos(d*x + c)^5 + 45*a^3*cos(d*x + c)^4 + 20*a^3*cos(d*x + c)^3 - 150*a^3*cos(d*x + c)^2 - 300*a^
3*cos(d*x + c) + 60*a^3*log(cos(d*x + c)) - 30*(6*a^3*cos(d*x + c) + a^3)/cos(d*x + c)^2)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (124) = 248\).

Time = 0.42 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.22 \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=\frac {60 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {30 \, {\left (15 \, a^{3} + \frac {14 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}} - \frac {399 \, a^{3} - \frac {1395 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {390 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {650 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {565 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {137 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{60 \, d} \]

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="giac")

[Out]

1/60*(60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + 30*(15*a^3 + 14*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^3*(cos(d*x + c) - 1)^2/(co
s(d*x + c) + 1)^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2 - (399*a^3 - 1395*a^3*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + 390*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 650*a^3*(cos(d*x + c) - 1)^3/(cos(d*x + c
) + 1)^3 - 565*a^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 137*a^3*(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 = 13.36 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx=-\frac {\frac {a^3\,{\cos \left (c+d\,x\right )}^3}{3}-5\,a^3\,\cos \left (c+d\,x\right )-\frac {5\,a^3\,{\cos \left (c+d\,x\right )}^2}{2}-\frac {3\,a^3\,\cos \left (c+d\,x\right )+\frac {a^3}{2}}{{\cos \left (c+d\,x\right )}^2}+\frac {3\,a^3\,{\cos \left (c+d\,x\right )}^4}{4}+\frac {a^3\,{\cos \left (c+d\,x\right )}^5}{5}+a^3\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

[In]

int(sin(c + d*x)^5*(a + a/cos(c + d*x))^3,x)

[Out]

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