\(\int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [77]

Optimal result

Integrand size = 21, antiderivative size = 55 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos ^4(c+d x)}{2 a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 55, 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, 45} \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos ^5(c+d x)}{5 a^2 d}+\frac {\cos ^4(c+d x)}{2 a^2 d}-\frac {\cos ^3(c+d x)}{3 a^2 d} \]

[In]

Int[Sin[c + d*x]^5/(a + a*Sec[c + d*x])^2,x]

[Out]

-1/3*Cos[c + d*x]^3/(a^2*d) + Cos[c + d*x]^4/(2*a^2*d) - Cos[c + d*x]^5/(5*a^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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 \frac {\cos ^2(c+d x) \sin ^5(c+d x)}{(-a-a \cos (c+d x))^2} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {(-a-x)^2 x^2}{a^2} \, dx,x,-a \cos (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int (-a-x)^2 x^2 \, dx,x,-a \cos (c+d x)\right )}{a^7 d} \\ & = \frac {\text {Subst}\left (\int \left (a^2 x^2+2 a x^3+x^4\right ) \, dx,x,-a \cos (c+d x)\right )}{a^7 d} \\ & = -\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos ^4(c+d x)}{2 a^2 d}-\frac {\cos ^5(c+d x)}{5 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {4 (4+3 \cos (c+d x)+3 \cos (2 (c+d x))) \sin ^6\left (\frac {1}{2} (c+d x)\right )}{15 a^2 d} \]

[In]

Integrate[Sin[c + d*x]^5/(a + a*Sec[c + d*x])^2,x]

[Out]

(4*(4 + 3*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]^6)/(15*a^2*d)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {-\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\)	\(39\)
default	\(\frac {-\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\)	\(39\)
parallelrisch	\(\frac {60 \cos \left (2 d x +2 c \right )-203-35 \cos \left (3 d x +3 c \right )-3 \cos \left (5 d x +5 c \right )-90 \cos \left (d x +c \right )+15 \cos \left (4 d x +4 c \right )}{240 a^{2} d}\)	\(63\)
risch	\(-\frac {3 \cos \left (d x +c \right )}{8 a^{2} d}-\frac {\cos \left (5 d x +5 c \right )}{80 d \,a^{2}}+\frac {\cos \left (4 d x +4 c \right )}{16 d \,a^{2}}-\frac {7 \cos \left (3 d x +3 c \right )}{48 d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right )}{4 d \,a^{2}}\)	\(84\)
norman	\(\frac {-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}-\frac {16}{15 a d}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d a}-\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a}\)	\(105\)

[In]

int(sin(d*x+c)^5/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{5} - 15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3}}{30 \, a^{2} d} \]

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(6*cos(d*x + c)^5 - 15*cos(d*x + c)^4 + 10*cos(d*x + c)^3)/(a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)**5/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{5} - 15 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3}}{30 \, a^{2} d} \]

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(6*cos(d*x + c)^5 - 15*cos(d*x + c)^4 + 10*cos(d*x + c)^3)/(a^2*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (49) = 98\).

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.16 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {8 \, {\left (\frac {10 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {20 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 2\right )}}{15 \, a^{2} d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}} \]

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-8/15*(10*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 20*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 15*(cos(d*x +
c) - 1)^3/(cos(d*x + c) + 1)^3 - 15*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 2)/(a^2*d*((cos(d*x + c) - 1)
/(cos(d*x + c) + 1) - 1)^5)

Mupad [B] (verification not implemented)

Time = 13.97 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {{\cos \left (c+d\,x\right )}^3\,\left (6\,{\cos \left (c+d\,x\right )}^2-15\,\cos \left (c+d\,x\right )+10\right )}{30\,a^2\,d} \]

[In]

int(sin(c + d*x)^5/(a + a/cos(c + d*x))^2,x)

[Out]

-(cos(c + d*x)^3*(6*cos(c + d*x)^2 - 15*cos(c + d*x) + 10))/(30*a^2*d)