\(\int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [95]

Optimal result

Integrand size = 21, antiderivative size = 89 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {5 \cos (c+d x)}{a^3 d}-\frac {3 \cos ^2(c+d x)}{2 a^3 d}+\frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {2}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {7 \log (1+\cos (c+d x))}{a^3 d} \]

[Out]

5*cos(d*x+c)/a^3/d-3/2*cos(d*x+c)^2/a^3/d+1/3*cos(d*x+c)^3/a^3/d-2/d/(a^3+a^3*cos(d*x+c))-7*ln(1+cos(d*x+c))/a
^3/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 89, 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, 78} \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {3 \cos ^2(c+d x)}{2 a^3 d}+\frac {5 \cos (c+d x)}{a^3 d}-\frac {2}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {7 \log (\cos (c+d x)+1)}{a^3 d} \]

[In]

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

[Out]

(5*Cos[c + d*x])/(a^3*d) - (3*Cos[c + d*x]^2)/(2*a^3*d) + Cos[c + d*x]^3/(3*a^3*d) - 2/(d*(a^3 + a^3*Cos[c + d
*x])) - (7*Log[1 + Cos[c + d*x]])/(a^3*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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 ^3(c+d x) \sin ^3(c+d x)}{(-a-a \cos (c+d x))^3} \, dx \\ & = \frac {\text {Subst}\left (\int \frac {(-a-x) x^3}{a^3 (-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {(-a-x) x^3}{(-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^6 d} \\ & = \frac {\text {Subst}\left (\int \left (-5 a^2-\frac {2 a^4}{(a-x)^2}+\frac {7 a^3}{a-x}-3 a x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a^6 d} \\ & = \frac {5 \cos (c+d x)}{a^3 d}-\frac {3 \cos ^2(c+d x)}{2 a^3 d}+\frac {\cos ^3(c+d x)}{3 a^3 d}-\frac {2}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {7 \log (1+\cos (c+d x))}{a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (389-184 \cos (2 (c+d x))+28 \cos (3 (c+d x))-4 \cos (4 (c+d x))+1344 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (-19+1344 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{24 a^3 d (1+\cos (c+d x))^3} \]

[In]

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

[Out]

-1/24*(Cos[(c + d*x)/2]^4*(389 - 184*Cos[2*(c + d*x)] + 28*Cos[3*(c + d*x)] - 4*Cos[4*(c + d*x)] + 1344*Log[Co
s[(c + d*x)/2]] + Cos[c + d*x]*(-19 + 1344*Log[Cos[(c + d*x)/2]])))/(a^3*d*(1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {3 \cos \left (d x +c \right )^{2}}{2}+5 \cos \left (d x +c \right )-7 \ln \left (\cos \left (d x +c \right )+1\right )-\frac {2}{\cos \left (d x +c \right )+1}}{d \,a^{3}}\)	\(60\)
default	\(\frac {\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {3 \cos \left (d x +c \right )^{2}}{2}+5 \cos \left (d x +c \right )-7 \ln \left (\cos \left (d x +c \right )+1\right )-\frac {2}{\cos \left (d x +c \right )+1}}{d \,a^{3}}\)	\(60\)
parallelrisch	\(\frac {-12 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+63 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )-9 \cos \left (2 d x +2 c \right )+84 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+121}{12 a^{3} d}\)	\(66\)
norman	\(\frac {\frac {34 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {41}{3 a d}+\frac {24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}+\frac {7 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a^{3} d}\)	\(109\)
risch	\(\frac {7 i x}{a^{3}}+\frac {21 \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{3} d}+\frac {21 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{3} d}+\frac {14 i c}{a^{3} d}-\frac {4 \,{\mathrm e}^{i \left (d x +c \right )}}{a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2}}-\frac {14 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{3} d}+\frac {\cos \left (3 d x +3 c \right )}{12 d \,a^{3}}-\frac {3 \cos \left (2 d x +2 c \right )}{4 d \,a^{3}}\)	\(137\)

[In]

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

[Out]

1/d/a^3*(1/3*cos(d*x+c)^3-3/2*cos(d*x+c)^2+5*cos(d*x+c)-7*ln(cos(d*x+c)+1)-2/(cos(d*x+c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {4 \, \cos \left (d x + c\right )^{4} - 14 \, \cos \left (d x + c\right )^{3} + 42 \, \cos \left (d x + c\right )^{2} - 84 \, {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 69 \, \cos \left (d x + c\right ) - 15}{12 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

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

[Out]

1/12*(4*cos(d*x + c)^4 - 14*cos(d*x + c)^3 + 42*cos(d*x + c)^2 - 84*(cos(d*x + c) + 1)*log(1/2*cos(d*x + c) +
1/2) + 69*cos(d*x + c) - 15)/(a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.81 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {12}{a^{3} \cos \left (d x + c\right ) + a^{3}} - \frac {2 \, \cos \left (d x + c\right )^{3} - 9 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right )}{a^{3}} + \frac {42 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(12/(a^3*cos(d*x + c) + a^3) - (2*cos(d*x + c)^3 - 9*cos(d*x + c)^2 + 30*cos(d*x + c))/a^3 + 42*log(cos(d
*x + c) + 1)/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {7 \, \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{a^{3} d} - \frac {2}{a^{3} d {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {2 \, a^{6} d^{5} \cos \left (d x + c\right )^{3} - 9 \, a^{6} d^{5} \cos \left (d x + c\right )^{2} + 30 \, a^{6} d^{5} \cos \left (d x + c\right )}{6 \, a^{9} d^{6}} \]

[In]

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

[Out]

-7*log(abs(-cos(d*x + c) - 1))/(a^3*d) - 2/(a^3*d*(cos(d*x + c) + 1)) + 1/6*(2*a^6*d^5*cos(d*x + c)^3 - 9*a^6*
d^5*cos(d*x + c)^2 + 30*a^6*d^5*cos(d*x + c))/(a^9*d^6)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {2}{a^3\,\cos \left (c+d\,x\right )+a^3}+\frac {7\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^3}-\frac {5\,\cos \left (c+d\,x\right )}{a^3}+\frac {3\,{\cos \left (c+d\,x\right )}^2}{2\,a^3}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^3}}{d} \]

[In]

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

[Out]

-(2/(a^3*cos(c + d*x) + a^3) + (7*log(cos(c + d*x) + 1))/a^3 - (5*cos(c + d*x))/a^3 + (3*cos(c + d*x)^2)/(2*a^
3) - cos(c + d*x)^3/(3*a^3))/d