\(\int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx\) [60]

Optimal result

Integrand size = 21, antiderivative size = 37 \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cos ^3(c+d x)}{3 a d}+\frac {\sin ^2(c+d x)}{2 a d} \]

[Out]

1/3*cos(d*x+c)^3/a/d+1/2*sin(d*x+c)^2/a/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3957, 2914, 2644, 30, 2645} \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\sin ^2(c+d x)}{2 a d}+\frac {\cos ^3(c+d x)}{3 a d} \]

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2914

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 (1+2 \cos (c+d x)) \sin ^4\left (\frac {1}{2} (c+d x)\right )}{3 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {\cos \left (d x +c \right )^{2}}{2}}{d a}\)	\(29\)
default	\(\frac {\frac {\cos \left (d x +c \right )^{3}}{3}-\frac {\cos \left (d x +c \right )^{2}}{2}}{d a}\)	\(29\)
parallelrisch	\(\frac {7+3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )-3 \cos \left (2 d x +2 c \right )}{12 d a}\)	\(39\)
risch	\(\frac {\cos \left (d x +c \right )}{4 a d}+\frac {\cos \left (3 d x +3 c \right )}{12 a d}-\frac {\cos \left (2 d x +2 c \right )}{4 a d}\)	\(50\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {2}{3 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2}}{6 \, a d} \]

[In]

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

[Out]

1/6*(2*cos(d*x + c)^3 - 3*cos(d*x + c)^2)/(a*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2}}{6 \, a d} \]

[In]

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

[Out]

1/6*(2*cos(d*x + c)^3 - 3*cos(d*x + c)^2)/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {2 \, \cos \left (d x + c\right )^{3}}{d} - \frac {3 \, \cos \left (d x + c\right )^{2}}{d}}{6 \, a} \]

[In]

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

[Out]

1/6*(2*cos(d*x + c)^3/d - 3*cos(d*x + c)^2/d)/a

Mupad [B] (verification not implemented)

Time = 13.50 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {{\cos \left (c+d\,x\right )}^2\,\left (2\,\cos \left (c+d\,x\right )-3\right )}{6\,a\,d} \]

[In]

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

[Out]

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