3.78 \(\int \frac{\sin ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=66 \[ \frac{\cos ^3(c+d x)}{3 a^2 d}-\frac{\cos ^2(c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d}-\frac{2 \log (\cos (c+d x)+1)}{a^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.163219, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2836, 12, 77} \[ \frac{\cos ^3(c+d x)}{3 a^2 d}-\frac{\cos ^2(c+d x)}{a^2 d}+\frac{2 \cos (c+d x)}{a^2 d}-\frac{2 \log (\cos (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2836

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*x)/b
)^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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 77

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]))))

Rubi steps

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

Mathematica [A]  time = 0.205094, size = 51, normalized size = 0.77 \[ \frac{27 \cos (c+d x)-6 \cos (2 (c+d x))+\cos (3 (c+d x))-48 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-22}{12 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a + a*Sec[c + d*x])^2,x]

[Out]

(-22 + 27*Cos[c + d*x] - 6*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] - 48*Log[Cos[(c + d*x)/2]])/(12*a^2*d)

________________________________________________________________________________________

Maple [A]  time = 0.083, size = 82, normalized size = 1.2 \begin{align*} -2\,{\frac{\ln \left ( 1+\sec \left ( dx+c \right ) \right ) }{d{a}^{2}}}+{\frac{1}{3\,d{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{d{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{1}{d{a}^{2}\sec \left ( dx+c \right ) }}+2\,{\frac{\ln \left ( \sec \left ( dx+c \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00854, size = 69, normalized size = 1.05 \begin{align*} \frac{\frac{\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right )}{a^{2}} - \frac{6 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.75959, size = 132, normalized size = 2. \begin{align*} \frac{\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) - 6 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**3/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.32994, size = 101, normalized size = 1.53 \begin{align*} -\frac{2 \, \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{a^{2} d} + \frac{a^{4} d^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{4} d^{2} \cos \left (d x + c\right )^{2} + 6 \, a^{4} d^{2} \cos \left (d x + c\right )}{3 \, a^{6} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-2*log(abs(-cos(d*x + c) - 1))/(a^2*d) + 1/3*(a^4*d^2*cos(d*x + c)^3 - 3*a^4*d^2*cos(d*x + c)^2 + 6*a^4*d^2*co
s(d*x + c))/(a^6*d^3)