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

Optimal. Leaf size=89 \[ \frac{\cos ^3(c+d x)}{3 a^3 d}-\frac{3 \cos ^2(c+d x)}{2 a^3 d}+\frac{5 \cos (c+d x)}{a^3 d}-\frac{2}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{7 \log (\cos (c+d x)+1)}{a^3 d} \]

[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)

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Rubi [A]  time = 0.183736, antiderivative size = 89, 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} \[ \frac{\cos ^3(c+d x)}{3 a^3 d}-\frac{3 \cos ^2(c+d x)}{2 a^3 d}+\frac{5 \cos (c+d x)}{a^3 d}-\frac{2}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{7 \log (\cos (c+d x)+1)}{a^3 d} \]

Antiderivative was successfully verified.

[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 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))^3} \, dx &=-\int \frac{\cos ^3(c+d x) \sin ^3(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{(-a-x) x^3}{(-a+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^6 d}\\ &=\frac{\operatorname{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]  time = 0.411911, size = 99, normalized size = 1.11 \[ -\frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (-184 \cos (2 (c+d x))+28 \cos (3 (c+d x))-4 \cos (4 (c+d x))+1344 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (1344 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-19\right )+389\right )}{24 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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[Cos[(c
+ d*x)/2]] + Cos[c + d*x]*(-19 + 1344*Log[Cos[(c + d*x)/2]])))/(24*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.097, size = 100, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d{a}^{3} \left ( 1+\sec \left ( dx+c \right ) \right ) }}-7\,{\frac{\ln \left ( 1+\sec \left ( dx+c \right ) \right ) }{d{a}^{3}}}+{\frac{1}{3\,d{a}^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{3}}}-{\frac{3}{2\,d{a}^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{2}}}+5\,{\frac{1}{d{a}^{3}\sec \left ( dx+c \right ) }}+7\,{\frac{\ln \left ( \sec \left ( dx+c \right ) \right ) }{d{a}^{3}}} \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))^3,x)

[Out]

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

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Maxima [A]  time = 0.976867, size = 97, normalized size = 1.09 \begin{align*} -\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} \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))^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

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Fricas [A]  time = 1.7707, size = 228, normalized size = 2.56 \begin{align*} \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 )}} \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))^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)

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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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29883, size = 127, normalized size = 1.43 \begin{align*} -\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}} \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))^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)