3.26 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=63 \[ -\frac{a^3 \sin ^3(c+d x)}{3 d}+\frac{4 a^3 \sin (c+d x)}{d}+\frac{3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{5 a^3 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0746903, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3791, 2637, 2635, 8, 2633} \[ -\frac{a^3 \sin ^3(c+d x)}{3 d}+\frac{4 a^3 \sin (c+d x)}{d}+\frac{3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{5 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.0614099, size = 44, normalized size = 0.7 \[ \frac{a^3 (45 \sin (c+d x)+9 \sin (2 (c+d x))+\sin (3 (c+d x))+30 c+30 d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(30*c + 30*d*x + 45*Sin[c + d*x] + 9*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.065, size = 74, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{3} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}}+3\,{a}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}\sin \left ( dx+c \right ) +{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1161, size = 96, normalized size = 1.52 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \,{\left (d x + c\right )} a^{3} - 36 \, a^{3} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 12*(d*x + c)*a^3 - 3
6*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.65539, size = 119, normalized size = 1.89 \begin{align*} \frac{15 \, a^{3} d x +{\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 22 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a^3*d*x + (2*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 22*a^3)*sin(d*x + c))/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(cos(d*x+c)**3*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.35408, size = 108, normalized size = 1.71 \begin{align*} \frac{15 \,{\left (d x + c\right )} a^{3} + \frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 33 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*(d*x + c)*a^3 + 2*(15*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*a^3*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d