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

Optimal. Leaf size=100 \[ \frac{a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} b x \left (3 a^2+2 b^2\right )+\frac{7 a^2 b \sin (c+d x) \cos (c+d x)}{6 d}+\frac{a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d} \]

[Out]

(b*(3*a^2 + 2*b^2)*x)/2 + (a*(2*a^2 + 9*b^2)*Sin[c + d*x])/(3*d) + (7*a^2*b*Cos[c + d*x]*Sin[c + d*x])/(6*d) +
(a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.149988, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3841, 4047, 2637, 4045, 8} \[ \frac{a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{3 d}+\frac{1}{2} b x \left (3 a^2+2 b^2\right )+\frac{7 a^2 b \sin (c+d x) \cos (c+d x)}{6 d}+\frac{a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*a^2 + 2*b^2)*x)/2 + (a*(2*a^2 + 9*b^2)*Sin[c + d*x])/(3*d) + (7*a^2*b*Cos[c + d*x]*Sin[c + d*x])/(6*d) +
(a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])*Sin[c + d*x])/(3*d)

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 2637

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 8

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

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) \left (7 a^2 b+a \left (2 a^2+9 b^2\right ) \sec (c+d x)+b \left (a^2+3 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) \left (7 a^2 b+b \left (a^2+3 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} \left (a \left (2 a^2+9 b^2\right )\right ) \int \cos (c+d x) \, dx\\ &=\frac{a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{3 d}+\frac{7 a^2 b \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{3 d}+\frac{1}{2} \left (b \left (3 a^2+2 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} b \left (3 a^2+2 b^2\right ) x+\frac{a \left (2 a^2+9 b^2\right ) \sin (c+d x)}{3 d}+\frac{7 a^2 b \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.120722, size = 80, normalized size = 0.8 \[ \frac{9 a \left (a^2+4 b^2\right ) \sin (c+d x)+9 a^2 b \sin (2 (c+d x))+18 a^2 b c+18 a^2 b d x+a^3 \sin (3 (c+d x))+12 b^3 c+12 b^3 d x}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18*a^2*b*c + 12*b^3*c + 18*a^2*b*d*x + 12*b^3*d*x + 9*a*(a^2 + 4*b^2)*Sin[c + d*x] + 9*a^2*b*Sin[2*(c + d*x)]
+ a^3*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.049, size = 76, normalized size = 0.8 \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}^{2}b \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,a{b}^{2}\sin \left ( dx+c \right ) +{b}^{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+b*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^2*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*a*b^2*sin(d*x+c)+
b^3*(d*x+c))

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Maxima [A]  time = 1.17445, size = 99, normalized size = 0.99 \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^{2} b - 12 \,{\left (d x + c\right )} b^{3} - 36 \, a b^{2} \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+b*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^2*b - 12*(d*x + c)*b^3 -
36*a*b^2*sin(d*x + c))/d

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Fricas [A]  time = 1.67069, size = 153, normalized size = 1.53 \begin{align*} \frac{3 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} d x +{\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{2} b \cos \left (d x + c\right ) + 4 \, a^{3} + 18 \, a b^{2}\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+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/6*(3*(3*a^2*b + 2*b^3)*d*x + (2*a^3*cos(d*x + c)^2 + 9*a^2*b*cos(d*x + c) + 4*a^3 + 18*a*b^2)*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+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28825, size = 230, normalized size = 2.3 \begin{align*} \frac{3 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a b^{2} \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+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/6*(3*(3*a^2*b + 2*b^3)*(d*x + c) + 2*(6*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*a*b
^2*tan(1/2*d*x + 1/2*c)^5 + 4*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^3*tan(1/2*d*x
+ 1/2*c) + 9*a^2*b*tan(1/2*d*x + 1/2*c) + 18*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d