3.327 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=117 \[ \frac{a^3 (B+3 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(B-2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{1}{2} a^3 x (7 B+6 C)+\frac{a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

[Out]

(a^3*(7*B + 6*C)*x)/2 + (a^3*(B + 3*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*B*Sin[c + d*x])/(2*d) + (a*B*Cos[c +
d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((B - 2*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.334513, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4072, 4017, 4018, 3996, 3770} \[ \frac{a^3 (B+3 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{(B-2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{1}{2} a^3 x (7 B+6 C)+\frac{a B \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(7*B + 6*C)*x)/2 + (a^3*(B + 3*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*B*Sin[c + d*x])/(2*d) + (a*B*Cos[c +
d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((B - 2*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(2*d)

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 4017

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 4018

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d*n
) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^2(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^2 (2 a (2 B+C)-a (B-2 C) \sec (c+d x)) \, dx\\ &=\frac{a B \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{(B-2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+a \sec (c+d x)) \left (5 a^2 B+2 a^2 (B+3 C) \sec (c+d x)\right ) \, dx\\ &=\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{(B-2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac{1}{2} \int \left (-a^3 (7 B+6 C)-2 a^3 (B+3 C) \sec (c+d x)\right ) \, dx\\ &=\frac{1}{2} a^3 (7 B+6 C) x+\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{(B-2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\left (a^3 (B+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^3 (7 B+6 C) x+\frac{a^3 (B+3 C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{5 a^3 B \sin (c+d x)}{2 d}+\frac{a B \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{(B-2 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 1.68187, size = 272, normalized size = 2.32 \[ \frac{1}{32} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (\frac{4 (3 B+C) \sin (c) \cos (d x)}{d}+\frac{4 (3 B+C) \cos (c) \sin (d x)}{d}-\frac{4 (B+3 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{4 (B+3 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{B \sin (2 c) \cos (2 d x)}{d}+\frac{B \cos (2 c) \sin (2 d x)}{d}+2 x (7 B+6 C)+\frac{4 C \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 C \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(2*(7*B + 6*C)*x - (4*(B + 3*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d
*x)/2]])/d + (4*(B + 3*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(3*B + C)*Cos[d*x]*Sin[c])/d + (B*C
os[2*d*x]*Sin[2*c])/d + (4*(3*B + C)*Cos[c]*Sin[d*x])/d + (B*Cos[2*c]*Sin[2*d*x])/d + (4*C*Sin[(d*x)/2])/(d*(C
os[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*C*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]))))/32

________________________________________________________________________________________

Maple [A]  time = 0.079, size = 145, normalized size = 1.2 \begin{align*}{\frac{B{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}Bx}{2}}+{\frac{7\,B{a}^{3}c}{2\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+3\,{\frac{B{a}^{3}\sin \left ( dx+c \right ) }{d}}+3\,{a}^{3}Cx+3\,{\frac{C{a}^{3}c}{d}}+3\,{\frac{{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}} \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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/2/d*B*a^3*sin(d*x+c)*cos(d*x+c)+7/2*a^3*B*x+7/2/d*B*a^3*c+a^3*C*sin(d*x+c)/d+3*a^3*B*sin(d*x+c)/d+3*a^3*C*x+
3/d*C*a^3*c+3/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+1/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+a^3*C*tan(d*x+c)/d

________________________________________________________________________________________

Maxima [A]  time = 0.94463, size = 189, normalized size = 1.62 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 12 \,{\left (d x + c\right )} B a^{3} + 12 \,{\left (d x + c\right )} C a^{3} + 2 \, B a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{3} \sin \left (d x + c\right ) + 4 \, C a^{3} \sin \left (d x + c\right ) + 4 \, C a^{3} \tan \left (d x + c\right )}{4 \, 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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 12*(d*x + c)*B*a^3 + 12*(d*x + c)*C*a^3 + 2*B*a^3*(log(sin(d*x +
c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^3*sin(d*x
+ c) + 4*C*a^3*sin(d*x + c) + 4*C*a^3*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 0.559321, size = 323, normalized size = 2.76 \begin{align*} \frac{{\left (7 \, B + 6 \, C\right )} a^{3} d x \cos \left (d x + c\right ) +{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (B a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, C a^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/2*((7*B + 6*C)*a^3*d*x*cos(d*x + c) + (B + 3*C)*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - (B + 3*C)*a^3*cos(d
*x + c)*log(-sin(d*x + c) + 1) + (B*a^3*cos(d*x + c)^2 + 2*(3*B + C)*a^3*cos(d*x + c) + 2*C*a^3)*sin(d*x + c))
/(d*cos(d*x + c))

________________________________________________________________________________________

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*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21781, size = 259, normalized size = 2.21 \begin{align*} -\frac{\frac{4 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} -{\left (7 \, B a^{3} + 6 \, C a^{3}\right )}{\left (d x + c\right )} - 2 \,{\left (B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \,{\left (B a^{3} + 3 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (5 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, C 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 )}^{2}}}{2 \, 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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/2*(4*C*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (7*B*a^3 + 6*C*a^3)*(d*x + c) - 2*(B*a^3 + 3
*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(B*a^3 + 3*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*B*a
^3*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*B*a^3*tan(1/2*d*x + 1/2*c) + 2*C*a^3*tan(1/2*d*
x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d