3.97 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=119 \[ \frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac{(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{1}{2} a^2 x (3 A+2 C)+\frac{2 a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.289675, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4087, 4018, 3996, 3770} \[ \frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac{(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac{1}{2} a^2 x (3 A+2 C)+\frac{2 a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4087

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

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

Mathematica [B]  time = 1.06006, size = 292, normalized size = 2.45 \[ -\frac{a^2 \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (4 \cos (d x) \left (3 A d x-4 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 C d x\right )+4 \cos (2 c+d x) \left (3 A d x-4 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 C d x\right )+A \sin (2 c+d x)+8 A \sin (c+2 d x)+8 A \sin (3 c+2 d x)+A \sin (2 c+3 d x)+A \sin (4 c+3 d x)+A \sin (d x)+16 C \sin (d x)\right )}{16 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right )-1\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right )+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Sec[(c + d*x)/2]^2*(4*Cos[d*x]*(3*A*d*x + 2*C*d*x - 4*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*C*L
og[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*Cos[2*c + d*x]*(3*A*d*x + 2*C*d*x - 4*C*Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] + 4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + A*Sin[d*x] + 16*C*Sin[d*x] + A*Sin[2*c + d*x
] + 8*A*Sin[c + 2*d*x] + 8*A*Sin[3*c + 2*d*x] + A*Sin[2*c + 3*d*x] + A*Sin[4*c + 3*d*x]))/(16*d*(Cos[c/2] - Si
n[c/2])*(Cos[c/2] + Sin[c/2])*(-1 + Tan[(c + d*x)/2])*(1 + Tan[(c + d*x)/2]))

________________________________________________________________________________________

Maple [A]  time = 0.079, size = 107, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}Ax}{2}}+{\frac{3\,{a}^{2}Ac}{2\,d}}+{a}^{2}Cx+{\frac{C{a}^{2}c}{d}}+2\,{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}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)^2*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.93985, size = 136, normalized size = 1.14 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \,{\left (d x + c\right )} A a^{2} + 4 \,{\left (d x + c\right )} C a^{2} + 4 \, C a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, C a^{2} \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)^2*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*A*a^2 + 4*(d*x + c)*C*a^2 + 4*C*a^2*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 8*A*a^2*sin(d*x + c) + 4*C*a^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 0.525422, size = 296, normalized size = 2.49 \begin{align*} \frac{{\left (3 \, A + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + 2 \, C a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, C a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (A a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\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)^2*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/2*((3*A + 2*C)*a^2*d*x*cos(d*x + c) + 2*C*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) - 2*C*a^2*cos(d*x + c)*log(
-sin(d*x + c) + 1) + (A*a^2*cos(d*x + c)^2 + 4*A*a^2*cos(d*x + c) + 2*C*a^2)*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)**2*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.22831, size = 193, normalized size = 1.62 \begin{align*} \frac{4 \, C a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, C a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{4 \, C a^{2} \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 (3 \, A a^{2} + 2 \, C a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, A a^{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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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