\(\int \cos ^2(c+d x) (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [89]

Optimal result

Integrand size = 31, antiderivative size = 58 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a (A+2 C) x+\frac {a C \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x)}{d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

1/2*a*(A+2*C)*x+a*C*arctanh(sin(d*x+c))/d+a*A*sin(d*x+c)/d+1/2*a*A*cos(d*x+c)*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4160, 4132, 8, 4130, 3855} \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a A \sin (c+d x)}{d}+\frac {a A \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} a x (A+2 C)+\frac {a C \text {arctanh}(\sin (c+d x))}{d} \]

[In]

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

[Out]

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

Rule 8

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

Rule 3855

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

Rule 4130

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 4132

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 4160

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(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[A*b*n + a*(C*n + A*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b,
d, e, f, A, C}, x] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {a A \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a A-a (A+2 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a A \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \int \cos (c+d x) \left (-2 a A-2 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{2} (a (A+2 C)) \int 1 \, dx \\ & = \frac {1}{2} a (A+2 C) x+\frac {a A \sin (c+d x)}{d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d}+(a C) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a (A+2 C) x+\frac {a C \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x)}{d}+\frac {a A \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (2 A c+2 A d x+4 C d x+4 C \text {arctanh}(\sin (c+d x))+4 A \sin (c+d x)+A \sin (2 (c+d x)))}{4 d} \]

[In]

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

[Out]

(a*(2*A*c + 2*A*d*x + 4*C*d*x + 4*C*ArcTanh[Sin[c + d*x]] + 4*A*Sin[c + d*x] + A*Sin[2*(c + d*x)]))/(4*d)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {a A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (d x +c \right )+a A \sin \left (d x +c \right )+C a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(65\)
default	\(\frac {a A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \left (d x +c \right )+a A \sin \left (d x +c \right )+C a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(65\)
parallelrisch	\(\frac {a \left (-2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A \sin \left (2 d x +2 c \right )}{2}+2 A \sin \left (d x +c \right )+\left (A +2 C \right ) x d \right )}{2 d}\)	\(67\)
risch	\(\frac {a A x}{2}+a x C -\frac {i a A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}\)	\(100\)
norman	\(\frac {\left (\frac {1}{2} a A +C a \right ) x +\left (\frac {1}{2} a A +C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-a A -2 C a \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {3 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}+\frac {C a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {C a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(192\)

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (A + 2 \, C\right )} a d x + C a \log \left (\sin \left (d x + c\right ) + 1\right ) - C a \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a \cos \left (d x + c\right ) + 2 \, A a\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

1/2*((A + 2*C)*a*d*x + C*a*log(sin(d*x + c) + 1) - C*a*log(-sin(d*x + c) + 1) + (A*a*cos(d*x + c) + 2*A*a)*sin
(d*x + c))/d

Sympy [F]

\[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \cos ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)

[Out]

a*(Integral(A*cos(c + d*x)**2, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(C*cos(c + d*x)**2*s
ec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.21 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a + 4 \, {\left (d x + c\right )} C a + 2 \, C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.71 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, C a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (A a + 2 \, C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 16.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.98 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^2*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)),x)

[Out]

(A*a*sin(c + d*x))/d + (A*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a*atan(sin(c/2 + (d*x)/2)/co
s(c/2 + (d*x)/2)))/d + (2*C*a*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a*sin(2*c + 2*d*x))/(4*d)