\(\int (a+a \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [5]

Optimal result

Integrand size = 31, antiderivative size = 42 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a C x+\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a C \sin (c+d x)}{d}+\frac {a A \tan (c+d x)}{d} \]

[Out]

a*C*x+a*A*arctanh(sin(d*x+c))/d+a*C*sin(d*x+c)/d+a*A*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3111, 3102, 2814, 3855} \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d}+\frac {a C \sin (c+d x)}{d}+a C x \]

[In]

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

[Out]

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

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3111

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m +
1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp
[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e +
f*x] + b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3855

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a C x+\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a C \cos (d x) \sin (c)}{d}+\frac {a C \cos (c) \sin (d x)}{d}+\frac {a A \tan (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a C \sin \left (d x +c \right )+a A \tan \left (d x +c \right )+a C \left (d x +c \right )}{d}\)	\(49\)
default	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a C \sin \left (d x +c \right )+a A \tan \left (d x +c \right )+a C \left (d x +c \right )}{d}\)	\(49\)
parts	\(\frac {a A \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a C \left (d x +c \right )}{d}+\frac {a C \sin \left (d x +c \right )}{d}\)	\(57\)
parallelrisch	\(\frac {a \left (2 d x C \cos \left (d x +c \right )-2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+2 A \sin \left (d x +c \right )+\sin \left (2 d x +2 c \right ) C \right )}{2 d \cos \left (d x +c \right )}\)	\(89\)
risch	\(a C x -\frac {i a C \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(100\)
norman	\(\frac {a C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a C x -2 a C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 a C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a \left (A -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (3 A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (3 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(209\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (42) = 84\).

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.05 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, C a d x \cos \left (d x + c\right ) + A a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - A a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a \cos \left (d x + c\right ) + A a\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.40 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} C a + A a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a \sin \left (d x + c\right ) + 2 \, A a \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (42) = 84\).

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.79 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {{\left (d x + c\right )} C a + A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - A a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.17 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

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