\(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [6]

Optimal result

Integrand size = 29, antiderivative size = 32 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=a B x+\frac {a (A+B) \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3047, 3100, 2814, 3855} \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {a (A+B) \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d}+a B x \]

[In]

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

[Out]

a*B*x + (a*(A + B)*ArcTanh[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 3047

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

Rule 3100

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=a B x+\frac {a A \text {arctanh}(\sin (c+d x))}{d}+\frac {a B \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56

method	result	size
parts	\(\frac {a A \tan \left (d x +c \right )}{d}+\frac {\left (a A +B a \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B a \left (d x +c \right )}{d}\)	\(50\)
derivativedivides	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \left (d x +c \right )+a A \tan \left (d x +c \right )+B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(57\)
default	\(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \left (d x +c \right )+a A \tan \left (d x +c \right )+B a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(57\)
parallelrisch	\(-\frac {\left (\left (A +B \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +B \right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-d x B \cos \left (d x +c \right )-A \sin \left (d x +c \right )\right ) a}{d \cos \left (d x +c \right )}\)	\(81\)
risch	\(a B x +\frac {2 i a A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(105\)
norman	\(\frac {a B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a B x -\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-a B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \left (A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(178\)

[In]

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

[Out]

a*A*tan(d*x+c)/d+(A*a+B*a)/d*ln(sec(d*x+c)+tan(d*x+c))+B*a/d*(d*x+c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (32) = 64\).

Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.47 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {2 \, B a d x \cos \left (d x + c\right ) + {\left (A + B\right )} a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + B\right )} a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, A a \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (32) = 64\).

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.28 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} B a + A a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + B a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*B*a + A*a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + B*a*(log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) + 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. 84 vs. \(2 (32) = 64\).

Time = 0.31 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.62 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {{\left (d x + c\right )} B a + {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.12 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {A\,a\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

[In]

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

[Out]

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