\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [14]

Optimal result

Integrand size = 29, antiderivative size = 82 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {1}{2} a^2 (4 A+3 B) x+\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d} \]

[Out]

1/2*a^2*(4*A+3*B)*x+a^2*A*arctanh(sin(d*x+c))/d+1/2*a^2*(2*A+3*B)*sin(d*x+c)/d+1/2*B*(a^2+a^2*cos(d*x+c))*sin(
d*x+c)/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3055, 3047, 3102, 2814, 3855} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (4 A+3 B)+\frac {B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d} \]

[In]

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

[Out]

(a^2*(4*A + 3*B)*x)/2 + (a^2*A*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*A + 3*B)*Sin[c + d*x])/(2*d) + (B*(a^2 + a^2
*Cos[c + d*x])*Sin[c + d*x])/(2*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 3055

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 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 {B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {1}{2} \int (a+a \cos (c+d x)) (2 a A+a (2 A+3 B) \cos (c+d x)) \sec (c+d x) \, dx \\ & = \frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^2 A+\left (2 a^2 A+a^2 (2 A+3 B)\right ) \cos (c+d x)+a^2 (2 A+3 B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {1}{2} \int \left (2 a^2 A+a^2 (4 A+3 B) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {1}{2} a^2 (4 A+3 B) x+\frac {a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\left (a^2 A\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^2 (4 A+3 B) x+\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a^2 \left (8 A d x+6 B d x-4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 (A+2 B) \sin (c+d x)+B \sin (2 (c+d x))\right )}{4 d} \]

[In]

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

[Out]

(a^2*(8*A*d*x + 6*B*d*x - 4*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*A*Log[Cos[(c + d*x)/2] + Sin[(c + d
*x)/2]] + 4*(A + 2*B)*Sin[c + d*x] + B*Sin[2*(c + d*x)]))/(4*d)

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(-\frac {\left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\sin \left (2 d x +2 c \right ) B}{4}+\left (-A -2 B \right ) \sin \left (d x +c \right )-2 d x \left (A +\frac {3 B}{4}\right )\right ) a^{2}}{d}\)	\(74\)
derivativedivides	\(\frac {A \,a^{2} \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} \left (d x +c \right )+2 B \,a^{2} \sin \left (d x +c \right )+A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )}{d}\)	\(96\)
default	\(\frac {A \,a^{2} \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} \left (d x +c \right )+2 B \,a^{2} \sin \left (d x +c \right )+A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )}{d}\)	\(96\)
parts	\(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(97\)
risch	\(2 a^{2} x A +\frac {3 a^{2} B x}{2}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{d}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{4 d}\)	\(153\)
norman	\(\frac {\left (2 A \,a^{2}+\frac {3}{2} B \,a^{2}\right ) x +\left (2 A \,a^{2}+\frac {3}{2} B \,a^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A \,a^{2}+\frac {9}{2} B \,a^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A \,a^{2}+\frac {9}{2} B \,a^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (2 A +3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (2 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{2} \left (A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(225\)

[In]

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

[Out]

-(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)-1/4*sin(2*d*x+2*c)*B+(-A-2*B)*sin(d*x+c)-2*d*x*(A+3/4*
B))*a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {{\left (4 \, A + 3 \, B\right )} a^{2} d x + A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{2} \cos \left (d x + c\right ) + 2 \, {\left (A + 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x), x) + Integral(A*cos(c + d*x)**2*se
c(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x), x) + In
tegral(B*cos(c + d*x)**3*sec(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {8 \, {\left (d x + c\right )} A a^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 4 \, {\left (d x + c\right )} B a^{2} + 4 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, B a^{2} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.77 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {2 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (4 \, A a^{2} + 3 \, B a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (4*A*a^2 + 3*B*
a^2)*(d*x + c) + 2*(2*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*A*a^2*tan(1/2*d*x + 1/
2*c) + 5*B*a^2*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 = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.72 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

(A*a^2*sin(c + d*x))/d + (2*B*a^2*sin(c + d*x))/d + (4*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(2*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (B*a^2*sin(2*c + 2*d*x))/(4*d)