\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [15]

Optimal result

Integrand size = 31, antiderivative size = 74 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=a^2 (A+2 B) x+\frac {a^2 (2 A+B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (A-B) \sin (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d} \]

[Out]

a^2*(A+2*B)*x+a^2*(2*A+B)*arctanh(sin(d*x+c))/d-a^2*(A-B)*sin(d*x+c)/d+A*(a^2+a^2*cos(d*x+c))*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 74, 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 = {3054, 3047, 3102, 2814, 3855} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {a^2 (2 A+B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (A-B) \sin (c+d x)}{d}+a^2 x (A+2 B)+\frac {A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d} \]

[In]

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

[Out]

a^2*(A + 2*B)*x + (a^2*(2*A + B)*ArcTanh[Sin[c + d*x]])/d - (a^2*(A - B)*Sin[c + d*x])/d + (A*(a^2 + a^2*Cos[c
+ d*x])*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 3054

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\int (a+a \cos (c+d x)) (a (2 A+B)-a (A-B) \cos (c+d x)) \sec (c+d x) \, dx \\ & = \frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\int \left (a^2 (2 A+B)+\left (-a^2 (A-B)+a^2 (2 A+B)\right ) \cos (c+d x)-a^2 (A-B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {a^2 (A-B) \sin (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\int \left (a^2 (2 A+B)+a^2 (A+2 B) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = a^2 (A+2 B) x-\frac {a^2 (A-B) \sin (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\left (a^2 (2 A+B)\right ) \int \sec (c+d x) \, dx \\ & = a^2 (A+2 B) x+\frac {a^2 (2 A+B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (A-B) \sin (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.93 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {a^2 \left (A c+2 B c+A d x+2 B d x-2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+B \sin (c+d x)+A \tan (c+d x)\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09

method	result	size
parts	\(\frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (d x +c \right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\sin \left (d x +c \right ) B \,a^{2}}{d}\)	\(81\)
derivativedivides	\(\frac {A \,a^{2} \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )+2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \,a^{2} \left (d x +c \right )+A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(88\)
default	\(\frac {A \,a^{2} \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )+2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \,a^{2} \left (d x +c \right )+A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(88\)
parallelrisch	\(-\frac {2 a^{2} \left (\cos \left (d x +c \right ) \left (A +\frac {B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\cos \left (d x +c \right ) \left (A +\frac {B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\sin \left (2 d x +2 c \right ) B}{4}-\frac {d x \left (A +2 B \right ) \cos \left (d x +c \right )}{2}-\frac {A \sin \left (d x +c \right )}{2}\right )}{d \cos \left (d x +c \right )}\)	\(103\)
risch	\(a^{2} x A +2 a^{2} B x -\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{2 d}+\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\)	\(163\)
norman	\(\frac {\left (-A \,a^{2}-2 B \,a^{2}\right ) x +\left (-2 A \,a^{2}-4 B \,a^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,a^{2}+2 B \,a^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A \,a^{2}+4 B \,a^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a^{2} \left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (3 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (3 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{2} \left (2 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (2 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(269\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.46 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {2 \, {\left (A + 2 \, B\right )} a^{2} d x \cos \left (d x + c\right ) + {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} A a^{2} + 4 \, {\left (d x + c\right )} B a^{2} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a^{2} \sin \left (d x + c\right ) + 2 \, A a^{2} \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (74) = 148\).

Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.09 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {{\left (A a^{2} + 2 \, B a^{2}\right )} {\left (d x + c\right )} + {\left (2 \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

[In]

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

[Out]

((A*a^2 + 2*B*a^2)*(d*x + c) + (2*A*a^2 + B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a^2 + B*a^2)*log(ab
s(tan(1/2*d*x + 1/2*c) - 1)) - 2*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - B*a^2*tan(1/2*d*x + 1/2*c)^3 + A*a^2*tan(1/2*
d*x + 1/2*c) + B*a^2*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 = 0.19 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.18 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx=\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

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