\(\int (a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [781]

Optimal result

Integrand size = 40, antiderivative size = 86 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} \left (4 a b B+2 a^2 C+b^2 C\right ) x+\frac {a^2 B \text {arctanh}(\sin (c+d x))}{d}+\frac {b (2 b B+3 a C) \sin (c+d x)}{2 d}+\frac {b C (a+b \cos (c+d x)) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3108, 3069, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^2 B \text {arctanh}(\sin (c+d x))}{d}+\frac {1}{2} x \left (2 a^2 C+4 a b B+b^2 C\right )+\frac {b (3 a C+2 b B) \sin (c+d x)}{2 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]

[In]

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

[Out]

((4*a*b*B + 2*a^2*C + b^2*C)*x)/2 + (a^2*B*ArcTanh[Sin[c + d*x]])/d + (b*(2*b*B + 3*a*C)*Sin[c + d*x])/(2*d) +
(b*C*(a + b*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 3069

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 - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[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 3108

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

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.40 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \left (4 a b B+2 a^2 C+b^2 C\right ) (c+d x)-4 a^2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b (b B+2 a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

(2*(4*a*b*B + 2*a^2*C + b^2*C)*(c + d*x) - 4*a^2*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*a^2*B*Log[Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*b*(b*B + 2*a*C)*Sin[c + d*x] + b^2*C*Sin[2*(c + d*x)])/(4*d)

Maple [A] (verified)

Time = 1.58 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \sin \left (d x +c \right ) C +2 B a b \left (d x +c \right )+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b^{2}}{d}\)	\(94\)
default	\(\frac {a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \sin \left (d x +c \right ) C +2 B a b \left (d x +c \right )+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \sin \left (d x +c \right ) b^{2}}{d}\)	\(94\)
parts	\(\frac {\left (B \,b^{2}+2 C a b \right ) \sin \left (d x +c \right )}{d}+\frac {\left (2 B a b +a^{2} C \right ) \left (d x +c \right )}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(95\)
parallelrisch	\(\frac {-4 B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+C \sin \left (2 d x +2 c \right ) b^{2}+\left (4 B \,b^{2}+8 C a b \right ) \sin \left (d x +c \right )+8 x \left (B a b +\frac {1}{2} a^{2} C +\frac {1}{4} b^{2} C \right ) d}{4 d}\)	\(97\)
risch	\(2 x B a b +a^{2} C x +\frac {b^{2} C x}{2}-\frac {i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i C a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C a b}{d}+\frac {B \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {\sin \left (2 d x +2 c \right ) b^{2} C}{4 d}\)	\(156\)
norman	\(\frac {\left (-2 B a b -a^{2} C -\frac {1}{2} b^{2} C \right ) x +\left (-6 B a b -3 a^{2} C -\frac {3}{2} b^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B a b +a^{2} C +\frac {1}{2} b^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 B a b +3 a^{2} C +\frac {3}{2} b^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-4 B a b -2 a^{2} C -b^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 B a b +2 a^{2} C +b^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (2 B b +4 a C -C b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 b \left (B b +2 a C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 b \left (B b +2 a C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {b \left (2 B b +4 a C +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 b^{2} C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {B \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(374\)

[In]

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

[Out]

1/d*(a^2*C*(d*x+c)+B*a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*b*sin(d*x+c)*C+2*B*a*b*(d*x+c)+b^2*C*(1/2*cos(d*x+c)*si
n(d*x+c)+1/2*d*x+1/2*c)+B*sin(d*x+c)*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {B a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} d x + {\left (C b^{2} \cos \left (d x + c\right ) + 4 \, C a b + 2 \, B b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((B + C*cos(c + d*x))*(a + b*cos(c + d*x))**2*cos(c + d*x)*sec(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a^{2} + 8 \, {\left (d x + c\right )} B a b + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, B b^{2} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*C*a^2 + 8*(d*x + c)*B*a*b + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^2 + 2*B*a^2*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 8*C*a*b*sin(d*x + c) + 4*B*b^2*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (80) = 160\).

Time = 0.34 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.07 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (2 \, C a^{2} + 4 \, B a b + C b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (2*C*a^2 + 4*B*
a*b + C*b^2)*(d*x + c) + 2*(4*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^2*tan(1/2*d*x + 1/2*c)^3 - C*b^2*tan(1/2*d*
x + 1/2*c)^3 + 4*C*a*b*tan(1/2*d*x + 1/2*c) + 2*B*b^2*tan(1/2*d*x + 1/2*c) + C*b^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 = 2.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.97 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {B\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

[In]

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

[Out]

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