\(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\) [951]

Optimal result

Integrand size = 41, antiderivative size = 118 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=b (b B+2 a C) x+\frac {\left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d}+\frac {a (A b+a B) \tan (c+d x)}{d}+\frac {A (a+b \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

b*(B*b+2*C*a)*x+1/2*(2*A*b^2+4*B*a*b+a^2*(A+2*C))*arctanh(sin(d*x+c))/d-1/2*b^2*(A-2*C)*sin(d*x+c)/d+a*(A*b+B*
a)*tan(d*x+c)/d+1/2*A*(a+b*cos(d*x+c))^2*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3126, 3110, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\left (a^2 (A+2 C)+4 a b B+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (a B+A b) \tan (c+d x)}{d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^2}{2 d}+b x (2 a C+b B)-\frac {b^2 (A-2 C) \sin (c+d x)}{2 d} \]

[In]

Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

b*(b*B + 2*a*C)*x + ((2*A*b^2 + 4*a*b*B + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*d) - (b^2*(A - 2*C)*Sin[c +
d*x])/(2*d) + (a*(A*b + a*B)*Tan[c + d*x])/d + (A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]*Tan[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 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 3110

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

Rule 3126

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(118)=236\).

Time = 3.91 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.35 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 b (b B+2 a C) (c+d x)-2 \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 A b^2+4 a b B+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (2 A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a (2 A b+a B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 b^2 C \sin (c+d x)}{4 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10

method	result	size
parts	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (2 a A b +B \,a^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (B \,b^{2}+2 a b C \right ) \left (d x +c \right )}{d}+\frac {\left (A \,b^{2}+2 B a b +a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} C \sin \left (d x +c \right )}{d}\)	\(130\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a A b \tan \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b C \left (d x +c \right )+A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{2} \left (d x +c \right )+C \sin \left (d x +c \right ) b^{2}}{d}\)	\(152\)
default	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a A b \tan \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b C \left (d x +c \right )+A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{2} \left (d x +c \right )+C \sin \left (d x +c \right ) b^{2}}{d}\)	\(152\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (2 A \,b^{2}+4 B a b +a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (2 A \,b^{2}+4 B a b +a^{2} \left (A +2 C \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 b d x \left (B b +2 C a \right ) \cos \left (2 d x +2 c \right )+\left (4 a A b +2 B \,a^{2}\right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{2}+\left (2 A \,a^{2}+C \,b^{2}\right ) \sin \left (d x +c \right )+2 b d x \left (B b +2 C a \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(201\)
risch	\(x B \,b^{2}+2 a b C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{2}}{2 d}-\frac {i a \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}-4 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-a A \,{\mathrm e}^{i \left (d x +c \right )}-4 A b -2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{d}\)	\(304\)
norman	\(\frac {\left (B \,b^{2}+2 a b C \right ) x +\left (-4 B \,b^{2}-8 a b C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}-2 a b C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,b^{2}-2 a b C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B \,b^{2}+2 a b C \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,b^{2}+4 a b C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,b^{2}+4 a b C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (A \,a^{2}-4 a A b -2 B \,a^{2}+2 C \,b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (A \,a^{2}+4 a A b +2 B \,a^{2}+2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (5 A \,a^{2}-12 a A b -6 B \,a^{2}+2 C \,b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (5 A \,a^{2}+12 a A b +6 B \,a^{2}+2 C \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (5 A \,a^{2}-4 a A b -2 B \,a^{2}-2 C \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (5 A \,a^{2}+4 a A b +2 B \,a^{2}-2 C \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (A \,a^{2}+2 A \,b^{2}+4 B a b +2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (A \,a^{2}+2 A \,b^{2}+4 B a b +2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(508\)

[In]

int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.40 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (2 \, C a b + B b^{2}\right )} d x \cos \left (d x + c\right )^{2} + {\left ({\left (A + 2 \, C\right )} a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 2 \, C\right )} a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.60 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {8 \, {\left (d x + c\right )} C a b + 4 \, {\left (d x + c\right )} B b^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C b^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right ) + 8 \, A a b \tan \left (d x + c\right )}{4 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (112) = 224\).

Time = 0.33 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.03 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\frac {4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (2 \, C a b + B b^{2}\right )} {\left (d x + c\right )} + {\left (A a^{2} + 2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{2} + 2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.85 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.18 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2\,\left (\frac {A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{4}+A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]

[In]

int(((a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^3,x)

[Out]

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