Integrand size = 41, antiderivative size = 126 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} \left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) x+\frac {a (2 A b+a B) \text {arctanh}(\sin (c+d x))}{d}-\frac {b (2 a A-b B-2 a C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d} \] Output:
1/2*(2*A*b^2+4*B*a*b+2*C*a^2+C*b^2)*x+a*(2*A*b+B*a)*arctanh(sin(d*x+c))/d- b*(2*A*a-B*b-2*C*a)*sin(d*x+c)/d-1/2*b^2*(2*A-C)*cos(d*x+c)*sin(d*x+c)/d+A *(a+b*cos(d*x+c))^2*tan(d*x+c)/d
Time = 4.14 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.23 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) (c+d x)-4 a (2 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a (2 A b+a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (4 a^2 A+b^2 C+4 b (b B+2 a C) \cos (c+d x)+b^2 C \cos (2 (c+d x))\right ) \tan (c+d x)}{4 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^2,x]
Output:
(2*(2*A*b^2 + 4*a*b*B + 2*a^2*C + b^2*C)*(c + d*x) - 4*a*(2*A*b + a*B)*Log [Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*a*(2*A*b + a*B)*Log[Cos[(c + d*x )/2] + Sin[(c + d*x)/2]] + (4*a^2*A + b^2*C + 4*b*(b*B + 2*a*C)*Cos[c + d* x] + b^2*C*Cos[2*(c + d*x)])*Tan[c + d*x])/(4*d)
Time = 0.92 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 3526, 3042, 3512, 3042, 3502, 3042, 3214, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \int (a+b \cos (c+d x)) \left (-b (2 A-C) \cos ^2(c+d x)+(b B+a C) \cos (c+d x)+2 A b+a B\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A b+a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}\) |
\(\Big \downarrow \) 3512 |
\(\displaystyle \frac {1}{2} \int \left (-2 b (2 a A-b B-2 a C) \cos ^2(c+d x)+\left (2 C a^2+4 b B a+2 A b^2+b^2 C\right ) \cos (c+d x)+2 a (2 A b+a B)\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {-2 b (2 a A-b B-2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (2 C a^2+4 b B a+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a (2 A b+a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {1}{2} \left (\int \left (2 a (2 A b+a B)+\left (2 C a^2+4 b B a+2 A b^2+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 b \sin (c+d x) (2 a A-2 a C-b B)}{d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\int \frac {2 a (2 A b+a B)+\left (2 C a^2+4 b B a+2 A b^2+b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b \sin (c+d x) (2 a A-2 a C-b B)}{d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {1}{2} \left (2 a (a B+2 A b) \int \sec (c+d x)dx+x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )-\frac {2 b \sin (c+d x) (2 a A-2 a C-b B)}{d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (2 a (a B+2 A b) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )-\frac {2 b \sin (c+d x) (2 a A-2 a C-b B)}{d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{2} \left (x \left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right )+\frac {2 a (a B+2 A b) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 b \sin (c+d x) (2 a A-2 a C-b B)}{d}\right )+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]
Output:
-1/2*(b^2*(2*A - C)*Cos[c + d*x]*Sin[c + d*x])/d + ((2*A*b^2 + 4*a*b*B + 2 *a^2*C + b^2*C)*x + (2*a*(2*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d - (2*b*(2* a*A - b*B - 2*a*C)*Sin[c + d*x])/d)/2 + (A*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/d
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
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] :> 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] + Simp[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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.68 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.96
method | result | size |
parts | \(\frac {A \,a^{2} \tan \left (d x +c \right )}{d}+\frac {\left (2 a A b +B \,a^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (B \,b^{2}+2 a b C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (A \,b^{2}+2 B a b +a^{2} C \right ) \left (d x +c \right )}{d}+\frac {C \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(121\) |
derivativedivides | \(\frac {A \tan \left (d x +c \right ) a^{2}+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (d x +c \right )+2 a A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \left (d x +c \right )+2 C \sin \left (d x +c \right ) a b +A \,b^{2} \left (d x +c \right )+B \sin \left (d x +c \right ) b^{2}+C \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(134\) |
default | \(\frac {A \tan \left (d x +c \right ) a^{2}+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (d x +c \right )+2 a A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \left (d x +c \right )+2 C \sin \left (d x +c \right ) a b +A \,b^{2} \left (d x +c \right )+B \sin \left (d x +c \right ) b^{2}+C \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(134\) |
parallelrisch | \(\frac {8 \left (-2 a A b -B \,a^{2}\right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \left (2 a A b +B \,a^{2}\right ) \cos \left (d x +c \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (B \,b^{2}+2 a b C \right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{2}+8 x d \left (\left (A +\frac {C}{2}\right ) b^{2}+2 B a b +a^{2} C \right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (8 A \,a^{2}+C \,b^{2}\right )}{8 d \cos \left (d x +c \right )}\) | \(163\) |
risch | \(x A \,b^{2}+2 x B a b +a^{2} C x +\frac {b^{2} C x}{2}-\frac {i C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,b^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a b C}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a b C}{d}+\frac {i C \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\) | \(247\) |
norman | \(\frac {\left (-A \,b^{2}-2 B a b -a^{2} C -\frac {1}{2} C \,b^{2}\right ) x +\left (-3 A \,b^{2}-6 B a b -3 a^{2} C -\frac {3}{2} C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (A \,b^{2}+2 B a b +a^{2} C +\frac {1}{2} C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (3 A \,b^{2}+6 B a b +3 a^{2} C +\frac {3}{2} C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-2 A \,b^{2}-4 B a b -2 a^{2} C -C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (2 A \,b^{2}+4 B a b +2 a^{2} C +C \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {2 \left (6 A \,a^{2}-C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 \left (2 A \,a^{2}-B \,b^{2}-2 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {4 \left (2 A \,a^{2}+B \,b^{2}+2 a b C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {\left (2 A \,a^{2}-2 B \,b^{2}-4 a b C +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (2 A \,a^{2}+2 B \,b^{2}+4 a b C +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {a \left (2 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \left (2 A b +B a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(465\) |
Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x,meth od=_RETURNVERBOSE)
Output:
A*a^2/d*tan(d*x+c)+(2*A*a*b+B*a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+(B*b^2+2*C* a*b)/d*sin(d*x+c)+(A*b^2+2*B*a*b+C*a^2)/d*(d*x+c)+C*b^2/d*(1/2*cos(d*x+c)* sin(d*x+c)+1/2*d*x+1/2*c)
Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.17 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {{\left (2 \, C a^{2} + 4 \, B a b + {\left (2 \, A + C\right )} b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C b^{2} \cos \left (d x + c\right )^{2} + 2 \, A a^{2} + 2 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2, x, algorithm="fricas")
Output:
1/2*((2*C*a^2 + 4*B*a*b + (2*A + C)*b^2)*d*x*cos(d*x + c) + (B*a^2 + 2*A*a *b)*cos(d*x + c)*log(sin(d*x + c) + 1) - (B*a^2 + 2*A*a*b)*cos(d*x + c)*lo g(-sin(d*x + c) + 1) + (C*b^2*cos(d*x + c)^2 + 2*A*a^2 + 2*(2*C*a*b + B*b^ 2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)* *2,x)
Output:
Integral((a + b*cos(c + d*x))**2*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)* sec(c + d*x)**2, x)
Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a^{2} + 8 \, {\left (d x + c\right )} B a b + 4 \, {\left (d x + c\right )} A b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, B b^{2} \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2, x, algorithm="maxima")
Output:
1/4*(4*(d*x + c)*C*a^2 + 8*(d*x + c)*B*a*b + 4*(d*x + c)*A*b^2 + (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)) + 4*A*a*b*(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) + 4*A*a^2*tan(d*x + c))/d
Time = 0.16 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.82 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {\frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2, x, algorithm="giac")
Output:
-1/2*(4*A*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (2*C*a^2 + 4*B*a*b + 2*A*b^2 + C*b^2)*(d*x + c) - 2*(B*a^2 + 2*A*a*b)*log(abs(tan( 1/2*d*x + 1/2*c) + 1)) + 2*(B*a^2 + 2*A*a*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 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*t an(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
Time = 1.13 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.17 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {B\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {C\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}-\frac {B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d}-\frac {B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \] Input:
int(((a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^2,x)
Output:
(B*b^2*sin(c + d*x))/d - (B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d* x)/2))*2i)/d - (A*b^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i )/d - (C*a^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (C* b^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d + (A*a^2*sin(c + d*x))/(d*cos(c + d*x)) + (2*C*a*b*sin(c + d*x))/d - (A*a*b*atan((sin(c/ 2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d - (B*a*b*atanh((sin(c/2 + (d*x) /2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (C*b^2*cos(c + d*x)*sin(c + d*x))/(2*d )
Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.33 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) b^{2} c -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b c +2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{3}+2 \cos \left (d x +c \right ) a^{2} c d x +6 \cos \left (d x +c \right ) a \,b^{2} d x +\cos \left (d x +c \right ) b^{2} c d x +2 \sin \left (d x +c \right ) a^{3}}{2 \cos \left (d x +c \right ) d} \] Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)
Output:
(cos(c + d*x)**2*sin(c + d*x)*b**2*c - 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b + 4*cos(c + d*x)*sin(c + d*x)*a*b*c + 2*cos(c + d*x)*sin(c + d*x)*b**3 + 2*cos(c + d* x)*a**2*c*d*x + 6*cos(c + d*x)*a*b**2*d*x + cos(c + d*x)*b**2*c*d*x + 2*si n(c + d*x)*a**3)/(2*cos(c + d*x)*d)