Integrand size = 31, antiderivative size = 58 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=a C x+\frac {a (A+2 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a A \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \]
Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.16 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=a C x+\frac {a A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a C \text {arctanh}(\sin (c+d x))}{d}+\frac {a A \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \]
a*C*x + (a*A*ArcTanh[Sin[c + d*x]])/(2*d) + (a*C*ArcTanh[Sin[c + d*x]])/d + (a*A*Tan[c + d*x])/d + (a*A*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.52 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 3511, 3042, 3500, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3511 |
| \(\displaystyle \frac {1}{2} \int \left (2 a C \cos ^2(c+d x)+a (A+2 C) \cos (c+d x)+2 a A\right ) \sec ^2(c+d x)dx+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {2 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a A}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{2} \left (\int (a (A+2 C)+2 a C \cos (c+d x)) \sec (c+d x)dx+\frac {2 a A \tan (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {a (A+2 C)+2 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a A \tan (c+d x)}{d}\right )+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a (A+2 C) \int \sec (c+d x)dx+\frac {2 a A \tan (c+d x)}{d}+2 a C x\right )+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a (A+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a A \tan (c+d x)}{d}+2 a C x\right )+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (A+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a A \tan (c+d x)}{d}+2 a C x\right )+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d}\) |
(a*A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a*C*x + (a*(A + 2*C)*ArcTanh[Si n[c + d*x]])/d + (2*a*A*Tan[c + d*x])/d)/2
3.1.6.3.1 Defintions of rubi rules used
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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-(b*c - a*d))*(A*b^2 + 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] + Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d )) - ((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, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 5.78 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28
| method | result | size |
| derivativedivides | \(\frac {a A \tan \left (d x +c \right )+a C \left (d x +c \right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(74\) |
| default | \(\frac {a A \tan \left (d x +c \right )+a C \left (d x +c \right )+a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(74\) |
| parts | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a C \left (d x +c \right )}{d}\) | \(82\) |
| parallelrisch | \(-\frac {\left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 d x C \cos \left (2 d x +2 c \right )-2 d x C -2 A \sin \left (d x +c \right )-2 \sin \left (2 d x +2 c \right ) A \right ) a}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(120\) |
| risch | \(a C x -\frac {i a A \left ({\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(139\) |
| norman | \(\frac {a C x +a C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {8 a A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-2 a C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(231\) |
1/d*(a*A*tan(d*x+c)+a*C*(d*x+c)+a*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec( d*x+c)+tan(d*x+c)))+a*C*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.74 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, C a d x \cos \left (d x + c\right )^{2} + {\left (A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, C\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a \cos \left (d x + c\right ) + A a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(4*C*a*d*x*cos(d*x + c)^2 + (A + 2*C)*a*cos(d*x + c)^2*log(sin(d*x + c ) + 1) - (A + 2*C)*a*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a*cos( d*x + c) + A*a)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)*sec(c + d*x)** 3, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**3, x))
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.64 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a - A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a \tan \left (d x + c\right )}{4 \, d} \]
1/4*(4*(d*x + c)*C*a - A*a*(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*(log(sin(d*x + c) + 1) - lo g(sin(d*x + c) - 1)) + 4*A*a*tan(d*x + c))/d
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.81 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} C a + {\left (A a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a + 2 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(2*(d*x + c)*C*a + (A*a + 2*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A*a + 2*C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(A*a*tan(1/2*d*x + 1/ 2*c)^3 - 3*A*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 1.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.21 \[ \int (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {A\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]