Integrand size = 21, antiderivative size = 66 \[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 \tan ^3(c+d x)}{3 d} \]
a^2*arctanh(sin(d*x+c))/d+2*a^2*tan(d*x+c)/d+a^2*sec(d*x+c)*tan(d*x+c)/d+1 /3*a^2*tan(d*x+c)^3/d
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 \tan ^3(c+d x)}{3 d} \]
(a^2*ArcTanh[Sin[c + d*x]])/d + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x] *Tan[c + d*x])/d + (a^2*Tan[c + d*x]^3)/(3*d)
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3236, 2009}
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.
\(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^2 \sec ^4(c+d x)+2 a^2 \sec ^3(c+d x)+a^2 \sec ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{d}\) |
(a^2*ArcTanh[Sin[c + d*x]])/d + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x] *Tan[c + d*x])/d + (a^2*Tan[c + d*x]^3)/(3*d)
3.1.21.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Time = 3.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {a^{2} \tan \left (d x +c \right )+2 a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(75\) |
default | \(\frac {a^{2} \tan \left (d x +c \right )+2 a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(75\) |
parts | \(-\frac {a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} \tan \left (d x +c \right )}{d}+\frac {a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(81\) |
risch | \(-\frac {2 i a^{2} \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-3 \,{\mathrm e}^{4 i \left (d x +c \right )}-12 \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}-5\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(111\) |
parallelrisch | \(-\frac {3 \left (\left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\cos \left (d x +c \right )-\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )-\frac {2 \sin \left (2 d x +2 c \right )}{3}-\frac {5 \sin \left (3 d x +3 c \right )}{9}\right ) a^{2}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(124\) |
norman | \(\frac {-\frac {6 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {20 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(166\) |
1/d*(a^2*tan(d*x+c)+2*a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan (d*x+c)))-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {3 \, a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (5 \, a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]
1/6*(3*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a^2*cos(d*x + c)^3*log (-sin(d*x + c) + 1) + 2*(5*a^2*cos(d*x + c)^2 + 3*a^2*cos(d*x + c) + a^2)* sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=a^{2} \left (\int 2 \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(2*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(cos(c + d*x)* *2*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**4, x))
Time = 0.24 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.29 \[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} \tan \left (d x + c\right )}{6 \, d} \]
1/6*(2*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 - 3*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)) + 6*a^2*t an(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.61 \[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]
1/3*(3*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 8*a^2*tan(1/2*d*x + 1/2 *c)^3 + 9*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 15.60 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.70 \[ \int (a+a \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+6\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]