Integrand size = 21, antiderivative size = 54 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
3/2*a^2*arctanh(sin(d*x+c))/d+2*a^2*tan(d*x+c)/d+1/2*a^2*sec(d*x+c)*tan(d* x+c)/d
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {a^2 \coth ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3,x]
Output:
(a^2*ArcCoth[Sin[c + d*x]])/d + (a^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a^2 *Tan[c + d*x])/d + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.28 (sec) , antiderivative size = 54, 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.
\(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^2 \sec ^3(c+d x)+2 a^2 \sec ^2(c+d x)+a^2 \sec (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 a^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\) |
Input:
Int[(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3,x]
Output:
(3*a^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)
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 = 3.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30
method | result | size |
derivativedivides | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} \tan \left (d x +c \right )+a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(70\) |
default | \(\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} \tan \left (d x +c \right )+a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(70\) |
parts | \(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} \tan \left (d x +c \right )}{d}\) | \(75\) |
risch | \(-\frac {i a^{2} \left ({\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}\) | \(99\) |
parallelrisch | \(\frac {a^{2} \left (3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (2 d x +2 c \right )-3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (2 d x +2 c \right )+4 \sin \left (2 d x +2 c \right )+2 \sin \left (d x +c \right )+3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )\right )}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(116\) |
norman | \(\frac {\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {7 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(148\) |
Input:
int((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a^2*tan(d*x+c)+a^2*(1/2*sec(d*x+c)*ta n(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {3 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="fricas")
Output:
1/4*(3*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*a^2*cos(d*x + c)^2*log (-sin(d*x + c) + 1) + 2*(4*a^2*cos(d*x + c) + a^2)*sin(d*x + c))/(d*cos(d* x + c)^2)
\[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=a^{2} \left (\int 2 \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*cos(d*x+c))**2*sec(d*x+c)**3,x)
Output:
a**2*(Integral(2*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(cos(c + d*x)* *2*sec(c + d*x)**3, x) + Integral(sec(c + d*x)**3, x))
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.63 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=-\frac {a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8 \, a^{2} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="maxima")
Output:
-1/4*(a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + l og(sin(d*x + c) - 1)) - 2*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 8*a^2*tan(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="giac")
Output:
1/2*(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)^3 - 5*a^2*tan(1/2*d*x + 1/2 *c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 44.76 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((a + a*cos(c + d*x))^2/cos(c + d*x)^3,x)
Output:
(3*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (3*a^2*tan(c/2 + (d*x)/2)^3 - 5*a^2* tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1) )
Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.91 \[ \int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {a^{2} \left (-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\cos \left (d x +c \right ) \sin \left (d x +c \right )+4 \sin \left (d x +c \right )^{3}-4 \sin \left (d x +c \right )\right )}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x)
Output:
(a**2*( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 3*cos (c + d*x)*log(tan((c + d*x)/2) - 1) + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1) - cos(c + d*x)*sin(c + d*x) + 4*sin(c + d*x)**3 - 4*sin(c + d*x)))/(2*cos(c + d*x)*d *(sin(c + d*x)**2 - 1))