Integrand size = 19, antiderivative size = 47 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \tan (c+d x)}{d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*a*arctanh(sin(d*x+c))/d+b*tan(d*x+c)/d+1/2*a*sec(d*x+c)*tan(d*x+c)/d
Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \tan (c+d x)}{d}+\frac {a \sec (c+d x) \tan (c+d x)}{2 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*Sec[c + d*x]^3,x]
Output:
(a*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Tan[c + d*x])/d + (a*Sec[c + d*x]*Tan [c + d*x])/(2*d)
Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3227, 3042, 4254, 24, 4255, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle a \int \sec ^3(c+d x)dx+b \int \sec ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {b \int 1d(-\tan (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {b \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle a \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b \tan (c+d x)}{d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle a \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b \tan (c+d x)}{d}\) |
Input:
Int[(a + b*Cos[c + d*x])*Sec[c + d*x]^3,x]
Output:
(b*Tan[c + d*x])/d + a*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d))
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.69 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b}{d}\) | \(47\) |
default | \(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b}{d}\) | \(47\) |
parts | \(\frac {a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b \tan \left (d x +c \right )}{d}\) | \(49\) |
parallelrisch | \(\frac {-a \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+a \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 a \sin \left (d x +c \right )+2 b \sin \left (2 d x +2 c \right )}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(92\) |
risch | \(-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )} a -2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}-2 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(98\) |
norman | \(\frac {\frac {\left (a -2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {\left (a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(125\) |
Input:
int((a+cos(d*x+c)*b)*sec(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+tan(d*x+c )*b)
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="fricas")
Output:
1/4*(a*cos(d*x + c)^2*log(sin(d*x + c) + 1) - a*cos(d*x + c)^2*log(-sin(d* x + c) + 1) + 2*(2*b*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)**3,x)
Output:
Integral((a + b*cos(c + d*x))*sec(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=-\frac {a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="maxima")
Output:
-1/4*(a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log (sin(d*x + c) - 1)) - 4*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (43) = 86\).
Time = 0.58 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.23 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="giac")
Output:
1/2*(a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(a*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1/2*c)^3 + a*tan( 1/2*d*x + 1/2*c) + 2*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^ 2)/d
Time = 40.45 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.72 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {\left (a-2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a+2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int((a + b*cos(c + d*x))/cos(c + d*x)^3,x)
Output:
(a*atanh(tan(c/2 + (d*x)/2)))/d + (tan(c/2 + (d*x)/2)^3*(a - 2*b) + tan(c/ 2 + (d*x)/2)*(a + 2*b))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1))
Time = 0.16 (sec) , antiderivative size = 159, normalized size of antiderivative = 3.38 \[ \int (a+b \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\cos \left (d x +c \right ) \sin \left (d x +c \right ) a +2 \sin \left (d x +c \right )^{3} b -2 \sin \left (d x +c \right ) b}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))*sec(d*x+c)^3,x)
Output:
( - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + cos(c + d*x )*log(tan((c + d*x)/2) - 1)*a + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin (c + d*x)**2*a - cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a - cos(c + d*x)*s in(c + d*x)*a + 2*sin(c + d*x)**3*b - 2*sin(c + d*x)*b)/(2*cos(c + d*x)*d* (sin(c + d*x)**2 - 1))