Integrand size = 19, antiderivative size = 63 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \tan (c+d x)}{d}+\frac {b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a \tan ^3(c+d x)}{3 d} \] Output:
1/2*b*arctanh(sin(d*x+c))/d+a*tan(d*x+c)/d+1/2*b*sec(d*x+c)*tan(d*x+c)/d+1 /3*a*tan(d*x+c)^3/d
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \] Input:
Integrate[(a + b*Cos[c + d*x])*Sec[c + d*x]^4,x]
Output:
(b*ArcTanh[Sin[c + d*x]])/(2*d) + (b*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a *(Tan[c + d*x] + Tan[c + d*x]^3/3))/d
Time = 0.40 (sec) , antiderivative size = 64, 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, 2009, 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.
\(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle a \int \sec ^4(c+d x)dx+b \int \sec ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {a \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle b \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
Input:
Int[(a + b*Cos[c + d*x])*Sec[c + d*x]^4,x]
Output:
b*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (a*( -Tan[c + d*x] - Tan[c + d*x]^3/3))/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 = 4.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(60\) |
default | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(60\) |
parts | \(-\frac {a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(62\) |
risch | \(-\frac {i \left (3 b \,{\mathrm e}^{5 i \left (d x +c \right )}-12 a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} b -4 a \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(99\) |
parallelrisch | \(\frac {-9 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 b \sin \left (2 d x +2 c \right )+4 a \left (\sin \left (3 d x +3 c \right )+3 \sin \left (d x +c \right )\right )}{6 d \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\) | \(126\) |
norman | \(\frac {-\frac {\left (2 a -3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {\left (2 a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (2 a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (3 b +2 a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(158\) |
Input:
int((a+cos(d*x+c)*b)*sec(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b*(1/2*sec(d*x+c)*tan(d*x+c)+1/ 2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.40 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {3 \, b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a \cos \left (d x + c\right )^{2} + 3 \, b \cos \left (d x + c\right ) + 2 \, a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")
Output:
1/12*(3*b*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*b*cos(d*x + c)^3*log(-s in(d*x + c) + 1) + 2*(4*a*cos(d*x + c)^2 + 3*b*cos(d*x + c) + 2*a)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)**4,x)
Output:
Integral((a + b*cos(c + d*x))*sec(c + d*x)**4, x)
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a - 3 \, b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{12 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="maxima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a - 3*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (57) = 114\).
Time = 0.55 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.94 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")
Output:
1/6*(3*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*b*log(abs(tan(1/2*d*x + 1/ 2*c) - 1)) - 2*(6*a*tan(1/2*d*x + 1/2*c)^5 - 3*b*tan(1/2*d*x + 1/2*c)^5 - 4*a*tan(1/2*d*x + 1/2*c)^3 + 6*a*tan(1/2*d*x + 1/2*c) + 3*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 41.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.76 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\left (2\,a+b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*cos(c + d*x))/cos(c + d*x)^4,x)
Output:
(b*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)^5*(2*a - b) + tan(c/ 2 + (d*x)/2)*(2*a + b) - (4*a*tan(c/2 + (d*x)/2)^3)/3)/(d*(3*tan(c/2 + (d* x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.56 \[ \int (a+b \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +4 \sin \left (d x +c \right )^{3} a -6 \sin \left (d x +c \right ) a}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))*sec(d*x+c)^4,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 3*cos( c + d*x)*sin(c + d*x)*b + 4*sin(c + d*x)**3*a - 6*sin(c + d*x)*a)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))