Integrand size = 21, antiderivative size = 80 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
a*b*arctanh(sin(d*x+c))/d+1/3*(2*a^2+3*b^2)*tan(d*x+c)/d+a*b*sec(d*x+c)*ta n(d*x+c)/d+1/3*a^2*sec(d*x+c)^2*tan(d*x+c)/d
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {a^2 \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4,x]
Output:
(a*b*ArcTanh[Sin[c + d*x]])/d + (b^2*Tan[c + d*x])/d + (a*b*Sec[c + d*x]*T an[c + d*x])/d + (a^2*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d
Time = 0.53 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3268, 3042, 3491, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3268 |
| \(\displaystyle \int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^4(c+d x)dx+2 a b \int \sec ^3(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
| \(\Big \downarrow \) 3491 |
| \(\displaystyle \frac {1}{3} \left (2 a^2+3 b^2\right ) \int \sec ^2(c+d x)dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (2 a^2+3 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {\left (2 a^2+3 b^2\right ) \int 1d(-\tan (c+d x))}{3 d}+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle 2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle 2 a b \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 a b \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}+2 a b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\) |
Input:
Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4,x]
Output:
((2*a^2 + 3*b^2)*Tan[c + d*x])/(3*d) + (a^2*Sec[c + d*x]^2*Tan[c + d*x])/( 3*d) + 2*a*b*(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 _)])^2, x_Symbol] :> Simp[2*c*(d/b) Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
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 = 5.42 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b^{2}}{d}\) | \(74\) |
| default | \(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\tan \left (d x +c \right ) b^{2}}{d}\) | \(74\) |
| parts | \(-\frac {a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{2} \tan \left (d x +c \right )}{d}+\frac {a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(79\) |
| risch | \(-\frac {2 i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-2 a^{2}-3 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(139\) |
| parallelrisch | \(\frac {-9 a \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 a \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (2 a^{2}+3 b^{2}\right ) \sin \left (3 d x +3 c \right )+6 a b \sin \left (2 d x +2 c \right )+\left (6 a^{2}+3 b^{2}\right ) \sin \left (d x +c \right )}{3 d \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\) | \(147\) |
| norman | \(\frac {-\frac {4 \left (a^{2}-3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {2 \left (a^{2}-a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {2 \left (a^{2}+a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a \left (2 a -3 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 a \left (3 b +2 a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}+\frac {a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(195\) |
Input:
int((a+cos(d*x+c)*b)^2*sec(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(-a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+2*a*b*(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^2)
Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x, algorithm="fricas")
Output:
1/6*(3*a*b*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a*b*cos(d*x + c)^3*log (-sin(d*x + c) + 1) + 2*(3*a*b*cos(d*x + c) + (2*a^2 + 3*b^2)*cos(d*x + c) ^2 + a^2)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**4,x)
Output:
Integral((a + b*cos(c + d*x))**2*sec(c + d*x)**4, x)
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, b^{2} \tan \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x, algorithm="maxima")
Output:
1/6*(2*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 - 3*a*b*(2*sin(d*x + c)/(sin( d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*b^2*t an(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).
Time = 0.53 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.22 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x, algorithm="giac")
Output:
1/3*(3*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2 *c)^5 + 3*b^2*tan(1/2*d*x + 1/2*c)^5 - 2*a^2*tan(1/2*d*x + 1/2*c)^3 - 6*b^ 2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 41.65 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.76 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {2\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,a^2-2\,a\,b+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,a^2}{3}-4\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+2\,a\,b+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*cos(c + d*x))^2/cos(c + d*x)^4,x)
Output:
(2*a*b*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)^5*(2*a^2 - 2*a*b + 2*b^2) - tan(c/2 + (d*x)/2)^3*((4*a^2)/3 + 4*b^2) + tan(c/2 + (d*x)/2)* (2*a*b + 2*a^2 + 2*b^2))/(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 = 0.17 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.42 \[ \int (a+b \cos (c+d x))^2 \sec ^4(c+d x) \, dx=\frac {-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a b -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +2 \sin \left (d x +c \right )^{3} a^{2}+3 \sin \left (d x +c \right )^{3} b^{2}-3 \sin \left (d x +c \right ) a^{2}-3 \sin \left (d x +c \right ) b^{2}}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))^2*sec(d*x+c)^4,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b + 3*cos(c + d*x)*log(tan((c + d*x)/2 ) + 1)*sin(c + d*x)**2*a*b - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b - 3*cos(c + d*x)*sin(c + d*x)*a*b + 2*sin(c + d*x)**3*a**2 + 3*sin(c + d*x )**3*b**2 - 3*sin(c + d*x)*a**2 - 3*sin(c + d*x)*b**2)/(3*cos(c + d*x)*d*( sin(c + d*x)**2 - 1))