Integrand size = 21, antiderivative size = 80 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
a*b*arctanh(sin(d*x+c))/d+1/3*(3*a^2+2*b^2)*tan(d*x+c)/d+a*b*sec(d*x+c)*ta n(d*x+c)/d+1/3*b^2*sec(d*x+c)^2*tan(d*x+c)/d
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \left (\tan (c+d x)+\frac {1}{3} \tan ^3(c+d x)\right )}{d} \]
(a*b*ArcTanh[Sin[c + d*x]])/d + (a^2*Tan[c + d*x])/d + (a*b*Sec[c + d*x]*T an[c + d*x])/d + (b^2*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d
Time = 0.55 (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, 4275, 3042, 4255, 3042, 4257, 4534, 3042, 4254, 24}
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.
| \(\displaystyle \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle \int \sec ^2(c+d x) \left (a^2+b^2 \sec ^2(c+d x)\right )dx+2 a b \int \sec ^3(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{3} \left (3 a^2+2 b^2\right ) \int \sec ^2(c+d x)dx+2 a b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 a^2+2 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+2 a b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {\left (3 a^2+2 b^2\right ) \int 1d(-\tan (c+d x))}{3 d}+2 a b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{3 d}+2 a b \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {b^2 \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
((3*a^2 + 2*b^2)*Tan[c + d*x])/(3*d) + (b^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))
3.5.58.3.1 Defintions of rubi rules used
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 1.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {\tan \left (d x +c \right ) a^{2}+2 a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(74\) |
| default | \(\frac {\tan \left (d x +c \right ) a^{2}+2 a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(74\) |
| parts | \(\frac {a^{2} \tan \left (d x +c \right )}{d}-\frac {b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(79\) |
| norman | \(\frac {\frac {4 \left (3 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {2 \left (a^{2}-a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {2 \left (a^{2}+a b +b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(132\) |
| risch | \(-\frac {2 i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-3 a^{2}-2 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(139\) |
| parallelrisch | \(\frac {-9 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (3 a^{2}+2 b^{2}\right ) \sin \left (3 d x +3 c \right )+6 a b \sin \left (2 d x +2 c \right )+3 \sin \left (d x +c \right ) \left (a^{2}+2 b^{2}\right )}{3 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(146\) |
1/d*(tan(d*x+c)*a^2+2*a*b*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan (d*x+c)))-b^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (3 \, a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]
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) + (3*a^2 + 2*b^2)*cos(d*x + c) ^2 + b^2)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{2} - 3 \, a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{2} \tan \left (d x + c\right )}{6 \, d} \]
1/6*(2*(tan(d*x + c)^3 + 3*tan(d*x + c))*b^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*a^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 = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.22 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]
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 - 6*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*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 = 15.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.76 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,a^2-2\,a\,b+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,a^2-\frac {4\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+2\,a\,b+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]