Integrand size = 21, antiderivative size = 73 \[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=-\frac {a^2 x}{2}+\frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^2 \tan (c+d x)}{d} \]
-1/2*a^2*x+2*a^2*arctanh(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d-1/2*a^2*cos(d*x+ c)*sin(d*x+c)/d+a^2*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(243\) vs. \(2(73)=146\).
Time = 1.22 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.33 \[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {1}{16} a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-2 x-\frac {8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {8 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {8 \cos (d x) \sin (c)}{d}-\frac {\cos (2 d x) \sin (2 c)}{d}-\frac {8 \cos (c) \sin (d x)}{d}-\frac {\cos (2 c) \sin (2 d x)}{d}+\frac {4 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \]
(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-2*x - (8*Log[Cos[(c + d*x)/ 2] - Sin[(c + d*x)/2]])/d + (8*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d - (8*Cos[d*x]*Sin[c])/d - (Cos[2*d*x]*Sin[2*c])/d - (8*Cos[c]*Sin[d*x])/d - (Cos[2*c]*Sin[2*d*x])/d + (4*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Co s[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[ c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/16
Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4359, 3042, 3188, 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.
\(\displaystyle \int \sin ^2(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 4359 |
\(\displaystyle \int \tan ^2(c+d x) (a (-\cos (c+d x))-a)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\tan \left (c+d x-\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3188 |
\(\displaystyle \frac {\int \left (-\cos ^2(c+d x) a^4+\sec ^2(c+d x) a^4-2 \cos (c+d x) a^4+2 \sec (c+d x) a^4\right )dx}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a^4 \sin (c+d x)}{d}+\frac {a^4 \tan (c+d x)}{d}-\frac {a^4 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {a^4 x}{2}}{a^2}\) |
(-1/2*(a^4*x) + (2*a^4*ArcTanh[Sin[c + d*x]])/d - (2*a^4*Sin[c + d*x])/d - (a^4*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a^4*Tan[c + d*x])/d)/a^2
3.1.32.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ ), x_Symbol] :> Simp[a^p Int[ExpandIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m _.), x_Symbol] :> Int[Cot[e + f*x]^p*(b + a*Sin[e + f*x])^m, x] /; FreeQ[{a , b, e, f, p}, x] && IntegerQ[m] && EqQ[m, p]
Time = 1.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(78\) |
default | \(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(78\) |
parts | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}+\frac {2 a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(83\) |
parallelrisch | \(-\frac {a^{2} \left (4 d x \cos \left (d x +c \right )+16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-7 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )+8 \sin \left (2 d x +2 c \right )\right )}{8 d \cos \left (d x +c \right )}\) | \(96\) |
risch | \(-\frac {a^{2} x}{2}+\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(144\) |
norman | \(\frac {\frac {a^{2} x}{2}+\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {6 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\frac {a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(186\) |
1/d*(a^2*(tan(d*x+c)-d*x-c)+2*a^2*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+ a^2*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))
Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.42 \[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=-\frac {a^{2} d x \cos \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{2} \cos \left (d x + c\right ) - 2 \, a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
-1/2*(a^2*d*x*cos(d*x + c) - 2*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) + 2* a^2*cos(d*x + c)*log(-sin(d*x + c) + 1) + (a^2*cos(d*x + c)^2 + 4*a^2*cos( d*x + c) - 2*a^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=a^{2} \left (\int 2 \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(2*sin(c + d*x)**2*sec(c + d*x), x) + Integral(sin(c + d*x)* *2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**2, x))
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11 \[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} + 4 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a^2 - 4*(d*x + c - tan(d*x + c))*a^2 + 4*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c))) /d
Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.75 \[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=-\frac {{\left (d x + c\right )} a^{2} - 4 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 4 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
-1/2*((d*x + c)*a^2 - 4*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 4*a^2*log (abs(tan(1/2*d*x + 1/2*c) - 1)) + 4*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 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 = 13.58 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.60 \[ \int (a+a \sec (c+d x))^2 \sin ^2(c+d x) \, dx=\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,x}{2}+\frac {5\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]