Integrand size = 19, antiderivative size = 51 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a x}{2}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \]
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a (c+d x)}{2 d}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin (c+d x)}{d}-\frac {a \sin (2 (c+d x))}{4 d} \]
(a*(c + d*x))/(2*d) + (b*ArcTanh[Sin[c + d*x]])/d - (b*Sin[c + d*x])/d - ( a*Sin[2*(c + d*x)])/(4*d)
Time = 0.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 4360, 25, 25, 3042, 3317, 3042, 3072, 262, 219, 3115, 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.
\(\displaystyle \int \sin ^2(c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -(\sin (c+d x) \tan (c+d x) (-a \cos (c+d x)-b))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -((b+a \cos (c+d x)) \sin (c+d x) \tan (c+d x))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \sin (c+d x) \tan (c+d x) (a \cos (c+d x)+b)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \sin ^2(c+d x)dx+b \int \sin (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin (c+d x)^2dx+b \int \sin (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle a \int \sin (c+d x)^2dx+\frac {b \int \frac {\sin ^2(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle a \int \sin (c+d x)^2dx+\frac {b \left (\int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)-\sin (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \int \sin (c+d x)^2dx+\frac {b (\text {arctanh}(\sin (c+d x))-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {b (\text {arctanh}(\sin (c+d x))-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {b (\text {arctanh}(\sin (c+d x))-\sin (c+d x))}{d}\) |
3.2.70.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 1.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(55\) |
default | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(55\) |
parts | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(57\) |
parallelrisch | \(\frac {2 a x d -4 \sin \left (d x +c \right ) b -a \sin \left (2 d x +2 c \right )+4 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\) | \(63\) |
risch | \(\frac {a x}{2}+\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(90\) |
norman | \(\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {\left (a -2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {a x}{2}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}-\frac {\left (a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(126\) |
1/d*(a*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+b*(-sin(d*x+c)+ln(sec(d* x+c)+tan(d*x+c))))
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a d x + b \log \left (\sin \left (d x + c\right ) + 1\right ) - b \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a \cos \left (d x + c\right ) + 2 \, b\right )} \sin \left (d x + c\right )}{2 \, d} \]
1/2*(a*d*x + b*log(sin(d*x + c) + 1) - b*log(-sin(d*x + c) + 1) - (a*cos(d *x + c) + 2*b)*sin(d*x + c))/d
\[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \sin ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a + 2 \, b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a + 2*b*(log(sin(d*x + c) + 1) - log (sin(d*x + c) - 1) - 2*sin(d*x + c)))/d
Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (47) = 94\).
Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.24 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (d x + c\right )} a + 2 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*((d*x + c)*a + 2*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*b*log(abs(ta n(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 = 13.93 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int (a+b \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {b\,\sin \left (c+d\,x\right )}{d} \]