Integrand size = 19, antiderivative size = 51 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a x}{2}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
1/2*a*x+a*arctanh(sin(d*x+c))/d-a*sin(d*x+c)/d-1/2*a*cos(d*x+c)*sin(d*x+c) /d
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a (c+d x)}{2 d}+\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin (2 (c+d x))}{4 d} \] Input:
Integrate[(a + a*Sec[c + d*x])*Sin[c + d*x]^2,x]
Output:
(a*(c + d*x))/(2*d) + (a*ArcTanh[Sin[c + d*x]])/d - (a*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.
\(\displaystyle \int \sin ^2(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -(\sin (c+d x) \tan (c+d x) (a (-\cos (c+d x))-a))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -((\cos (c+d x) a+a) \sin (c+d x) \tan (c+d x))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \sin (c+d x) \tan (c+d x) (a \cos (c+d x)+a)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \sin ^2(c+d x)dx+a \int \sin (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin (c+d x)^2dx+a \int \sin (c+d x) \tan (c+d x)dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {a \int \frac {\sin ^2(c+d x)}{1-\sin ^2(c+d x)}d\sin (c+d x)}{d}+a \int \sin (c+d x)^2dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {a \left (\int \frac {1}{1-\sin ^2(c+d x)}d\sin (c+d x)-\sin (c+d x)\right )}{d}+a \int \sin (c+d x)^2dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \int \sin (c+d x)^2dx+\frac {a (\text {arctanh}(\sin (c+d x))-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {a (\text {arctanh}(\sin (c+d x))-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {a (\text {arctanh}(\sin (c+d x))-\sin (c+d x))}{d}+a \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\) |
Input:
Int[(a + a*Sec[c + d*x])*Sin[c + d*x]^2,x]
Output:
(a*(ArcTanh[Sin[c + d*x]] - Sin[c + d*x]))/d + a*(x/2 - (Cos[c + d*x]*Sin[ c + d*x])/(2*d))
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 = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(\frac {a \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(55\) |
default | \(\frac {a \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(55\) |
parallelrisch | \(-\frac {a \left (-2 d x -4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \sin \left (d x +c \right )+\sin \left (2 d x +2 c \right )\right )}{4 d}\) | \(57\) |
parts | \(\frac {a \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(57\) |
risch | \(\frac {a x}{2}+\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(90\) |
norman | \(\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {a x}{2}-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(119\) |
Input:
int((a+a*sec(d*x+c))*sin(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a*(-1/2*cos(d*x+c)*sin(d*x+ c)+1/2*d*x+1/2*c))
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a d x + a \log \left (\sin \left (d x + c\right ) + 1\right ) - a \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a \cos \left (d x + c\right ) + 2 \, a\right )} \sin \left (d x + c\right )}{2 \, d} \] Input:
integrate((a+a*sec(d*x+c))*sin(d*x+c)^2,x, algorithm="fricas")
Output:
1/2*(a*d*x + a*log(sin(d*x + c) + 1) - a*log(-sin(d*x + c) + 1) - (a*cos(d *x + c) + 2*a)*sin(d*x + c))/d
\[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=a \left (\int \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))*sin(d*x+c)**2,x)
Output:
a*(Integral(sin(c + d*x)**2*sec(c + d*x), x) + Integral(sin(c + d*x)**2, x ))
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a + 2 \, a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \] Input:
integrate((a+a*sec(d*x+c))*sin(d*x+c)^2,x, algorithm="maxima")
Output:
1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a + 2*a*(log(sin(d*x + c) + 1) - log (sin(d*x + c) - 1) - 2*sin(d*x + c)))/d
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.73 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {{\left (d x + c\right )} a + 2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:
integrate((a+a*sec(d*x+c))*sin(d*x+c)^2,x, algorithm="giac")
Output:
1/2*((d*x + c)*a + 2*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*a*log(abs(ta n(1/2*d*x + 1/2*c) - 1)) - 2*(a*tan(1/2*d*x + 1/2*c)^3 + 3*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 10.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.57 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a\,x}{2}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \] Input:
int(sin(c + d*x)^2*(a + a/cos(c + d*x)),x)
Output:
(a*x)/2 - (3*a*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^3)/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) + (2*a*atanh(tan(c/2 + (d*x)/2)) )/d
Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18 \[ \int (a+a \sec (c+d x)) \sin ^2(c+d x) \, dx=\frac {a \left (-\cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \sin \left (d x +c \right )+d x \right )}{2 d} \] Input:
int((a+a*sec(d*x+c))*sin(d*x+c)^2,x)
Output:
(a*( - cos(c + d*x)*sin(c + d*x) - 2*log(tan((c + d*x)/2) - 1) + 2*log(tan ((c + d*x)/2) + 1) - 2*sin(c + d*x) + d*x))/(2*d)