Integrand size = 27, antiderivative size = 48 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}-\frac {a \cot (c+d x)}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \] Output:
-a*arctanh(cos(d*x+c))/d-a*cot(d*x+c)/d+a*sec(d*x+c)/d+a*tan(d*x+c)/d
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.42 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cot (c+d x)}{d}-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \] Input:
Integrate[Csc[c + d*x]^2*Sec[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
-((a*Cot[c + d*x])/d) - (a*Log[Cos[(c + d*x)/2]])/d + (a*Log[Sin[(c + d*x) /2]])/d + (a*Sec[c + d*x])/d + (a*Tan[c + d*x])/d
Time = 0.41 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3317, 3042, 3100, 244, 2009, 3102, 25, 262, 219}
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.
\(\displaystyle \int \csc ^2(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \sin (c+d x)+a}{\sin (c+d x)^2 \cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \csc ^2(c+d x) \sec ^2(c+d x)dx+a \int \csc (c+d x) \sec ^2(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc (c+d x) \sec (c+d x)^2dx+a \int \csc (c+d x)^2 \sec (c+d x)^2dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {a \int \cot ^2(c+d x) \left (\tan ^2(c+d x)+1\right )d\tan (c+d x)}{d}+a \int \csc (c+d x) \sec (c+d x)^2dx\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {a \int \left (\cot ^2(c+d x)+1\right )d\tan (c+d x)}{d}+a \int \csc (c+d x) \sec (c+d x)^2dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \csc (c+d x) \sec (c+d x)^2dx+\frac {a (\tan (c+d x)-\cot (c+d x))}{d}\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {a \int -\frac {\sec ^2(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{d}+\frac {a (\tan (c+d x)-\cot (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {a (\tan (c+d x)-\cot (c+d x))}{d}-\frac {a \int \frac {\sec ^2(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {a \left (\sec (c+d x)-\int \frac {1}{1-\sec ^2(c+d x)}d\sec (c+d x)\right )}{d}+\frac {a (\tan (c+d x)-\cot (c+d x))}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a (\sec (c+d x)-\text {arctanh}(\sec (c+d x)))}{d}+\frac {a (\tan (c+d x)-\cot (c+d x))}{d}\) |
Input:
Int[Csc[c + d*x]^2*Sec[c + d*x]^2*(a + a*Sin[c + d*x]),x]
Output:
(a*(-ArcTanh[Sec[c + d*x]] + Sec[c + d*x]))/d + (a*(-Cot[c + d*x] + Tan[c + d*x]))/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] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 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[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[1/f Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] , x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, 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]
Time = 0.60 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.27
method | result | size |
derivativedivides | \(\frac {a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )}{d}\) | \(61\) |
default | \(\frac {a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a \left (\frac {1}{\sin \left (d x +c \right ) \cos \left (d x +c \right )}-2 \cot \left (d x +c \right )\right )}{d}\) | \(61\) |
parallelrisch | \(\frac {\left (\left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-6\right ) a}{2 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(66\) |
risch | \(\frac {-4 a -2 i a \,{\mathrm e}^{i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\) | \(97\) |
norman | \(\frac {\frac {a}{2 d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {5 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 d}+\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(149\) |
Input:
int(csc(d*x+c)^2*sec(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+a*(1/sin(d*x+c)/cos(d*x+c) -2*cot(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (48) = 96\).
Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 3.44 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + {\left (a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (a \cos \left (d x + c\right )^{2} + {\left (a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - 2 \, a}{2 \, {\left (d \cos \left (d x + c\right )^{2} + {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/2*(4*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) + (a*cos(d*x + c)^2 + (a*cos(d *x + c) + a)*sin(d*x + c) - a)*log(1/2*cos(d*x + c) + 1/2) - (a*cos(d*x + c)^2 + (a*cos(d*x + c) + a)*sin(d*x + c) - a)*log(-1/2*cos(d*x + c) + 1/2) - 2*(2*a*cos(d*x + c) + a)*sin(d*x + c) - 2*a)/(d*cos(d*x + c)^2 + (d*cos (d*x + c) + d)*sin(d*x + c) - d)
\[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=a \left (\int \csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(csc(d*x+c)**2*sec(d*x+c)**2*(a+a*sin(d*x+c)),x)
Output:
a*(Integral(csc(c + d*x)**2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)*cs c(c + d*x)**2*sec(c + d*x)**2, x))
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, a {\left (\frac {1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )}}{2 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/2*(a*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 2*a*(1/tan(d*x + c) - tan(d*x + c)))/d
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.81 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {2 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \] Input:
integrate(csc(d*x+c)^2*sec(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/2*(2*a*log(abs(tan(1/2*d*x + 1/2*c))) + a*tan(1/2*d*x + 1/2*c) - (a*tan( 1/2*d*x + 1/2*c)^2 + 4*a*tan(1/2*d*x + 1/2*c) - a)/(tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c)))/d
Time = 31.65 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.60 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a-5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )} \] Input:
int((a + a*sin(c + d*x))/(cos(c + d*x)^2*sin(c + d*x)^2),x)
Output:
(a*tan(c/2 + (d*x)/2))/(2*d) + (a*log(tan(c/2 + (d*x)/2)))/d - (a - 5*a*ta n(c/2 + (d*x)/2))/(d*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^2))
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.08 \[ \int \csc ^2(c+d x) \sec ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )}{2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:
int(csc(d*x+c)^2*sec(d*x+c)^2*(a+a*sin(d*x+c)),x)
Output:
(a*(2*log(tan((c + d*x)/2))*tan((c + d*x)/2)**2 - 2*log(tan((c + d*x)/2))* tan((c + d*x)/2) + tan((c + d*x)/2)**3 - 6*tan((c + d*x)/2)**2 + 1))/(2*ta n((c + d*x)/2)*d*(tan((c + d*x)/2) - 1))