Integrand size = 19, antiderivative size = 75 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos (c+d x)}{a^3 d}-\frac {1}{2 a d (a+a \cos (c+d x))^2}+\frac {3}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {3 \log (1+\cos (c+d x))}{a^3 d} \] Output:
-cos(d*x+c)/a^3/d-1/2/a/d/(a+a*cos(d*x+c))^2+3/d/(a^3+a^3*cos(d*x+c))+3*ln (1+cos(d*x+c))/a^3/d
Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.37 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (21-2 \cos (3 (c+d x))+72 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\cos (2 (c+d x)) \left (-5+24 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos (c+d x) \left (22+96 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{4 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[Sin[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
(Cos[(c + d*x)/2]^2*(21 - 2*Cos[3*(c + d*x)] + 72*Log[Cos[(c + d*x)/2]] + Cos[2*(c + d*x)]*(-5 + 24*Log[Cos[(c + d*x)/2]]) + Cos[c + d*x]*(22 + 96*L og[Cos[(c + d*x)/2]])))/(4*a^3*d*(1 + Cos[c + d*x])^3)
Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 4360, 25, 25, 3042, 25, 3312, 27, 49, 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.
\(\displaystyle \int \frac {\sin (c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\sin (c+d x) \cos ^3(c+d x)}{(a (-\cos (c+d x))-a)^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos ^3(c+d x) \sin (c+d x)}{(\cos (c+d x) a+a)^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sin (c+d x) \cos ^3(c+d x)}{(a \cos (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \cos \left (c+d x+\frac {\pi }{2}\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right ) \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}{\left (\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^3}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle -\frac {\int \frac {\cos ^3(c+d x)}{(\cos (c+d x) a+a)^3}d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^3 \cos ^3(c+d x)}{(\cos (c+d x) a+a)^3}d(a \cos (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {\int \left (-\frac {a^3}{(\cos (c+d x) a+a)^3}+\frac {3 a^2}{(\cos (c+d x) a+a)^2}-\frac {3 a}{\cos (c+d x) a+a}+1\right )d(a \cos (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {a^3}{2 (a \cos (c+d x)+a)^2}-\frac {3 a^2}{a \cos (c+d x)+a}+a \cos (c+d x)-3 a \log (a \cos (c+d x)+a)}{a^4 d}\) |
Input:
Int[Sin[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
-((a*Cos[c + d*x] + a^3/(2*(a + a*Cos[c + d*x])^2) - (3*a^2)/(a + a*Cos[c + d*x]) - 3*a*Log[a + a*Cos[c + d*x]])/(a^4*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, 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 = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79
method | result | size |
parallelrisch | \(\frac {-29-24 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+12 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-8 \cos \left (d x +c \right )}{8 d \,a^{3}}\) | \(59\) |
derivativedivides | \(\frac {-\frac {1}{2 \left (1+\sec \left (d x +c \right )\right )^{2}}-\frac {2}{1+\sec \left (d x +c \right )}+3 \ln \left (1+\sec \left (d x +c \right )\right )-\frac {1}{\sec \left (d x +c \right )}-3 \ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}}\) | \(63\) |
default | \(\frac {-\frac {1}{2 \left (1+\sec \left (d x +c \right )\right )^{2}}-\frac {2}{1+\sec \left (d x +c \right )}+3 \ln \left (1+\sec \left (d x +c \right )\right )-\frac {1}{\sec \left (d x +c \right )}-3 \ln \left (\sec \left (d x +c \right )\right )}{d \,a^{3}}\) | \(63\) |
norman | \(\frac {\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d a}-\frac {13}{4 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2}}-\frac {3 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}\) | \(90\) |
risch | \(-\frac {3 i x}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}-\frac {6 i c}{d \,a^{3}}+\frac {6 \,{\mathrm e}^{3 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}}{a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4}}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) | \(128\) |
Input:
int(sin(d*x+c)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8*(-29-24*ln(sec(1/2*d*x+1/2*c)^2)-sec(1/2*d*x+1/2*c)^4+12*sec(1/2*d*x+1 /2*c)^2-8*cos(d*x+c))/d/a^3
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.28 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - 6 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, \cos \left (d x + c\right ) - 5}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(sin(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
-1/2*(2*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - 6*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - 4*cos(d*x + c) - 5)/(a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sin(d*x+c)/(a+a*sec(d*x+c))**3,x)
Output:
Integral(sin(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x ) + 1), x)/a**3
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {6 \, \cos \left (d x + c\right ) + 5}{a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \cos \left (d x + c\right ) + a^{3}} - \frac {2 \, \cos \left (d x + c\right )}{a^{3}} + \frac {6 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \] Input:
integrate(sin(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
1/2*((6*cos(d*x + c) + 5)/(a^3*cos(d*x + c)^2 + 2*a^3*cos(d*x + c) + a^3) - 2*cos(d*x + c)/a^3 + 6*log(cos(d*x + c) + 1)/a^3)/d
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos \left (d x + c\right )}{a^{3} d} + \frac {3 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {6 \, \cos \left (d x + c\right ) + 5}{2 \, a^{3} d {\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \] Input:
integrate(sin(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
-cos(d*x + c)/(a^3*d) + 3*log(abs(cos(d*x + c) + 1))/(a^3*d) + 1/2*(6*cos( d*x + c) + 5)/(a^3*d*(cos(d*x + c) + 1)^2)
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {3\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^3\,d}-\frac {\cos \left (c+d\,x\right )}{a^3\,d}+\frac {3\,\cos \left (c+d\,x\right )+\frac {5}{2}}{a^3\,d\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \] Input:
int(sin(c + d*x)/(a + a/cos(c + d*x))^3,x)
Output:
(3*log(cos(c + d*x) + 1))/(a^3*d) - cos(c + d*x)/(a^3*d) + (3*cos(c + d*x) + 5/2)/(a^3*d*(cos(c + d*x) + 1)^2)
Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.41 \[ \int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:
int(sin(d*x+c)/(a+a*sec(d*x+c))^3,x)
Output:
( - 24*log(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)**2 - 24*log(tan((c + d*x)/2)**2 + 1) - tan((c + d*x)/2)**6 + 9*tan((c + d*x)/2)**4 + 26*tan((c + d*x)/2)**2)/(8*a**3*d*(tan((c + d*x)/2)**2 + 1))