Integrand size = 19, antiderivative size = 71 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {5 a \log (1-\sin (c+d x))}{4 d}-\frac {a \log (1+\sin (c+d x))}{4 d}+\frac {a \sin (c+d x)}{d}+\frac {a^2}{2 d (a-a \sin (c+d x))} \] Output:
5/4*a*ln(1-sin(d*x+c))/d-1/4*a*ln(1+sin(d*x+c))/d+a*sin(d*x+c)/d+1/2*a^2/d /(a-a*sin(d*x+c))
Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {3 a \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a \left (2 \log (\cos (c+d x))+\sec ^2(c+d x)\right )}{2 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{2 d}-\frac {a \sin (c+d x) \tan ^2(c+d x)}{d} \] Input:
Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^3,x]
Output:
(-3*a*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*Log[Cos[c + d*x]] + Sec[c + d*x ]^2))/(2*d) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(2*d) - (a*Sin[c + d*x]*Tan[ c + d*x]^2)/d
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 99, 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.
\(\displaystyle \int \tan ^3(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {a^2}{2 (a-a \sin (c+d x))^2}-\frac {5 a}{4 (a-a \sin (c+d x))}-\frac {a}{4 (\sin (c+d x) a+a)}+1\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2}{2 (a-a \sin (c+d x))}+a \sin (c+d x)+\frac {5}{4} a \log (a-a \sin (c+d x))-\frac {1}{4} a \log (a \sin (c+d x)+a)}{d}\) |
Input:
Int[(a + a*Sin[c + d*x])*Tan[c + d*x]^3,x]
Output:
((5*a*Log[a - a*Sin[c + d*x]])/4 - (a*Log[a + a*Sin[c + d*x]])/4 + a*Sin[c + d*x] + a^2/(2*(a - a*Sin[c + d*x])))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.62 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(81\) |
default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a \left (\frac {\tan \left (d x +c \right )^{2}}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(81\) |
parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(89\) |
risch | \(-i a x -\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 i a c}{d}-\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(115\) |
Input:
int((a+a*sin(d*x+c))*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/2*sin(d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x+c)-3/2* ln(sec(d*x+c)+tan(d*x+c)))+a*(1/2*tan(d*x+c)^2+ln(cos(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{2} + {\left (a \sin \left (d x + c\right ) - a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, {\left (a \sin \left (d x + c\right ) - a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a \sin \left (d x + c\right ) - 2 \, a}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^3,x, algorithm="fricas")
Output:
-1/4*(4*a*cos(d*x + c)^2 + (a*sin(d*x + c) - a)*log(sin(d*x + c) + 1) - 5* (a*sin(d*x + c) - a)*log(-sin(d*x + c) + 1) + 4*a*sin(d*x + c) - 2*a)/(d*s in(d*x + c) - d)
\[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)**3,x)
Output:
a*(Integral(sin(c + d*x)*tan(c + d*x)**3, x) + Integral(tan(c + d*x)**3, x ))
Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {a \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, a \sin \left (d x + c\right ) + \frac {2 \, a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^3,x, algorithm="maxima")
Output:
-1/4*(a*log(sin(d*x + c) + 1) - 5*a*log(sin(d*x + c) - 1) - 4*a*sin(d*x + c) + 2*a/(sin(d*x + c) - 1))/d
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=-\frac {1}{4} \, a {\left (\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {4 \, \sin \left (d x + c\right )}{d} + \frac {2}{d {\left (\sin \left (d x + c\right ) - 1\right )}}\right )} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^3,x, algorithm="giac")
Output:
-1/4*a*(log(abs(sin(d*x + c) + 1))/d - 5*log(abs(sin(d*x + c) - 1))/d - 4* sin(d*x + c)/d + 2/(d*(sin(d*x + c) - 1)))
Time = 17.05 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.17 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {5\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{2\,d}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{2\,d}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int(tan(c + d*x)^3*(a + a*sin(c + d*x)),x)
Output:
(5*a*log(tan(c/2 + (d*x)/2) - 1))/(2*d) - (a*log(tan(c/2 + (d*x)/2) + 1))/ (2*d) + (3*a*tan(c/2 + (d*x)/2) - 4*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^3)/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4 + 1)) - (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.31 \[ \int (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx=\frac {a \left (-\cos \left (d x +c \right ) \tan \left (d x +c \right )-\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right ) \tan \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )+\tan \left (d x +c \right )^{2}\right )}{2 d} \] Input:
int((a+a*sin(d*x+c))*tan(d*x+c)^3,x)
Output:
(a*( - cos(c + d*x)*tan(c + d*x) - log(tan(c + d*x)**2 + 1) + 3*log(tan((c + d*x)/2) - 1) - 3*log(tan((c + d*x)/2) + 1) + sin(c + d*x)*tan(c + d*x)* *2 + 4*sin(c + d*x) + tan(c + d*x)**2))/(2*d)