Integrand size = 21, antiderivative size = 91 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {7 a^3 \log (1-\sin (c+d x))}{d}+\frac {5 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin ^2(c+d x)}{2 d}+\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {2 a^4}{d (a-a \sin (c+d x))} \] Output:
7*a^3*ln(1-sin(d*x+c))/d+5*a^3*sin(d*x+c)/d+3/2*a^3*sin(d*x+c)^2/d+1/3*a^3 *sin(d*x+c)^3/d+2*a^4/d/(a-a*sin(d*x+c))
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {a^3 \left (42 \log (1-\sin (c+d x))+\frac {12}{1-\sin (c+d x)}+30 \sin (c+d x)+9 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )}{6 d} \] Input:
Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^3,x]
Output:
(a^3*(42*Log[1 - Sin[c + d*x]] + 12/(1 - Sin[c + d*x]) + 30*Sin[c + d*x] + 9*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3))/(6*d)
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 86, 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.
| \(\displaystyle \int \tan ^3(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^3 (a \sin (c+d x)+a)^3dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x) (\sin (c+d x) a+a)}{(a-a \sin (c+d x))^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (\frac {2 a^4}{(a-a \sin (c+d x))^2}-\frac {7 a^3}{a-a \sin (c+d x)}+\sin ^2(c+d x) a^2+3 \sin (c+d x) a^2+5 a^2\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^4}{a-a \sin (c+d x)}+\frac {1}{3} a^3 \sin ^3(c+d x)+\frac {3}{2} a^3 \sin ^2(c+d x)+5 a^3 \sin (c+d x)+7 a^3 \log (a-a \sin (c+d x))}{d}\) |
Input:
Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^3,x]
Output:
(7*a^3*Log[a - a*Sin[c + d*x]] + 5*a^3*Sin[c + d*x] + (3*a^3*Sin[c + d*x]^ 2)/2 + (a^3*Sin[c + d*x]^3)/3 + (2*a^4)/(a - a*Sin[c + d*x]))/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 5.91 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {3 \sin \left (d x +c \right )^{2}}{2}+5 \sin \left (d x +c \right )-\frac {2}{\sin \left (d x +c \right )-1}+7 \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(60\) |
| default | \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {3 \sin \left (d x +c \right )^{2}}{2}+5 \sin \left (d x +c \right )-\frac {2}{\sin \left (d x +c \right )-1}+7 \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(60\) |
| risch | \(-7 i a^{3} x -\frac {21 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {21 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {14 i a^{3} c}{d}-\frac {4 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d}+\frac {14 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}-\frac {3 a^{3} \cos \left (2 d x +2 c \right )}{4 d}\) | \(142\) |
| parts | \(\frac {a^{3} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{7}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{2}+\frac {5 \sin \left (d x +c \right )^{3}}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \left (\frac {\sin \left (d x +c \right )^{6}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(218\) |
Input:
int((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
a^3/d*(1/3*sin(d*x+c)^3+3/2*sin(d*x+c)^2+5*sin(d*x+c)-2/(sin(d*x+c)-1)+7*l n(sin(d*x+c)-1))
Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {4 \, a^{3} \cos \left (d x + c\right )^{4} - 50 \, a^{3} \cos \left (d x + c\right )^{2} + 31 \, a^{3} + 84 \, {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (14 \, a^{3} \cos \left (d x + c\right )^{2} + 55 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="fricas")
Output:
1/12*(4*a^3*cos(d*x + c)^4 - 50*a^3*cos(d*x + c)^2 + 31*a^3 + 84*(a^3*sin( d*x + c) - a^3)*log(-sin(d*x + c) + 1) - (14*a^3*cos(d*x + c)^2 + 55*a^3)* sin(d*x + c))/(d*sin(d*x + c) - d)
\[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))**3*tan(d*x+c)**3,x)
Output:
a**3*(Integral(3*sin(c + d*x)*tan(c + d*x)**3, x) + Integral(3*sin(c + d*x )**2*tan(c + d*x)**3, x) + Integral(sin(c + d*x)**3*tan(c + d*x)**3, x) + Integral(tan(c + d*x)**3, x))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 42 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, a^{3} \sin \left (d x + c\right ) - \frac {12 \, a^{3}}{\sin \left (d x + c\right ) - 1}}{6 \, d} \] Input:
integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="maxima")
Output:
1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 42*a^3*log(sin(d*x + c) - 1) + 30*a^3*sin(d*x + c) - 12*a^3/(sin(d*x + c) - 1))/d
Time = 0.23 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {7 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, a^{3}}{d {\left (\sin \left (d x + c\right ) - 1\right )}} + \frac {2 \, a^{3} d^{2} \sin \left (d x + c\right )^{3} + 9 \, a^{3} d^{2} \sin \left (d x + c\right )^{2} + 30 \, a^{3} d^{2} \sin \left (d x + c\right )}{6 \, d^{3}} \] Input:
integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="giac")
Output:
7*a^3*log(abs(sin(d*x + c) - 1))/d - 2*a^3/(d*(sin(d*x + c) - 1)) + 1/6*(2 *a^3*d^2*sin(d*x + c)^3 + 9*a^3*d^2*sin(d*x + c)^2 + 30*a^3*d^2*sin(d*x + c))/d^3
Time = 18.15 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.88 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {14\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}+\frac {14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {98\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {100\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {98\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+14\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {7\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int(tan(c + d*x)^3*(a + a*sin(c + d*x))^3,x)
Output:
(14*a^3*log(tan(c/2 + (d*x)/2) - 1))/d + ((98*a^3*tan(c/2 + (d*x)/2)^3)/3 - 14*a^3*tan(c/2 + (d*x)/2)^2 - (100*a^3*tan(c/2 + (d*x)/2)^4)/3 + (98*a^3 *tan(c/2 + (d*x)/2)^5)/3 - 14*a^3*tan(c/2 + (d*x)/2)^6 + 14*a^3*tan(c/2 + (d*x)/2)^7 + 14*a^3*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 - 2*tan (c/2 + (d*x)/2) - 6*tan(c/2 + (d*x)/2)^3 + 6*tan(c/2 + (d*x)/2)^4 - 6*tan( c/2 + (d*x)/2)^5 + 4*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^7 + tan(c /2 + (d*x)/2)^8 + 1)) - (7*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.18 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.67 \[ \int (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {a^{3} \left (-3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )-36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+78 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-78 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \sin \left (d x +c \right )^{5}+9 \sin \left (d x +c \right )^{4}+28 \sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}-18 \sin \left (d x +c \right )^{2}-42 \sin \left (d x +c \right )-3 \tan \left (d x +c \right )^{2}\right )}{6 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*sin(d*x+c))^3*tan(d*x+c)^3,x)
Output:
(a**3*( - 3*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2 + 3*log(tan(c + d*x)* *2 + 1) - 36*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 + 36*log(tan((c + d*x)/2)**2 + 1) + 78*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 78*log( tan((c + d*x)/2) - 1) - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 6*lo g(tan((c + d*x)/2) + 1) + 2*sin(c + d*x)**5 + 9*sin(c + d*x)**4 + 28*sin(c + d*x)**3 + 3*sin(c + d*x)**2*tan(c + d*x)**2 - 18*sin(c + d*x)**2 - 42*s in(c + d*x) - 3*tan(c + d*x)**2))/(6*d*(sin(c + d*x)**2 - 1))