Integrand size = 29, antiderivative size = 82 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {3 a^3 \log (1-\sin (c+d x))}{d}-\frac {a^3 \sin (c+d x)}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {3 a^5}{d \left (a^2-a^2 \sin (c+d x)\right )} \] Output:
-3*a^3*ln(1-sin(d*x+c))/d-a^3*sin(d*x+c)/d+1/2*a^5/d/(a-a*sin(d*x+c))^2-3* a^5/d/(a^2-a^2*sin(d*x+c))
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {a^3 \left (6 \log (1-\sin (c+d x))+\frac {5-6 \sin (c+d x)}{(-1+\sin (c+d x))^2}+2 \sin (c+d x)\right )}{2 d} \] Input:
Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^3,x]
Output:
-1/2*(a^3*(6*Log[1 - Sin[c + d*x]] + (5 - 6*Sin[c + d*x])/(-1 + Sin[c + d* x])^2 + 2*Sin[c + d*x]))/d
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 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.
\(\displaystyle \int \tan ^3(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 (a \sin (c+d x)+a)^3}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\sin ^3(c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^2 \int \left (\frac {a^3}{(a-a \sin (c+d x))^3}-\frac {3 a^2}{(a-a \sin (c+d x))^2}+\frac {3 a}{a-a \sin (c+d x)}-1\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 \left (\frac {a^3}{2 (a-a \sin (c+d x))^2}-\frac {3 a^2}{a-a \sin (c+d x)}-a \sin (c+d x)-3 a \log (a-a \sin (c+d x))\right )}{d}\) |
Input:
Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^3,x]
Output:
(a^2*(-3*a*Log[a - a*Sin[c + d*x]] - a*Sin[c + d*x] + a^3/(2*(a - a*Sin[c + d*x])^2) - (3*a^2)/(a - a*Sin[c + d*x])))/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_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 3.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(-\frac {a^{3} \left (\sin \left (d x +c \right )-\frac {3}{\sin \left (d x +c \right )-1}-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+3 \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(51\) |
default | \(-\frac {a^{3} \left (\sin \left (d x +c \right )-\frac {3}{\sin \left (d x +c \right )-1}-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+3 \ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(51\) |
risch | \(3 i a^{3} x +\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {6 i a^{3} c}{d}+\frac {2 i \left (-3 a^{3} {\mathrm e}^{i \left (d x +c \right )}-5 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {6 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(140\) |
Input:
int(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
-a^3/d*(sin(d*x+c)-3/(sin(d*x+c)-1)-1/2/(sin(d*x+c)-1)^2+3*ln(sin(d*x+c)-1 ))
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {4 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} - 6 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="frica s")
Output:
1/2*(4*a^3*cos(d*x + c)^2 + a^3 - 6*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^3)*log(-sin(d*x + c) + 1) - 2*(a^3*cos(d*x + c)^2 + a^3)*sin(d*x + c))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**2*(a+a*sin(d*x+c))**3*tan(d*x+c)**3,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {6 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 2 \, a^{3} \sin \left (d x + c\right ) - \frac {6 \, a^{3} \sin \left (d x + c\right ) - 5 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="maxim a")
Output:
-1/2*(6*a^3*log(sin(d*x + c) - 1) + 2*a^3*sin(d*x + c) - (6*a^3*sin(d*x + c) - 5*a^3)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=-\frac {1}{2} \, a^{3} {\left (\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} + \frac {2 \, \sin \left (d x + c\right )}{d} - \frac {6 \, \sin \left (d x + c\right ) - 5}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*tan(d*x+c)^3,x, algorithm="giac" )
Output:
-1/2*a^3*(6*log(abs(sin(d*x + c) - 1))/d + 2*sin(d*x + c)/d - (6*sin(d*x + c) - 5)/(d*(sin(d*x + c) - 1)^2))
Time = 32.96 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.50 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {3\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {6\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}-\frac {6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-18\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-18\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )} \] Input:
int((tan(c + d*x)^3*(a + a*sin(c + d*x))^3)/cos(c + d*x)^2,x)
Output:
(3*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (6*a^3*log(tan(c/2 + (d*x)/2) - 1))/d - (20*a^3*tan(c/2 + (d*x)/2)^3 - 18*a^3*tan(c/2 + (d*x)/2)^2 - 18*a^ 3*tan(c/2 + (d*x)/2)^4 + 6*a^3*tan(c/2 + (d*x)/2)^5 + 6*a^3*tan(c/2 + (d*x )/2))/(d*(7*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 8*tan(c/2 + (d*x )/2)^3 + 7*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^5 + tan(c/2 + (d*x) /2)^6 + 1))
Time = 0.20 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.05 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^3(c+d x) \, dx=\frac {a^{3} \left (6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}-3\right )}{2 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*tan(d*x+c)^3,x)
Output:
(a**3*(6*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 12*log(tan((c + d* x)/2)**2 + 1)*sin(c + d*x) + 6*log(tan((c + d*x)/2)**2 + 1) - 12*log(tan(( c + d*x)/2) - 1)*sin(c + d*x)**2 + 24*log(tan((c + d*x)/2) - 1)*sin(c + d* x) - 12*log(tan((c + d*x)/2) - 1) - 2*sin(c + d*x)**3 + 6*sin(c + d*x)**2 - 3))/(2*d*(sin(c + d*x)**2 - 2*sin(c + d*x) + 1))