Integrand size = 29, antiderivative size = 68 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {2 a^5}{d \left (a^2-a^2 \sin (c+d x)\right )} \] Output:
-a^3*ln(1-sin(d*x+c))/d+1/2*a^5/d/(a-a*sin(d*x+c))^2-2*a^5/d/(a^2-a^2*sin( d*x+c))
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {a^3 \left (2 \log (1-\sin (c+d x))+\frac {3-4 \sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{2 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^2,x]
Output:
-1/2*(a^3*(2*Log[1 - Sin[c + d*x]] + (3 - 4*Sin[c + d*x])/(-1 + Sin[c + d* x])^2))/d
Time = 0.32 (sec) , antiderivative size = 58, 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.
\(\displaystyle \int \tan ^2(c+d x) \sec ^3(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 (a \sin (c+d x)+a)^3}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\sin ^2(c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {a^2 \sin ^2(c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^3 \int \left (\frac {a^2}{(a-a \sin (c+d x))^3}-\frac {2 a}{(a-a \sin (c+d x))^2}+\frac {1}{a-a \sin (c+d x)}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (\frac {a^2}{2 (a-a \sin (c+d x))^2}-\frac {2 a}{a-a \sin (c+d x)}-\log (a-a \sin (c+d x))\right )}{d}\) |
Input:
Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^2,x]
Output:
(a^3*(-Log[a - a*Sin[c + d*x]] + a^2/(2*(a - a*Sin[c + d*x])^2) - (2*a)/(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 = 0.51 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(-\frac {a^{3} \left (-\frac {2}{\sin \left (d x +c \right )-1}-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+\ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(43\) |
default | \(-\frac {a^{3} \left (-\frac {2}{\sin \left (d x +c \right )-1}-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+\ln \left (\sin \left (d x +c \right )-1\right )\right )}{d}\) | \(43\) |
risch | \(i a^{3} x +\frac {2 i a^{3} c}{d}+\frac {2 i a^{3} \left (-3 i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4}}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(98\) |
Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^3*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-a^3/d*(-2/(sin(d*x+c)-1)-1/2/(sin(d*x+c)-1)^2+ln(sin(d*x+c)-1))
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {4 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3} + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="frica s")
Output:
-1/2*(4*a^3*sin(d*x + c) - 3*a^3 + 2*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^3)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))**3*tan(d*x+c)**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {2 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {4 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="maxim a")
Output:
-1/2*(2*a^3*log(sin(d*x + c) - 1) - (4*a^3*sin(d*x + c) - 3*a^3)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {1}{2} \, a^{3} {\left (\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {4 \, \sin \left (d x + c\right ) - 3}{d {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))^3*tan(d*x+c)^2,x, algorithm="giac" )
Output:
-1/2*a^3*(2*log(abs(sin(d*x + c) - 1))/d - (4*sin(d*x + c) - 3)/(d*(sin(d* x + c) - 1)^2))
Time = 32.67 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.60 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^3\,\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )+a^3\,\left (4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )-8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )+2\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^3\,\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )+a^3\,\left (4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )-8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )+2\right )\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (6\,a^3\,\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )+a^3\,\left (6\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )-12\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )+6\right )\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^4}-\frac {a^3\,\left (2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )}{d} \] Input:
int((tan(c + d*x)^2*(a + a*sin(c + d*x))^3)/cos(c + d*x)^3,x)
Output:
- (tan(c/2 + (d*x)/2)*(4*a^3*(2*log(tan(c/2 + (d*x)/2) - 1) - log(tan(c/2 + (d*x)/2)^2 + 1)) + a^3*(4*log(tan(c/2 + (d*x)/2)^2 + 1) - 8*log(tan(c/2 + (d*x)/2) - 1) + 2)) + tan(c/2 + (d*x)/2)^3*(4*a^3*(2*log(tan(c/2 + (d*x) /2) - 1) - log(tan(c/2 + (d*x)/2)^2 + 1)) + a^3*(4*log(tan(c/2 + (d*x)/2)^ 2 + 1) - 8*log(tan(c/2 + (d*x)/2) - 1) + 2)) - tan(c/2 + (d*x)/2)^2*(6*a^3 *(2*log(tan(c/2 + (d*x)/2) - 1) - log(tan(c/2 + (d*x)/2)^2 + 1)) + a^3*(6* log(tan(c/2 + (d*x)/2)^2 + 1) - 12*log(tan(c/2 + (d*x)/2) - 1) + 6)))/(d*( tan(c/2 + (d*x)/2) - 1)^4) - (a^3*(2*log(tan(c/2 + (d*x)/2) - 1) - log(tan (c/2 + (d*x)/2)^2 + 1)))/d
Time = 0.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.32 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {a^{3} \left (2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \sin \left (d x +c \right )^{2}-1\right )}{2 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^3*(a+a*sin(d*x+c))^3*tan(d*x+c)^2,x)
Output:
(a**3*(2*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 - 4*log(tan((c + d*x )/2)**2 + 1)*sin(c + d*x) + 2*log(tan((c + d*x)/2)**2 + 1) - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 4*log(tan((c + d*x)/2) - 1) + 2*sin(c + d*x)**2 - 1))/(2*d*(sin(c + d*x) **2 - 2*sin(c + d*x) + 1))