Integrand size = 27, antiderivative size = 68 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {a^5}{4 d \left (a^3-a^3 \sin (c+d x)\right )} \] Output:
-1/4*a^2*arctanh(sin(d*x+c))/d+1/4*a^4/d/(a-a*sin(d*x+c))^2-1/4*a^5/d/(a^3 -a^3*sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.53 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^2 \left (\text {arctanh}(\sin (c+d x))-\frac {\sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{4 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*Tan[c + d*x],x]
Output:
-1/4*(a^2*(ArcTanh[Sin[c + d*x]] - Sin[c + d*x]/(-1 + Sin[c + d*x])^2))/d
Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3315, 27, 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.
\(\displaystyle \int \tan (c+d x) \sec ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) (a \sin (c+d x)+a)^2}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a^5 \int \frac {\sin (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^4 \int \frac {a \sin (c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {a^4 \int \left (\frac {1}{2 (a-a \sin (c+d x))^3}-\frac {1}{4 \left (a^2-a^2 \sin ^2(c+d x)\right ) a}-\frac {1}{4 (a-a \sin (c+d x))^2 a}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^4 \left (-\frac {\text {arctanh}(\sin (c+d x))}{4 a^2}-\frac {1}{4 a (a-a \sin (c+d x))}+\frac {1}{4 (a-a \sin (c+d x))^2}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2*Tan[c + d*x],x]
Output:
(a^4*(-1/4*ArcTanh[Sin[c + d*x]]/a^2 + 1/(4*(a - a*Sin[c + d*x])^2) - 1/(4 *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_))*((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[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]
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35
method | result | size |
risch | \(\frac {i \left (a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-a^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}\) | \(92\) |
derivativedivides | \(\frac {\frac {a^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(106\) |
default | \(\frac {\frac {a^{2} \sin \left (d x +c \right )^{4}}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(106\) |
Input:
int(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*tan(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/2*I*(a^2*exp(3*I*(d*x+c))-a^2*exp(I*(d*x+c)))/(exp(I*(d*x+c))-I)^4/d-1/4 *a^2/d*ln(exp(I*(d*x+c))+I)+1/4*a^2/d*ln(exp(I*(d*x+c))-I)
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.76 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, a^{2} \sin \left (d x + c\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*tan(d*x+c),x, algorithm="fricas" )
Output:
-1/8*(2*a^2*sin(d*x + c) + (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^ 2)*log(sin(d*x + c) + 1) - (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^ 2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
\[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=a^{2} \left (\int \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**2*tan(d*x+c),x)
Output:
a**2*(Integral(tan(c + d*x)*sec(c + d*x)**4, x) + Integral(2*sin(c + d*x)* tan(c + d*x)*sec(c + d*x)**4, x) + Integral(sin(c + d*x)**2*tan(c + d*x)*s ec(c + d*x)**4, x))
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, a^{2} \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*tan(d*x+c),x, algorithm="maxima" )
Output:
-1/8*(a^2*log(sin(d*x + c) + 1) - a^2*log(sin(d*x + c) - 1) - 2*a^2*sin(d* x + c)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.31 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {1}{16} \, a^{2} {\left (\frac {\log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right )}{d} - \frac {\log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right )}{d} + \frac {\frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 6}{d {\left (\frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2\right )}}\right )} \] Input:
integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*tan(d*x+c),x, algorithm="giac")
Output:
-1/16*a^2*(log(abs(1/sin(d*x + c) + sin(d*x + c) + 2))/d - log(abs(1/sin(d *x + c) + sin(d*x + c) - 2))/d + (1/sin(d*x + c) + sin(d*x + c) - 6)/(d*(1 /sin(d*x + c) + sin(d*x + c) - 2)))
Time = 33.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.56 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d} \] Input:
int((tan(c + d*x)*(a + a*sin(c + d*x))^2)/cos(c + d*x)^4,x)
Output:
((a^2*tan(c/2 + (d*x)/2)^3)/2 + (a^2*tan(c/2 + (d*x)/2))/2)/(d*(6*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + ( d*x)/2)^4 + 1)) - (a^2*atanh(tan(c/2 + (d*x)/2)))/(2*d)
Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.21 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {a^{2} \left (2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right )^{2}+1\right )}{8 d \left (\sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)^4*(a+a*sin(d*x+c))^2*tan(d*x+c),x)
Output:
(a**2*(2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 4*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x) + 2*log(tan((c + d*x)/2) - 1) - 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 2*log(t an((c + d*x)/2) + 1) + sin(c + d*x)**2 + 1))/(8*d*(sin(c + d*x)**2 - 2*sin (c + d*x) + 1))