Integrand size = 25, antiderivative size = 74 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
-1/8*b*arctanh(sin(d*x+c))/d+1/4*a*sec(d*x+c)^4/d-1/8*b*sec(d*x+c)*tan(d*x +c)/d+1/4*b*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \sec ^4(c+d x)}{4 d}-\frac {b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])*Tan[c + d*x],x]
Output:
-1/8*(b*ArcTanh[Sin[c + d*x]])/d + (a*Sec[c + d*x]^4)/(4*d) - (b*Sec[c + d *x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3313, 3042, 3086, 15, 3091, 3042, 4255, 3042, 4257}
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.
| \(\displaystyle \int \tan (c+d x) \sec ^4(c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x) (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \sec ^4(c+d x) \tan (c+d x)dx+b \int \sec ^3(c+d x) \tan ^2(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sec (c+d x)^4 \tan (c+d x)dx+b \int \sec (c+d x)^3 \tan (c+d x)^2dx\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \frac {a \int \sec ^3(c+d x)d\sec (c+d x)}{d}+b \int \sec (c+d x)^3 \tan (c+d x)^2dx\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle b \int \sec (c+d x)^3 \tan (c+d x)^2dx+\frac {a \sec ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle b \left (\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {1}{4} \int \sec ^3(c+d x)dx\right )+\frac {a \sec ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a \sec ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {1}{4} \left (-\frac {1}{2} \int \sec (c+d x)dx-\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a \sec ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {1}{4} \left (-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a \sec ^4(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a \sec ^4(c+d x)}{4 d}+b \left (\frac {1}{4} \left (-\frac {\text {arctanh}(\sin (c+d x))}{2 d}-\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])*Tan[c + d*x],x]
Output:
(a*Sec[c + d*x]^4)/(4*d) + b*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (-1/2* ArcTanh[Sin[c + d*x]]/d - (Sec[c + d*x]*Tan[c + d*x])/(2*d))/4)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.63 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {\frac {a}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(80\) |
| default | \(\frac {\frac {a}{4 \cos \left (d x +c \right )^{4}}+b \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(80\) |
| risch | \(\frac {i \left (b \,{\mathrm e}^{7 i \left (d x +c \right )}-16 i a \,{\mathrm e}^{4 i \left (d x +c \right )}-7 b \,{\mathrm e}^{5 i \left (d x +c \right )}+7 b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{8 d}\) | \(120\) |
Input:
int(sec(d*x+c)^4*(a+b*sin(d*x+c))*tan(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*(1/4/cos(d*x+c)^4*a+b*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/ cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))*tan(d*x+c),x, algorithm="fricas")
Output:
-1/16*(b*cos(d*x + c)^4*log(sin(d*x + c) + 1) - b*cos(d*x + c)^4*log(-sin( d*x + c) + 1) + 2*(b*cos(d*x + c)^2 - 2*b)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)^4)
\[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right ) \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*sin(d*x+c))*tan(d*x+c),x)
Output:
Integral((a + b*sin(c + d*x))*tan(c + d*x)*sec(c + d*x)**4, x)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \log \left (\sin \left (d x + c\right ) + 1\right ) - b \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (b \sin \left (d x + c\right )^{3} + b \sin \left (d x + c\right ) + 2 \, a\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))*tan(d*x+c),x, algorithm="maxima")
Output:
-1/16*(b*log(sin(d*x + c) + 1) - b*log(sin(d*x + c) - 1) - 2*(b*sin(d*x + c)^3 + b*sin(d*x + c) + 2*a)/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=-\frac {b \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} + \frac {b \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {b \sin \left (d x + c\right )^{3} + b \sin \left (d x + c\right ) + 2 \, a}{8 \, {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2} d} \] Input:
integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))*tan(d*x+c),x, algorithm="giac")
Output:
-1/16*b*log(abs(sin(d*x + c) + 1))/d + 1/16*b*log(abs(sin(d*x + c) - 1))/d + 1/8*(b*sin(d*x + c)^3 + b*sin(d*x + c) + 2*a)/((sin(d*x + c)^2 - 1)^2*d )
Time = 21.41 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.14 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \] Input:
int((tan(c + d*x)*(a + b*sin(c + d*x)))/cos(c + d*x)^4,x)
Output:
((b*tan(c/2 + (d*x)/2))/4 + 2*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x) /2)^6 + (7*b*tan(c/2 + (d*x)/2)^3)/4 + (7*b*tan(c/2 + (d*x)/2)^5)/4 + (b*t an(c/2 + (d*x)/2)^7)/4)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^ 2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (b*atanh(tan(c/2 + (d*x)/2)))/(4*d)
Time = 0.19 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.54 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x)) \tan (c+d x) \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -2 \sin \left (d x +c \right )^{4} a +\sin \left (d x +c \right )^{3} b +4 \sin \left (d x +c \right )^{2} a +\sin \left (d x +c \right ) b}{8 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^4*(a+b*sin(d*x+c))*tan(d*x+c),x)
Output:
(log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b - 2*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2*b + log(tan((c + d*x)/2) - 1)*b - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - log (tan((c + d*x)/2) + 1)*b - 2*sin(c + d*x)**4*a + sin(c + d*x)**3*b + 4*sin (c + d*x)**2*a + sin(c + d*x)*b)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))