Integrand size = 27, antiderivative size = 55 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^3(c+d x)}{3 a d}+\frac {\sec ^5(c+d x)}{5 a d}-\frac {\tan ^5(c+d x)}{5 a d} \] Output:
-1/3*sec(d*x+c)^3/a/d+1/5*sec(d*x+c)^5/a/d-1/5*tan(d*x+c)^5/a/d
Time = 1.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.93 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^3(c+d x) (40+66 \cos (c+d x)-192 \cos (2 (c+d x))+22 \cos (3 (c+d x))+24 \cos (4 (c+d x))+16 \sin (c+d x)+22 \sin (2 (c+d x))-48 \sin (3 (c+d x))+11 \sin (4 (c+d x)))}{960 a d (1+\sin (c+d x))} \] Input:
Integrate[(Sec[c + d*x]*Tan[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(Sec[c + d*x]^3*(40 + 66*Cos[c + d*x] - 192*Cos[2*(c + d*x)] + 22*Cos[3*(c + d*x)] + 24*Cos[4*(c + d*x)] + 16*Sin[c + d*x] + 22*Sin[2*(c + d*x)] - 4 8*Sin[3*(c + d*x)] + 11*Sin[4*(c + d*x)]))/(960*a*d*(1 + Sin[c + d*x]))
Time = 0.43 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3318, 3042, 3086, 25, 244, 2009, 3087, 15}
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.
| \(\displaystyle \int \frac {\tan ^3(c+d x) \sec (c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^4 (a \sin (c+d x)+a)}dx\) |
| \(\Big \downarrow \) 3318 |
| \(\displaystyle \frac {\int \sec ^3(c+d x) \tan ^3(c+d x)dx}{a}-\frac {\int \sec ^2(c+d x) \tan ^4(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^3 \tan (c+d x)^3dx}{a}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3086 |
| \(\displaystyle \frac {\int -\sec ^2(c+d x) \left (1-\sec ^2(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \sec ^2(c+d x) \left (1-\sec ^2(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle -\frac {\int \left (\sec ^2(c+d x)-\sec ^4(c+d x)\right )d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^5(c+d x)-\frac {1}{3} \sec ^3(c+d x)}{a d}-\frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^5(c+d x)-\frac {1}{3} \sec ^3(c+d x)}{a d}-\frac {\int \tan ^4(c+d x)d\tan (c+d x)}{a d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\frac {1}{5} \sec ^5(c+d x)-\frac {1}{3} \sec ^3(c+d x)}{a d}-\frac {\tan ^5(c+d x)}{5 a d}\) |
Input:
Int[(Sec[c + d*x]*Tan[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
Output:
(-1/3*Sec[c + d*x]^3 + Sec[c + d*x]^5/5)/(a*d) - Tan[c + d*x]^5/(5*a*d)
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.98
| method | result | size |
| risch | \(-\frac {2 i \left (8 i {\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 i {\mathrm e}^{i \left (d x +c \right )}-5 \,{\mathrm e}^{4 i \left (d x +c \right )}-3+10 i {\mathrm e}^{5 i \left (d x +c \right )}+15 \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d a}\) | \(109\) |
| derivativedivides | \(\frac {\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {16}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-128}}{d a}\) | \(115\) |
| default | \(\frac {\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {16}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-128}}{d a}\) | \(115\) |
Input:
int(sec(d*x+c)*tan(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-2/15*I*(8*I*exp(3*I*(d*x+c))+9*exp(2*I*(d*x+c))+6*I*exp(I*(d*x+c))-5*exp( 4*I*(d*x+c))-3+10*I*exp(5*I*(d*x+c))+15*exp(6*I*(d*x+c)))/(exp(I*(d*x+c))+ I)^5/(exp(I*(d*x+c))-I)^3/d/a
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.36 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 4}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
1/15*(3*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 1)*sin(d*x + c) + 4)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)
\[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**3*sec(c + d*x)/(sin(c + d*x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (49) = 98\).
Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 4.25 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {4 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-4/15*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 15*sin(d*x + c)^4/(cos(d* x + c) + 1)^4 + 1)/((a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6 *a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a*sin(d*x + c)^8/(cos(d *x + c) + 1)^8)*d)
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (49) = 98\).
Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.18 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {5 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(sec(d*x+c)*tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/120*(5*(3*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 5)/(a*(tan( 1/2*d*x + 1/2*c) - 1)^3) - (15*tan(1/2*d*x + 1/2*c)^4 + 60*tan(1/2*d*x + 1 /2*c)^3 + 10*tan(1/2*d*x + 1/2*c)^2 + 20*tan(1/2*d*x + 1/2*c) + 7)/(a*(tan (1/2*d*x + 1/2*c) + 1)^5))/d
Time = 32.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {4\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{15\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \] Input:
int(tan(c + d*x)^3/(cos(c + d*x)*(a + a*sin(c + d*x))),x)
Output:
-(4*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) + 6*tan(c/2 + (d*x)/2)^ 3 + 15*tan(c/2 + (d*x)/2)^4 - 1))/(15*a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan( c/2 + (d*x)/2) + 1)^5)
Time = 0.17 (sec) , antiderivative size = 326, normalized size of antiderivative = 5.93 \[ \int \frac {\sec (c+d x) \tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \sin \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}-10 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \sin \left (d x +c \right )^{3}+5 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}-10 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \sin \left (d x +c \right )^{2}-5 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \sin \left (d x +c \right ) \tan \left (d x +c \right )^{2}+10 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+10 \cos \left (d x +c \right ) \sec \left (d x +c \right )+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right )-3 \sin \left (d x +c \right )^{4}+12 \sin \left (d x +c \right )^{3}+12 \sin \left (d x +c \right )^{2}-8 \sin \left (d x +c \right )-8}{15 \cos \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)*tan(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(5*cos(c + d*x)*sec(c + d*x)*sin(c + d*x)**3*tan(c + d*x)**2 - 10*cos(c + d*x)*sec(c + d*x)*sin(c + d*x)**3 + 5*cos(c + d*x)*sec(c + d*x)*sin(c + d* x)**2*tan(c + d*x)**2 - 10*cos(c + d*x)*sec(c + d*x)*sin(c + d*x)**2 - 5*c os(c + d*x)*sec(c + d*x)*sin(c + d*x)*tan(c + d*x)**2 + 10*cos(c + d*x)*se c(c + d*x)*sin(c + d*x) - 5*cos(c + d*x)*sec(c + d*x)*tan(c + d*x)**2 + 10 *cos(c + d*x)*sec(c + d*x) + 8*cos(c + d*x)*sin(c + d*x)**3 + 8*cos(c + d* x)*sin(c + d*x)**2 - 8*cos(c + d*x)*sin(c + d*x) - 8*cos(c + d*x) - 3*sin( c + d*x)**4 + 12*sin(c + d*x)**3 + 12*sin(c + d*x)**2 - 8*sin(c + d*x) - 8 )/(15*cos(c + d*x)*a*d*(sin(c + d*x)**3 + sin(c + d*x)**2 - sin(c + d*x) - 1))