Integrand size = 21, antiderivative size = 82 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{8 a d}+\frac {3 \sec (c+d x) \tan (c+d x)}{8 a d}-\frac {\sec (c+d x) \tan ^3(c+d x)}{4 a d}+\frac {\tan ^4(c+d x)}{4 a d} \] Output:
-3/8*arctanh(sin(d*x+c))/a/d+3/8*sec(d*x+c)*tan(d*x+c)/a/d-1/4*sec(d*x+c)* tan(d*x+c)^3/a/d+1/4*tan(d*x+c)^4/a/d
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))+\frac {1}{-1+\sin (c+d x)}-\frac {1}{(1+\sin (c+d x))^2}+\frac {4}{1+\sin (c+d x)}}{8 a d} \] Input:
Integrate[Tan[c + d*x]^3/(a + a*Sin[c + d*x]),x]
Output:
-1/8*(3*ArcTanh[Sin[c + d*x]] + (-1 + Sin[c + d*x])^(-1) - (1 + Sin[c + d* x])^(-2) + 4/(1 + Sin[c + d*x]))/(a*d)
Time = 0.51 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3185, 3042, 3087, 15, 3091, 3042, 3091, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{a \sin (c+d x)+a}dx\) |
| \(\Big \downarrow \) 3185 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) \tan ^3(c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^4(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^3dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \tan ^3(c+d x)d\tan (c+d x)}{a d}-\frac {\int \sec (c+d x) \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {\tan ^4(c+d x)}{4 a d}-\frac {\int \sec (c+d x) \tan (c+d x)^4dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^4(c+d x)}{4 a d}-\frac {\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^4(c+d x)}{4 a d}-\frac {\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx}{a}\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^4(c+d x)}{4 a d}-\frac {\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^4(c+d x)}{4 a d}-\frac {\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )}{a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan ^4(c+d x)}{4 a d}-\frac {\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )}{a}\) |
Input:
Int[Tan[c + d*x]^3/(a + a*Sin[c + d*x]),x]
Output:
Tan[c + d*x]^4/(4*a*d) - ((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1/2*A rcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)/a
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -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[((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[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.42 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{16}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{16}}{d a}\) | \(67\) |
| default | \(\frac {-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{16}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{16}}{d a}\) | \(67\) |
| risch | \(-\frac {i \left (2 i {\mathrm e}^{4 i \left (d x +c \right )}-2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 i {\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{5 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 a d}\) | \(139\) |
Input:
int(tan(d*x+c)^3/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(-1/8/(sin(d*x+c)-1)+3/16*ln(sin(d*x+c)-1)+1/8/(1+sin(d*x+c))^2-1/2/ (1+sin(d*x+c))-3/16*ln(1+sin(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.52 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {10 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) - 6}{16 \, {\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/16*(10*cos(d*x + c)^2 + 3*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2 )*log(sin(d*x + c) + 1) - 3*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2) *log(-sin(d*x + c) + 1) - 2*sin(d*x + c) - 6)/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d*cos(d*x + c)^2)
\[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(tan(d*x+c)**3/(a+a*sin(d*x+c)),x)
Output:
Integral(tan(c + d*x)**3/(sin(c + d*x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 2\right )}}{a \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} + \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/16*(2*(5*sin(d*x + c)^2 + sin(d*x + c) - 2)/(a*sin(d*x + c)^3 + a*sin(d *x + c)^2 - a*sin(d*x + c) - a) + 3*log(sin(d*x + c) + 1)/a - 3*log(sin(d* x + c) - 1)/a)/d
Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, a d} + \frac {3 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, a d} - \frac {5 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 2}{8 \, a d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-3/16*log(abs(sin(d*x + c) + 1))/(a*d) + 3/16*log(abs(sin(d*x + c) - 1))/( a*d) - 1/8*(5*sin(d*x + c)^2 + sin(d*x + c) - 2)/(a*d*(sin(d*x + c) + 1)^2 *(sin(d*x + c) - 1))
Time = 18.85 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.10 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a\,d} \] Input:
int(tan(c + d*x)^3/(a + a*sin(c + d*x)),x)
Output:
((3*tan(c/2 + (d*x)/2))/4 + (3*tan(c/2 + (d*x)/2)^2)/2 - tan(c/2 + (d*x)/2 )^3/2 + (3*tan(c/2 + (d*x)/2)^4)/2 + (3*tan(c/2 + (d*x)/2)^5)/4)/(d*(a + 2 *a*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^2 - 4*a*tan(c/2 + (d*x)/2)^3 - a*tan(c/2 + (d*x)/2)^4 + 2*a*tan(c/2 + (d*x)/2)^5 + a*tan(c/2 + (d*x)/2) ^6)) - (3*atanh(tan(c/2 + (d*x)/2)))/(4*a*d)
Time = 0.19 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.61 \[ \int \frac {\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}-3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )^{3}-6 \sin \left (d x +c \right )^{2}+3}{8 a d \left (\sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )-1\right )} \] Input:
int(tan(d*x+c)^3/(a+a*sin(d*x+c)),x)
Output:
(3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 3*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2 - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 3*log(tan((c + d*x)/2) - 1) - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 3*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)**2 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d* x) + 3*log(tan((c + d*x)/2) + 1) - sin(c + d*x)**3 - 6*sin(c + d*x)**2 + 3 )/(8*a*d*(sin(c + d*x)**3 + sin(c + d*x)**2 - sin(c + d*x) - 1))