\(\int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx\) [54]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec (c+d x)}{a d}+\frac {2 \sec ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan ^5(c+d x)}{5 a d} \] Output:

-sec(d*x+c)/a/d+2/3*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
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^3(c+d x) (200-534 \cos (c+d x)+288 \cos (2 (c+d x))-178 \cos (3 (c+d x))+24 \cos (4 (c+d x))-64 \sin (c+d x)-178 \sin (2 (c+d x))+192 \sin (3 (c+d x))-89 \sin (4 (c+d x)))}{960 a d (1+\sin (c+d x))} \] Input:

Integrate[Tan[c + d*x]^4/(a + a*Sin[c + d*x]),x]
 

Output:

-1/960*(Sec[c + d*x]^3*(200 - 534*Cos[c + d*x] + 288*Cos[2*(c + d*x)] - 17 
8*Cos[3*(c + d*x)] + 24*Cos[4*(c + d*x)] - 64*Sin[c + d*x] - 178*Sin[2*(c 
+ d*x)] + 192*Sin[3*(c + d*x)] - 89*Sin[4*(c + d*x)]))/(a*d*(1 + Sin[c + d 
*x]))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3185, 3042, 3086, 210, 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.

\(\displaystyle \int \frac {\tan ^4(c+d x)}{a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^4}{a \sin (c+d x)+a}dx\)

\(\Big \downarrow \) 3185

\(\displaystyle \frac {\int \sec ^2(c+d x) \tan ^4(c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^5(c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^5dx}{a}\)

\(\Big \downarrow \) 3086

\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\int \left (\sec ^2(c+d x)-1\right )^2d\sec (c+d x)}{a d}\)

\(\Big \downarrow \) 210

\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\int \left (\sec ^4(c+d x)-2 \sec ^2(c+d x)+1\right )d\sec (c+d x)}{a d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\frac {1}{5} \sec ^5(c+d x)-\frac {2}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\)

\(\Big \downarrow \) 3087

\(\displaystyle \frac {\int \tan ^4(c+d x)d\tan (c+d x)}{a d}-\frac {\frac {1}{5} \sec ^5(c+d x)-\frac {2}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {\tan ^5(c+d x)}{5 a d}-\frac {\frac {1}{5} \sec ^5(c+d x)-\frac {2}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\)

Input:

Int[Tan[c + d*x]^4/(a + a*Sin[c + d*x]),x]
 

Output:

-((Sec[c + d*x] - (2*Sec[c + d*x]^3)/3 + Sec[c + d*x]^5/5)/(a*d)) + Tan[c 
+ d*x]^5/(5*a*d)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 210

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 
)^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3086

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])
 

rule 3087

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])
 

rule 3185

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]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.74

method	result	size
risch	\(-\frac {2 \left (25 i {\mathrm e}^{4 i \left (d x +c \right )}+5 \,{\mathrm e}^{5 i \left (d x +c \right )}+21 i {\mathrm e}^{2 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 i {\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{7 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}+3 i\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} a d}\)	\(120\)
derivativedivides	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a d}\)	\(130\)
default	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a d}\)	\(130\)

Input:

int(tan(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(25*I*exp(4*I*(d*x+c))+5*exp(5*I*(d*x+c))+21*I*exp(2*I*(d*x+c))+13*e 
xp(3*I*(d*x+c))+15*I*exp(6*I*(d*x+c))+15*exp(7*I*(d*x+c))-9*exp(I*(d*x+c)) 
+3*I)/(exp(I*(d*x+c))+I)^5/(exp(I*(d*x+c))-I)^3/a/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate(tan(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")
 

Output:

-1/15*(3*cos(d*x + c)^4 + 6*cos(d*x + c)^2 + 4*(3*cos(d*x + c)^2 - 1)*sin( 
d*x + c) - 1)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)
 

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\tan ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate(tan(d*x+c)**4/(a+a*sin(d*x+c)),x)
 

Output:

Integral(tan(c + d*x)**4/(sin(c + d*x) + 1), x)/a
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (63) = 126\).

Time = 0.04 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.10 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {16 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \] Input:

integrate(tan(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")
 

Output:

-16/15*(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 + 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)
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.74 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {5 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 490 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 73}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \] Input:

integrate(tan(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/120*(5*(9*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 11)/(a*(tan 
(1/2*d*x + 1/2*c) - 1)^3) - (45*tan(1/2*d*x + 1/2*c)^4 + 240*tan(1/2*d*x + 
 1/2*c)^3 + 490*tan(1/2*d*x + 1/2*c)^2 + 320*tan(1/2*d*x + 1/2*c) + 73)/(a 
*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
 

Mupad [B] (verification not implemented)

Time = 17.81 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {16\,\left (-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{15\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \] Input:

int(tan(c + d*x)^4/(a + a*sin(c + d*x)),x)
 

Output:

(16*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^2 - 6*tan(c/2 + (d*x)/2)^ 
3 + 1))/(15*a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^5)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 302, normalized size of antiderivative = 4.38 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} \tan \left (d x +c \right )^{3}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} \tan \left (d x +c \right )+7 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{3}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )+7 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \tan \left (d x +c \right )^{3}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \tan \left (d x +c \right )-7 \cos \left (d x +c \right ) \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3}+15 \cos \left (d x +c \right ) \tan \left (d x +c \right )-7 \cos \left (d x +c \right )+23 \sin \left (d x +c \right )^{4}+8 \sin \left (d x +c \right )^{3}-27 \sin \left (d x +c \right )^{2}-7 \sin \left (d x +c \right )+8}{15 \cos \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{3}+\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )-1\right )} \] Input:

int(tan(d*x+c)^4/(a+a*sin(d*x+c)),x)
 

Output:

(5*cos(c + d*x)*sin(c + d*x)**3*tan(c + d*x)**3 - 15*cos(c + d*x)*sin(c + 
d*x)**3*tan(c + d*x) + 7*cos(c + d*x)*sin(c + d*x)**3 + 5*cos(c + d*x)*sin 
(c + d*x)**2*tan(c + d*x)**3 - 15*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x 
) + 7*cos(c + d*x)*sin(c + d*x)**2 - 5*cos(c + d*x)*sin(c + d*x)*tan(c + d 
*x)**3 + 15*cos(c + d*x)*sin(c + d*x)*tan(c + d*x) - 7*cos(c + d*x)*sin(c 
+ d*x) - 5*cos(c + d*x)*tan(c + d*x)**3 + 15*cos(c + d*x)*tan(c + d*x) - 7 
*cos(c + d*x) + 23*sin(c + d*x)**4 + 8*sin(c + d*x)**3 - 27*sin(c + d*x)** 
2 - 7*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))