\(\int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [84]

Optimal result

Integrand size = 19, antiderivative size = 105 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{16 a^4 d}+\frac {1}{8 d (a+a \sin (c+d x))^4}-\frac {1}{12 a d (a+a \sin (c+d x))^3}-\frac {1}{16 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {1}{16 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:

1/16*arctanh(sin(d*x+c))/a^4/d+1/8/d/(a+a*sin(d*x+c))^4-1/12/a/d/(a+a*sin( 
d*x+c))^3-1/16/d/(a^2+a^2*sin(d*x+c))^2-1/16/d/(a^4+a^4*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.59 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {3 \text {arctanh}(\sin (c+d x))-\frac {4+19 \sin (c+d x)+12 \sin ^2(c+d x)+3 \sin ^3(c+d x)}{(1+\sin (c+d x))^4}}{48 a^4 d} \] Input:

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

Output:

(3*ArcTanh[Sin[c + d*x]] - (4 + 19*Sin[c + d*x] + 12*Sin[c + d*x]^2 + 3*Si 
n[c + d*x]^3)/(1 + Sin[c + d*x])^4)/(48*a^4*d)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3186

\(\displaystyle \frac {\int \frac {a \sin (c+d x)}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {\int \left (-\frac {1}{2 (\sin (c+d x) a+a)^5}+\frac {1}{4 (\sin (c+d x) a+a)^4 a}+\frac {1}{8 (\sin (c+d x) a+a)^3 a^2}+\frac {1}{16 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^3}+\frac {1}{16 (\sin (c+d x) a+a)^2 a^3}\right )d(a \sin (c+d x))}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\text {arctanh}(\sin (c+d x))}{16 a^4}-\frac {1}{16 a^3 (a \sin (c+d x)+a)}-\frac {1}{16 a^2 (a \sin (c+d x)+a)^2}-\frac {1}{12 a (a \sin (c+d x)+a)^3}+\frac {1}{8 (a \sin (c+d x)+a)^4}}{d}\)

Input:

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

Output:

(ArcTanh[Sin[c + d*x]]/(16*a^4) + 1/(8*(a + a*Sin[c + d*x])^4) - 1/(12*a*( 
a + a*Sin[c + d*x])^3) - 1/(16*a^2*(a + a*Sin[c + d*x])^2) - 1/(16*a^3*(a 
+ a*Sin[c + d*x])))/d
 

Defintions of rubi rules used

rule 86

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]))))
 

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 3186

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) 
^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
 

Maple [A] (verified)

Time = 5.35 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{32}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{32}}{d \,a^{4}}\)	\(79\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{32}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{32}}{d \,a^{4}}\)	\(79\)
risch	\(-\frac {i \left (-3 \,{\mathrm e}^{i \left (d x +c \right )}+85 \,{\mathrm e}^{3 i \left (d x +c \right )}+24 i {\mathrm e}^{2 i \left (d x +c \right )}-80 i {\mathrm e}^{4 i \left (d x +c \right )}-85 \,{\mathrm e}^{5 i \left (d x +c \right )}+24 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{24 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{16 a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{16 a^{4} d}\)	\(148\)

Input:

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

Output:

1/d/a^4*(-1/32*ln(sin(d*x+c)-1)+1/8/(1+sin(d*x+c))^4-1/12/(1+sin(d*x+c))^3 
-1/16/(1+sin(d*x+c))^2-1/16/(1+sin(d*x+c))+1/32*ln(1+sin(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (95) = 190\).

Time = 0.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.89 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {24 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{4} - 8 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 8\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 22\right )} \sin \left (d x + c\right ) - 32}{96 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 8 \, a^{4} d \cos \left (d x + c\right )^{2} + 8 \, a^{4} d - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \] Input:

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

Output:

1/96*(24*cos(d*x + c)^2 + 3*(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d* 
x + c)^2 - 2)*sin(d*x + c) + 8)*log(sin(d*x + c) + 1) - 3*(cos(d*x + c)^4 
- 8*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)*log(-sin(d*x 
 + c) + 1) + 2*(3*cos(d*x + c)^2 - 22)*sin(d*x + c) - 32)/(a^4*d*cos(d*x + 
 c)^4 - 8*a^4*d*cos(d*x + c)^2 + 8*a^4*d - 4*(a^4*d*cos(d*x + c)^2 - 2*a^4 
*d)*sin(d*x + c))
 

Sympy [F]

\[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\tan {\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

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

Output:

Integral(tan(c + d*x)/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x 
)**2 + 4*sin(c + d*x) + 1), x)/a**4
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} + 12 \, \sin \left (d x + c\right )^{2} + 19 \, \sin \left (d x + c\right ) + 4\right )}}{a^{4} \sin \left (d x + c\right )^{4} + 4 \, a^{4} \sin \left (d x + c\right )^{3} + 6 \, a^{4} \sin \left (d x + c\right )^{2} + 4 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4}}}{96 \, d} \] Input:

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

Output:

-1/96*(2*(3*sin(d*x + c)^3 + 12*sin(d*x + c)^2 + 19*sin(d*x + c) + 4)/(a^4 
*sin(d*x + c)^4 + 4*a^4*sin(d*x + c)^3 + 6*a^4*sin(d*x + c)^2 + 4*a^4*sin( 
d*x + c) + a^4) - 3*log(sin(d*x + c) + 1)/a^4 + 3*log(sin(d*x + c) - 1)/a^ 
4)/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{32 \, a^{4} d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{32 \, a^{4} d} - \frac {3 \, \sin \left (d x + c\right )^{3} + 12 \, \sin \left (d x + c\right )^{2} + 19 \, \sin \left (d x + c\right ) + 4}{48 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{4}} \] Input:

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

Output:

1/32*log(abs(sin(d*x + c) + 1))/(a^4*d) - 1/32*log(abs(sin(d*x + c) - 1))/ 
(a^4*d) - 1/48*(3*sin(d*x + c)^3 + 12*sin(d*x + c)^2 + 19*sin(d*x + c) + 4 
)/(a^4*d*(sin(d*x + c) + 1)^4)
 

Mupad [B] (verification not implemented)

Time = 19.83 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.29 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^4\,d}+\frac {-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{8}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {43\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+28\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+56\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+70\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+56\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+28\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^4\right )} \] Input:

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

Output:

atanh(tan(c/2 + (d*x)/2))/(8*a^4*d) + (tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d 
*x)/2)/8 + (43*tan(c/2 + (d*x)/2)^3)/24 + (10*tan(c/2 + (d*x)/2)^4)/3 + (4 
3*tan(c/2 + (d*x)/2)^5)/24 + tan(c/2 + (d*x)/2)^6 - tan(c/2 + (d*x)/2)^7/8 
)/(d*(28*a^4*tan(c/2 + (d*x)/2)^2 + 56*a^4*tan(c/2 + (d*x)/2)^3 + 70*a^4*t 
an(c/2 + (d*x)/2)^4 + 56*a^4*tan(c/2 + (d*x)/2)^5 + 28*a^4*tan(c/2 + (d*x) 
/2)^6 + 8*a^4*tan(c/2 + (d*x)/2)^7 + a^4*tan(c/2 + (d*x)/2)^8 + a^4 + 8*a^ 
4*tan(c/2 + (d*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.67 \[ \int \frac {\tan (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}-72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}+48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+19 \sin \left (d x +c \right )^{4}+64 \sin \left (d x +c \right )^{3}+66 \sin \left (d x +c \right )^{2}+3}{192 a^{4} d \left (\sin \left (d x +c \right )^{4}+4 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2}+4 \sin \left (d x +c \right )+1\right )} \] Input:

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

Output:

( - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 - 48*log(tan((c + d*x)/2) 
 - 1)*sin(c + d*x)**3 - 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 48* 
log(tan((c + d*x)/2) - 1)*sin(c + d*x) - 12*log(tan((c + d*x)/2) - 1) + 12 
*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 48*log(tan((c + d*x)/2) + 1)* 
sin(c + d*x)**3 + 72*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 48*log(ta 
n((c + d*x)/2) + 1)*sin(c + d*x) + 12*log(tan((c + d*x)/2) + 1) + 19*sin(c 
 + d*x)**4 + 64*sin(c + d*x)**3 + 66*sin(c + d*x)**2 + 3)/(192*a**4*d*(sin 
(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1) 
)