\(\int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [194]

Optimal result

Integrand size = 19, antiderivative size = 149 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b)^3 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b)^3 d}+\frac {a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {a}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {a^2+b^2}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \] Output:

-1/2*ln(1-sin(d*x+c))/(a+b)^3/d-1/2*ln(1+sin(d*x+c))/(a-b)^3/d+a*(a^2+3*b^ 
2)*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d-1/2*a/(a^2-b^2)/d/(a+b*sin(d*x+c))^2-( 
a^2+b^2)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))
 

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.43 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {-\frac {\log (1-\sin (c+d x))}{(a+b)^2}+\frac {\log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 a b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {2 b}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}+a \left (\frac {\log (1-\sin (c+d x))}{(a+b)^3}-\frac {\log (1+\sin (c+d x))}{(a-b)^3}+\frac {b \left (2 \left (3 a^2+b^2\right ) \log (a+b \sin (c+d x))+\frac {\left (a^2-b^2\right ) \left (-5 a^2+b^2-4 a b \sin (c+d x)\right )}{(a+b \sin (c+d x))^2}\right )}{\left (a^2-b^2\right )^3}\right )}{2 b d} \] Input:

Integrate[Tan[c + d*x]/(a + b*Sin[c + d*x])^3,x]
 

Output:

(-(Log[1 - Sin[c + d*x]]/(a + b)^2) + Log[1 + Sin[c + d*x]]/(a - b)^2 - (4 
*a*b*Log[a + b*Sin[c + d*x]])/(a^2 - b^2)^2 + (2*b)/((a^2 - b^2)*(a + b*Si 
n[c + d*x])) + a*(Log[1 - Sin[c + d*x]]/(a + b)^3 - Log[1 + Sin[c + d*x]]/ 
(a - b)^3 + (b*(2*(3*a^2 + b^2)*Log[a + b*Sin[c + d*x]] + ((a^2 - b^2)*(-5 
*a^2 + b^2 - 4*a*b*Sin[c + d*x]))/(a + b*Sin[c + d*x])^2))/(a^2 - b^2)^3)) 
/(2*b*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3200, 594, 27, 657, 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.

\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3200

\(\displaystyle \frac {\int \frac {b \sin (c+d x)}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 594

\(\displaystyle \frac {\frac {\int -\frac {2 \left (b^2-a b \sin (c+d x)\right )}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )}-\frac {a}{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {b^2-a b \sin (c+d x)}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{a^2-b^2}-\frac {a}{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}}{d}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {-\frac {\int \left (\frac {b-a}{2 (a+b)^2 (b-b \sin (c+d x))}-\frac {a \left (a^2+3 b^2\right )}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {a+b}{2 (a-b)^2 (\sin (c+d x) b+b)}+\frac {-a^2-b^2}{(a-b) (a+b) (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{a^2-b^2}-\frac {a}{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {a}{2 \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {\frac {a^2+b^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {a \left (a^2+3 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac {(a-b) \log (b-b \sin (c+d x))}{2 (a+b)^2}+\frac {(a+b) \log (b \sin (c+d x)+b)}{2 (a-b)^2}}{a^2-b^2}}{d}\)

Input:

Int[Tan[c + d*x]/(a + b*Sin[c + d*x])^3,x]
 

Output:

(-1/2*a/((a^2 - b^2)*(a + b*Sin[c + d*x])^2) - (((a - b)*Log[b - b*Sin[c + 
 d*x]])/(2*(a + b)^2) - (a*(a^2 + 3*b^2)*Log[a + b*Sin[c + d*x]])/(a^2 - b 
^2)^2 + ((a + b)*Log[b + b*Sin[c + d*x]])/(2*(a - b)^2) + (a^2 + b^2)/((a^ 
2 - b^2)*(a + b*Sin[c + d*x])))/(a^2 - b^2))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

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 3200

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)/(b^2 - x^2)^((p + 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b 
^2, 0] && IntegerQ[(p + 1)/2]
 

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {a}{2 \left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a^{2}+b^{2}}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a \left (a^{2}+3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\)	\(134\)
default	\(\frac {-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {a}{2 \left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a^{2}+b^{2}}{\left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a \left (a^{2}+3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\)	\(134\)
risch	\(\frac {i x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {i c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i a^{3} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {2 i a^{3} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {6 i a \,b^{2} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {6 i a \,b^{2} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {2 \left (-i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-i {\mathrm e}^{3 i \left (d x +c \right )} b^{3}+i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+i {\mathrm e}^{i \left (d x +c \right )} b^{3}+3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )} a \,b^{2}\right )}{\left (-i {\mathrm e}^{2 i \left (d x +c \right )} b +i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} \left (a^{2}-b^{2}\right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(601\)

Input:

int(tan(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/2/(a-b)^3*ln(1+sin(d*x+c))-1/2/(a+b)^3*ln(sin(d*x+c)-1)-1/2*a/(a+b 
)/(a-b)/(a+b*sin(d*x+c))^2-(a^2+b^2)/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))+a*(a 
^2+3*b^2)/(a-b)^3/(a+b)^3*ln(a+b*sin(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (143) = 286\).

Time = 0.20 (sec) , antiderivative size = 462, normalized size of antiderivative = 3.10 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4} - 2 \, {\left (a^{5} + 4 \, a^{3} b^{2} + 3 \, a b^{4} - {\left (a^{3} b^{2} + 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{5} + 3 \, a^{4} b + 4 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{5} - 3 \, a^{4} b + 4 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5} - {\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{4} b - b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} d\right )}} \] Input:

integrate(tan(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/2*(3*a^5 - 2*a^3*b^2 - a*b^4 - 2*(a^5 + 4*a^3*b^2 + 3*a*b^4 - (a^3*b^2 + 
 3*a*b^4)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^2*b^3)*sin(d*x + c))*log(b*sin(d 
*x + c) + a) + (a^5 + 3*a^4*b + 4*a^3*b^2 + 4*a^2*b^3 + 3*a*b^4 + b^5 - (a 
^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*cos(d*x + c)^2 + 2*(a^4*b + 3*a^3*b^2 
+ 3*a^2*b^3 + a*b^4)*sin(d*x + c))*log(sin(d*x + c) + 1) + (a^5 - 3*a^4*b 
+ 4*a^3*b^2 - 4*a^2*b^3 + 3*a*b^4 - b^5 - (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - 
 b^5)*cos(d*x + c)^2 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*sin(d*x + 
 c))*log(-sin(d*x + c) + 1) + 2*(a^4*b - b^5)*sin(d*x + c))/((a^6*b^2 - 3* 
a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3 
*b^5 - a*b^7)*d*sin(d*x + c) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*d)
 

Sympy [F]

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

integrate(tan(d*x+c)/(a+b*sin(d*x+c))**3,x)
 

Output:

Integral(tan(c + d*x)/(a + b*sin(c + d*x))**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.53 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (a^{3} + 3 \, a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {3 \, a^{3} + a b^{2} + 2 \, {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{2 \, d} \] Input:

integrate(tan(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/2*(2*(a^3 + 3*a*b^2)*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^ 
4 - b^6) - (3*a^3 + a*b^2 + 2*(a^2*b + b^3)*sin(d*x + c))/(a^6 - 2*a^4*b^2 
 + a^2*b^4 + (a^4*b^2 - 2*a^2*b^4 + b^6)*sin(d*x + c)^2 + 2*(a^5*b - 2*a^3 
*b^3 + a*b^5)*sin(d*x + c)) - log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b 
^2 - b^3) - log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.41 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {{\left (a^{3} b + 3 \, a b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {\log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} - \frac {\log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} - \frac {3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4} + 2 \, {\left (a^{4} b - b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d} \] Input:

integrate(tan(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")
 

Output:

(a^3*b + 3*a*b^3)*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3* 
a^2*b^5*d - b^7*d) - 1/2*log(abs(-sin(d*x + c) + 1))/(a^3*d + 3*a^2*b*d + 
3*a*b^2*d + b^3*d) - 1/2*log(abs(-sin(d*x + c) - 1))/(a^3*d - 3*a^2*b*d + 
3*a*b^2*d - b^3*d) - 1/2*(3*a^5 - 2*a^3*b^2 - a*b^4 + 2*(a^4*b - b^5)*sin( 
d*x + c))/((b*sin(d*x + c) + a)^2*(a + b)^3*(a - b)^3*d)
 

Mupad [B] (verification not implemented)

Time = 18.17 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.04 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^2\,b^2+b^4\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,{\left (a+b\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,{\left (a-b\right )}^3}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^3+3\,a\,b^2\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \] Input:

int(tan(c + d*x)/(a + b*sin(c + d*x))^3,x)
 

Output:

((2*tan(c/2 + (d*x)/2)^2*(b^4 + 3*a^2*b^2))/(a*(a^4 + b^4 - 2*a^2*b^2)) + 
(4*a^2*b*tan(c/2 + (d*x)/2))/(a^4 + b^4 - 2*a^2*b^2) + (4*a^2*b*tan(c/2 + 
(d*x)/2)^3)/(a^4 + b^4 - 2*a^2*b^2))/(d*(tan(c/2 + (d*x)/2)^2*(2*a^2 + 4*b 
^2) + a^2*tan(c/2 + (d*x)/2)^4 + a^2 + 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b* 
tan(c/2 + (d*x)/2))) - log(tan(c/2 + (d*x)/2) - 1)/(d*(a + b)^3) - log(tan 
(c/2 + (d*x)/2) + 1)/(d*(a - b)^3) + (log(a + 2*b*tan(c/2 + (d*x)/2) + a*t 
an(c/2 + (d*x)/2)^2)*(3*a*b^2 + a^3))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^ 
2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1005, normalized size of antiderivative = 6.74 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(tan(d*x+c)/(a+b*sin(d*x+c))^3,x)
 

Output:

( - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b**2 + 6*log(tan((c + 
 d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**3 - 6*log(tan((c + d*x)/2) - 1)*sin( 
c + d*x)**2*a**2*b**4 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**5 
 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**5*b + 12*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)*a**4*b**2 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
*a**3*b**3 + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b**4 - 2*log(ta 
n((c + d*x)/2) - 1)*a**6 + 6*log(tan((c + d*x)/2) - 1)*a**5*b - 6*log(tan( 
(c + d*x)/2) - 1)*a**4*b**2 + 2*log(tan((c + d*x)/2) - 1)*a**3*b**3 - 2*lo 
g(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4*b**2 - 6*log(tan((c + d*x)/2) 
 + 1)*sin(c + d*x)**2*a**3*b**3 - 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x) 
**2*a**2*b**4 - 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**5 - 4*log 
(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**5*b - 12*log(tan((c + d*x)/2) + 1)* 
sin(c + d*x)*a**4*b**2 - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**3*b* 
*3 - 4*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a**2*b**4 - 2*log(tan((c + d 
*x)/2) + 1)*a**6 - 6*log(tan((c + d*x)/2) + 1)*a**5*b - 6*log(tan((c + d*x 
)/2) + 1)*a**4*b**2 - 2*log(tan((c + d*x)/2) + 1)*a**3*b**3 + 2*log(tan((c 
 + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a**4*b**2 + 6* 
log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a**2 
*b**4 + 4*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d* 
x)*a**5*b + 12*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*si...