\(\int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [183]

Optimal result

Integrand size = 19, antiderivative size = 109 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b)^2 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b)^2 d}+\frac {\left (a^2+b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {a}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.49 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {a \left ((a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (1+\sin (c+d x))-2 \left (-a^2+b^2+\left (a^2+b^2\right ) \log (a+b \sin (c+d x))\right )\right )+b \left ((a-b)^2 \log (1-\sin (c+d x))+(a+b)^2 \log (1+\sin (c+d x))-2 \left (a^2+b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)}{2 (a-b)^2 (a+b)^2 d (a+b \sin (c+d x))} \] Input:

Integrate[Tan[c + d*x]/(a + b*Sin[c + d*x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.17, 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, 25, 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.

\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3200

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

\(\Big \downarrow \) 594

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 657

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]/(a + b*Sin[c + d*x])^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 = 0.70 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}}{d}\)	\(98\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {a}{\left (a +b \right ) \left (a -b \right ) \left (a +b \sin \left (d x +c \right )\right )}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}}{d}\)	\(98\)
risch	\(\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {i c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i x}{a^{2}+2 a b +b^{2}}+\frac {i c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 i a^{2} x}{a^{4}-2 b^{2} a^{2}+b^{4}}-\frac {2 i a^{2} c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {2 i b^{2} x}{a^{4}-2 b^{2} a^{2}+b^{4}}-\frac {2 i b^{2} c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\left (a^{2}+2 a b +b^{2}\right ) d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) a^{2}}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\)	\(401\)

Input:

int(tan(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.79 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, a^{3} - 2 \, a b^{2} - 2 \, {\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \] Input:

integrate(tan(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(tan(d*x+c)/(a+b*sin(d*x+c))**2,x)
 

Output:

Integral(tan(c + d*x)/(a + b*sin(c + d*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, a}{a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \] Input:

integrate(tan(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.37 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\left (a^{2} b + b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b d - 2 \, a^{2} b^{3} d + b^{5} d} - \frac {\log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} - \frac {\log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} - \frac {a^{3} - a b^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )} {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d} \] Input:

integrate(tan(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 18.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.45 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2+b^2\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,{\left (a-b\right )}^2}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,{\left (a+b\right )}^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2-b^2\right )\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )} \] Input:

int(tan(c + d*x)/(a + b*sin(c + d*x))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 459, normalized size of antiderivative = 4.21 \[ \int \frac {\tan (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a^{2} b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a \,b^{2}-a^{3}+a \,b^{2}}{d \left (\sin \left (d x +c \right ) a^{4} b -2 \sin \left (d x +c \right ) a^{2} b^{3}+\sin \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int(tan(d*x+c)/(a+b*sin(d*x+c))^2,x)
 

Output:

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