\(\int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [177]

Optimal result

Integrand size = 21, antiderivative size = 96 \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 a^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d} \] Output:

-2*a^2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)/d- 
b*sec(d*x+c)/(a^2-b^2)/d+a*tan(d*x+c)/(a^2-b^2)/d
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.58 \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 a^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) \cos (c+d x)+\sqrt {a^2-b^2} (-b+b \cos (c+d x)+a \sin (c+d x))}{(a-b) (a+b) \sqrt {a^2-b^2} d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

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

Output:

(-2*a^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*Cos[c + d*x] + Sq 
rt[a^2 - b^2]*(-b + b*Cos[c + d*x] + a*Sin[c + d*x]))/((a - b)*(a + b)*Sqr 
t[a^2 - b^2]*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + S 
in[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3206, 3042, 3086, 24, 3139, 1083, 217, 4254, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3206

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3086

\(\displaystyle -\frac {a^2 \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2-b^2}-\frac {b \int 1d\sec (c+d x)}{d \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {a^2 \int \frac {1}{a+b \sin (c+d x)}dx}{a^2-b^2}+\frac {a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2-b^2}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2-b^2}-\frac {2 a^2 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4254

\(\displaystyle -\frac {a \int 1d(-\tan (c+d x))}{d \left (a^2-b^2\right )}-\frac {2 a^2 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {2 a^2 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

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 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3206

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[a/(a^2 - b^2)   Int[(g*Tan[e + f*x])^p/Sin[e + f*x] 
^2, x], x] + (-Simp[b*(g/(a^2 - b^2))   Int[(g*Tan[e + f*x])^(p - 1)/Cos[e 
+ f*x], x], x] - Simp[a^2*(g^2/(a^2 - b^2))   Int[(g*Tan[e + f*x])^(p - 2)/ 
(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2 
, 0] && IntegersQ[2*p] && GtQ[p, 1]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {2 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}-\frac {8}{\left (8 a +8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{\left (8 a -8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(112\)
default	\(\frac {-\frac {2 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}-\frac {8}{\left (8 a +8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{\left (8 a -8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(112\)
risch	\(\frac {-2 i a +2 \,{\mathrm e}^{i \left (d x +c \right )} b}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(208\)

Input:

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

Output:

1/d*(-2*a^2/(a-b)/(a+b)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c) 
+2*b)/(a^2-b^2)^(1/2))-8/(8*a+8*b)/(tan(1/2*d*x+1/2*c)-1)-8/(8*a-8*b)/(tan 
(1/2*d*x+1/2*c)+1))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.18 \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a^{2} \cos \left (d x + c\right ) \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}, \frac {\sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{2} b + b^{3} + {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )}\right ] \] Input:

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

Output:

[1/2*(sqrt(-a^2 + b^2)*a^2*cos(d*x + c)*log(((2*a^2 - b^2)*cos(d*x + c)^2 
- 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos( 
d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 
 - b^2)) - 2*a^2*b + 2*b^3 + 2*(a^3 - a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b 
^2 + b^4)*d*cos(d*x + c)), (sqrt(a^2 - b^2)*a^2*arctan(-(a*sin(d*x + c) + 
b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c) - a^2*b + b^3 + (a^3 - a*b 
^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*d*cos(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.11 \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}\right )}}{d} \] Input:

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

Output:

-2*((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2 
*c) + b)/sqrt(a^2 - b^2)))*a^2/(a^2 - b^2)^(3/2) + (a*tan(1/2*d*x + 1/2*c) 
 - b)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1)))/d
 

Mupad [B] (verification not implemented)

Time = 17.92 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2\,b}{a^2-b^2}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a^2\,\mathrm {atan}\left (\frac {\frac {a^2\,\left (2\,a^2\,b-2\,b^3\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}}{2\,a^2}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:

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

Output:

((2*b)/(a^2 - b^2) - (2*a*tan(c/2 + (d*x)/2))/(a^2 - b^2))/(d*(tan(c/2 + ( 
d*x)/2)^2 - 1)) - (2*a^2*atan(((a^2*(2*a^2*b - 2*b^3))/((a + b)^(3/2)*(a - 
 b)^(3/2)) + (2*a^3*tan(c/2 + (d*x)/2)*(a^2 - b^2))/((a + b)^(3/2)*(a - b) 
^(3/2)))/(2*a^2)))/(d*(a + b)^(3/2)*(a - b)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.99 \[ \int \frac {\tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{3}+\cos \left (d x +c \right ) \tan \left (d x +c \right ) a^{4}-2 \cos \left (d x +c \right ) \tan \left (d x +c \right ) a^{2} b^{2}+\cos \left (d x +c \right ) \tan \left (d x +c \right ) b^{4}+\cos \left (d x +c \right ) a^{3} b -\cos \left (d x +c \right ) a \,b^{3}+\sin \left (d x +c \right ) a^{2} b^{2}-\sin \left (d x +c \right ) b^{4}-a^{3} b +a \,b^{3}}{\cos \left (d x +c \right ) a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:

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

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*c 
os(c + d*x)*a**3 + cos(c + d*x)*tan(c + d*x)*a**4 - 2*cos(c + d*x)*tan(c + 
 d*x)*a**2*b**2 + cos(c + d*x)*tan(c + d*x)*b**4 + cos(c + d*x)*a**3*b - c 
os(c + d*x)*a*b**3 + sin(c + d*x)*a**2*b**2 - sin(c + d*x)*b**4 - a**3*b + 
 a*b**3)/(cos(c + d*x)*a*d*(a**4 - 2*a**2*b**2 + b**4))