\(\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1460]

Optimal result

Integrand size = 29, antiderivative size = 222 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {x}{b^2}-\frac {2 a^5 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac {4 a^3 \left (a^2-2 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac {a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.82 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {2 a b^3+a^4 (c+d x)-2 a^2 b^2 (c+d x)+b^4 (c+d x)}{\left (-a^2 b+b^3\right )^2}+\frac {2 a^3 \left (a^2-4 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2}}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {a^4 \cos (c+d x)}{(a-b)^2 b (a+b)^2 (a+b \sin (c+d x))}}{d} \] Input:

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

Output:

(-((2*a*b^3 + a^4*(c + d*x) - 2*a^2*b^2*(c + d*x) + b^4*(c + d*x))/(-(a^2* 
b) + b^3)^2) + (2*a^3*(a^2 - 4*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a 
^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)) + Sin[(c + d*x)/2]/((a + b)^2*(Cos[(c 
+ d*x)/2] - Sin[(c + d*x)/2])) + Sin[(c + d*x)/2]/((a - b)^2*(Cos[(c + d*x 
)/2] + Sin[(c + d*x)/2])) - (a^4*Cos[c + d*x])/((a - b)^2*b*(a + b)^2*(a + 
 b*Sin[c + d*x])))/d
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3376, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^4}{\cos (c+d x)^2 (a+b \sin (c+d x))^2}dx\)

\(\Big \downarrow \) 3376

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^5 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}-\frac {a^4 \cos (c+d x)}{b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {4 a^3 \left (a^2-2 b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}-\frac {x}{b^2}\)

Input:

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

Output:

-(x/b^2) - (2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*( 
a^2 - b^2)^(5/2)*d) + (4*a^3*(a^2 - 2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2]) 
/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)*d) + Cos[c + d*x]/(2*(a + b)^2*d 
*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) - ( 
a^4*Cos[c + d*x])/(b*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))
 

Defintions of rubi rules used

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 3376

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) 
 + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ 
e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr 
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( 
LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\frac {2 a^{3} \left (\frac {-b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}-4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{b^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(184\)
default	\(\frac {\frac {2 a^{3} \left (\frac {-b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a b}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}-4 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{b^{2} \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\)	\(184\)
risch	\(-\frac {x}{b^{2}}+\frac {2 i \left (3 i a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-i b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+2 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{2} b^{3}+i b^{5}+2 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}+i a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+a^{5} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{4} b +{\mathrm e}^{i \left (d x +c \right )} a^{5}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \left (a^{2}-b^{2}\right )^{2}}-\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{2}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,b^{2}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(532\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 724, normalized size of antiderivative = 3.26 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/2*(2*a^4*b^3 - 4*a^2*b^5 + 2*b^7 + 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a* 
b^6)*d*x*cos(d*x + c) + 2*(a^6*b - b^7)*cos(d*x + c)^2 - ((a^5*b - 4*a^3*b 
^3)*cos(d*x + c)*sin(d*x + c) + (a^6 - 4*a^4*b^2)*cos(d*x + c))*sqrt(-a^2 
+ b^2)*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^5*b^2 - 2*a^3*b^ 
4 + a*b^6 - (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*x*cos(d*x + c))*sin(d* 
x + c))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d*cos(d*x + c)*sin(d*x + 
c) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*cos(d*x + c)), -(a^4*b^3 
- 2*a^2*b^5 + b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*x*cos(d*x + c) 
 + (a^6*b - b^7)*cos(d*x + c)^2 + ((a^5*b - 4*a^3*b^3)*cos(d*x + c)*sin(d* 
x + c) + (a^6 - 4*a^4*b^2)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d* 
x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (a^5*b^2 - 2*a^3*b^4 + a*b^6 
 - (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*x*cos(d*x + c))*sin(d*x + c))/( 
(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d*cos(d*x + c)*sin(d*x + c) + (a^7 
*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*cos(d*x + c))]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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 = 0.47 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.19 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (2 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{4} - 2 \, a^{2} b^{2}\right )}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}} - \frac {d x + c}{b^{2}}}{d} \] Input:

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

Output:

(2*(a^5 - 4*a^3*b^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^4*b^2 - 2*a^2*b^4 + b^6)*s 
qrt(a^2 - b^2)) - 2*(2*a^3*b*tan(1/2*d*x + 1/2*c)^3 + a*b^3*tan(1/2*d*x + 
1/2*c)^3 + a^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^4*tan(1/2*d*x + 1/2*c)^2 - 3*a 
*b^3*tan(1/2*d*x + 1/2*c) - a^4 - 2*a^2*b^2)/((a^4*b - 2*a^2*b^3 + b^5)*(a 
*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x + 1 
/2*c) - a)) - (d*x + c)/b^2)/d
 

Mupad [B] (verification not implemented)

Time = 24.57 (sec) , antiderivative size = 4797, normalized size of antiderivative = 21.61 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (2*tan 
(c/2 + (d*x)/2)^2*(a^4 + 2*b^4))/(b*(a^4 + b^4 - 2*a^2*b^2)) - (2*a^2*(a^2 
 + 2*b^2))/(b*(a^2 - b^2)^2) - (6*a*b^2*tan(c/2 + (d*x)/2))/(a^2 - b^2)^2) 
/(d*(a + 2*b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^4 - 2*b*tan(c/2 + ( 
d*x)/2)^3)) - (2*atan((64*a*b^23*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3* 
b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 
1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21* 
b^3) - (704*a^3*b^21*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520* 
a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^ 
11 + 5440*a^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) + (3520 
*a^5*b^19*tan(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 
9536*a^7*b^17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a 
^15*b^9 - 4544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) - (9536*a^7*b^17*t 
an(c/2 + (d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^ 
17 + 14464*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4 
544*a^17*b^7 + 1536*a^19*b^5 - 192*a^21*b^3) + (14464*a^9*b^15*tan(c/2 + ( 
d*x)/2))/(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464 
*a^9*b^15 - 11072*a^11*b^13 + 1024*a^13*b^11 + 5440*a^15*b^9 - 4544*a^17*b 
^7 + 1536*a^19*b^5 - 192*a^21*b^3) - (11072*a^11*b^13*tan(c/2 + (d*x)/2))/ 
(64*a*b^23 - 704*a^3*b^21 + 3520*a^5*b^19 - 9536*a^7*b^17 + 14464*a^9*b...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 863, normalized size of antiderivative = 3.89 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos( 
c + d*x)*sin(c + d*x)*a**5*b**2 - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 
2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)*a**3*b**4 + 2*sqrt( 
a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*cos(c + d*x) 
*a**6*b - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b* 
*2))*cos(c + d*x)*a**4*b**3 + cos(c + d*x)*sin(c + d*x)*tan(c + d*x)*a**6* 
b**2 - 3*cos(c + d*x)*sin(c + d*x)*tan(c + d*x)*a**4*b**4 + 3*cos(c + d*x) 
*sin(c + d*x)*tan(c + d*x)*a**2*b**6 - cos(c + d*x)*sin(c + d*x)*tan(c + d 
*x)*b**8 - cos(c + d*x)*sin(c + d*x)*a**7*b - cos(c + d*x)*sin(c + d*x)*a* 
*6*b**2*d*x + 4*cos(c + d*x)*sin(c + d*x)*a**5*b**3 + 3*cos(c + d*x)*sin(c 
 + d*x)*a**4*b**4*d*x - 5*cos(c + d*x)*sin(c + d*x)*a**3*b**5 - 3*cos(c + 
d*x)*sin(c + d*x)*a**2*b**6*d*x + 2*cos(c + d*x)*sin(c + d*x)*a*b**7 + cos 
(c + d*x)*sin(c + d*x)*b**8*d*x + cos(c + d*x)*tan(c + d*x)*a**7*b - 3*cos 
(c + d*x)*tan(c + d*x)*a**5*b**3 + 3*cos(c + d*x)*tan(c + d*x)*a**3*b**5 - 
 cos(c + d*x)*tan(c + d*x)*a*b**7 - cos(c + d*x)*a**8 - cos(c + d*x)*a**7* 
b*d*x + 4*cos(c + d*x)*a**6*b**2 + 3*cos(c + d*x)*a**5*b**3*d*x - 5*cos(c 
+ d*x)*a**4*b**4 - 3*cos(c + d*x)*a**3*b**5*d*x + 2*cos(c + d*x)*a**2*b**6 
 + cos(c + d*x)*a*b**7*d*x + 3*sin(c + d*x)**2*a**4*b**4 - 3*sin(c + d*x)* 
*2*a**2*b**6 - sin(c + d*x)*a**7*b + 4*sin(c + d*x)*a**5*b**3 - 5*sin(c + 
d*x)*a**3*b**5 + 2*sin(c + d*x)*a*b**7 - a**6*b**2 - a**4*b**4 + 2*a**2...