\(\int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1350]

Optimal result

Integrand size = 29, antiderivative size = 165 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a^2 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \] Output:

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

Mathematica [A] (verified)

Time = 1.89 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.21 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {48 a^2 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(c+d x) \left (-4 a^2 b-8 b^3+3 b \left (5 a^2+b^2\right ) \cos (c+d x)-12 a^2 b \cos (2 (c+d x))+5 a^2 b \cos (3 (c+d x))+b^3 \cos (3 (c+d x))-6 a^3 \sin (c+d x)+12 a b^2 \sin (c+d x)+2 a^3 \sin (3 (c+d x))+4 a b^2 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}}{24 d} \] Input:

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

Output:

((48*a^2*b^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2) 
^(5/2) - (Sec[c + d*x]^3*(-4*a^2*b - 8*b^3 + 3*b*(5*a^2 + b^2)*Cos[c + d*x 
] - 12*a^2*b*Cos[2*(c + d*x)] + 5*a^2*b*Cos[3*(c + d*x)] + b^3*Cos[3*(c + 
d*x)] - 6*a^3*Sin[c + d*x] + 12*a*b^2*Sin[c + d*x] + 2*a^3*Sin[3*(c + d*x) 
] + 4*a*b^2*Sin[3*(c + d*x)]))/((a - b)^2*(a + b)^2))/(24*d)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3381, 3042, 3086, 15, 3175, 27, 3042, 3139, 1083, 217, 4254, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3381

\(\displaystyle \frac {a \int \sec ^4(c+d x)dx}{a^2-b^2}-\frac {b \int \sec ^3(c+d x) \tan (c+d x)dx}{a^2-b^2}-\frac {a^2 \int \frac {\sec ^2(c+d x)}{a+b \sin (c+d x)}dx}{a^2-b^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3086

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 3175

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {a \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{a^2-b^2}-\frac {a^2 \left (-\frac {2 b^2 \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {\sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )}\right )}{a^2-b^2}-\frac {b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

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 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 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 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 3175

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^ 
(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2* 
(a^2 - b^2)*(p + 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m* 
(a^2*(p + 2) - b^2*(m + p + 2) + a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; F 
reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegersQ 
[2*m, 2*p]
 

rule 3381

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a*(d^2/(a^2 
- b^2))   Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-Simp[ 
b*(d/(a^2 - b^2))   Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] 
 - Simp[a^2*(d^2/(g^2*(a^2 - b^2)))   Int[(g*Cos[e + f*x])^(p + 2)*((d*Sin[ 
e + f*x])^(n - 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g 
}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 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 = 1.28 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {\frac {2 a^{2} b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (8 a -8 b \right )}+\frac {4}{\left (8 a -8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (8 a +8 b \right )}-\frac {4}{\left (8 a +8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(207\)
default	\(\frac {\frac {2 a^{2} b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (8 a -8 b \right )}+\frac {4}{\left (8 a -8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (8 a +8 b \right )}-\frac {4}{\left (8 a +8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(207\)
risch	\(\frac {-2 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+2 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-4 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+\frac {4 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {8 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{3}-\frac {2 i a^{3}}{3}-\frac {4 i a \,b^{2}}{3}+2 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b^{2} a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b^{2} a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(302\)

Input:

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

Output:

1/d*(2*a^2*b^2/(a-b)^2/(a+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))-8/3/(tan(1/2*d*x+1/2*c)+1)^3/(8*a-8*b)+4/(8* 
a-8*b)/(tan(1/2*d*x+1/2*c)+1)^2+1/2*b/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)-8/3/( 
tan(1/2*d*x+1/2*c)-1)^3/(8*a+8*b)-4/(8*a+8*b)/(tan(1/2*d*x+1/2*c)-1)^2-1/2 
*b/(a+b)^2/(tan(1/2*d*x+1/2*c)-1))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 470, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{2} b^{2} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} a^{2} b^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)^2*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 = 0.41 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2} b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{2} b - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 21.71 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2\,a^2\,b^2\,\mathrm {atan}\left (\frac {\frac {a^2\,b^2\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,a^2\,b^2}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,\left (2\,a^2\,b+b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^3+a\,b^2\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.55 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{2} b^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{4} b -3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b^{3}+\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{5}-2 \cos \left (d x +c \right ) a^{4} b +3 \cos \left (d x +c \right ) a^{2} b^{3}-\cos \left (d x +c \right ) b^{5}-\sin \left (d x +c \right )^{3} a^{5}-\sin \left (d x +c \right )^{3} a^{3} b^{2}+2 \sin \left (d x +c \right )^{3} a \,b^{4}+3 \sin \left (d x +c \right )^{2} a^{4} b -3 \sin \left (d x +c \right )^{2} a^{2} b^{3}+3 \sin \left (d x +c \right ) a^{3} b^{2}-3 \sin \left (d x +c \right ) a \,b^{4}-2 a^{4} b +a^{2} b^{3}+b^{5}}{3 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2} a^{6}-3 \sin \left (d x +c \right )^{2} a^{4} b^{2}+3 \sin \left (d x +c \right )^{2} a^{2} b^{4}-\sin \left (d x +c \right )^{2} b^{6}-a^{6}+3 a^{4} b^{2}-3 a^{2} b^{4}+b^{6}\right )} \] Input:

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

Output:

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