\(\int \frac {\sin ^2(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [62]

Optimal result

Integrand size = 23, antiderivative size = 184 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+6 b) x}{2 a^4}-\frac {\sqrt {b} \left (15 a^2+40 a b+24 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^4 (a+b)^{3/2} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {3 b \tan (e+f x)}{4 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (11 a+12 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

1/2*(a+6*b)*x/a^4-1/8*b^(1/2)*(15*a^2+40*a*b+24*b^2)*arctan(b^(1/2)*tan(f* 
x+e)/(a+b)^(1/2))/a^4/(a+b)^(3/2)/f-1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*t 
an(f*x+e)^2)^2-3/4*b*tan(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2-1/8*b*(11*a+1 
2*b)*tan(f*x+e)/a^3/(a+b)/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 14.06 (sec) , antiderivative size = 1915, normalized size of antiderivative = 10.41 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[Sin[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(5*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(((3*a^2 + 8*a*b + 8*b^ 
2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) - (a*Sqrt[b]* 
(3*a^2 + 16*a*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*x) 
])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(8192*b^(5/2)*f*(a + b*S 
ec[e + f*x]^2)^3) + ((a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((-3* 
a*(a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) + (S 
qrt[b]*(3*a^3 + 14*a^2*b + 24*a*b^2 + 16*b^3 + a*(3*a^2 + 4*a*b + 4*b^2)*C 
os[2*(e + f*x)])*Sin[2*(e + f*x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x) 
])^2)))/(2048*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3) - ((a + 2*b + a*Cos[2*e 
+ 2*f*x])^3*Sec[e + f*x]^6*((2*(3*a^5 - 10*a^4*b + 80*a^3*b^2 + 480*a^2*b^ 
3 + 640*a*b^4 + 256*b^5)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 
2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[ 
e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^ 
4]) + (Sec[2*e]*(256*b^2*(a + b)^2*(3*a^2 + 8*a*b + 8*b^2)*f*x*Cos[2*e] + 
512*a*b^2*(a + b)^2*(a + 2*b)*f*x*Cos[2*f*x] + 128*a^4*b^2*f*x*Cos[2*(e + 
2*f*x)] + 256*a^3*b^3*f*x*Cos[2*(e + 2*f*x)] + 128*a^2*b^4*f*x*Cos[2*(e + 
2*f*x)] + 512*a^4*b^2*f*x*Cos[4*e + 2*f*x] + 2048*a^3*b^3*f*x*Cos[4*e + 2* 
f*x] + 2560*a^2*b^4*f*x*Cos[4*e + 2*f*x] + 1024*a*b^5*f*x*Cos[4*e + 2*f*x] 
 + 128*a^4*b^2*f*x*Cos[6*e + 4*f*x] + 256*a^3*b^3*f*x*Cos[6*e + 4*f*x] + 1 
28*a^2*b^4*f*x*Cos[6*e + 4*f*x] - 9*a^6*Sin[2*e] + 12*a^5*b*Sin[2*e] + ...
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4620, 373, 402, 27, 402, 397, 216, 218}

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.

\(\displaystyle \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {\frac {\int \frac {-5 b \tan ^2(e+f x)+a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {2 (a+b) \left (-9 b \tan ^2(e+f x)+2 a+3 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a (a+b)}-\frac {3 b \tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {-9 b \tan ^2(e+f x)+2 a+3 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}-\frac {3 b \tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {4 a^2+17 b a+12 b^2-b (11 a+12 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}-\frac {b (11 a+12 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {3 b \tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {\frac {4 (a+b) (a+6 b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {b \left (15 a^2+40 a b+24 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a (a+b)}-\frac {b (11 a+12 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {3 b \tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {\frac {4 (a+b) (a+6 b) \arctan (\tan (e+f x))}{a}-\frac {b \left (15 a^2+40 a b+24 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a (a+b)}-\frac {b (11 a+12 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {3 b \tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\frac {4 (a+b) (a+6 b) \arctan (\tan (e+f x))}{a}-\frac {\sqrt {b} \left (15 a^2+40 a b+24 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a (a+b)}-\frac {b (11 a+12 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {3 b \tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )^2}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\)

Input:

Int[Sin[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(-1/2*Tan[e + f*x]/(a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)^2) + 
 ((-3*b*Tan[e + f*x])/(2*a*(a + b + b*Tan[e + f*x]^2)^2) + (((4*(a + b)*(a 
 + 6*b)*ArcTan[Tan[e + f*x]])/a - (Sqrt[b]*(15*a^2 + 40*a*b + 24*b^2)*ArcT 
an[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a*(a + b)) - ( 
b*(11*a + 12*b)*Tan[e + f*x])/(2*a*(a + b)*(a + b + b*Tan[e + f*x]^2)))/(2 
*a))/(2*a))/f
 

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 216

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

rule 218

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

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4620

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, 
 x] && IntegerQ[m/2] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 3.03 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\frac {\frac {a b \left (7 a +8 b \right ) \tan \left (f x +e \right )^{3}}{8 a +8 b}+\frac {a \left (9 a +8 b \right ) \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+40 a b +24 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}+\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (1+\tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +6 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{4}}}{f}\)	\(155\)
default	\(\frac {-\frac {b \left (\frac {\frac {a b \left (7 a +8 b \right ) \tan \left (f x +e \right )^{3}}{8 a +8 b}+\frac {a \left (9 a +8 b \right ) \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+40 a b +24 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{4}}+\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (1+\tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +6 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{4}}}{f}\)	\(155\)
risch	\(\frac {x}{2 a^{3}}+\frac {3 x b}{a^{4}}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{3} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{3} f}-\frac {i b \left (9 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+32 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+24 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+102 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+152 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+80 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+80 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b +56 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}+9 a^{3}+10 a^{2} b \right )}{4 a^{4} \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 \left (a +b \right )^{2} f \,a^{2}}-\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{2 \left (a +b \right )^{2} f \,a^{3}}-\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{2} f \,a^{4}}+\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 \left (a +b \right )^{2} f \,a^{2}}+\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{2 \left (a +b \right )^{2} f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{2} f \,a^{4}}\)	\(592\)

Input:

int(sin(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/a^4*b*((1/8*a*b*(7*a+8*b)/(a+b)*tan(f*x+e)^3+1/8*a*(9*a+8*b)*tan(f 
*x+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(15*a^2+40*a*b+24*b^2)/(a+b)/((a+b)*b)^( 
1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))+1/a^4*(-1/2*a*tan(f*x+e)/(1+tan 
(f*x+e)^2)+1/2*(a+6*b)*arctan(tan(f*x+e))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (166) = 332\).

Time = 0.16 (sec) , antiderivative size = 815, normalized size of antiderivative = 4.43 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(sin(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[1/32*(16*(a^4 + 7*a^3*b + 6*a^2*b^2)*f*x*cos(f*x + e)^4 + 32*(a^3*b + 7*a 
^2*b^2 + 6*a*b^3)*f*x*cos(f*x + e)^2 + 16*(a^2*b^2 + 7*a*b^3 + 6*b^4)*f*x 
+ ((15*a^4 + 40*a^3*b + 24*a^2*b^2)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 
 + 24*b^4 + 2*(15*a^3*b + 40*a^2*b^2 + 24*a*b^3)*cos(f*x + e)^2)*sqrt(-b/( 
a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos( 
f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x 
 + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*co 
s(f*x + e)^2 + b^2)) - 4*(4*(a^4 + a^3*b)*cos(f*x + e)^5 + (17*a^3*b + 18* 
a^2*b^2)*cos(f*x + e)^3 + (11*a^2*b^2 + 12*a*b^3)*cos(f*x + e))*sin(f*x + 
e))/((a^7 + a^6*b)*f*cos(f*x + e)^4 + 2*(a^6*b + a^5*b^2)*f*cos(f*x + e)^2 
 + (a^5*b^2 + a^4*b^3)*f), 1/16*(8*(a^4 + 7*a^3*b + 6*a^2*b^2)*f*x*cos(f*x 
 + e)^4 + 16*(a^3*b + 7*a^2*b^2 + 6*a*b^3)*f*x*cos(f*x + e)^2 + 8*(a^2*b^2 
 + 7*a*b^3 + 6*b^4)*f*x + ((15*a^4 + 40*a^3*b + 24*a^2*b^2)*cos(f*x + e)^4 
 + 15*a^2*b^2 + 40*a*b^3 + 24*b^4 + 2*(15*a^3*b + 40*a^2*b^2 + 24*a*b^3)*c 
os(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*s 
qrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) - 2*(4*(a^4 + a^3*b)*cos(f*x 
 + e)^5 + (17*a^3*b + 18*a^2*b^2)*cos(f*x + e)^3 + (11*a^2*b^2 + 12*a*b^3) 
*cos(f*x + e))*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x + e)^4 + 2*(a^6*b + 
a^5*b^2)*f*cos(f*x + e)^2 + (a^5*b^2 + a^4*b^3)*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**2/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.48 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 40 \, a b^{2} + 24 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + a^{4} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (11 \, a b^{2} + 12 \, b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (17 \, a^{2} b + 40 \, a b^{2} + 24 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (4 \, a^{3} + 21 \, a^{2} b + 29 \, a b^{2} + 12 \, b^{3}\right )} \tan \left (f x + e\right )}{{\left (a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (f x + e\right )^{6} + a^{6} + 3 \, a^{5} b + 3 \, a^{4} b^{2} + a^{3} b^{3} + {\left (2 \, a^{5} b + 5 \, a^{4} b^{2} + 3 \, a^{3} b^{3}\right )} \tan \left (f x + e\right )^{4} + {\left (a^{6} + 5 \, a^{5} b + 7 \, a^{4} b^{2} + 3 \, a^{3} b^{3}\right )} \tan \left (f x + e\right )^{2}} - \frac {4 \, {\left (f x + e\right )} {\left (a + 6 \, b\right )}}{a^{4}}}{8 \, f} \] Input:

integrate(sin(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

-1/8*((15*a^2*b + 40*a*b^2 + 24*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b) 
)/((a^5 + a^4*b)*sqrt((a + b)*b)) + ((11*a*b^2 + 12*b^3)*tan(f*x + e)^5 + 
(17*a^2*b + 40*a*b^2 + 24*b^3)*tan(f*x + e)^3 + (4*a^3 + 21*a^2*b + 29*a*b 
^2 + 12*b^3)*tan(f*x + e))/((a^4*b^2 + a^3*b^3)*tan(f*x + e)^6 + a^6 + 3*a 
^5*b + 3*a^4*b^2 + a^3*b^3 + (2*a^5*b + 5*a^4*b^2 + 3*a^3*b^3)*tan(f*x + e 
)^4 + (a^6 + 5*a^5*b + 7*a^4*b^2 + 3*a^3*b^3)*tan(f*x + e)^2) - 4*(f*x + e 
)*(a + 6*b)/a^4)/f
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.13 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 40 \, a b^{2} + 24 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + a^{4} b\right )} \sqrt {a b + b^{2}}} + \frac {7 \, a b^{2} \tan \left (f x + e\right )^{3} + 8 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) + 17 \, a b^{2} \tan \left (f x + e\right ) + 8 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + a^{3} b\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac {4 \, {\left (f x + e\right )} {\left (a + 6 \, b\right )}}{a^{4}} + \frac {4 \, \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} a^{3}}}{8 \, f} \] Input:

integrate(sin(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

-1/8*((15*a^2*b + 40*a*b^2 + 24*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) 
+ arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^5 + a^4*b)*sqrt(a*b + b^2)) 
+ (7*a*b^2*tan(f*x + e)^3 + 8*b^3*tan(f*x + e)^3 + 9*a^2*b*tan(f*x + e) + 
17*a*b^2*tan(f*x + e) + 8*b^3*tan(f*x + e))/((a^4 + a^3*b)*(b*tan(f*x + e) 
^2 + a + b)^2) - 4*(f*x + e)*(a + 6*b)/a^4 + 4*tan(f*x + e)/((tan(f*x + e) 
^2 + 1)*a^3))/f
 

Mupad [B] (verification not implemented)

Time = 15.71 (sec) , antiderivative size = 2628, normalized size of antiderivative = 14.28 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(sin(e + f*x)^2/(a + b/cos(e + f*x)^2)^3,x)
 

Output:

(atan(((((tan(e + f*x)*(3264*a*b^6 + 1152*b^7 + 3296*a^2*b^5 + 1424*a^3*b^ 
4 + 241*a^4*b^3))/(32*(2*a^7*b + a^8 + a^6*b^2)) - (((6*a^8*b^5 + (29*a^9* 
b^4)/2 + (21*a^10*b^3)/2 + 2*a^11*b^2)/(2*a^10*b + a^11 + a^9*b^2) - (tan( 
e + f*x)*(a*1i + b*6i)*(512*a^8*b^5 + 1280*a^9*b^4 + 1024*a^10*b^3 + 256*a 
^11*b^2))/(128*a^4*(2*a^7*b + a^8 + a^6*b^2)))*(a*1i + b*6i))/(4*a^4))*(a* 
1i + b*6i)*1i)/(4*a^4) + (((tan(e + f*x)*(3264*a*b^6 + 1152*b^7 + 3296*a^2 
*b^5 + 1424*a^3*b^4 + 241*a^4*b^3))/(32*(2*a^7*b + a^8 + a^6*b^2)) + (((6* 
a^8*b^5 + (29*a^9*b^4)/2 + (21*a^10*b^3)/2 + 2*a^11*b^2)/(2*a^10*b + a^11 
+ a^9*b^2) + (tan(e + f*x)*(a*1i + b*6i)*(512*a^8*b^5 + 1280*a^9*b^4 + 102 
4*a^10*b^3 + 256*a^11*b^2))/(128*a^4*(2*a^7*b + a^8 + a^6*b^2)))*(a*1i + b 
*6i))/(4*a^4))*(a*1i + b*6i)*1i)/(4*a^4))/(((297*a*b^6)/4 + 27*b^7 + (279* 
a^2*b^5)/4 + (805*a^3*b^4)/32 + (165*a^4*b^3)/64)/(2*a^10*b + a^11 + a^9*b 
^2) - (((tan(e + f*x)*(3264*a*b^6 + 1152*b^7 + 3296*a^2*b^5 + 1424*a^3*b^4 
 + 241*a^4*b^3))/(32*(2*a^7*b + a^8 + a^6*b^2)) - (((6*a^8*b^5 + (29*a^9*b 
^4)/2 + (21*a^10*b^3)/2 + 2*a^11*b^2)/(2*a^10*b + a^11 + a^9*b^2) - (tan(e 
 + f*x)*(a*1i + b*6i)*(512*a^8*b^5 + 1280*a^9*b^4 + 1024*a^10*b^3 + 256*a^ 
11*b^2))/(128*a^4*(2*a^7*b + a^8 + a^6*b^2)))*(a*1i + b*6i))/(4*a^4))*(a*1 
i + b*6i))/(4*a^4) + (((tan(e + f*x)*(3264*a*b^6 + 1152*b^7 + 3296*a^2*b^5 
 + 1424*a^3*b^4 + 241*a^4*b^3))/(32*(2*a^7*b + a^8 + a^6*b^2)) + (((6*a^8* 
b^5 + (29*a^9*b^4)/2 + (21*a^10*b^3)/2 + 2*a^11*b^2)/(2*a^10*b + a^11 +...
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1738, normalized size of antiderivative = 9.45 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sin(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*sin(e + f*x)**4*a**4 - 40*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*ta 
n((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b - 24*sqrt(b)*sqr 
t(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f* 
x)**4*a**2*b**2 + 30*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2 
) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**4 + 110*sqrt(b)*sqrt(a + b)*atan( 
(sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3*b + 
 128*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqr 
t(b))*sin(e + f*x)**2*a**2*b**2 + 48*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b) 
*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**3 - 15*sqrt(b)* 
sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 - 
70*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*a**3*b - 119*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - 
sqrt(a))/sqrt(b))*a**2*b**2 - 88*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan 
((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b**3 - 24*sqrt(b)*sqrt(a + b)*atan((sq 
rt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**4 - 15*sqrt(b)*sqrt(a + 
b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4* 
a**4 - 40*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a) 
)/sqrt(b))*sin(e + f*x)**4*a**3*b - 24*sqrt(b)*sqrt(a + b)*atan((sqrt(a + 
b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b**2 + 30*...