\(\int \frac {\sin ^2(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [49]

Optimal result

Integrand size = 23, antiderivative size = 130 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+4 b) x}{2 a^3}-\frac {\sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 \sqrt {a+b} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]

[Out]

1/2*(a+4*b)*x/a^3-1/2*(3*a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^(1/2)/a^3/f/(a+b)^(1/2)-1/2*cos(f*x+e
)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)-b*tan(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4217, 482, 541, 536, 209, 211} \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 f \sqrt {a+b}}+\frac {x (a+4 b)}{2 a^3}-\frac {b \tan (e+f x)}{a^2 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[In]

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

[Out]

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

Rule 209

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

Rule 211

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

Rule 482

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

Rule 536

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

Rule 541

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

Rule 4217

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {a+b-3 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{2 a f} \\ & = -\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {2 (a+b) (a+2 b)-4 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 a^2 (a+b) f} \\ & = -\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(a+4 b) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}-\frac {(b (3 a+4 b)) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f} \\ & = \frac {(a+4 b) x}{2 a^3}-\frac {\sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 \sqrt {a+b} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 8.73 (sec) , antiderivative size = 699, normalized size of antiderivative = 5.38 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^2 \sec ^4(e+f x) \left (-\frac {2 \left (16 x+\frac {\left (-a^3+6 a^2 b+24 a b^2+16 b^3\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\left (a^2+8 a b+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (a+2 b+a \cos (2 (e+f x))) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{a^2}-\frac {-64 (a+2 b) x+\frac {\left (a^4-16 a^3 b-144 a^2 b^2-256 a b^3-128 b^4\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {16 a \cos (2 f x) \sin (2 e)}{f}+\frac {16 a \cos (2 e) \sin (2 f x)}{f}-\frac {\left (a^3+18 a^2 b+48 a b^2+32 b^3\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (a+2 b+a \cos (2 (e+f x))) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}}{a^3}+\frac {2 \left (\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}\right )}{b^{3/2} f}+\frac {-\frac {a \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\frac {\sqrt {b} (a+2 b) \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}}{b^{3/2} f}\right )}{256 \left (a+b \sec ^2(e+f x)\right )^2} \]

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*((-2*(16*x + ((-a^3 + 6*a^2*b + 24*a*b^2 + 16*b^3)*ArcTan[(Se
c[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*S
in[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a^2 + 8*a*b + 8*b^2
)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] +
Sin[e]))))/a^2 - (-64*(a + 2*b)*x + ((a^4 - 16*a^3*b - 144*a^2*b^2 - 256*a*b^3 - 128*b^4)*ArcTan[(Sec[f*x]*(C
os[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]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (16*a*Cos[2*f*x]*Sin[2*e])/f +
(16*a*Cos[2*e]*Sin[2*f*x])/f - ((a^3 + 18*a^2*b + 48*a*b^2 + 32*b^3)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*
(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + Sin[e])))/a^3 + (2*(((a + 2*b)*ArcTan[(Sq
rt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (a*Sqrt[b]*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e
+ f*x)]))))/(b^(3/2)*f) + (-((a*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2)) + (Sqrt[b]*(a + 2*b
)*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)])))/(b^(3/2)*f)))/(256*(a + b*Sec[e + f*x]^2)^2)

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}+\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (1+\tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +4 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{3}}}{f}\)	\(109\)
default	\(\frac {-\frac {b \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}+\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (1+\tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +4 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{3}}}{f}\)	\(109\)
risch	\(\frac {x}{2 a^{2}}+\frac {2 x b}{a^{3}}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f}-\frac {i b \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}-\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right ) f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{\left (a +b \right ) f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right ) f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{\left (a +b \right ) f \,a^{3}}\)	\(340\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.39 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{2} + 4 \, a b\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a b + 4 \, b^{2}\right )} f x + {\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 4 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left (a^{2} \cos \left (f x + e\right )^{3} + 2 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}, \frac {2 \, {\left (a^{2} + 4 \, a b\right )} f x \cos \left (f x + e\right )^{2} + 2 \, {\left (a b + 4 \, b^{2}\right )} f x + {\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 4 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left (a^{2} \cos \left (f x + e\right )^{3} + 2 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}\right ] \]

[In]

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

[Out]

[1/8*(4*(a^2 + 4*a*b)*f*x*cos(f*x + e)^2 + 4*(a*b + 4*b^2)*f*x + ((3*a^2 + 4*a*b)*cos(f*x + e)^2 + 3*a*b + 4*b
^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*cos(f*x + e)^2 + b^2)) - 4*(a^2*cos(f*x + e)^3 + 2*a*b*cos(f*x + e))*sin(f*x + e))/(a^4*f*cos(f*x
+ e)^2 + a^3*b*f), 1/4*(2*(a^2 + 4*a*b)*f*x*cos(f*x + e)^2 + 2*(a*b + 4*b^2)*f*x + ((3*a^2 + 4*a*b)*cos(f*x +
e)^2 + 3*a*b + 4*b^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x +
e)*sin(f*x + e))) - 2*(a^2*cos(f*x + e)^3 + 2*a*b*cos(f*x + e))*sin(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f)
]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.97 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {2 \, b \tan \left (f x + e\right )^{3} + {\left (a + 2 \, b\right )} \tan \left (f x + e\right )}{a^{2} b \tan \left (f x + e\right )^{4} + a^{3} + a^{2} b + {\left (a^{3} + 2 \, a^{2} b\right )} \tan \left (f x + e\right )^{2}} - \frac {{\left (f x + e\right )} {\left (a + 4 \, b\right )}}{a^{3}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}}}{2 \, f} \]

[In]

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

[Out]

-1/2*((2*b*tan(f*x + e)^3 + (a + 2*b)*tan(f*x + e))/(a^2*b*tan(f*x + e)^4 + a^3 + a^2*b + (a^3 + 2*a^2*b)*tan(
f*x + e)^2) - (f*x + e)*(a + 4*b)/a^3 + (3*a*b + 4*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b
)*a^3))/f

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (f x + e\right )} {\left (a + 4 \, b\right )}}{a^{3}} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a b + 4 \, b^{2}\right )}}{\sqrt {a b + b^{2}} a^{3}} - \frac {2 \, b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + 2 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )} a^{2}}}{2 \, f} \]

[In]

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

[Out]

1/2*((f*x + e)*(a + 4*b)/a^3 - (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*
(3*a*b + 4*b^2)/(sqrt(a*b + b^2)*a^3) - (2*b*tan(f*x + e)^3 + a*tan(f*x + e) + 2*b*tan(f*x + e))/((b*tan(f*x +
e)^4 + a*tan(f*x + e)^2 + 2*b*tan(f*x + e)^2 + a + b)*a^2))/f

Mupad [B] (verification not implemented)

Time = 18.50 (sec) , antiderivative size = 816, normalized size of antiderivative = 6.28 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+2\,b\right )}{2\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^3}{a^2}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (a+2\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atan}\left (\frac {b^2\,\mathrm {tan}\left (e+f\,x\right )}{4\,\left (\frac {b^2}{4}+\frac {b^3}{a}\right )}+\frac {b^3\,\mathrm {tan}\left (e+f\,x\right )}{b^3+\frac {a\,b^2}{4}}\right )\,\left (a\,1{}\mathrm {i}+b\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^3\,f}+\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}+\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^4+b\,a^3}+\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}-\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^4+b\,a^3}}{\frac {\frac {3\,a^2\,b^3}{2}+8\,a\,b^4+8\,b^5}{a^6}+\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}+\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}-\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}-\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}\,2{}\mathrm {i}}{f\,\left (a^4+b\,a^3\right )} \]

[In]

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

[Out]

(atan(((((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3))/a^4 + (((4*a^6*b^3 + 2*a^7*b^2)/a^6 + (tan(e + f*x)*(1
6*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2)
)/(a^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/2)*1i)/(a^3*b + a^4) + (((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a
^2*b^3))/a^4 - (((4*a^6*b^3 + 2*a^7*b^2)/a^6 - (tan(e + f*x)*(16*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*(-b*(a + b
))^(1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/
2)*1i)/(a^3*b + a^4))/((8*a*b^4 + 8*b^5 + (3*a^2*b^3)/2)/a^6 + (((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3)
)/a^4 + (((4*a^6*b^3 + 2*a^7*b^2)/a^6 + (tan(e + f*x)*(16*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*(-b*(a + b))^(1/2
))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^
3*b + a^4) - (((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3))/a^4 - (((4*a^6*b^3 + 2*a^7*b^2)/a^6 - (tan(e + f
*x)*(16*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))
^(1/2))/(a^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2)*2i)/
(f*(a^3*b + a^4)) - (atan((b^2*tan(e + f*x))/(4*(b^2/4 + b^3/a)) + (b^3*tan(e + f*x))/((a*b^2)/4 + b^3))*(a*1i
+ b*4i)*1i)/(2*a^3*f) - ((tan(e + f*x)*(a + 2*b))/(2*a^2) + (b*tan(e + f*x)^3)/a^2)/(f*(a + b + b*tan(e + f*x
)^4 + tan(e + f*x)^2*(a + 2*b)))