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

Optimal. Leaf size=184 \[ -\frac{\sqrt{b} \left (15 a^2+40 a b+24 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^4 f (a+b)^{3/2}}-\frac{b (11 a+12 b) \tan (e+f x)}{8 a^3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}-\frac{3 b \tan (e+f x)}{4 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac{x (a+6 b)}{2 a^4}-\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

((a + 6*b)*x)/(2*a^4) - (Sqrt[b]*(15*a^2 + 40*a*b + 24*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^4
*(a + b)^(3/2)*f) - (Cos[e + f*x]*Sin[e + f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)^2) - (3*b*Tan[e + f*x])/(4*a
^2*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(11*a + 12*b)*Tan[e + f*x])/(8*a^3*(a + b)*f*(a + b + b*Tan[e + f*x]^2
))

________________________________________________________________________________________

Rubi [A]  time = 0.278306, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4132, 471, 527, 522, 203, 205} \[ -\frac{\sqrt{b} \left (15 a^2+40 a b+24 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^4 f (a+b)^{3/2}}-\frac{b (11 a+12 b) \tan (e+f x)}{8 a^3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}-\frac{3 b \tan (e+f x)}{4 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac{x (a+6 b)}{2 a^4}-\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 6*b)*x)/(2*a^4) - (Sqrt[b]*(15*a^2 + 40*a*b + 24*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^4
*(a + b)^(3/2)*f) - (Cos[e + f*x]*Sin[e + f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)^2) - (3*b*Tan[e + f*x])/(4*a
^2*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(11*a + 12*b)*Tan[e + f*x])/(8*a^3*(a + b)*f*(a + b + b*Tan[e + f*x]^2
))

Rule 4132

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]

Rule 471

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 527

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 522

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 203

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^3} \, 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 )^2}+\frac{\operatorname{Subst}\left (\int \frac{a+b-5 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, 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 )^2}-\frac{3 b \tan (e+f x)}{4 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{2 (a+b) (2 a+3 b)-18 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 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 )^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 )}+\frac{\operatorname{Subst}\left (\int \frac{2 (a+b) \left (4 a^2+17 a b+12 b^2\right )-2 b (a+b) (11 a+12 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{16 a^3 (a+b)^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 )}+\frac{(a+6 b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 a^4 f}-\frac{\left (b \left (15 a^2+40 a b+24 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^4 (a+b) f}\\ &=\frac{(a+6 b) x}{2 a^4}-\frac{\sqrt{b} \left (15 a^2+40 a b+24 b^2\right ) \tan ^{-1}\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 )}\\ \end{align*}

Mathematica [C]  time = 19.2405, size = 1915, normalized size = 10.41 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(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])/Sqr
t[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*Sec[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)
+ (Sqrt[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)*Cos[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*Co
s[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)*ArcTa
n[(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*C
os[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] + 128*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] + 684*a^4*b^2*Sin[2*e] + 2880*a^3*b^3*Sin[2*e] + 5280*a^2*b^4*Sin
[2*e] + 4608*a*b^5*Sin[2*e] + 1536*b^6*Sin[2*e] + 9*a^6*Sin[2*f*x] - 14*a^5*b*Sin[2*f*x] - 608*a^4*b^2*Sin[2*f
*x] - 2112*a^3*b^3*Sin[2*f*x] - 2560*a^2*b^4*Sin[2*f*x] - 1024*a*b^5*Sin[2*f*x] + 3*a^6*Sin[2*(e + 2*f*x)] - 1
2*a^5*b*Sin[2*(e + 2*f*x)] - 204*a^4*b^2*Sin[2*(e + 2*f*x)] - 384*a^3*b^3*Sin[2*(e + 2*f*x)] - 192*a^2*b^4*Sin
[2*(e + 2*f*x)] - 3*a^6*Sin[4*e + 2*f*x] + 10*a^5*b*Sin[4*e + 2*f*x] + 304*a^4*b^2*Sin[4*e + 2*f*x] + 1056*a^3
*b^3*Sin[4*e + 2*f*x] + 1280*a^2*b^4*Sin[4*e + 2*f*x] + 512*a*b^5*Sin[4*e + 2*f*x]))/(a + 2*b + a*Cos[2*(e + f
*x)])^2))/(4096*a^3*b^2*(a + b)^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*((-6*a^2*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]) + (a*Se
c[2*e]*((-9*a^4 - 16*a^3*b + 48*a^2*b^2 + 128*a*b^3 + 64*b^4)*Sin[2*f*x] + a*(-3*a^3 + 2*a^2*b + 24*a*b^2 + 16
*b^3)*Sin[2*(e + 2*f*x)] + (3*a^4 - 64*a^2*b^2 - 128*a*b^3 - 64*b^4)*Sin[4*e + 2*f*x]) + (9*a^5 + 18*a^4*b - 6
4*a^3*b^2 - 256*a^2*b^3 - 320*a*b^4 - 128*b^5)*Tan[2*e])/(a^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(4096*b^2*(a
+ b)^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*(-1536*(a + 2*b)*x - (3
*(a^6 - 8*a^5*b + 120*a^4*b^2 + 1280*a^3*b^3 + 3200*a^2*b^4 + 3072*a*b^5 + 1024*b^6)*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])]*(Co
s[2*e] - I*Sin[2*e]))/(b^2*(a + b)^(5/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (4*(a^4 + 32*a^3*b + 160*a^2*b^2 +
256*a*b^3 + 128*b^4)*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)]
)^2) + (256*a*Sin[2*(e + f*x)])/f + (a*(-3*a^5 + 26*a^4*b + 736*a^3*b^2 + 2624*a^2*b^3 + 3200*a*b^4 + 1280*b^5
)*Sec[2*e]*Sin[2*f*x] + (3*a^6 - 24*a^5*b - 920*a^4*b^2 - 4864*a^3*b^3 - 10112*a^2*b^4 - 9216*a*b^5 - 3072*b^6
)*Tan[2*e])/(b^2*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)]))))/(8192*a^4*(a + b*Sec[e + f*x]^2)^3)

________________________________________________________________________________________

Maple [A]  time = 0.101, size = 314, normalized size = 1.7 \begin{align*} -{\frac{\tan \left ( fx+e \right ) }{2\,f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{2\,f{a}^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{f{a}^{4}}}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( a+b \right ) }}-{\frac{{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{f{a}^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( a+b \right ) }}-{\frac{9\,b\tan \left ( fx+e \right ) }{8\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{b}^{2}\tan \left ( fx+e \right ) }{f{a}^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,b}{8\,f{a}^{2} \left ( a+b \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-5\,{\frac{{b}^{2}}{f{a}^{3} \left ( a+b \right ) \sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{b\tan \left ( fx+e \right ) }{\sqrt{ \left ( a+b \right ) b}}} \right ) }-3\,{\frac{{b}^{3}}{f{a}^{4} \left ( a+b \right ) \sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{b\tan \left ( fx+e \right ) }{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f/a^3*tan(f*x+e)/(tan(f*x+e)^2+1)+1/2/f/a^3*arctan(tan(f*x+e))+3/f/a^4*arctan(tan(f*x+e))*b-7/8/f/a^2*b^2
/(a+b+b*tan(f*x+e)^2)^2/(a+b)*tan(f*x+e)^3-1/f/a^3*b^3/(a+b+b*tan(f*x+e)^2)^2/(a+b)*tan(f*x+e)^3-9/8*b*tan(f*x
+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2-1/f*b^2/a^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)-15/8/f/a^2*b/(a+b)/((a+b)*b)^(1
/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-5/f/a^3*b^2/(a+b)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2)
)-3/f/a^4*b^3/(a+b)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 0.753736, size = 1872, normalized size = 10.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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*cos(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), 1/1
6*(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 + 2
4*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))*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*(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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.29612, size = 296, normalized size = 1.61 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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