3.147 \(\int \frac{\sin ^2(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=181 \[ \frac{(a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f (a-b)^{7/2}}-\frac{b (13 a+2 b) \tan (e+f x)}{6 a f (a-b)^3 \sqrt{a+b \tan ^2(e+f x)}}-\frac{5 b \tan (e+f x)}{6 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\sin (e+f x) \cos (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
[Out]
((a + 4*b)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*(a - b)^(7/2)*f) - (Cos[e + f*x]*
Sin[e + f*x])/(2*(a - b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (5*b*Tan[e + f*x])/(6*(a - b)^2*f*(a + b*Tan[e + f*
x]^2)^(3/2)) - (b*(13*a + 2*b)*Tan[e + f*x])/(6*a*(a - b)^3*f*Sqrt[a + b*Tan[e + f*x]^2])
________________________________________________________________________________________
Rubi [A] time = 0.208875, antiderivative size = 181, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{3663, 471, 527, 12, 377, 203} \[ \frac{(a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f (a-b)^{7/2}}-\frac{b (13 a+2 b) \tan (e+f x)}{6 a f (a-b)^3 \sqrt{a+b \tan ^2(e+f x)}}-\frac{5 b \tan (e+f x)}{6 f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\sin (e+f x) \cos (e+f x)}{2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sin[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]
[Out]
((a + 4*b)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*(a - b)^(7/2)*f) - (Cos[e + f*x]*
Sin[e + f*x])/(2*(a - b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - (5*b*Tan[e + f*x])/(6*(a - b)^2*f*(a + b*Tan[e + f*
x]^2)^(3/2)) - (b*(13*a + 2*b)*Tan[e + f*x])/(6*a*(a - b)^3*f*Sqrt[a + b*Tan[e + f*x]^2])
Rule 3663
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
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])
Rubi steps
\begin{align*} \int \frac{\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a-4 b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{2 (a-b) f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a (3 a+2 b)-10 a b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{6 a (a-b)^2 f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+2 b) \tan (e+f x)}{6 a (a-b)^3 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2 (a+4 b)}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 a^2 (a-b)^3 f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+2 b) \tan (e+f x)}{6 a (a-b)^3 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(a+4 b) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 (a-b)^3 f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+2 b) \tan (e+f x)}{6 a (a-b)^3 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(a+4 b) \operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^3 f}\\ &=\frac{(a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{7/2} f}-\frac{\cos (e+f x) \sin (e+f x)}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+2 b) \tan (e+f x)}{6 a (a-b)^3 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 4.28657, size = 309, normalized size = 1.71 \[ -\frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (-\frac{3 a b (a+4 b) \sin (2 (e+f x)) \sin ^2(e+f x) \left (\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}\right )^{3/2} \left (2 (a-b) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )-2 a \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{\sqrt{2}}-(a-b) \sin (2 (e+f x)) \left (8 a b^2-4 b (6 a+b) ((a-b) \cos (2 (e+f x))+a+b)-3 a ((a-b) \cos (2 (e+f x))+a+b)^2\right )\right )}{12 \sqrt{2} a f (a-b)^4 ((a-b) \cos (2 (e+f x))+a+b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]
[Out]
-(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-((a - b)*(8*a*b^2 - 4*b*(6*a + b)*(a + b + (a - b)
*Cos[2*(e + f*x)]) - 3*a*(a + b + (a - b)*Cos[2*(e + f*x)])^2)*Sin[2*(e + f*x)]) - (3*a*b*(a + 4*b)*(((a + b +
(a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)^(3/2)*(2*(a - b)*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(
e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - 2*a*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*
(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sin[e + f*x]^2*Sin[2*(e + f*x)])/Sqrt[2]))/(12*Sqrt[2]*a*(a - b)^
4*f*(a + b + (a - b)*Cos[2*(e + f*x)])^2)
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Maple [B] time = 0.217, size = 2511, normalized size = 13.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)
[Out]
-1/3/f*(-1+cos(2*f*x+2*e))*(-3*cos(2*f*x+2*e)*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f
*x+2*e)+1))^(3/2)*(b^4*(a-b))^(1/2)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*
cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^2+6*cos(2*f*x+2*e)^2*a^3*b^3-14*cos(2*f*x+2*
e)^2*a^2*b^4+10*cos(2*f*x+2*e)^2*a*b^5-2*cos(2*f*x+2*e)^2*b^6-3*(b^4*(a-b))^(1/2)*arctan(b^2*(a-b)/(b^4*(a-b))
^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*
sin(2*f*x+2*e)*a^2*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)+10*cos(2*f*x+2*e)*a^2*b^
4-14*cos(2*f*x+2*e)*a*b^5+4*cos(2*f*x+2*e)*b^6-6*a^3*b^3+4*a^2*b^4+4*a*b^5-2*b^6)/sin(2*f*x+2*e)^3/(a-b)^3/a^2
/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)/b^2-1/12/f*(-1+cos(2*f*x+2*e))*(6*cos(2*f*
x+2*e)*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-
1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^5*b+16*
cos(2*f*x+2*e)*(a-b)^(3/2)*b^6-24*cos(2*f*x+2*e)*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(
2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)
+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^4*b^2-9*cos(2*f*x+2*e)^3*(a-b)^(3/2)*a^4*b^2+3*cos(2*f*x+2*e)^2*(a-b)^(3/2)
*a^5*b+3*cos(2*f*x+2*e)^2*(a-b)^(3/2)*a^4*b^2+3*cos(2*f*x+2*e)^3*(a-b)^(3/2)*a^5*b-3*(a-b)^(3/2)*a^5*b*cos(2*f
*x+2*e)+9*(a-b)^(3/2)*a^4*b^2*cos(2*f*x+2*e)+9*cos(2*f*x+2*e)^3*(a-b)^(3/2)*a^3*b^3-3*cos(2*f*x+2*e)^3*(a-b)^(
3/2)*a^2*b^4+21*cos(2*f*x+2*e)^2*(a-b)^(3/2)*a^3*b^3-71*cos(2*f*x+2*e)^2*(a-b)^(3/2)*a^2*b^4+52*cos(2*f*x+2*e)
^2*(a-b)^(3/2)*a*b^5+3*cos(2*f*x+2*e)*(a-b)^(3/2)*a^3*b^3+55*cos(2*f*x+2*e)*(a-b)^(3/2)*a^2*b^4-80*cos(2*f*x+2
*e)*(a-b)^(3/2)*a*b^5-33*(a-b)^(3/2)*a^3*b^3+19*(a-b)^(3/2)*a^2*b^4+28*(a-b)^(3/2)*a*b^5-12*cos(2*f*x+2*e)*sin
(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*
x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^2*b^4-24*(b^4*(a-
b))^(1/2)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*cos(2*f*x
+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*sin(2*f*x+2*e)*a^2*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2
*e)+1))^(3/2)*(a-b)^(3/2)-24*cos(2*f*x+2*e)*sin(2*f*x+2*e)*(a-b)^(3/2)*(b^4*(a-b))^(1/2)*((a*cos(2*f*x+2*e)-b*
cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x
+2*e)/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^2+6*sin(2*f*x+2*e)*((a*cos(2*f*x+2
*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*
cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^5*b-24*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*c
os(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f
*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^4*b^2+30*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*
f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*
e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^3*b^3-12*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2
*e)+a+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*
cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^2*b^4-3*(a-b)^(3/2)*a^5*b-3*(a-b)^(3/2)*a^4*b^2-8*(a-b)^(3/2)
*b^6-8*cos(2*f*x+2*e)^2*(a-b)^(3/2)*b^6+30*cos(2*f*x+2*e)*sin(2*f*x+2*e)*((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a
+b)/(cos(2*f*x+2*e)+1))^(3/2)*arctan((a-b)^(1/2)*(-1+cos(2*f*x+2*e))/sin(2*f*x+2*e)/((a*cos(2*f*x+2*e)-b*cos(2
*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+1))^(1/2))*a^3*b^3)/b/((a*cos(2*f*x+2*e)-b*cos(2*f*x+2*e)+a+b)/(cos(2*f*x+2*e)+
1))^(3/2)/a^2/(a-b)^(11/2)/sin(2*f*x+2*e)^3
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
integrate(sin(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(5/2), x)
________________________________________________________________________________________
Fricas [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*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
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*tan(f*x+e)**2)**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
integrate(sin(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(5/2), x)