3.127 \(\int \frac{\sin ^2(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac{(a+5 b) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{2 a^{7/2} f}-\frac{b (13 a+15 b) \tan (e+f x)}{6 a^3 f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[Out]

((a + 5*b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*a^(7/2)*f) - (Cos[e + f*x]*Sin[e
+ f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (5*b*Tan[e + f*x])/(6*a^2*f*(a + b + b*Tan[e + f*x]^2)^(3/2
)) - (b*(13*a + 15*b)*Tan[e + f*x])/(6*a^3*(a + b)*f*Sqrt[a + b + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.201752, antiderivative size = 167, 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 = {4132, 471, 527, 12, 377, 203} \[ \frac{(a+5 b) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{2 a^{7/2} f}-\frac{b (13 a+15 b) \tan (e+f x)}{6 a^3 f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}-\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 5*b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*a^(7/2)*f) - (Cos[e + f*x]*Sin[e
+ f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)^(3/2)) - (5*b*Tan[e + f*x])/(6*a^2*f*(a + b + b*Tan[e + f*x]^2)^(3/2
)) - (b*(13*a + 15*b)*Tan[e + f*x])/(6*a^3*(a + b)*f*Sqrt[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 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 \sec ^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+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 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a+b-4 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{5/2}} \, 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 )^{3/2}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{(a+b) (3 a+5 b)-10 b (a+b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{6 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 )^{3/2}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+15 b) \tan (e+f x)}{6 a^3 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 (a+b)^2 (a+5 b)}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 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 )^{3/2}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+15 b) \tan (e+f x)}{6 a^3 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{(a+5 b) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+15 b) \tan (e+f x)}{6 a^3 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{(a+5 b) \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{2 a^3 f}\\ &=\frac{(a+5 b) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{2 a^{7/2} f}-\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{5 b \tan (e+f x)}{6 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac{b (13 a+15 b) \tan (e+f x)}{6 a^3 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [B]  time = 10.7456, size = 983, normalized size = 5.89 \[ -\frac{(\cos (2 e+2 f x) a+a+2 b)^{5/2} \csc (e+f x) \left (-\frac{12 \sin ^4(e+f x)}{a+b}+\frac{(\cos (2 (e+f x)) a+a+2 b) \sin ^2(e+f x)}{(a+b)^2}+\frac{\sin ^2(e+f x)}{a+b}+\frac{16 \left (-a \sin ^2(e+f x)+a+b\right ) \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right ) \left (\frac{a^2 (a+b) \sin ^4(e+f x)}{\left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac{6 a (a+b) \sin ^2(e+f x)}{\cos (2 (e+f x)) a+a+2 b}+\frac{3 \sqrt{a} \sqrt{a+b} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right ) \sin (e+f x)}{\sqrt{\frac{-a \sin ^2(e+f x)+a+b}{a+b}}}\right )}{a^3}\right ) \sec ^5(e+f x)}{256 \sqrt{2} f \left (b \sec ^2(e+f x)+a\right )^{5/2} \left (-a \sin ^2(e+f x)+a+b\right )^{3/2}}-\frac{(\cos (2 e+2 f x) a+a+2 b)^{5/2} \csc (e+f x) \left (\frac{96 \sin ^6(e+f x)}{a}-\frac{24 \sin ^4(e+f x)}{a+b}+\frac{(\cos (2 (e+f x)) a+a+2 b) \sin ^2(e+f x)}{(a+b)^2}+\frac{\sin ^2(e+f x)}{a+b}+\frac{80 \left (-a \sin ^2(e+f x)+a+b\right ) \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right ) \left (\frac{a^2 (a+b) \sin ^4(e+f x)}{\left (-a \sin ^2(e+f x)+a+b\right )^2}-\frac{6 a (a+b) \sin ^2(e+f x)}{\cos (2 (e+f x)) a+a+2 b}+\frac{3 \sqrt{a} \sqrt{a+b} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right ) \sin (e+f x)}{\sqrt{\frac{-a \sin ^2(e+f x)+a+b}{a+b}}}\right )}{a^3}-\frac{160 \left (-a \sin ^2(e+f x)+a+b\right ) \left (1-\frac{a \sin ^2(e+f x)}{a+b}\right ) \left (\frac{a^2 \sin ^4(e+f x)}{\left (\frac{a \sin ^2(e+f x)}{a+b}-1\right )^2}-\frac{6 a (a+b)^2 \sin ^2(e+f x)}{\cos (2 (e+f x)) a+a+2 b}+\frac{3 \sqrt{a} (a+b)^{3/2} \sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right ) \sin (e+f x)}{\sqrt{\frac{-a \sin ^2(e+f x)+a+b}{a+b}}}\right )}{a^4}\right ) \sec ^5(e+f x)}{768 \sqrt{2} f \left (b \sec ^2(e+f x)+a\right )^{5/2} \left (-a \sin ^2(e+f x)+a+b\right )^{3/2}}+\frac{5 (\cos (2 (e+f x)) a+2 a+3 b) (\cos (2 e+2 f x) a+a+2 b)^{5/2} \tan (e+f x) \sec ^4(e+f x)}{384 (a+b)^2 f (\cos (2 (e+f x)) a+a+2 b)^{3/2} \left (b \sec ^2(e+f x)+a\right )^{5/2}}-\frac{(b+(3 a+2 b) \cos (2 (e+f x))) (\cos (2 e+2 f x) a+a+2 b)^{5/2} \tan (e+f x) \sec ^4(e+f x)}{384 (a+b)^2 f (\cos (2 (e+f x)) a+a+2 b)^{3/2} \left (b \sec ^2(e+f x)+a\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*(Sin[e + f*x]^2/(a + b) + ((a + 2*b + a*Cos
[2*(e + f*x)])*Sin[e + f*x]^2)/(a + b)^2 - (12*Sin[e + f*x]^4)/(a + b) + (16*(a + b - a*Sin[e + f*x]^2)*(1 - (
a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e
+ f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin
[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3))/(256*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^(5/2)*(a + b
- a*Sin[e + f*x]^2)^(3/2)) - ((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Csc[e + f*x]*Sec[e + f*x]^5*(Sin[e + f*x]^
2/(a + b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/(a + b)^2 - (24*Sin[e + f*x]^4)/(a + b) + (96*Sin[
e + f*x]^6)/a + (80*(a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)
/(a + 2*b + a*Cos[2*(e + f*x)]) + (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[
a + b]*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3
- (160*(a + b - a*Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)^2*Sin[e + f*x]^2)/(a + 2*b
+ a*Cos[2*(e + f*x)]) + (3*Sqrt[a]*(a + b)^(3/2)*ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt
[(a + b - a*Sin[e + f*x]^2)/(a + b)] + (a^2*Sin[e + f*x]^4)/(-1 + (a*Sin[e + f*x]^2)/(a + b))^2))/a^4))/(768*S
qrt[2]*f*(a + b*Sec[e + f*x]^2)^(5/2)*(a + b - a*Sin[e + f*x]^2)^(3/2)) + (5*(2*a + 3*b + a*Cos[2*(e + f*x)])*
(a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^4*Tan[e + f*x])/(384*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x
)])^(3/2)*(a + b*Sec[e + f*x]^2)^(5/2)) - ((b + (3*a + 2*b)*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*e + 2*f*x])^(
5/2)*Sec[e + f*x]^4*Tan[e + f*x])/(384*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)*(a + b*Sec[e + f*x]^2)
^(5/2))

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Maple [C]  time = 0.582, size = 3158, normalized size = 18.9 \begin{align*} \text{output 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*sec(f*x+e)^2)^(5/2),x)

[Out]

-1/6/f/a^3/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*sin(f*x+e)*(b+a*cos(f*x+e)^2)*(3*cos(f*x+e)^2*sin(f*x+e)*2^
(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)
*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e
))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2
)/(a+b)^2)^(1/2))*a^3+18*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1
/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+
e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*
I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*b+15*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*
(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*co
s(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2
*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b
)^2)^(1/2))*a*b^2-6*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a
*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)
/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a
^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^
3-36*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/
(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))
)^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+
a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b-30*cos(f*x+
e)^2*sin(f*x+e)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)
))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*Ellip
ticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-
(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^2+3*cos(f*x+e)^5*((2*I*a^(
1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3+3*cos(f*x+e)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+3*2^(1/2)*(1/(
a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*
x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)
^(1/2))*a^2*b*sin(f*x+e)+18*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(
1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))
^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I
*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^2*sin(f*x+e)+15*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(
1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)
*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/
2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^3*sin(f*x+e)-6*2^(1/
2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I
*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))
*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-
a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b*sin(f*x+e)-36*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*
a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2
)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^
(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^2*sin(f*x+e)-30*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*
b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(
f*x+e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),
-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^
(1/2))*b^3*sin(f*x+e)-3*cos(f*x+e)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^3-3*cos(f*x+e)^4*((2*I*a^(1/2)*
b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+18*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+20*cos(f*x+e)^3*
((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-18*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-20
*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+13*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1
/2)*a*b^2+15*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3-13*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)
*a*b^2-15*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3)/(a+b)/(-1+cos(f*x+e))/cos(f*x+e)^5/((b+a*cos(f*x+e)^2)/
cos(f*x+e)^2)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{2}}{{\left (b \sec \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*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(5/2), x)

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Fricas [B]  time = 7.39595, size = 2037, normalized size = 12.2 \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)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(3*((a^4 + 6*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 6*a*b^3 + 5*b^4 + 2*(a^3*b + 6*a^2*b^2 + 5*a
*b^3)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*
a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b
^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b
+ 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)
/cos(f*x + e)^2)*sin(f*x + e)) + 8*(3*(a^4 + a^3*b)*cos(f*x + e)^5 + 2*(9*a^3*b + 10*a^2*b^2)*cos(f*x + e)^3 +
(13*a^2*b^2 + 15*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*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), -1/24*(3*((a^4 + 6*a^3*b +
5*a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 6*a*b^3 + 5*b^4 + 2*(a^3*b + 6*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*sqrt(
a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x
+ e)^2)*sin(f*x + e))) + 4*(3*(a^4 + a^3*b)*cos(f*x + e)^5 + 2*(9*a^3*b + 10*a^2*b^2)*cos(f*x + e)^3 + (13*a^2
*b^2 + 15*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*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)]

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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)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{2}}{{\left (b \sec \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*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^2/(b*sec(f*x + e)^2 + a)^(5/2), x)