\(\int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [123]

Optimal result

Integrand size = 25, antiderivative size = 93 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {a \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{3/2} f}-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) f} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 2.55 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.90 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left ((a-b) (a+b+(a-b) \cos (2 (e+f x)))+\sqrt {2} a (-a+b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )+\sqrt {2} a^2 \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \sec ^2(e+f x) \sin (2 (e+f x))}{4 \sqrt {2} (a-b)^2 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \] Input:

Integrate[Sin[e + f*x]^2/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

-1/4*(((a - b)*(a + b + (a - b)*Cos[2*(e + f*x)]) + Sqrt[2]*a*(-a + b)*Sqr 
t[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[ 
Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] + 
 Sqrt[2]*a^2*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*E 
llipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Cs 
c[e + f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x)])/(Sqrt[2]*( 
a - b)^2*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4146, 373, 27, 291, 216}

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.

\(\displaystyle \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^2}{\sqrt {a+b \tan (e+f x)^2}}dx\)

\(\Big \downarrow \) 4146

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

\(\Big \downarrow \) 373

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {a \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{3/2}}-\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}}{f}\)

Input:

Int[Sin[e + f*x]^2/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

((a*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*(a - 
 b)^(3/2)) - (Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*(a - b)*(1 + Tan 
[e + f*x]^2)))/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 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

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 3042

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

rule 4146

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]}, Sim 
p[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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(176\) vs. \(2(81)=162\).

Time = 12.20 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.90

method	result	size
default	\(-\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )}{\sqrt {a -b}\, \left (\cos \left (f x +e \right )-1\right )}\right ) \left (-1-\sec \left (f x +e \right )\right )+\sqrt {a -b}\, \sin \left (f x +e \right ) \cos \left (f x +e \right ) a +\sqrt {a -b}\, b \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )}{2 f \left (a -b \right )^{\frac {3}{2}} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(177\)

Input:

int(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (81) = 162\).

Time = 0.29 (sec) , antiderivative size = 696, normalized size of antiderivative = 7.48 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-1/16*(8*(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f* 
x + e)*sin(f*x + e) - a*sqrt(-a + b)*log(128*(a^4 - 4*a^3*b + 6*a^2*b^2 - 
4*a*b^3 + b^4)*cos(f*x + e)^8 - 256*(a^4 - 5*a^3*b + 9*a^2*b^2 - 7*a*b^3 + 
 2*b^4)*cos(f*x + e)^6 + 32*(5*a^4 - 34*a^3*b + 77*a^2*b^2 - 72*a*b^3 + 24 
*b^4)*cos(f*x + e)^4 + a^4 - 32*a^3*b + 160*a^2*b^2 - 256*a*b^3 + 128*b^4 
- 32*(a^4 - 11*a^3*b + 34*a^2*b^2 - 40*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8* 
(16*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 24*(a^3 - 4*a^2*b + 5 
*a*b^2 - 2*b^3)*cos(f*x + e)^5 + 2*(5*a^3 - 29*a^2*b + 48*a*b^2 - 24*b^3)* 
cos(f*x + e)^3 - (a^3 - 10*a^2*b + 24*a*b^2 - 16*b^3)*cos(f*x + e))*sqrt(- 
a + b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)))/(( 
a^2 - 2*a*b + b^2)*f), -1/8*(4*(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/c 
os(f*x + e)^2)*cos(f*x + e)*sin(f*x + e) - sqrt(a - b)*a*arctan(-1/4*(8*(a 
^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 8*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^3 
+ (a^2 - 8*a*b + 8*b^2)*cos(f*x + e))*sqrt(a - b)*sqrt(((a - b)*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2)/((2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e) 
^4 - a^2*b + 3*a*b^2 - 2*b^3 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*cos(f*x + 
 e)^2)*sin(f*x + e))))/((a^2 - 2*a*b + b^2)*f)]
 

Sympy [F]

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

integrate(sin(f*x+e)**2/(a+b*tan(f*x+e)**2)**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \] Input:

integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sin(f*x + e)^2/sqrt(b*tan(f*x + e)^2 + a), x)
 

Giac [F]

\[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \] Input:

integrate(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sin(f*x + e)^2/sqrt(b*tan(f*x + e)^2 + a), x)
 

Mupad [F(-1)]

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

int(sin(e + f*x)^2/(a + b*tan(e + f*x)^2)^(1/2),x)
 

Output:

int(sin(e + f*x)^2/(a + b*tan(e + f*x)^2)^(1/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \sin \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} b +a}d x \] Input:

int(sin(f*x+e)^2/(a+b*tan(f*x+e)^2)^(1/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*sin(e + f*x)**2)/(tan(e + f*x)**2*b + a), 
x)