Integrand size = 25, antiderivative size = 134 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+2 b) \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{5/2} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 (a-b) f \sqrt {a+b \tan ^2(e+f x)}}-\frac {3 b \tan (e+f x)}{2 (a-b)^2 f \sqrt {a+b \tan ^2(e+f x)}} \]
1/2*(a+2*b)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^ (5/2)/f-1/2*cos(f*x+e)*sin(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)^(1/2)-3/2*b*t an(f*x+e)/(a-b)^2/f/(a+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 4.36 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.10 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left ((a-b) (a+5 b+(a-b) \cos (2 (e+f x)))-\sqrt {2} \left (a^2+a b-2 b^2\right ) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )+\sqrt {2} a (a+2 b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \sec ^2(e+f x) \sin (2 (e+f x))}{4 \sqrt {2} (a-b)^3 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
-1/4*(((a - b)*(a + 5*b + (a - b)*Cos[2*(e + f*x)]) - Sqrt[2]*(a^2 + a*b - 2*b^2)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Ellipt icF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqr t[2]], 1] + Sqrt[2]*a*(a + 2*b)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*C sc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*C os[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x)])/(Sqrt[2]*(a - b)^3*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[ e + f*x]^2])
Time = 0.33 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4146, 373, 402, 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.
| \(\displaystyle \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^2}{\left (a+b \tan (e+f x)^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\int \frac {a-2 b \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^{3/2}}d\tan (e+f x)}{2 (a-b)}-\frac {\tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a (a+2 b)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a (a-b)}-\frac {3 b \tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}-\frac {\tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {(a+2 b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a-b}-\frac {3 b \tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}-\frac {\tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {(a+2 b) \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{a-b}-\frac {3 b \tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}-\frac {\tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {(a+2 b) \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{(a-b)^{3/2}}-\frac {3 b \tan (e+f x)}{(a-b) \sqrt {a+b \tan ^2(e+f x)}}}{2 (a-b)}-\frac {\tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)}}}{f}\) |
(-1/2*Tan[e + f*x]/((a - b)*(1 + Tan[e + f*x]^2)*Sqrt[a + b*Tan[e + f*x]^2 ]) + (((a + 2*b)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x] ^2]])/(a - b)^(3/2) - (3*b*Tan[e + f*x])/((a - b)*Sqrt[a + b*Tan[e + f*x]^ 2]))/(2*(a - b)))/f
3.2.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1596\) vs. \(2(118)=236\).
Time = 2.92 (sec) , antiderivative size = 1597, normalized size of antiderivative = 11.92
1/f/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b* tan(f*x+e)^2)^(1/2)*tan(f*x+e))-b*tan(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^ (1/2)+1/f/b/(a-b)^3*(b^4*(a-b))^(1/2)*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/( b*(-cos(2*f*x+2*e)+1)^2*csc(2*f*x+2*e)^2+a)^(1/2)*(csc(2*f*x+2*e)-cot(2*f* x+2*e)))-1/2/f/(a-b)^2*a/(b*(-cos(2*f*x+2*e)+1)^2*csc(2*f*x+2*e)^2+a)^(1/2 )/(1/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^ 2*b+b*csc(2*f*x+2*e)^2+a)*a*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)^2-2/(cos(2*f*x +2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x +2*e)^2+a)*a*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)+1/(cos(2*f*x+2*e)^2*csc(2*f*x +2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*a*csc( 2*f*x+2*e)^2-b/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2 *f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)^2+2*b/ (cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b *csc(2*f*x+2*e)^2+a)*csc(2*f*x+2*e)^2*cos(2*f*x+2*e)-b/(cos(2*f*x+2*e)^2*c sc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a )*csc(2*f*x+2*e)^2+1)*csc(2*f*x+2*e)+1/2/f/(a-b)^2*a/(b*(-cos(2*f*x+2*e)+1 )^2*csc(2*f*x+2*e)^2+a)^(1/2)/(1/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*co s(2*f*x+2*e)*csc(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*a*csc(2*f*x+2*e)^2*c os(2*f*x+2*e)^2-2/(cos(2*f*x+2*e)^2*csc(2*f*x+2*e)^2*b-2*cos(2*f*x+2*e)*cs c(2*f*x+2*e)^2*b+b*csc(2*f*x+2*e)^2+a)*a*csc(2*f*x+2*e)^2*cos(2*f*x+2*e...
Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (118) = 236\).
Time = 3.78 (sec) , antiderivative size = 908, normalized size of antiderivative = 6.78 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (a^{2} + a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {-a + b} \log \left (128 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - 5 \, a^{3} b + 9 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 34 \, a^{3} b + 77 \, a^{2} b^{2} - 72 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 32 \, a^{3} b + 160 \, a^{2} b^{2} - 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{4} - 11 \, a^{3} b + 34 \, a^{2} b^{2} - 40 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - 4 \, a^{2} b + 5 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 29 \, a^{2} b + 48 \, a b^{2} - 24 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 10 \, a^{2} b + 24 \, a b^{2} - 16 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}, \frac {{\left ({\left (a^{2} + a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {a - b} \arctan \left (-\frac {{\left (8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{2} b + 3 \, a b^{2} - 2 \, b^{3} - {\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) - 4 \, {\left ({\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}\right ] \]
[-1/16*(((a^2 + a*b - 2*b^2)*cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(-a + b)*lo g(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)) + 8*((a^2 - 2*a*b + b^2)*cos(f*x + e)^3 + 3*(a*b - b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e) ^2)*sin(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f), 1/8*(((a^2 + a*b - 2*b^2)* cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(a - b)*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))) - 4*((a^2 - 2*a*b + b^2)*cos(f*x + e)^3 + 3*(a*b - b^2)*cos(f*x + e ))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^...
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sin ^{2}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]