Integrand size = 25, antiderivative size = 93 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {a \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{3/2} f}-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) f} \]
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
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 4.13 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.90 \[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left ((a-b) (a+b+(a-b) \cos (2 (e+f x)))+\sqrt {2} a (-a+b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )+\sqrt {2} a^2 \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \sec ^2(e+f x) \sin (2 (e+f x))}{4 \sqrt {2} (a-b)^2 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
-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])
Time = 0.28 (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 |
| \(\displaystyle \frac {\frac {a \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{3/2}}-\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}}{f}\) |
((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
3.2.23.3.1 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[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. \(279\) vs. \(2(81)=162\).
Time = 4.40 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.01
| method | result | size |
| default | \(-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {a -b}\, a -\sqrt {a -b}\, b \cos \left (f x +e \right ) \sin \left (f x +e \right )+\arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a +\sqrt {a -b}\, b \tan \left (f x +e \right )+\arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a \sec \left (f x +e \right )}{2 f \left (a -b \right )^{\frac {3}{2}} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(280\) |
-1/2/f/(a-b)^(3/2)/(a+b*tan(f*x+e)^2)^(1/2)*(sin(f*x+e)*cos(f*x+e)*(a-b)^( 1/2)*a-(a-b)^(1/2)*b*cos(f*x+e)*sin(f*x+e)+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)*(cot(f*x+e)+csc(f*x+e)))*(( a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a+(a-b)^(1/2)*b*t an(f*x+e)+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)*(cot(f*x+e)+csc(f*x+e)))*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/ (cos(f*x+e)+1)^2)^(1/2)*a*sec(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (81) = 162\).
Time = 0.47 (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=\left [-\frac {8 \, {\left (a - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a \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 )}{16 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}, -\frac {4 \, {\left (a - b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - \sqrt {a - b} a \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 )}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}\right ] \]
[-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)]
\[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\sin ^{2}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^2}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]