Integrand size = 25, antiderivative size = 146 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {3 a^2 \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 (a-b)^{5/2} f}-\frac {(5 a-2 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 (a-b) f} \]
3/8*a^2*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2 )/f-1/8*(5*a-2*b)*cos(f*x+e)*sin(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/(a-b)^2/f +1/4*cos(f*x+e)^3*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 = 5.16 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.15 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left ((a-b) \left (7 a^2+8 a b-3 b^2+2 \left (3 a^2-5 a b+2 b^2\right ) \cos (2 (e+f x))-(a-b)^2 \cos (4 (e+f x))\right )+6 \sqrt {2} a^2 (-a+b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )+6 \sqrt {2} a^3 \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )\right ) \sec ^2(e+f x) \sin (2 (e+f x))}{32 \sqrt {2} (a-b)^3 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \]
-1/32*(((a - b)*(7*a^2 + 8*a*b - 3*b^2 + 2*(3*a^2 - 5*a*b + 2*b^2)*Cos[2*( e + f*x)] - (a - b)^2*Cos[4*(e + f*x)]) + 6*Sqrt[2]*a^2*(-a + b)*Sqrt[((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] + 6*Sqr t[2]*a^3*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Ellip ticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[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 = 0.34 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4146, 372, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{\sqrt {a+b \tan (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\int \frac {a-2 (2 a-b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {3 a^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {3 a^2 \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {3 a^2 \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}-\frac {\frac {(5 a-2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {3 a^2 \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 (a-b)^{3/2}}}{4 (a-b)}}{f}\) |
((Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(4*(a - b)*(1 + Tan[e + f*x]^2) ^2) - ((-3*a^2*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2 ]])/(2*(a - b)^(3/2)) + ((5*a - 2*b)*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^ 2])/(2*(a - b)*(1 + Tan[e + f*x]^2)))/(4*(a - b)))/f
3.2.22.3.1 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[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[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. \(410\) vs. \(2(130)=260\).
Time = 7.36 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.82
| method | result | size |
| default | \(-\frac {-2 \cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) \sqrt {a -b}\, a^{2}+4 \cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) \sqrt {a -b}\, a b -2 \cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) \sqrt {a -b}\, b^{2}+5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {a -b}\, a^{2}-9 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {a -b}\, a b +4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {a -b}\, b^{2}+3 \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{2}+3 \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a^{2} \sec \left (f x +e \right )+5 \sqrt {a -b}\, a b \tan \left (f x +e \right )-2 \sqrt {a -b}\, b^{2} \tan \left (f x +e \right )}{8 f \left (a -b \right )^{\frac {5}{2}} \sqrt {a +b \tan \left (f x +e \right )^{2}}}\) | \(411\) |
-1/8/f/(a-b)^(5/2)/(a+b*tan(f*x+e)^2)^(1/2)*(-2*cos(f*x+e)^3*sin(f*x+e)*(a -b)^(1/2)*a^2+4*cos(f*x+e)^3*sin(f*x+e)*(a-b)^(1/2)*a*b-2*cos(f*x+e)^3*sin (f*x+e)*(a-b)^(1/2)*b^2+5*cos(f*x+e)*sin(f*x+e)*(a-b)^(1/2)*a^2-9*cos(f*x+ e)*sin(f*x+e)*(a-b)^(1/2)*a*b+4*cos(f*x+e)*sin(f*x+e)*(a-b)^(1/2)*b^2+3*ar ctan(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^2+3*arctan(1/(a-b)^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(c os(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^2*sec(f*x+e)+5*(a-b)^(1/2)*a*b*tan(f*x+e )-2*(a-b)^(1/2)*b^2*tan(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (130) = 260\).
Time = 2.25 (sec) , antiderivative size = 788, normalized size of antiderivative = 5.40 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {3 \, a^{2} \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 (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 7 \, a b + 2 \, 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 )}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, \frac {3 \, \sqrt {a - b} a^{2} \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 (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 7 \, a b + 2 \, 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 )}{32 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \]
[-1/64*(3*a^2*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)*co s(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)*sqr t(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 8*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 7*a*b + 2*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^3 - 3*a^2*b + 3*a*b^2 - b^3)*f), 1/32*(3*sqrt(a - b)*a^2*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)/c os(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*(2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 7*a*b + 2* 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^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)]
\[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\sin ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]