Integrand size = 25, antiderivative size = 125 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f} \]
-1/2*(a+2*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/b^(3/2)/ f+arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/(a-b)^(1/2)+1/ 2*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 6.37 (sec) , antiderivative size = 713, normalized size of antiderivative = 5.70 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {-\frac {b (a+b) \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}+\frac {4 b^2 \sqrt {1+\cos (2 (e+f x))} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (1+\cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {1+\cos (2 (e+f x))} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{b f}+\frac {\sqrt {\frac {a+b+a \cos (2 (e+f x))-b \cos (2 (e+f x))}{1+\cos (2 (e+f x))}} \tan (e+f x)}{2 b f} \]
-((-((b*(a + b)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f* x)])]*Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b] *Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)] )*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(a*(a + b + (a - b)*Cos[ 2*(e + f*x)]))) + (4*b^2*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(a + b + (a - b)* Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e + f*x]^2)/b)]* Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b )*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[S qrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Si n[e + f*x]^4)/(4*a*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*( e + f*x)]]) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x )])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f *x]^2)/b]*Csc[2*(e + f*x)]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/ (2*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x) ]])))/Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])/(b*f)) + (Sqrt[(a + b + a*Co s[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*Tan[e + f*x]) /(2*b*f)
Time = 0.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4153, 381, 398, 224, 219, 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.
| \(\displaystyle \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^4}{\sqrt {a+b \tan (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 381 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\int \frac {(a+2 b) \tan ^2(e+f x)+a}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}}{f}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {(a+2 b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-2 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {(a+2 b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-2 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-2 b \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{2 b}}{f}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-2 b \int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}}{2 b}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b}-\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {b}}-\frac {2 b \arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}}{2 b}}{f}\) |
(-1/2*((-2*b*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]] )/Sqrt[a - b] + ((a + 2*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/Sqrt[b])/b + (Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*b)) /f
3.4.27.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 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^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q }, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 , p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}+\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}}{f}\) | \(157\) |
| default | \(\frac {-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}+\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}}{f}\) | \(157\) |
1/f*(-ln(b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))/b^(1/2)+1/2*tan(f*x+ e)/b*(a+b*tan(f*x+e)^2)^(1/2)-1/2*a/b^(3/2)*ln(b^(1/2)*tan(f*x+e)+(a+b*tan (f*x+e)^2)^(1/2))+(b^4*(a-b))^(1/2)/b^2/(a-b)*arctan(b^2*(a-b)/(b^4*(a-b)) ^(1/2)/(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)))
Time = 0.73 (sec) , antiderivative size = 647, normalized size of antiderivative = 5.18 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {2 \, \sqrt {-a + b} b^{2} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a^{2} + a b - 2 \, b^{2}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{4 \, {\left (a b^{2} - b^{3}\right )} f}, -\frac {\sqrt {-a + b} b^{2} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a^{2} + a b - 2 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) - \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f}, \frac {4 \, \sqrt {a - b} b^{2} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + {\left (a^{2} + a b - 2 \, b^{2}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{4 \, {\left (a b^{2} - b^{3}\right )} f}, \frac {2 \, \sqrt {a - b} b^{2} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + {\left (a^{2} + a b - 2 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) + \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f}\right ] \]
[-1/4*(2*sqrt(-a + b)*b^2*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f* x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - (a^2 + a*b - 2*b^2)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) - 2*sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*ta n(f*x + e))/((a*b^2 - b^3)*f), -1/2*(sqrt(-a + b)*b^2*log(-((a - 2*b)*tan( f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/( tan(f*x + e)^2 + 1)) - (a^2 + a*b - 2*b^2)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) - sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*tan(f*x + e))/((a*b^2 - b^3)*f), 1/4*(4*sqrt(a - b)*b^2*arctan(-sqrt (b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + (a^2 + a*b - 2*b^2)*s qrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f *x + e) + a) + 2*sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*tan(f*x + e))/((a* b^2 - b^3)*f), 1/2*(2*sqrt(a - b)*b^2*arctan(-sqrt(b*tan(f*x + e)^2 + a)/( sqrt(a - b)*tan(f*x + e))) + (a^2 + a*b - 2*b^2)*sqrt(-b)*arctan(sqrt(b*ta n(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + sqrt(b*tan(f*x + e)^2 + a)* (a*b - b^2)*tan(f*x + e))/((a*b^2 - b^3)*f)]
\[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \]