\(\int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [327]

Optimal result

Integrand size = 25, antiderivative size = 125 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f} \] Output:

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

Mathematica [C] (verified)

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

Time = 4.36 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.17 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\left (\sqrt {2} a (-a+b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )-2 \sqrt {2} a b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \operatorname {EllipticPi}\left (-\frac {b}{a-b},\arcsin \left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right ),1\right )+(a-b) (a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)\right ) \tan (e+f x)}{2 \sqrt {2} b (-a+b) f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \] Input:

Integrate[Tan[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

-1/2*((Sqrt[2]*a*(-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] - 2*Sqrt[2]*a*b*Sqrt[((a + b + (a - b)*Cos[2*( 
e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b 
 + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] + (a - b)*(a 
+ b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2)*Tan[e + f*x])/(Sqrt[2]*b*( 
-a + b)*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.53 (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

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

Input:

Int[Tan[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

(-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
 

Defintions of rubi rules used

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 219

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])
 

rule 224

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]
 

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 381

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]
 

rule 398

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]
 

rule 3042

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

rule 4153

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]))
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}}{f}\)	\(157\)
default	\(\frac {\frac {\tan \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{2 b^{\frac {3}{2}}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\right )}{b^{2} \left (a -b \right )}-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \tan \left (f x +e \right )^{2}}\right )}{\sqrt {b}}}{f}\)	\(157\)

Input:

int(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(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)*ta 
n(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))-ln(b^(1/2)*ta 
n(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))/b^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 631, normalized size of antiderivative = 5.05 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx =\text {Too large to display} \] Input:

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[-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)/sqrt(b*tan(f*x + e)^2 + a)) - 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(a - 
b)*tan(f*x + e)/sqrt(b*tan(f*x + e)^2 + a)) + (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)*tan(f*x + e))/((a*b^2 - b 
^3)*f), 1/2*(2*sqrt(a - b)*b^2*arctan(sqrt(a - b)*tan(f*x + e)/sqrt(b*tan( 
f*x + e)^2 + a)) + (a^2 + a*b - 2*b^2)*sqrt(-b)*arctan(sqrt(-b)*tan(f*x + 
e)/sqrt(b*tan(f*x + e)^2 + a)) + sqrt(b*tan(f*x + e)^2 + a)*(a*b - b^2)*ta 
n(f*x + e))/((a*b^2 - b^3)*f)]
 

Sympy [F]

\[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \] Input:

integrate(tan(f*x+e)**4/(a+b*tan(f*x+e)**2)**(1/2),x)
 

Output:

Integral(tan(e + f*x)**4/sqrt(a + b*tan(e + f*x)**2), x)
 

Maxima [F]

\[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{4}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \] Input:

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(tan(f*x + e)^4/sqrt(b*tan(f*x + e)^2 + a), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \] Input:

int(tan(e + f*x)^4/(a + b*tan(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(tan(e + f*x)^4/(a + b*tan(e + f*x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{2} b +a}d x \] Input:

int(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x)
 

Output:

int((sqrt(tan(e + f*x)**2*b + a)*tan(e + f*x)**4)/(tan(e + f*x)**2*b + a), 
x)