\(\int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [320]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} f}-\frac {(a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2 f} \] Output:

-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/(a-b)^(1/2)/f-(a+b)*(a+b*ta 
n(f*x+e)^2)^(1/2)/b^2/f+1/3*(a+b*tan(f*x+e)^2)^(3/2)/b^2/f
 

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \left (2 a+3 b-b \tan ^2(e+f x)\right ) \sqrt {a+b \tan ^2(e+f x)}}{3 b^2}}{2 f} \] Input:

Integrate[Tan[e + f*x]^5/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

-1/2*((2*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/Sqrt[a - b] + (2 
*(2*a + 3*b - b*Tan[e + f*x]^2)*Sqrt[a + b*Tan[e + f*x]^2])/(3*b^2))/f
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4153, 354, 99, 2009}

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.

\(\displaystyle \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^5}{\sqrt {a+b \tan (e+f x)^2}}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {\int \frac {\tan ^5(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (\frac {-a-b}{b \sqrt {b \tan ^2(e+f x)+a}}+\frac {\sqrt {b \tan ^2(e+f x)+a}}{b}+\frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}\right )d\tan ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {2 \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 b^2}-\frac {2 (a+b) \sqrt {a+b \tan ^2(e+f x)}}{b^2}}{2 f}\)

Input:

Int[Tan[e + f*x]^5/Sqrt[a + b*Tan[e + f*x]^2],x]
 

Output:

((-2*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/Sqrt[a - b] - (2*(a 
+ b)*Sqrt[a + b*Tan[e + f*x]^2])/b^2 + (2*(a + b*Tan[e + f*x]^2)^(3/2))/(3 
*b^2))/(2*f)
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 = 0.83 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b}-\frac {2 a \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b^{2}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}}{f}\)	\(103\)
default	\(\frac {\frac {\tan \left (f x +e \right )^{2} \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b}-\frac {2 a \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 b^{2}}+\frac {\arctan \left (\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{b}}{f}\)	\(103\)

Input:

int(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/3*tan(f*x+e)^2/b*(a+b*tan(f*x+e)^2)^(1/2)-2/3*a/b^2*(a+b*tan(f*x+e) 
^2)^(1/2)+1/(-a+b)^(1/2)*arctan((a+b*tan(f*x+e)^2)^(1/2)/(-a+b)^(1/2))-1/b 
*(a+b*tan(f*x+e)^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 340, normalized size of antiderivative = 3.58 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [\frac {3 \, \sqrt {a - b} b^{2} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - a b + 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left (a b^{2} - b^{3}\right )} f}, -\frac {3 \, \sqrt {-a + b} b^{2} \arctan \left (-\frac {{\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{2 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} - a b\right )}}\right ) - 2 \, {\left ({\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - a b + 3 \, b^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left (a b^{2} - b^{3}\right )} f}\right ] \] Input:

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

[1/12*(3*sqrt(a - b)*b^2*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan( 
f*x + e)^2 - 4*(b*tan(f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqr 
t(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 
 4*((a*b - b^2)*tan(f*x + e)^2 - 2*a^2 - a*b + 3*b^2)*sqrt(b*tan(f*x + e)^ 
2 + a))/((a*b^2 - b^3)*f), -1/6*(3*sqrt(-a + b)*b^2*arctan(-1/2*(b*tan(f*x 
 + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/((a*b - b^2)*ta 
n(f*x + e)^2 + a^2 - a*b)) - 2*((a*b - b^2)*tan(f*x + e)^2 - 2*a^2 - a*b + 
 3*b^2)*sqrt(b*tan(f*x + e)^2 + a))/((a*b^2 - b^3)*f)]
 

Sympy [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \] Input:

integrate(tan(f*x+e)**5/(a+b*tan(f*x+e)**2)**(1/2),x)
 

Output:

Integral(tan(e + f*x)**5/sqrt(a + b*tan(e + f*x)**2), x)
 

Maxima [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \tan \left (f x + e\right )^{2} + a}} \,d x } \] Input:

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(tan(f*x + e)^5/sqrt(b*tan(f*x + e)^2 + a), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 9.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{3\,b^2\,f}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sqrt {a-b}}\right )}{f\,\sqrt {a-b}}-\left (\frac {2\,a}{b^2\,f}-\frac {a-b}{b^2\,f}\right )\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \] Input:

int(tan(e + f*x)^5/(a + b*tan(e + f*x)^2)^(1/2),x)
 

Output:

(a + b*tan(e + f*x)^2)^(3/2)/(3*b^2*f) - atanh((a + b*tan(e + f*x)^2)^(1/2 
)/(a - b)^(1/2))/(f*(a - b)^(1/2)) - ((2*a)/(b^2*f) - (a - b)/(b^2*f))*(a 
+ b*tan(e + f*x)^2)^(1/2)
 

Reduce [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{2} b -2 \sqrt {\tan \left (f x +e \right )^{2} b +a}\, a -3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2} b +a}d x \right ) b^{2} f}{3 b^{2} f} \] Input:

int(tan(f*x+e)^5/(a+b*tan(f*x+e)^2)^(1/2),x)
 

Output:

(sqrt(tan(e + f*x)**2*b + a)*tan(e + f*x)**2*b - 2*sqrt(tan(e + f*x)**2*b 
+ a)*a - 3*int((sqrt(tan(e + f*x)**2*b + a)*tan(e + f*x)**3)/(tan(e + f*x) 
**2*b + a),x)*b**2*f)/(3*b**2*f)