\(\int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [402]

Optimal result

Integrand size = 25, antiderivative size = 89 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f}-\frac {(a+2 b) \sqrt {a+b \sec ^2(e+f x)}}{b^2 f}+\frac {\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f} \] Output:

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.20 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {-2 a (a+3 b)-b (a+6 b) \sec ^2(e+f x)+b^2 \sec ^4(e+f x)-3 b^2 \text {arctanh}\left (\sqrt {1+\frac {b \sec ^2(e+f x)}{a}}\right ) \sqrt {1+\frac {b \sec ^2(e+f x)}{a}}}{3 b^2 f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

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

Output:

(-2*a*(a + 3*b) - b*(a + 6*b)*Sec[e + f*x]^2 + b^2*Sec[e + f*x]^4 - 3*b^2* 
ArcTanh[Sqrt[1 + (b*Sec[e + f*x]^2)/a]]*Sqrt[1 + (b*Sec[e + f*x]^2)/a])/(3 
*b^2*f*Sqrt[a + b*Sec[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 87, 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, 4627, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 99

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-2*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/Sqrt[a] - (2*(a + 2*b)*S 
qrt[a + b*Sec[e + f*x]^2])/b^2 + (2*(a + b*Sec[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 4627

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si 
mp[1/f   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] 
, x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( 
m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers 
Q[2*n, p])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(77)=154\).

Time = 11.88 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.17

method	result	size
default	\(\frac {\left (-\frac {\sec \left (f x +e \right )^{2}}{3}-2\right ) b \,a^{\frac {3}{2}}+\left (\frac {\sec \left (f x +e \right )^{4}}{3}-2 \sec \left (f x +e \right )^{2}\right ) b^{2} \sqrt {a}-\frac {2 a^{\frac {5}{2}}}{3}-\frac {\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) b^{2} \left (3+3 \sec \left (f x +e \right )\right )}{3}}{f \,b^{2} \sqrt {a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(193\)

Input:

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

Output:

1/f*((-1/3*sec(f*x+e)^2-2)*b*a^(3/2)+(1/3*sec(f*x+e)^4-2*sec(f*x+e)^2)*b^2 
*a^(1/2)-2/3*a^(5/2)-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4* 
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)*cos(f*x+e)+4*a^(1/2)*( 
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)+4*cos(f*x+e)*a)*b^2*(3+3*sec(f* 
x+e)))/b^2/a^(1/2)/(a+b*sec(f*x+e)^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (77) = 154\).

Time = 0.47 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.61 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) - 8 \, {\left (2 \, {\left (a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, a b^{2} f \cos \left (f x + e\right )^{2}}, \frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} - 4 \, {\left (2 \, {\left (a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, a b^{2} f \cos \left (f x + e\right )^{2}}\right ] \] Input:

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

Output:

[1/24*(3*sqrt(a)*b^2*cos(f*x + e)^2*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b 
*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b 
^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x 
 + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
+ e)^2)) - 8*(2*(a^2 + 3*a*b)*cos(f*x + e)^2 - a*b)*sqrt((a*cos(f*x + e)^2 
 + b)/cos(f*x + e)^2))/(a*b^2*f*cos(f*x + e)^2), 1/12*(3*sqrt(-a)*b^2*arct 
an(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt(( 
a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos( 
f*x + e)^2 + a*b^2))*cos(f*x + e)^2 - 4*(2*(a^2 + 3*a*b)*cos(f*x + e)^2 - 
a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*b^2*f*cos(f*x + e)^2) 
]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 765 vs. \(2 (77) = 154\).

Time = 0.69 (sec) , antiderivative size = 765, normalized size of antiderivative = 8.60 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

2/3*(3*arctan(-1/2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f* 
x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b 
*tan(1/2*f*x + 1/2*e)^2 + a + b) + sqrt(a + b))/sqrt(-a))/sqrt(-a) - 2*(3* 
(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*ta 
n(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2* 
e)^2 + a + b))^5 - 21*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2 
*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 
2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^4*sqrt(a + b) + 2*(sqrt(a + b)*tan(1/ 
2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^ 
4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b))^3*(3 
*a - 17*b) + 6*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 
1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan 
(1/2*f*x + 1/2*e)^2 + a + b))^2*(5*a + 9*b)*sqrt(a + b) - 3*(sqrt(a + b)*t 
an(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f*x + 1/ 
2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)) 
*(3*a^2 - 18*a*b - 37*b^2) - (9*a^2 + 10*a*b - 47*b^2)*sqrt(a + b))/((sqrt 
(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2 
*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 
+ a + b))^2 - 2*(sqrt(a + b)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 
 1/2*e)^4 + b*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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