\(\int \frac {\tan ^5(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [415]

Optimal result

Integrand size = 25, antiderivative size = 88 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(a+b)^2}{a b^2 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\sqrt {a+b \sec ^2(e+f x)}}{b^2 f} \] Output:

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

Mathematica [A] (verified)

Time = 3.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.38 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {(a+2 b+a \cos (2 (e+f x))) \left (2 a^2+4 a b+b^2+\left (2 a^2+2 a b+b^2\right ) \cos (2 (e+f x))\right ) \sec ^4(e+f x)}{4 a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \] Input:

Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, 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, 98, 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.

\(\displaystyle \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^5}{\left (a+b \sec (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int \frac {\cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 98

\(\displaystyle \frac {\int \left (-\frac {(a+b)^2}{a b \left (b \sec ^2(e+f x)+a\right )^{3/2}}+\frac {\cos (e+f x)}{a \sqrt {b \sec ^2(e+f x)+a}}+\frac {1}{b \sqrt {b \sec ^2(e+f x)+a}}\right )d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 98

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x 
_)), x_] :> Int[ExpandIntegrand[(e + f*x)^FractionalPart[p], (c + d*x)^n*(( 
e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] 
&& IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]
 

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. \(336\) vs. \(2(78)=156\).

Time = 7.12 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.83

method	result	size
default	\(\frac {2 a^{\frac {9}{2}}+a^{\frac {7}{2}} b \left (2+3 \sec \left (f x +e \right )^{2}\right )+a^{\frac {5}{2}} b^{2} \left (1+2 \sec \left (f x +e \right )^{2}+\sec \left (f x +e \right )^{4}\right )+a^{\frac {3}{2}} b^{3} \sec \left (f x +e \right )^{2}+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) a^{2} b^{2} \left (-1-\sec \left (f x +e \right )\right )+\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}\, \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+4 \cos \left (f x +e \right ) a \right ) a \,b^{3} \left (-\sec \left (f x +e \right )^{2}-\sec \left (f x +e \right )^{3}\right )}{f \,b^{2} a^{\frac {5}{2}} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\)	\(337\)

Input:

int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/f/b^2/a^(5/2)/(a+b*sec(f*x+e)^2)^(3/2)*(2*a^(9/2)+a^(7/2)*b*(2+3*sec(f*x 
+e)^2)+a^(5/2)*b^2*(1+2*sec(f*x+e)^2+sec(f*x+e)^4)+a^(3/2)*b^3*sec(f*x+e)^ 
2+((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+c 
os(f*x+e))^2)^(1/2)+4*cos(f*x+e)*a)*a^2*b^2*(-1-sec(f*x+e))+((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)*a*b^3*(-sec(f*x+e)^2-sec(f*x+e)^3))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (78) = 156\).

Time = 0.63 (sec) , antiderivative size = 458, normalized size of antiderivative = 5.20 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b^{2} \cos \left (f x + e\right )^{2} + b^{3}\right )} \sqrt {a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, {\left (a^{2} b + {\left (2 \, a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{2} + a^{2} b^{3} f\right )}}, \frac {{\left (a b^{2} \cos \left (f x + e\right )^{2} + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) + 4 \, {\left (a^{2} b + {\left (2 \, a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{2} + a^{2} b^{3} f\right )}}\right ] \] Input:

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*((a*b^2*cos(f*x + e)^2 + b^3)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 25 
6*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)/c 
os(f*x + e)^2)) + 8*(a^2*b + (2*a^3 + 2*a^2*b + a*b^2)*cos(f*x + e)^2)*sqr 
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^3*b^2*f*cos(f*x + e)^2 + a^2* 
b^3*f), 1/4*((a*b^2*cos(f*x + e)^2 + b^3)*sqrt(-a)*arctan(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) 
) + 4*(a^2*b + (2*a^3 + 2*a^2*b + a*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + 
 e)^2 + b)/cos(f*x + e)^2))/(a^3*b^2*f*cos(f*x + e)^2 + a^2*b^3*f)]
 

Sympy [F]

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

integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**2)**(3/2),x)
 

Output:

Integral(tan(e + f*x)**5/(a + b*sec(e + f*x)**2)**(3/2), x)
 

Maxima [F]

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

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(tan(f*x + e)^5/(b*sec(f*x + e)^2 + a)^(3/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 560 vs. \(2 (78) = 156\).

Time = 0.99 (sec) , antiderivative size = 560, normalized size of antiderivative = 6.36 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

-(((a^4*b*sgn(cos(f*x + e)) + 2*a^3*b^2*sgn(cos(f*x + e)) + a^2*b^3*sgn(co 
s(f*x + e)))*tan(1/2*f*x + 1/2*e)^2/(a^3*b^3) - (a^4*b*sgn(cos(f*x + e)) + 
 2*a^3*b^2*sgn(cos(f*x + e)) + a^2*b^3*sgn(cos(f*x + e)))/(a^3*b^3))/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*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)*a*sgn(cos(f*x + e))) - 4*(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 + 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*tan(1/2*f*x + 1/2*e)^2 + a + b))*sqrt(a + b) + a - 3* 
b)*b*sgn(cos(f*x + e))))/f
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:

int(tan(e + f*x)^5/(a + b/cos(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^5/(a + b/cos(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{5}}{\sec \left (f x +e \right )^{4} b^{2}+2 \sec \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(3/2),x)
 

Output:

int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**5)/(sec(e + f*x)**4*b**2 + 
2*sec(e + f*x)**2*a*b + a**2),x)