\(\int \frac {\tan ^5(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [428]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {(a+b)^2}{3 a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\frac {1}{a^2}-\frac {1}{b^2}}{f \sqrt {a+b \sec ^2(e+f x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 5.57 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.93 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {4 (a+b) \operatorname {AppellF1}\left (3,\frac {1}{2},\frac {5}{2},4,\sin ^2(e+f x),\frac {a \sin ^2(e+f x)}{a+b}\right ) \tan ^6(e+f x)}{3 f \left (a+b \sec ^2(e+f x)\right )^{5/2} \left (8 (a+b) \operatorname {AppellF1}\left (3,\frac {1}{2},\frac {5}{2},4,\sin ^2(e+f x),\frac {a \sin ^2(e+f x)}{a+b}\right )+\left (5 a \operatorname {AppellF1}\left (4,\frac {1}{2},\frac {7}{2},5,\sin ^2(e+f x),\frac {a \sin ^2(e+f x)}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (4,\frac {3}{2},\frac {5}{2},5,\sin ^2(e+f x),\frac {a \sin ^2(e+f x)}{a+b}\right )\right ) \sin ^2(e+f x)\right )} \] Input:

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

Output:

(4*(a + b)*AppellF1[3, 1/2, 5/2, 4, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a 
+ b)]*Tan[e + f*x]^6)/(3*f*(a + b*Sec[e + f*x]^2)^(5/2)*(8*(a + b)*AppellF 
1[3, 1/2, 5/2, 4, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a*Appel 
lF1[4, 1/2, 7/2, 5, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (a + b)* 
AppellF1[4, 3/2, 5/2, 5, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[ 
e + f*x]^2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4627

\(\displaystyle \frac {\int \frac {\cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {\int \frac {\cos (e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 98

\(\displaystyle \frac {\int \left (-\frac {(a+b)^2}{a b \left (b \sec ^2(e+f x)+a\right )^{5/2}}+\frac {\cos (e+f x)}{a^2 \sqrt {b \sec ^2(e+f x)+a}}+\frac {a^2-b^2}{a^2 b \left (b \sec ^2(e+f x)+a\right )^{3/2}}\right )d\sec ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 \left (\frac {1}{a^2}-\frac {1}{b^2}\right )}{\sqrt {a+b \sec ^2(e+f x)}}+\frac {2 (a+b)^2}{3 a b^2 \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{2 f}\)

Input:

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

Output:

((-2*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/a^(5/2) + (2*(a + b)^2)/ 
(3*a*b^2*(a + b*Sec[e + f*x]^2)^(3/2)) + (2*(a^(-2) - b^(-2)))/Sqrt[a + b* 
Sec[e + f*x]^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. \(1569\) vs. \(2(85)=170\).

Time = 5.46 (sec) , antiderivative size = 1570, normalized size of antiderivative = 16.19

method	result	size
default	\(\text {Expression too large to display}\)	\(1570\)

Input:

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

Output:

-1/3/f/a^(5/2)/((-a*b)^(1/2)-a)^2/((-a*b)^(1/2)+a)^2/b^2/(cos(f*x+e)^2*a^3 
+(2*cos(f*x+e)^2+1)*a^2*b+(cos(f*x+e)^2+2)*b^2*a+b^3)/(a+b*sec(f*x+e)^2)^( 
5/2)*(3*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+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^9*b^2*(cos(f*x+e)^2+cos(f*x+e) 
)+3*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+a* 
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^8*b^3*(3+4*cos(f*x+e)^2+4*cos(f*x+ 
e)+3*sec(f*x+e))+9*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+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^7*b^4*(4+2*cos(f*x+ 
e)^2+2*cos(f*x+e)+4*sec(f*x+e)+sec(f*x+e)^2+sec(f*x+e)^3)+3*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+a*cos(f*x+e)^2)/(1+c 
os(f*x+e))^2)^(1/2)*a^6*b^5*(18+4*cos(f*x+e)^2+4*cos(f*x+e)+18*sec(f*x+e)+ 
12*sec(f*x+e)^2+12*sec(f*x+e)^3+sec(f*x+e)^4+sec(f*x+e)^5)+3*ln(4*((b+a*co 
s(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+a*cos(f*x+e)^2)/(1+ 
cos(f*x+e))^2)^(1/2)*a^5*b^6*(4*sec(f*x+e)^5+4*sec(f*x+e)^4+18*sec(f*x+e)^ 
3+18*sec(f*x+e)^2+cos(f*x+e)^2+12*sec(f*x+e)+cos(f*x+e)+12)+9*ln(4*((b+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (85) = 170\).

Time = 0.79 (sec) , antiderivative size = 564, normalized size of antiderivative = 5.81 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + 2 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4}\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 (2 \, {\left (a^{4} - a^{3} b - 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (a^{3} b - a b^{3}\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}}}}{24 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{3} f \cos \left (f x + e\right )^{2} + a^{3} b^{4} f\right )}}, \frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + 2 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4}\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 (2 \, {\left (a^{4} - a^{3} b - 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (a^{3} b - a b^{3}\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}}}}{12 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{3} f \cos \left (f x + e\right )^{2} + a^{3} b^{4} f\right )}}\right ] \] Input:

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

Output:

[1/24*(3*(a^2*b^2*cos(f*x + e)^4 + 2*a*b^3*cos(f*x + e)^2 + b^4)*sqrt(a)*l 
og(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^4 - a^3*b - 2*a^2* 
b^2)*cos(f*x + e)^4 + 3*(a^3*b - a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2))/(a^5*b^2*f*cos(f*x + e)^4 + 2*a^4*b^3*f*cos(f*x 
 + e)^2 + a^3*b^4*f), 1/12*(3*(a^2*b^2*cos(f*x + e)^4 + 2*a*b^3*cos(f*x + 
e)^2 + b^4)*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*(2*(a^4 - a^3*b - 2*a^2* 
b^2)*cos(f*x + e)^4 + 3*(a^3*b - a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + 
e)^2 + b)/cos(f*x + e)^2))/(a^5*b^2*f*cos(f*x + e)^4 + 2*a^4*b^3*f*cos(f*x 
 + e)^2 + a^3*b^4*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (85) = 170\).

Time = 1.56 (sec) , antiderivative size = 466, normalized size of antiderivative = 4.80 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*(((((2*a^11*sgn(cos(f*x + e)) + a^10*b*sgn(cos(f*x + e)) - 4*a^9*b^2*s 
gn(cos(f*x + e)) - 3*a^8*b^3*sgn(cos(f*x + e)))*tan(1/2*f*x + 1/2*e)^2/(a^ 
10*b^2) - 3*(2*a^11*sgn(cos(f*x + e)) - 3*a^10*b*sgn(cos(f*x + e)) - 4*a^9 
*b^2*sgn(cos(f*x + e)) + a^8*b^3*sgn(cos(f*x + e)))/(a^10*b^2))*tan(1/2*f* 
x + 1/2*e)^2 + 3*(2*a^11*sgn(cos(f*x + e)) - 3*a^10*b*sgn(cos(f*x + e)) - 
4*a^9*b^2*sgn(cos(f*x + e)) + a^8*b^3*sgn(cos(f*x + e)))/(a^10*b^2))*tan(1 
/2*f*x + 1/2*e)^2 - (2*a^11*sgn(cos(f*x + e)) + a^10*b*sgn(cos(f*x + e)) - 
 4*a^9*b^2*sgn(cos(f*x + e)) - 3*a^8*b^3*sgn(cos(f*x + e)))/(a^10*b^2))/(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)^(3/2) + 6*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^2*sgn(cos(f*x + e))))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{5}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

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

Output:

int((sqrt(sec(e + f*x)**2*b + a)*tan(e + f*x)**5)/(sec(e + f*x)**6*b**3 + 
3*sec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)