\(\int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [185]

Optimal result

Integrand size = 23, antiderivative size = 108 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} \sqrt {a+b} f}+\frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\sin ^5(e+f x)}{5 a f} \] Output:

-b^3*arctanh(a^(1/2)*sin(f*x+e)/(a+b)^(1/2))/a^(7/2)/(a+b)^(1/2)/f+(a^2-a* 
b+b^2)*sin(f*x+e)/a^3/f-1/3*(2*a-b)*sin(f*x+e)^3/a^2/f+1/5*sin(f*x+e)^5/a/ 
f
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {120 b^3 \left (\log \left (\sqrt {a+b}-\sqrt {a} \sin (e+f x)\right )-\log \left (\sqrt {a+b}+\sqrt {a} \sin (e+f x)\right )\right )}{\sqrt {a+b}}+30 \sqrt {a} \left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x)+5 a^{3/2} (5 a-4 b) \sin (3 (e+f x))+3 a^{5/2} \sin (5 (e+f x))}{240 a^{7/2} f} \] Input:

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

Output:

((120*b^3*(Log[Sqrt[a + b] - Sqrt[a]*Sin[e + f*x]] - Log[Sqrt[a + b] + Sqr 
t[a]*Sin[e + f*x]]))/Sqrt[a + b] + 30*Sqrt[a]*(5*a^2 - 6*a*b + 8*b^2)*Sin[ 
e + f*x] + 5*a^(3/2)*(5*a - 4*b)*Sin[3*(e + f*x)] + 3*a^(5/2)*Sin[5*(e + f 
*x)])/(240*a^(7/2)*f)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4635, 300, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4635

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

\(\Big \downarrow \) 300

\(\displaystyle \frac {\int \left (\frac {\sin ^4(e+f x)}{a}-\frac {(2 a-b) \sin ^2(e+f x)}{a^2}+\frac {a^2-b a+b^2}{a^3}-\frac {b^3}{a^3 \left (-a \sin ^2(e+f x)+a+b\right )}\right )d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} \sqrt {a+b}}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2}+\frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3}+\frac {\sin ^5(e+f x)}{5 a}}{f}\)

Input:

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

Output:

(-((b^3*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(a^(7/2)*Sqrt[a + b]) 
) + ((a^2 - a*b + b^2)*Sin[e + f*x])/a^3 - ((2*a - b)*Sin[e + f*x]^3)/(3*a 
^2) + Sin[e + f*x]^5/(5*a))/f
 

Defintions of rubi rules used

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

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 4635

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
 Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m 
+ n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In 
tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+\frac {a b \sin \left (f x +e \right )^{3}}{3}+\sin \left (f x +e \right ) a^{2}-\sin \left (f x +e \right ) a b +\sin \left (f x +e \right ) b^{2}}{a^{3}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}}{f}\)	\(110\)
default	\(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+\frac {a b \sin \left (f x +e \right )^{3}}{3}+\sin \left (f x +e \right ) a^{2}-\sin \left (f x +e \right ) a b +\sin \left (f x +e \right ) b^{2}}{a^{3}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}}{f}\)	\(110\)
risch	\(-\frac {5 i {\mathrm e}^{i \left (f x +e \right )}}{16 a f}+\frac {3 i {\mathrm e}^{i \left (f x +e \right )} b}{8 a^{2} f}-\frac {i {\mathrm e}^{i \left (f x +e \right )} b^{2}}{2 f \,a^{3}}+\frac {5 i {\mathrm e}^{-i \left (f x +e \right )}}{16 a f}-\frac {3 i {\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{2} f}+\frac {i {\mathrm e}^{-i \left (f x +e \right )} b^{2}}{2 f \,a^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, f \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, f \,a^{3}}+\frac {\sin \left (5 f x +5 e \right )}{80 a f}+\frac {5 \sin \left (3 f x +3 e \right )}{48 a f}-\frac {\sin \left (3 f x +3 e \right ) b}{12 a^{2} f}\)	\(282\)

Input:

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

Output:

1/f*(1/a^3*(1/5*a^2*sin(f*x+e)^5-2/3*a^2*sin(f*x+e)^3+1/3*a*b*sin(f*x+e)^3 
+sin(f*x+e)*a^2-sin(f*x+e)*a*b+sin(f*x+e)*b^2)-b^3/a^3/(a*(a+b))^(1/2)*arc 
tanh(a*sin(f*x+e)/(a*(a+b))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.82 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {15 \, \sqrt {a^{2} + a b} b^{3} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3} + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{5} + a^{4} b\right )} f}, \frac {15 \, \sqrt {-a^{2} - a b} b^{3} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) + {\left (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3} + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{5} + a^{4} b\right )} f}\right ] \] Input:

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

Output:

[1/30*(15*sqrt(a^2 + a*b)*b^3*log(-(a*cos(f*x + e)^2 + 2*sqrt(a^2 + a*b)*s 
in(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(3*(a^4 + a^3*b)*cos(f* 
x + e)^4 + 8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3 + (4*a^4 - a^3*b - 5*a^2 
*b^2)*cos(f*x + e)^2)*sin(f*x + e))/((a^5 + a^4*b)*f), 1/15*(15*sqrt(-a^2 
- a*b)*b^3*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (3*(a^4 + a^3*b 
)*cos(f*x + e)^4 + 8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3 + (4*a^4 - a^3*b 
 - 5*a^2*b^2)*cos(f*x + e)^2)*sin(f*x + e))/((a^5 + a^4*b)*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, b^{3} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {2 \, {\left (3 \, a^{2} \sin \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - a b + b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{3}}}{30 \, f} \] Input:

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

Output:

1/30*(15*b^3*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt 
((a + b)*a)))/(sqrt((a + b)*a)*a^3) + 2*(3*a^2*sin(f*x + e)^5 - 5*(2*a^2 - 
 a*b)*sin(f*x + e)^3 + 15*(a^2 - a*b + b^2)*sin(f*x + e))/a^3)/f
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, b^{3} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} a^{3}} + \frac {3 \, a^{4} \sin \left (f x + e\right )^{5} - 10 \, a^{4} \sin \left (f x + e\right )^{3} + 5 \, a^{3} b \sin \left (f x + e\right )^{3} + 15 \, a^{4} \sin \left (f x + e\right ) - 15 \, a^{3} b \sin \left (f x + e\right ) + 15 \, a^{2} b^{2} \sin \left (f x + e\right )}{a^{5}}}{15 \, f} \] Input:

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

Output:

1/15*(15*b^3*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*a^3 
) + (3*a^4*sin(f*x + e)^5 - 10*a^4*sin(f*x + e)^3 + 5*a^3*b*sin(f*x + e)^3 
 + 15*a^4*sin(f*x + e) - 15*a^3*b*sin(f*x + e) + 15*a^2*b^2*sin(f*x + e))/ 
a^5)/f
 

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sin \left (e+f\,x\right )\,\left (\frac {3}{a}+\frac {\left (a+b\right )\,\left (\frac {a+b}{a^2}-\frac {3}{a}\right )}{a}\right )}{f}+\frac {{\sin \left (e+f\,x\right )}^5}{5\,a\,f}+\frac {{\sin \left (e+f\,x\right )}^3\,\left (\frac {a+b}{3\,a^2}-\frac {1}{a}\right )}{f}-\frac {b^3\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )}{a^{7/2}\,f\,\sqrt {a+b}} \] Input:

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

Output:

(sin(e + f*x)*(3/a + ((a + b)*((a + b)/a^2 - 3/a))/a))/f + sin(e + f*x)^5/ 
(5*a*f) + (sin(e + f*x)^3*((a + b)/(3*a^2) - 1/a))/f - (b^3*atanh((a^(1/2) 
*sin(e + f*x))/(a + b)^(1/2)))/(a^(7/2)*f*(a + b)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {15 \sqrt {a}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{3}-15 \sqrt {a}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{3}+6 \sin \left (f x +e \right )^{5} a^{4}+6 \sin \left (f x +e \right )^{5} a^{3} b -20 \sin \left (f x +e \right )^{3} a^{4}-10 \sin \left (f x +e \right )^{3} a^{3} b +10 \sin \left (f x +e \right )^{3} a^{2} b^{2}+30 \sin \left (f x +e \right ) a^{4}+30 \sin \left (f x +e \right ) a \,b^{3}}{30 a^{4} f \left (a +b \right )} \] Input:

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

Output:

(15*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) 
- 2*sqrt(a)*tan((e + f*x)/2))*b**3 - 15*sqrt(a)*sqrt(a + b)*log(sqrt(a + b 
)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*b**3 + 6 
*sin(e + f*x)**5*a**4 + 6*sin(e + f*x)**5*a**3*b - 20*sin(e + f*x)**3*a**4 
 - 10*sin(e + f*x)**3*a**3*b + 10*sin(e + f*x)**3*a**2*b**2 + 30*sin(e + f 
*x)*a**4 + 30*sin(e + f*x)*a*b**3)/(30*a**4*f*(a + b))