\(\int \frac {\sec ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [193]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\text {arctanh}(\sin (e+f x))}{b^2 f}-\frac {\sqrt {a} (2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 b^2 (a+b)^{3/2} f}-\frac {a \sin (e+f x)}{2 b (a+b) f \left (a+b-a \sin ^2(e+f x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.24 (sec) , antiderivative size = 980, normalized size of antiderivative = 9.61 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*((-2*I)*a*(2*a + 3*b)*ArcTa 
n[(2*Sin[e]*(I*a + I*b + I*(a + b)*Cos[2*e] + Sqrt[a]*Sqrt[a + b]*Cos[f*x] 
*Sqrt[(Cos[e] - I*Sin[e])^2] - Sqrt[a]*Sqrt[a + b]*Cos[2*e + f*x]*Sqrt[(Co 
s[e] - I*Sin[e])^2] + a*Sin[2*e] + b*Sin[2*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt 
[(Cos[e] - I*Sin[e])^2]*Sin[f*x] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I* 
Sin[e])^2]*Sin[2*e + f*x]))/(I*(a + 3*b)*Cos[e] + I*(a + b)*Cos[3*e] + I*a 
*Cos[e + 2*f*x] + I*a*Cos[3*e + 2*f*x] + 3*a*Sin[e] + b*Sin[e] + a*Sin[3*e 
] + b*Sin[3*e] + a*Sin[e + 2*f*x] - a*Sin[3*e + 2*f*x])]*(a + 2*b + a*Cos[ 
2*(e + f*x)])*Sec[e + f*x]*(Cos[e] - I*Sin[e]) - a*(2*a + 3*b)*(a + 2*b + 
a*Cos[2*(e + f*x)])*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x)] - (2*I 
)*a*Sin[2*e] - (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*S 
in[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin 
[2*e + f*x]]*Sec[e + f*x]*(Cos[e] - I*Sin[e]) + a*(2*a + 3*b)*(a + 2*b + a 
*Cos[2*(e + f*x)])*Log[-a - 2*(a + b)*Cos[2*e] + a*Cos[2*(e + f*x)] + (2*I 
)*a*Sin[2*e] + (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*S 
in[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin 
[2*e + f*x]]*Sec[e + f*x]*(Cos[e] - I*Sin[e]) - 8*Sqrt[a]*(a + b)^(3/2)*(a 
 + 2*b + a*Cos[2*(e + f*x)])*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sec[ 
e + f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] + 8*Sqrt[a]*(a + b)^(3/2)*(a + 2*b + 
a*Cos[2*(e + f*x)])*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sec[e + f*...
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4635, 316, 25, 397, 219, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4635

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

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 = 0.88 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {a \left (\frac {b \sin \left (f x +e \right )}{2 \left (a +b \right ) \left (-a -b +a \sin \left (f x +e \right )^{2}\right )}-\frac {\left (2 a +3 b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{2}}-\frac {\ln \left (\sin \left (f x +e \right )-1\right )}{2 b^{2}}+\frac {\ln \left (\sin \left (f x +e \right )+1\right )}{2 b^{2}}}{f}\)	\(110\)
default	\(\frac {\frac {a \left (\frac {b \sin \left (f x +e \right )}{2 \left (a +b \right ) \left (-a -b +a \sin \left (f x +e \right )^{2}\right )}-\frac {\left (2 a +3 b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}\right )}{b^{2}}-\frac {\ln \left (\sin \left (f x +e \right )-1\right )}{2 b^{2}}+\frac {\ln \left (\sin \left (f x +e \right )+1\right )}{2 b^{2}}}{f}\)	\(110\)
risch	\(\frac {i a \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{b \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{b^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{b^{2} f}+\frac {\sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a \left (a +b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a}-1\right ) a}{2 \left (a +b \right )^{2} f \,b^{2}}+\frac {3 \sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a \left (a +b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a}-1\right )}{4 \left (a +b \right )^{2} f b}-\frac {\sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a \left (a +b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a}-1\right ) a}{2 \left (a +b \right )^{2} f \,b^{2}}-\frac {3 \sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a \left (a +b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a}-1\right )}{4 \left (a +b \right )^{2} f b}\)	\(336\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 392, normalized size of antiderivative = 3.84 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [-\frac {2 \, a b \sin \left (f x + e\right ) - {\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + 3 \, b^{2}\right )} \sqrt {\frac {a}{a + b}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {a}{a + b}} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{4 \, {\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a b^{3} + b^{4}\right )} f\right )}}, -\frac {a b \sin \left (f x + e\right ) - {\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + 3 \, b^{2}\right )} \sqrt {-\frac {a}{a + b}} \arctan \left (\sqrt {-\frac {a}{a + b}} \sin \left (f x + e\right )\right ) - {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a b^{3} + b^{4}\right )} f\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.43 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (2 \, a + 3 \, b\right )} a \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {2 \, a \sin \left (f x + e\right )}{a^{2} b + 2 \, a b^{2} + b^{3} - {\left (a^{2} b + a b^{2}\right )} \sin \left (f x + e\right )^{2}} + \frac {2 \, \log \left (\sin \left (f x + e\right ) + 1\right )}{b^{2}} - \frac {2 \, \log \left (\sin \left (f x + e\right ) - 1\right )}{b^{2}}}{4 \, f} \] Input:

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

Output:

1/4*((2*a + 3*b)*a*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) 
+ sqrt((a + b)*a)))/((a*b^2 + b^3)*sqrt((a + b)*a)) - 2*a*sin(f*x + e)/(a^ 
2*b + 2*a*b^2 + b^3 - (a^2*b + a*b^2)*sin(f*x + e)^2) + 2*log(sin(f*x + e) 
 + 1)/b^2 - 2*log(sin(f*x + e) - 1)/b^2)/f
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.24 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (2 \, a^{2} + 3 \, a b\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt {-a^{2} - a b}} + \frac {a \sin \left (f x + e\right )}{{\left (a \sin \left (f x + e\right )^{2} - a - b\right )} {\left (a b + b^{2}\right )}} + \frac {\log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right )}{b^{2}} - \frac {\log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right )}{b^{2}}}{2 \, f} \] Input:

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

Output:

1/2*((2*a^2 + 3*a*b)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a*b^2 + b^3 
)*sqrt(-a^2 - a*b)) + a*sin(f*x + e)/((a*sin(f*x + e)^2 - a - b)*(a*b + b^ 
2)) + log(abs(sin(f*x + e) + 1))/b^2 - log(abs(sin(f*x + e) - 1))/b^2)/f
 

Mupad [B] (verification not implemented)

Time = 16.89 (sec) , antiderivative size = 2039, normalized size of antiderivative = 19.99 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((((4*a^2*b^6 + 6*a^3*b^5 + 2*a^4*b^4)/(2*(2*a*b^4 + b^5 + a^2*b^3) 
) - (sin(e + f*x)*(16*a^2*b^7 + 64*a^3*b^6 + 80*a^4*b^5 + 32*a^5*b^4))/(8* 
b^2*(2*a*b^3 + b^4 + a^2*b^2)))*1i)/(2*b^2) + (sin(e + f*x)*(20*a^4*b + 8* 
a^5 + 13*a^3*b^2)*1i)/(4*(2*a*b^3 + b^4 + a^2*b^2)))/b^2 - ((((4*a^2*b^6 + 
 6*a^3*b^5 + 2*a^4*b^4)/(2*(2*a*b^4 + b^5 + a^2*b^3)) + (sin(e + f*x)*(16* 
a^2*b^7 + 64*a^3*b^6 + 80*a^4*b^5 + 32*a^5*b^4))/(8*b^2*(2*a*b^3 + b^4 + a 
^2*b^2)))*1i)/(2*b^2) - (sin(e + f*x)*(20*a^4*b + 8*a^5 + 13*a^3*b^2)*1i)/ 
(4*(2*a*b^3 + b^4 + a^2*b^2)))/b^2)/((((4*a^2*b^6 + 6*a^3*b^5 + 2*a^4*b^4) 
/(2*(2*a*b^4 + b^5 + a^2*b^3)) - (sin(e + f*x)*(16*a^2*b^7 + 64*a^3*b^6 + 
80*a^4*b^5 + 32*a^5*b^4))/(8*b^2*(2*a*b^3 + b^4 + a^2*b^2)))/(2*b^2) + (si 
n(e + f*x)*(20*a^4*b + 8*a^5 + 13*a^3*b^2))/(4*(2*a*b^3 + b^4 + a^2*b^2))) 
/b^2 + (((4*a^2*b^6 + 6*a^3*b^5 + 2*a^4*b^4)/(2*(2*a*b^4 + b^5 + a^2*b^3)) 
 + (sin(e + f*x)*(16*a^2*b^7 + 64*a^3*b^6 + 80*a^4*b^5 + 32*a^5*b^4))/(8*b 
^2*(2*a*b^3 + b^4 + a^2*b^2)))/(2*b^2) - (sin(e + f*x)*(20*a^4*b + 8*a^5 + 
 13*a^3*b^2))/(4*(2*a*b^3 + b^4 + a^2*b^2)))/b^2 - ((3*a^3*b)/2 + a^4)/(2* 
a*b^4 + b^5 + a^2*b^3)))*1i)/(b^2*f) - (atan(((((sin(e + f*x)*(20*a^4*b + 
8*a^5 + 13*a^3*b^2))/(2*(2*a*b^3 + b^4 + a^2*b^2)) + ((a*(a + b)^3)^(1/2)* 
((4*a^2*b^6 + 6*a^3*b^5 + 2*a^4*b^4)/(2*a*b^4 + b^5 + a^2*b^3) - (sin(e + 
f*x)*(a*(a + b)^3)^(1/2)*(2*a + 3*b)*(16*a^2*b^7 + 64*a^3*b^6 + 80*a^4*b^5 
 + 32*a^5*b^4))/(8*(2*a*b^3 + b^4 + a^2*b^2)*(3*a*b^4 + b^5 + 3*a^2*b^3...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 879, normalized size of antiderivative = 8.62 \[ \int \frac {\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*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))*sin(e + f*x)**2*a**2 + 3*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))*sin(e + f*x)**2*a*b - 2*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))*a**2 - 5*sqrt(a)*s 
qrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*t 
an((e + f*x)/2))*a*b - 3*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**2 - 2*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))*sin(e + f*x)**2*a**2 - 3*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*t 
an((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x 
)**2*a*b + 2*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqr 
t(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*a**2 + 5*sqrt(a)*sqrt(a + b)*log(sq 
rt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))* 
a*b + 3*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**2 - 4*log(tan((e + f*x)/2) - 1)*sin(e 
 + f*x)**2*a**3 - 8*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a**2*b - 4*l 
og(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a*b**2 + 4*log(tan((e + f*x)/2) - 
 1)*a**3 + 12*log(tan((e + f*x)/2) - 1)*a**2*b + 12*log(tan((e + f*x)/2) - 
 1)*a*b**2 + 4*log(tan((e + f*x)/2) - 1)*b**3 + 4*log(tan((e + f*x)/2) ...