\(\int \sec ^5(e+f x) (5-6 \sec ^2(e+f x)) \, dx\) [26]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=-\frac {\sec ^5(e+f x) \tan (e+f x)}{f} \]

[Out]

-sec(f*x+e)^5*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {4128} \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan (e+f x) \sec ^5(e+f x)}{f} \]

[In]

Int[Sec[e + f*x]^5*(5 - 6*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]^5*Tan[e + f*x])/f)

Rule 4128

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] /; FreeQ[{b, e, f, A, C, m}, x] && EqQ[C*m + A*(m + 1), 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sec ^5(e+f x) \tan (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=-\frac {\sec ^5(e+f x) \tan (e+f x)}{f} \]

[In]

Integrate[Sec[e + f*x]^5*(5 - 6*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]^5*Tan[e + f*x])/f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16

method	result	size
risch	\(\frac {32 i \left ({\mathrm e}^{7 i \left (f x +e \right )}-{\mathrm e}^{5 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6}}\)	\(41\)
parallelrisch	\(-\frac {32 \sin \left (f x +e \right )}{f \left (\cos \left (6 f x +6 e \right )+6 \cos \left (4 f x +4 e \right )+15 \cos \left (2 f x +2 e \right )+10\right )}\)	\(47\)
derivativedivides	\(\frac {-5 \left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+6 \left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )}{f}\)	\(70\)
default	\(\frac {-5 \left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+6 \left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )}{f}\)	\(70\)
parts	\(\frac {-5 \left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {15 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}}{f}-\frac {6 \left (-\left (-\frac {\sec \left (f x +e \right )^{5}}{6}-\frac {5 \sec \left (f x +e \right )^{3}}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )+\frac {5 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16}\right )}{f}\)	\(110\)
norman	\(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}-\frac {20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}-\frac {20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}-\frac {10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{6}}\)	\(112\)

[In]

int(sec(f*x+e)^5*(5-6*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

32*I/f/(exp(2*I*(f*x+e))+1)^6*(exp(7*I*(f*x+e))-exp(5*I*(f*x+e)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{6}} \]

[In]

integrate(sec(f*x+e)^5*(5-6*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-sin(f*x + e)/(f*cos(f*x + e)^6)

Sympy [F]

\[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=- \int \left (- 5 \sec ^{5}{\left (e + f x \right )}\right )\, dx - \int 6 \sec ^{7}{\left (e + f x \right )}\, dx \]

[In]

integrate(sec(f*x+e)**5*(5-6*sec(f*x+e)**2),x)

[Out]

-Integral(-5*sec(e + f*x)**5, x) - Integral(6*sec(e + f*x)**7, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21 \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=\frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1\right )} f} \]

[In]

integrate(sec(f*x+e)^5*(5-6*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

sin(f*x + e)/((sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f*x + e)^2 - 1)*f)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=\frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{3} f} \]

[In]

integrate(sec(f*x+e)^5*(5-6*sec(f*x+e)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21 \[ \int \sec ^5(e+f x) \left (5-6 \sec ^2(e+f x)\right ) \, dx=\frac {\sin \left (e+f\,x\right )}{f\,\left ({\sin \left (e+f\,x\right )}^6-3\,{\sin \left (e+f\,x\right )}^4+3\,{\sin \left (e+f\,x\right )}^2-1\right )} \]

[In]

int(-(6/cos(e + f*x)^2 - 5)/cos(e + f*x)^5,x)

[Out]

sin(e + f*x)/(f*(3*sin(e + f*x)^2 - 3*sin(e + f*x)^4 + sin(e + f*x)^6 - 1))