\(\int \sec ^3(e+f x) (3-4 \sec ^2(e+f x)) \, dx\) [28]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sec ^3(e+f x) \tan (e+f x)}{f} \]

[Out]

-sec(f*x+e)^3*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} \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\tan (e+f x) \sec ^3(e+f x)}{f} \]

[In]

Int[Sec[e + f*x]^3*(3 - 4*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]^3*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 ^3(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 \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sec ^3(e+f x) \tan (e+f x)}{f} \]

[In]

Integrate[Sec[e + f*x]^3*(3 - 4*Sec[e + f*x]^2),x]

[Out]

-((Sec[e + f*x]^3*Tan[e + f*x])/f)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89

method	result	size
parallelrisch	\(-\frac {8 \sin \left (f x +e \right )}{f \left (\cos \left (4 f x +4 e \right )+4 \cos \left (2 f x +2 e \right )+3\right )}\)	\(36\)
risch	\(\frac {8 i \left ({\mathrm e}^{5 i \left (f x +e \right )}-{\mathrm e}^{3 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}\)	\(41\)
derivativedivides	\(\frac {\frac {3 \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+4 \left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )}{f}\)	\(47\)
default	\(\frac {\frac {3 \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+4 \left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )}{f}\)	\(47\)
norman	\(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}-\frac {6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}\)	\(80\)
parts	\(\frac {3 \sec \left (f x +e \right ) \tan \left (f x +e \right )}{2 f}+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2 f}-\frac {4 \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}\)	\(87\)

[In]

int(sec(f*x+e)^3*(3-4*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

-8/f*sin(f*x+e)/(cos(4*f*x+4*e)+4*cos(2*f*x+2*e)+3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{4}} \]

[In]

integrate(sec(f*x+e)^3*(3-4*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-sin(f*x + e)/(f*cos(f*x + e)^4)

Sympy [F]

\[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=- \int \left (- 3 \sec ^{3}{\left (e + f x \right )}\right )\, dx - \int 4 \sec ^{5}{\left (e + f x \right )}\, dx \]

[In]

integrate(sec(f*x+e)**3*(3-4*sec(f*x+e)**2),x)

[Out]

-Integral(-3*sec(e + f*x)**3, x) - Integral(4*sec(e + f*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1\right )} f} \]

[In]

integrate(sec(f*x+e)^3*(3-4*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2} f} \]

[In]

integrate(sec(f*x+e)^3*(3-4*sec(f*x+e)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \sec ^3(e+f x) \left (3-4 \sec ^2(e+f x)\right ) \, dx=-\frac {\sin \left (e+f\,x\right )}{f\,{\left ({\sin \left (e+f\,x\right )}^2-1\right )}^2} \]

[In]

int(-(4/cos(e + f*x)^2 - 3)/cos(e + f*x)^3,x)

[Out]

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