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

Optimal. Leaf size=19 \[ -\frac {\tan (e+f x) \sec ^5(e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {4043} \[ -\frac {\tan (e+f x) \sec ^5(e+f x)}{f} \]

Antiderivative was successfully verified.

[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 4043

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*} \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}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 19, normalized size = 1.00 \[ -\frac {\tan (e+f x) \sec ^5(e+f x)}{f} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.41, size = 19, normalized size = 1.00 \[ -\frac {\sin \left (f x + e\right )}{f \cos \left (f x + e\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 24, normalized size = 1.26 \[ \frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 1.56, size = 70, normalized size = 3.68 \[ \frac {-5 \left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+6 \left (-\frac {\left (\sec ^{5}\left (f x +e \right )\right )}{6}-\frac {5 \left (\sec ^{3}\left (f x +e \right )\right )}{24}-\frac {5 \sec \left (f x +e \right )}{16}\right ) \tan \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-5*(-1/4*sec(f*x+e)^3-3/8*sec(f*x+e))*tan(f*x+e)+6*(-1/6*sec(f*x+e)^5-5/24*sec(f*x+e)^3-5/16*sec(f*x+e))*
tan(f*x+e))

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maxima [B]  time = 0.63, size = 42, normalized size = 2.21 \[ \frac {\sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.46, size = 42, normalized size = 2.21 \[ \frac {\sin \left (e+f\,x\right )}{f\,\left ({\sin \left (e+f\,x\right )}^6-3\,{\sin \left (e+f\,x\right )}^4+3\,{\sin \left (e+f\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- 5 \sec ^{5}{\left (e + f x \right )}\right )\, dx - \int 6 \sec ^{7}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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