3.1.36 \(\int \cos ^5(e+f x) (-5+4 \sec ^2(e+f x)) \, dx\) [36]

Optimal. Leaf size=19 \[ -\frac {\cos ^4(e+f x) \sin (e+f x)}{f} \]

[Out]

-cos(f*x+e)^4*sin(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 = {4128} \begin {gather*} -\frac {\sin (e+f x) \cos ^4(e+f x)}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^5*(-5 + 4*Sec[e + f*x]^2),x]

[Out]

-((Cos[e + f*x]^4*Sin[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*} \int \cos ^5(e+f x) \left (-5+4 \sec ^2(e+f x)\right ) \, dx &=-\frac {\cos ^4(e+f x) \sin (e+f x)}{f}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(19)=38\).
time = 0.02, size = 59, normalized size = 3.11 \begin {gather*} \frac {7 \sin (e+f x)}{8 f}-\frac {4 \sin ^3(e+f x)}{3 f}-\frac {25 \sin (3 (e+f x))}{48 f}-\frac {\sin (5 (e+f x))}{16 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^5*(-5 + 4*Sec[e + f*x]^2),x]

[Out]

(7*Sin[e + f*x])/(8*f) - (4*Sin[e + f*x]^3)/(3*f) - (25*Sin[3*(e + f*x)])/(48*f) - Sin[5*(e + f*x)]/(16*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(51\) vs. \(2(19)=38\).
time = 0.57, size = 52, normalized size = 2.74

method	result	size
risch	\(-\frac {\sin \left (f x +e \right )}{8 f}-\frac {\sin \left (5 f x +5 e \right )}{16 f}-\frac {3 \sin \left (3 f x +3 e \right )}{16 f}\)	\(41\)
derivativedivides	\(\frac {-\left (\frac {8}{3}+\cos ^{4}\left (f x +e \right )+\frac {4 \left (\cos ^{2}\left (f x +e \right )\right )}{3}\right ) \sin \left (f x +e \right )+\frac {4 \left (2+\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{3}}{f}\)	\(52\)
default	\(\frac {-\left (\frac {8}{3}+\cos ^{4}\left (f x +e \right )+\frac {4 \left (\cos ^{2}\left (f x +e \right )\right )}{3}\right ) \sin \left (f x +e \right )+\frac {4 \left (2+\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{3}}{f}\)	\(52\)
norman	\(\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {10 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {20 \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {20 \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {10 \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 33, normalized size = 1.74 \begin {gather*} -\frac {\sin \left (f x + e\right )^{5} - 2 \, \sin \left (f x + e\right )^{3} + \sin \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

-(sin(f*x + e)^5 - 2*sin(f*x + e)^3 + sin(f*x + e))/f

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Fricas [A]
time = 2.73, size = 21, normalized size = 1.11 \begin {gather*} -\frac {\cos \left (f x + e\right )^{4} \sin \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

-cos(f*x + e)^4*sin(f*x + e)/f

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (4 \sec ^{2}{\left (e + f x \right )} - 5\right ) \cos ^{5}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**5*(-5+4*sec(f*x+e)**2),x)

[Out]

Integral((4*sec(e + f*x)**2 - 5)*cos(e + f*x)**5, x)

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Giac [A]
time = 0.45, size = 30, normalized size = 1.58 \begin {gather*} -\frac {\sin \left (f x + e\right )^{5} - 2 \, \sin \left (f x + e\right )^{3} + \sin \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(-5+4*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-(sin(f*x + e)^5 - 2*sin(f*x + e)^3 + sin(f*x + e))/f

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Mupad [B]
time = 2.42, size = 23, normalized size = 1.21 \begin {gather*} -\frac {\sin \left (e+f\,x\right )\,{\left ({\sin \left (e+f\,x\right )}^2-1\right )}^2}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^5*(4/cos(e + f*x)^2 - 5),x)

[Out]

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

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