3.3 \(\int \sin ^4(e+f x) (5-6 \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=18 \[ \frac {\sin ^5(e+f x) \cos (e+f x)}{f} \]

[Out]

cos(f*x+e)*sin(f*x+e)^5/f

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Rubi [A]  time = 0.02, antiderivative size = 18, 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 = {3011} \[ \frac {\sin ^5(e+f x) \cos (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4*(5 - 6*Sin[e + f*x]^2),x]

[Out]

(Cos[e + f*x]*Sin[e + f*x]^5)/f

Rule 3011

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

Rubi steps

\begin {align*} \int \sin ^4(e+f x) \left (5-6 \sin ^2(e+f x)\right ) \, dx &=\frac {\cos (e+f x) \sin ^5(e+f x)}{f}\\ \end {align*}

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Mathematica [B]  time = 0.11, size = 39, normalized size = 2.17 \[ \frac {5 \sin (2 (e+f x))-4 \sin (4 (e+f x))+\sin (6 (e+f x))+24 e}{32 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4*(5 - 6*Sin[e + f*x]^2),x]

[Out]

(24*e + 5*Sin[2*(e + f*x)] - 4*Sin[4*(e + f*x)] + Sin[6*(e + f*x)])/(32*f)

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fricas [A]  time = 0.43, size = 35, normalized size = 1.94 \[ \frac {{\left (\cos \left (f x + e\right )^{5} - 2 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(5-6*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.16, size = 46, normalized size = 2.56 \[ \frac {\sin \left (6 \, f x + 6 \, e\right )}{32 \, f} - \frac {\sin \left (4 \, f x + 4 \, e\right )}{8 \, f} + \frac {5 \, \sin \left (2 \, f x + 2 \, e\right )}{32 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(5-6*sin(f*x+e)^2),x, algorithm="giac")

[Out]

1/32*sin(6*f*x + 6*e)/f - 1/8*sin(4*f*x + 4*e)/f + 5/32*sin(2*f*x + 2*e)/f

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maple [B]  time = 0.43, size = 65, normalized size = 3.61 \[ \frac {\left (\sin ^{5}\left (f x +e \right )+\frac {5 \left (\sin ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )-\frac {5 \left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4*(5-6*sin(f*x+e)^2),x)

[Out]

1/f*((sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)-5/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e))

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maxima [B]  time = 0.65, size = 44, normalized size = 2.44 \[ \frac {\tan \left (f x + e\right )^{5}}{{\left (\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(5-6*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

tan(f*x + e)^5/((tan(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f*x + e)^2 + 1)*f)

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mupad [B]  time = 13.28, size = 43, normalized size = 2.39 \[ \frac {5\,\sin \left (2\,e+2\,f\,x\right )}{32\,f}-\frac {\sin \left (4\,e+4\,f\,x\right )}{8\,f}+\frac {\sin \left (6\,e+6\,f\,x\right )}{32\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-sin(e + f*x)^4*(6*sin(e + f*x)^2 - 5),x)

[Out]

(5*sin(2*e + 2*f*x))/(32*f) - sin(4*e + 4*f*x)/(8*f) + sin(6*e + 6*f*x)/(32*f)

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sympy [A]  time = 4.83, size = 236, normalized size = 13.11 \[ \begin {cases} - \frac {15 x \sin ^{6}{\left (e + f x \right )}}{8} - \frac {45 x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{8} + \frac {15 x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {45 x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{8} + \frac {15 x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac {15 x \cos ^{6}{\left (e + f x \right )}}{8} + \frac {15 x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {33 \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {5 \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{f} - \frac {25 \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {15 \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{8 f} - \frac {15 \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text {for}\: f \neq 0 \\x \left (5 - 6 \sin ^{2}{\relax (e )}\right ) \sin ^{4}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4*(5-6*sin(f*x+e)**2),x)

[Out]

Piecewise((-15*x*sin(e + f*x)**6/8 - 45*x*sin(e + f*x)**4*cos(e + f*x)**2/8 + 15*x*sin(e + f*x)**4/8 - 45*x*si
n(e + f*x)**2*cos(e + f*x)**4/8 + 15*x*sin(e + f*x)**2*cos(e + f*x)**2/4 - 15*x*cos(e + f*x)**6/8 + 15*x*cos(e
+ f*x)**4/8 + 33*sin(e + f*x)**5*cos(e + f*x)/(8*f) + 5*sin(e + f*x)**3*cos(e + f*x)**3/f - 25*sin(e + f*x)**
3*cos(e + f*x)/(8*f) + 15*sin(e + f*x)*cos(e + f*x)**5/(8*f) - 15*sin(e + f*x)*cos(e + f*x)**3/(8*f), Ne(f, 0)
), (x*(5 - 6*sin(e)**2)*sin(e)**4, True))

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