3.1 \(\int \sin ^m(e+f x) (1+m-(2+m) \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=20 \[ \frac {\cos (e+f x) \sin ^{m+1}(e+f x)}{f} \]

[Out]

cos(f*x+e)*sin(f*x+e)^(1+m)/f

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {3011} \[ \frac {\cos (e+f x) \sin ^{m+1}(e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^m*(1 + m - (2 + m)*Sin[e + f*x]^2),x]

[Out]

(Cos[e + f*x]*Sin[e + f*x]^(1 + m))/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 ^m(e+f x) \left (1+m-(2+m) \sin ^2(e+f x)\right ) \, dx &=\frac {\cos (e+f x) \sin ^{1+m}(e+f x)}{f}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 107, normalized size = 5.35 \[ \frac {\cos (e+f x) \sin ^{m+1}(e+f x) \left ((m+3) \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(e+f x)\right )-(m+2) \sin ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {m+3}{2};\frac {m+5}{2};\sin ^2(e+f x)\right )\right )}{f (m+3) \sqrt {\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^m*(1 + m - (2 + m)*Sin[e + f*x]^2),x]

[Out]

(Cos[e + f*x]*Sin[e + f*x]^(1 + m)*((3 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[e + f*x]^2] - (2
+ m)*Hypergeometric2F1[1/2, (3 + m)/2, (5 + m)/2, Sin[e + f*x]^2]*Sin[e + f*x]^2))/(f*(3 + m)*Sqrt[Cos[e + f*x
]^2])

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fricas [A]  time = 0.43, size = 24, normalized size = 1.20 \[ \frac {\sin \left (f x + e\right )^{m} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^m*(1+m-(2+m)*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -{\left ({\left (m + 2\right )} \sin \left (f x + e\right )^{2} - m - 1\right )} \sin \left (f x + e\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^m*(1+m-(2+m)*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate(-((m + 2)*sin(f*x + e)^2 - m - 1)*sin(f*x + e)^m, x)

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maple [F]  time = 5.20, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{m}\left (f x +e \right )\right ) \left (1+m -\left (2+m \right ) \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^m*(1+m-(2+m)*sin(f*x+e)^2),x)

[Out]

int(sin(f*x+e)^m*(1+m-(2+m)*sin(f*x+e)^2),x)

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maxima [B]  time = 1.41, size = 248, normalized size = 12.40 \[ -\frac {{\left (\left (-1\right )^{\frac {1}{2} \, m} e^{\left (\frac {1}{2} \, m \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) + \frac {1}{2} \, m \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )\right )} \sin \left (-{\left (f x + e\right )} {\left (m + 2\right )} + m \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - m \arctan \left (\sin \left (f x + e\right ), -\cos \left (f x + e\right ) + 1\right )\right ) - \left (-1\right )^{\frac {1}{2} \, m} e^{\left (\frac {1}{2} \, m \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) + \frac {1}{2} \, m \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )\right )} \sin \left (-{\left (f x + e\right )} {\left (m - 2\right )} + m \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - m \arctan \left (\sin \left (f x + e\right ), -\cos \left (f x + e\right ) + 1\right )\right )\right )} 2^{-m - 2}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^m*(1+m-(2+m)*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

-((-1)^(1/2*m)*e^(1/2*m*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) + 1/2*m*log(cos(f*x + e)^2 +
sin(f*x + e)^2 - 2*cos(f*x + e) + 1))*sin(-(f*x + e)*(m + 2) + m*arctan2(sin(f*x + e), cos(f*x + e) + 1) - m*
arctan2(sin(f*x + e), -cos(f*x + e) + 1)) - (-1)^(1/2*m)*e^(1/2*m*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(
f*x + e) + 1) + 1/2*m*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*cos(f*x + e) + 1))*sin(-(f*x + e)*(m - 2) + m*ar
ctan2(sin(f*x + e), cos(f*x + e) + 1) - m*arctan2(sin(f*x + e), -cos(f*x + e) + 1)))*2^(-m - 2)/f

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mupad [B]  time = 13.74, size = 22, normalized size = 1.10 \[ \frac {{\sin \left (e+f\,x\right )}^m\,\sin \left (2\,e+2\,f\,x\right )}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^m*(m - sin(e + f*x)^2*(m + 2) + 1),x)

[Out]

(sin(e + f*x)^m*sin(2*e + 2*f*x))/(2*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**m*(1+m-(2+m)*sin(f*x+e)**2),x)

[Out]

Timed out

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