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[e + f*x]*Sin[e + f*x]^(1 + m))/f

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Rubi [A]  time = 0.0284205, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, 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*}

Mathematica [C]  time = 0.175617, 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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Maple [F]  time = 1.375, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{m} \left ( 1+m- \left ( 2+m \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

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 = 2.33524, size = 335, normalized size = 16.75 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.73164, size = 59, normalized size = 2.95 \begin{align*} \frac{\sin \left (f x + e\right )^{m} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{f} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)