3.5.13 \(\int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx\) [413]

Optimal. Leaf size=46 \[ \frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]

[Out]

2*c*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2817} \begin {gather*} \frac {2 c \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(2*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

Rule 2817

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx &=\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 85, normalized size = 1.85 \begin {gather*} \frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)}}{f (1+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c*Sin[e + f*x]])/(f*(1 + 2*m)*(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [F]
time = 0.18, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c -c \sin \left (f x +e \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (47) = 94\).
time = 0.50, size = 124, normalized size = 2.70 \begin {gather*} -\frac {2 \, {\left (a^{m} \sqrt {c} + \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{f {\left (2 \, m + 1\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a^m*sqrt(c) + a^m*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1))*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)
- m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/(f*(2*m + 1)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))

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Fricas [A]
time = 0.37, size = 82, normalized size = 1.78 \begin {gather*} \frac {2 \, \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{2 \, f m + {\left (2 \, f m + f\right )} \cos \left (f x + e\right ) - {\left (2 \, f m + f\right )} \sin \left (f x + e\right ) + f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m*(cos(f*x + e) + sin(f*x + e) + 1)/(2*f*m + (2*f*m + f)*cos(
f*x + e) - (2*f*m + f)*sin(f*x + e) + f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(e + f*x) + 1))**m*sqrt(-c*(sin(e + f*x) - 1)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, x)

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Mupad [B]
time = 0.45, size = 53, normalized size = 1.15 \begin {gather*} -\frac {2\,\cos \left (e+f\,x\right )\,{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}}{f\,\left (2\,m+1\right )\,\left (\sin \left (e+f\,x\right )-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*cos(e + f*x)*(a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2))/(f*(2*m + 1)*(sin(e + f*x) - 1))

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