3.412 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=100 \[ \frac {8 c^2 \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (4 m^2+8 m+3\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f (2 m+3)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2740, 2738} \[ \frac {8 c^2 \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (4 m^2+8 m+3\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(8*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(3 + 8*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e + f*x]
*(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]])/(f*(3 + 2*m))

Rule 2738

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)]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx &=\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)}}{f (3+2 m)}+\frac {(4 c) \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx}{3+2 m}\\ &=\frac {8 c^2 \cos (e+f x) (a+a \sin (e+f x))^m}{f \left (3+8 m+4 m^2\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)}}{f (3+2 m)}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 110, normalized size = 1.10 \[ -\frac {2 c \sqrt {c-c \sin (e+f x)} ((2 m+1) \sin (e+f x)-2 m-5) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m}{f (2 m+1) (2 m+3) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

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

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fricas [A]  time = 0.47, size = 145, normalized size = 1.45 \[ \frac {2 \, {\left ({\left (2 \, c m + c\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c m + 5 \, c\right )} \cos \left (f x + e\right ) - {\left ({\left (2 \, c m + c\right )} \cos \left (f x + e\right ) - 4 \, c\right )} \sin \left (f x + e\right ) + 4 \, c\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{4 \, f m^{2} + 8 \, f m + {\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \cos \left (f x + e\right ) - {\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \sin \left (f x + e\right ) + 3 \, f} \]

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))^(3/2),x, algorithm="fricas")

[Out]

2*((2*c*m + c)*cos(f*x + e)^2 + (2*c*m + 5*c)*cos(f*x + e) - ((2*c*m + c)*cos(f*x + e) - 4*c)*sin(f*x + e) + 4
*c)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(4*f*m^2 + 8*f*m + (4*f*m^2 + 8*f*m + 3*f)*cos(f*x + e) -
(4*f*m^2 + 8*f*m + 3*f)*sin(f*x + e) + 3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

integrate((-c*sin(f*x + e) + c)^(3/2)*(a*sin(f*x + e) + a)^m, x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

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))^(3/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3/2),x)

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maxima [B]  time = 0.44, size = 193, normalized size = 1.93 \[ -\frac {2 \, {\left (a^{m} c^{\frac {3}{2}} {\left (2 \, m + 5\right )} - \frac {a^{m} c^{\frac {3}{2}} {\left (2 \, m - 3\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {a^{m} c^{\frac {3}{2}} {\left (2 \, m - 3\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{m} c^{\frac {3}{2}} {\left (2 \, m + 5\right )} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + 3\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \]

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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.19, size = 94, normalized size = 0.94 \[ -\frac {c\,{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (10\,\cos \left (e+f\,x\right )-\sin \left (2\,e+2\,f\,x\right )+4\,m\,\cos \left (e+f\,x\right )-2\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (4\,m^2+8\,m+3\right )} \]

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))^(3/2),x)

[Out]

-(c*(a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2)*(10*cos(e + f*x) - sin(2*e + 2*f*x) + 4*m*cos(e + f
*x) - 2*m*sin(2*e + 2*f*x)))/(f*(sin(e + f*x) - 1)*(8*m + 4*m^2 + 3))

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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((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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