3.411 \(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=160 \[ \frac{16 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f \left (4 m^2+16 m+15\right )}+\frac{64 c^3 \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+5) \left (4 m^2+8 m+3\right ) \sqrt{c-c \sin (e+f x)}}+\frac{2 c \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{f (2 m+5)} \]

[Out]

(64*c^3*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(5 + 2*m)*(3 + 8*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]) + (16*c^
2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]])/(f*(15 + 16*m + 4*m^2)) + (2*c*Cos[e + f*x]*(a
+ a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(f*(5 + 2*m))

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Rubi [A]  time = 0.251838, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {2740, 2738} \[ \frac{16 c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f \left (4 m^2+16 m+15\right )}+\frac{64 c^3 \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+5) \left (4 m^2+8 m+3\right ) \sqrt{c-c \sin (e+f x)}}+\frac{2 c \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(64*c^3*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(5 + 2*m)*(3 + 8*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]) + (16*c^
2*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]])/(f*(15 + 16*m + 4*m^2)) + (2*c*Cos[e + f*x]*(a
+ a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(f*(5 + 2*m))

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])

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)]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx &=\frac{2 c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}+\frac{(8 c) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx}{5+2 m}\\ &=\frac{16 c^2 \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)}}{f \left (15+16 m+4 m^2\right )}+\frac{2 c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}+\frac{\left (32 c^2\right ) \int (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)} \, dx}{15+16 m+4 m^2}\\ &=\frac{64 c^3 \cos (e+f x) (a+a \sin (e+f x))^m}{f \left (15+46 m+36 m^2+8 m^3\right ) \sqrt{c-c \sin (e+f x)}}+\frac{16 c^2 \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{c-c \sin (e+f x)}}{f \left (15+16 m+4 m^2\right )}+\frac{2 c \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{f (5+2 m)}\\ \end{align*}

Mathematica [A]  time = 2.54694, size = 149, normalized size = 0.93 \[ -\frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a (\sin (e+f x)+1))^m \left (4 \left (4 m^2+16 m+7\right ) \sin (e+f x)+\left (4 m^2+8 m+3\right ) \cos (2 (e+f x))-12 m^2-56 m-89\right )}{f (2 m+1) (2 m+3) (2 m+5) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(5/2),x]

[Out]

-((c^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]]*(-89 - 56*m - 1
2*m^2 + (3 + 8*m + 4*m^2)*Cos[2*(e + f*x)] + 4*(7 + 16*m + 4*m^2)*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2
*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])))

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Maple [F]  time = 4.983, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

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))^(5/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x)

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Maxima [A]  time = 1.98713, size = 392, normalized size = 2.45 \begin{align*} -\frac{2 \,{\left ({\left (4 \, m^{2} + 24 \, m + 43\right )} a^{m} c^{\frac{5}{2}} - \frac{{\left (12 \, m^{2} + 40 \, m - 15\right )} a^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \,{\left (4 \, m^{2} + 8 \, m + 35\right )} a^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{2 \,{\left (4 \, m^{2} + 8 \, m + 35\right )} a^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{{\left (12 \, m^{2} + 40 \, m - 15\right )} a^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{{\left (4 \, m^{2} + 24 \, m + 43\right )} a^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\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 (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} f{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{5}{2}}} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

-2*((4*m^2 + 24*m + 43)*a^m*c^(5/2) - (12*m^2 + 40*m - 15)*a^m*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*(4*
m^2 + 8*m + 35)*a^m*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*(4*m^2 + 8*m + 35)*a^m*c^(5/2)*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 - (12*m^2 + 40*m - 15)*a^m*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + (4*m^2 + 24*
m + 43)*a^m*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*l
og(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^3 + 36*m^2 + 46*m + 15)*f*(sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 + 1)^(5/2))

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Fricas [A]  time = 1.15751, size = 630, normalized size = 3.94 \begin{align*} -\frac{2 \,{\left ({\left (4 \, c^{2} m^{2} + 8 \, c^{2} m + 3 \, c^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (4 \, c^{2} m^{2} + 24 \, c^{2} m + 11 \, c^{2}\right )} \cos \left (f x + e\right )^{2} - 32 \, c^{2} - 2 \,{\left (4 \, c^{2} m^{2} + 16 \, c^{2} m + 23 \, c^{2}\right )} \cos \left (f x + e\right ) +{\left ({\left (4 \, c^{2} m^{2} + 8 \, c^{2} m + 3 \, c^{2}\right )} \cos \left (f x + e\right )^{2} - 32 \, c^{2} + 2 \,{\left (4 \, c^{2} m^{2} + 16 \, c^{2} m + 7 \, c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m +{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) -{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \end{align*}

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))^(5/2),x, algorithm="fricas")

[Out]

-2*((4*c^2*m^2 + 8*c^2*m + 3*c^2)*cos(f*x + e)^3 - (4*c^2*m^2 + 24*c^2*m + 11*c^2)*cos(f*x + e)^2 - 32*c^2 - 2
*(4*c^2*m^2 + 16*c^2*m + 23*c^2)*cos(f*x + e) + ((4*c^2*m^2 + 8*c^2*m + 3*c^2)*cos(f*x + e)^2 - 32*c^2 + 2*(4*
c^2*m^2 + 16*c^2*m + 7*c^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f*
m^3 + 36*f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f*m^3 + 36*f*m^2 + 46*f*m + 1
5*f)*sin(f*x + e) + 15*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((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [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((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out