3.4.59 \(\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx\) [359]

Optimal. Leaf size=134 \[ -\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \]

[Out]

-1/5*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f-2/15*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f
/(a+a*sin(f*x+e))^(1/2)-1/5*a^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2819, 2817} \begin {gather*} -\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)]

Rule 2819

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-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] &&
EqQ[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))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx &=-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}+\frac {1}{5} (4 a) \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}+\frac {1}{5} \left (2 a^2\right ) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx\\ &=-\frac {2 a^3 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 77, normalized size = 0.57 \begin {gather*} \frac {a^2 c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (150 \sin (e+f x)+25 \sin (3 (e+f x))+3 \sin (5 (e+f x)))}{240 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(150*Sin[e + f*x] + 25*Sin[3*(e + f*
x)] + 3*Sin[5*(e + f*x)]))/(240*f)

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Maple [A]
time = 16.94, size = 67, normalized size = 0.50

method	result	size
default	\(\frac {\left (3 \left (\cos ^{4}\left (f x +e \right )\right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+8\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{15 f \cos \left (f x +e \right )^{5}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/f*(3*cos(f*x+e)^4+4*cos(f*x+e)^2+8)*(-c*(sin(f*x+e)-1))^(5/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(5/2)/cos(f*x
+e)^5

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 91, normalized size = 0.68 \begin {gather*} \frac {{\left (3 \, a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} c^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/15*(3*a^2*c^2*cos(f*x + e)^4 + 4*a^2*c^2*cos(f*x + e)^2 + 8*a^2*c^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*
x + e) + c)*sin(f*x + e)/(f*cos(f*x + e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.51, size = 165, normalized size = 1.23 \begin {gather*} -\frac {16 \, {\left (6 \, a^{2} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 15 \, a^{2} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 10 \, a^{2} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*(6*a^2*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*
f*x + 1/2*e)) - 15*a^2*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*p
i + 1/2*f*x + 1/2*e)) + 10*a^2*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(si
n(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)*sqrt(c)/f

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Mupad [B]
time = 1.50, size = 83, normalized size = 0.62 \begin {gather*} \frac {a^2\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )\right )}{240\,f\,\left (\cos \left (2\,e+2\,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))^(5/2)*(c - c*sin(e + f*x))^(5/2),x)

[Out]

(a^2*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(175*sin(2*e + 2*f*x) + 28*sin(4*e + 4*f*x
) + 3*sin(6*e + 6*f*x)))/(240*f*(cos(2*e + 2*f*x) + 1))

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