3.1.12 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx\) [12]

Optimal. Leaf size=92 \[ \frac {c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{4 a f} \]

[Out]

1/6*c*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f/(c-c*sin(f*x+e))^(1/2)+1/4*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*
sin(f*x+e))^(1/2)/a/f

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Rubi [A]
time = 0.26, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2920, 2819, 2817} \begin {gather*} \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 a f}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 a f \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6*a*f*Sqrt[c - c*Sin[e + f*x]]) + (Cos[e + f*x]*(a + a*Sin[e + f*
x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])/(4*a*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])

Rule 2920

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rubi steps

\begin {align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx &=\frac {\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{4 a f}+\frac {\int (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx}{2 a}\\ &=\frac {c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{4 a f}\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 83, normalized size = 0.90 \begin {gather*} \frac {a \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-12 \cos (2 (e+f x))-3 \cos (4 (e+f x))+8 (9 \sin (e+f x)+\sin (3 (e+f x))))}{96 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-12*Cos[2*(e + f*x)] - 3*Cos[4*(e + f*x)]
+ 8*(9*Sin[e + f*x] + Sin[3*(e + f*x)])))/(96*f)

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Maple [A]
time = 0.16, size = 90, normalized size = 0.98

method	result	size
default	\(-\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-3 \left (\cos ^{4}\left (f x +e \right )\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-4 \left (\cos ^{2}\left (f x +e \right )\right )+5 \sin \left (f x +e \right )-5\right )}{12 f \cos \left (f x +e \right )^{3}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/12/f*(-c*(sin(f*x+e)-1))^(1/2)*sin(f*x+e)*(a*(1+sin(f*x+e)))^(3/2)*(-3*cos(f*x+e)^4+cos(f*x+e)^2*sin(f*x+e)
-4*cos(f*x+e)^2+5*sin(f*x+e)-5)/cos(f*x+e)^3

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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(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 81, normalized size = 0.88 \begin {gather*} -\frac {{\left (3 \, a \cos \left (f x + e\right )^{4} - 4 \, {\left (a \cos \left (f x + e\right )^{2} + 2 \, a\right )} \sin \left (f x + e\right ) - 3 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/12*(3*a*cos(f*x + e)^4 - 4*(a*cos(f*x + e)^2 + 2*a)*sin(f*x + e) - 3*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*si
n(f*x + e) + c)/(f*cos(f*x + e))

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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 )^{\frac {3}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \cos ^{2}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(3/2)*sqrt(-c*(sin(e + f*x) - 1))*cos(e + f*x)**2, x)

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Giac [A]
time = 0.54, size = 104, normalized size = 1.13 \begin {gather*} \frac {4 \, {\left (3 \, a \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 ) - 4 \, a \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}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

4/3*(3*a*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*pi + 1/2*f*x + 1/2*
e)) - 4*a*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(sin(-1/4*pi + 1/2*f*x + 1/2
*e)))*sqrt(a)*sqrt(c)/f

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Mupad [B]
time = 9.81, size = 97, normalized size = 1.05 \begin {gather*} -\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )-80\,\sin \left (2\,e+2\,f\,x\right )-8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,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(cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(1/2),x)

[Out]

-(a*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(12*cos(e + f*x) + 15*cos(3*e + 3*f*x) + 3*cos(
5*e + 5*f*x) - 80*sin(2*e + 2*f*x) - 8*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))

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