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

Optimal. Leaf size=140 \[ -\frac {2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 c f} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2841, 2740, 2738} \[ -\frac {2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{5 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 2841

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} (c-c \sin (e+f x))^{3/2} \, dx &=\frac {\int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}+\frac {4 \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx}{5 c}\\ &=-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}+\frac {(2 a) \int \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2} \, dx}{5 c}\\ &=-\frac {2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 82, normalized size = 0.59 \[ -\frac {c (\sin (e+f x)-1) (150 \sin (e+f x)+25 \sin (3 (e+f x))+3 \sin (5 (e+f x))) \sec ^3(e+f x) (a (\sin (e+f x)+1))^{3/2} \sqrt {c-c \sin (e+f x)}}{240 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 73, normalized size = 0.52 \[ \frac {{\left (3 \, a c \cos \left (f x + e\right )^{4} + 4 \, a c \cos \left (f x + e\right )^{2} + 8 \, a c\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/15*(3*a*c*cos(f*x + e)^4 + 4*a*c*cos(f*x + e)^2 + 8*a*c)*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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)sqrt(2*a)*sqrt(2*c)*(-80*a*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(f
*x+exp(1))/(16*f)^2-480*a*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(3*f*x+
3*exp(1))/(96*f)^2-160*a*c*f*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(5*f*x+5
*exp(1))/(160*f)^2)

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maple [A]  time = 0.34, size = 67, normalized size = 0.48 \[ \frac {\left (3 \left (\cos ^{4}\left (f x +e \right )\right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+8\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \sin \left (f x +e \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}}{15 f \cos \left (f x +e \right )^{3}} \]

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))^(3/2),x)

[Out]

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

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

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))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.48, size = 79, normalized size = 0.56 \[ \frac {a\,c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )\right )}{240\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

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))^(3/2),x)

[Out]

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

[Out]

Timed out

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