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

Optimal. Leaf size=168 \[ \frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f^2}+\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}+\frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4604, 4422, 2644} \[ \frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f^2}+\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}+\frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/f^2 - (3*c*x*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[
c + c*Sin[e + f*x]])/(4*f) + (c*Sin[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/(4*f^2) + (x*S
ec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(5/2))/(2*c*f)

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rule 4422

Int[Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol]
:> Simp[((e + f*x)^m*(a + b*Sin[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*d*(n + 1)), Int[(e + f*x
)^(m - 1)*(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 4604

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_
)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^F
racPart[m])/Cos[e + f*x]^(2*FracPart[m]), Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x],
x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ
[2*m] && IGeQ[n - m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.64, size = 73, normalized size = 0.43 \[ \frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c (\sin (e+f x)+1)} (\sin (e+f x)+4 f x \tan (e+f x)-f x \cos (2 (e+f x)) \sec (e+f x)+4)}{4 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(4 - f*x*Cos[2*(e + f*x)]*Sec[e + f*x] + Sin[e + f*x] +
4*f*x*Tan[e + f*x]))/(4*f^2)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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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(x*(c+c*sin(f*x+e))^(3/2)*(a-a*sin(f*x+e))^(1/2),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)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
)*(-1/16*c*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*sin(2*f*x+2*exp(1))/f^2-1/2*c
*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(f*x+exp(1))/f^2-1/2*c*x*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))/f+1/8*c*x*sign(sin(1/2*(f*x+exp(1))
-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(2*f*x+2*exp(1))/f)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int x \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.21, size = 123, normalized size = 0.73 \[ -\frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-16\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+8\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+2\,f\,x\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+2\,f\,x\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )+16\right )}{8\,f^2\,\left (2\,{\sin \left (e+f\,x\right )}^2-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*(-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(sin(e + f*x) + sin(3*e + 3*f*x) - 16*sin(e + f
*x)^2 + 8*f*x*sin(2*e + 2*f*x) + 2*f*x*(2*sin(e/2 + (f*x)/2)^2 - 1) + 2*f*x*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1)
+ 16))/(8*f^2*(2*sin(e + f*x)^2 - 2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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