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

Optimal. Leaf size=155 \[ -\frac {6 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^4}-\frac {6 x \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^3}+\frac {3 x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}+\frac {x^3 \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f} \]

[Out]

-6*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^4+3*x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2-6
*x*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)*tan(f*x+e)/f^3+x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1
/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.20, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4604, 3296, 2638} \[ \frac {3 x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}-\frac {6 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^4}-\frac {6 x \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^3}+\frac {x^3 \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/f^4 + (3*x^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e
+ f*x]])/f^2 - (6*x*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*Tan[e + f*x])/f^3 + (x^3*Sqrt[a - a*Sin[
e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*Tan[e + f*x])/f

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

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Mathematica [A]  time = 0.51, size = 61, normalized size = 0.39 \[ \frac {\left (f x \left (f^2 x^2-6\right ) \tan (e+f x)+3 f^2 x^2-6\right ) \sqrt {a-a \sin (e+f x)} \sqrt {c (\sin (e+f x)+1)}}{f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(-6 + 3*f^2*x^2 + f*x*(-6 + f^2*x^2)*Tan[e + f*x]))/f^4

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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^3*(a-a*sin(f*x+e))^(1/2)*(c+c*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^3*(a-a*sin(f*x+e))^(1/2)*(c+c*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
)*(1/2*(6*sign(sin(1/2*(f*x+exp(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-3*f^2*x^2*sign(sin(1/2*(f*x+ex
p(1))-1/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(f*x+exp(1))/f^4-1/2*(-6*f*x*sign(sin(1/2*(f*x+exp(1))-1
/4*pi))*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+f^3*x^3*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^4)

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int x^{3} \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {c +c \sin \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.05, size = 111, normalized size = 0.72 \[ -\frac {\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (6\,\cos \left (2\,e+2\,f\,x\right )-3\,f^2\,x^2+6\,f\,x\,\sin \left (2\,e+2\,f\,x\right )-3\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )-f^3\,x^3\,\sin \left (2\,e+2\,f\,x\right )+6\right )}{f^4\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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