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

Optimal. Leaf size=118 \[ \frac{2 x \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^2}-\frac{2 \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^3}+\frac{x^2 \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f} \]

[Out]

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

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Rubi [A]  time = 0.176069, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4604, 3296, 2637} \[ \frac{2 x \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^2}-\frac{2 \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f^3}+\frac{x^2 \tan (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.332446, size = 54, normalized size = 0.46 \[ \frac{\left (\left (f^2 x^2-2\right ) \tan (e+f x)+2 f x\right ) \sqrt{a-a \sin (e+f x)} \sqrt{c (\sin (e+f x)+1)}}{f^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*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]]*(2*f*x + (-2 + f^2*x^2)*Tan[e + f*x]))/f^3

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Maple [F]  time = 0.082, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{c+c\sin \left ( fx+e \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a \sin \left (f x + e\right ) + a} \sqrt{c \sin \left (f x + e\right ) + c} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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^2, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{c \left (\sin{\left (e + f x \right )} + 1\right )} \sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a \sin \left (f x + e\right ) + a} \sqrt{c \sin \left (f x + e\right ) + c} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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