3.168 \(\int \frac{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}{x} \, dx\)

Optimal. Leaf size=86 \[ \cos (e) \text{CosIntegral}(f x) \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}-\sin (e) \text{Si}(f x) \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c} \]

[Out]

Cos[e]*CosIntegral[f*x]*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]] - Sec[e + f*x]*Sin[e]*S
qrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*SinIntegral[f*x]

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Rubi [A]  time = 0.183241, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4604, 3303, 3299, 3302} \[ \cos (e) \text{CosIntegral}(f x) \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}-\sin (e) \text{Si}(f x) \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[e]*CosIntegral[f*x]*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]] - Sec[e + f*x]*Sin[e]*S
qrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*SinIntegral[f*x]

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 3303

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.222457, size = 52, normalized size = 0.6 \[ \sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{c (\sin (e+f x)+1)} (\cos (e) \text{CosIntegral}(f x)-\sin (e) \text{Si}(f x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sec[e + f*x]*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(Cos[e]*CosIntegral[f*x] - Sin[e]*SinIntegral
[f*x])

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Maple [F]  time = 0.083, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\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((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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