∫ ∘ _________ a−a sin(e+fx) ∘ _________ c+c sin(e+fx)

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

Optimal. Leaf size=123 \[ -f \sin (e) \text {Ci}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-f \cos (e) \text {Si}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-\frac {\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{x} \]

[Out]

-(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/x-f*cos(e)*sec(f*x+e)*Si(f*x)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f

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

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Rubi [A] time = 0.20, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4604, 3297, 3303, 3299, 3302} \[ -f \sin (e) \text {CosIntegral}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-f \cos (e) \text {Si}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-\frac {\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{x} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/x) - f*CosIntegral[f*x]*Sec[e + f*x]*Sin[e]*Sqrt[a - a*S

in[e + f*x]]*Sqrt[c + c*Sin[e + f*x]] - f*Cos[e]*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]

]*SinIntegral[f*x]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(

m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[

m, -1]

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]

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

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Mathematica [A] time = 0.24, size = 65, normalized size = 0.53 \[ -\frac {\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c (\sin (e+f x)+1)} (f x \sin (e) \text {Ci}(f x)+f x \cos (e) \text {Si}(f x)+\cos (e+f x))}{x} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

] + f*x*Cos[e]*SinIntegral[f*x]))/x)

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/