∫ ∘ _________ a−a sin(e+fx) (c+c sin(e+fx))3∕2 ______________________________________________________

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

Optimal. Leaf size=273 \[ -c f \sin (e) \text {Ci}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}+c f \cos (2 e) \text {Ci}(2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-c f \sin (2 e) \text {Si}(2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-c f \cos (e) \text {Si}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{x}-\frac {c \sin (2 e+2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{2 x} \]

[Out]

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

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

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

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

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Rubi [A] time = 0.66, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4604, 6741, 12, 6742, 3297, 3303, 3299, 3302} \[ -c f \sin (e) \text {CosIntegral}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}+c f \cos (2 e) \text {CosIntegral}(2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-c f \sin (2 e) \text {Si}(2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-c f \cos (e) \text {Si}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}-\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{x}-\frac {c \sin (2 e+2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{2 x} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((c*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])/x) + c*f*Cos[2*e]*CosIntegral[2*f*x]*Sec[e + f*x]*Sqrt

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

x]]*Sqrt[c + c*Sin[e + f*x]] - (c*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*Sin[2*e + 2*f

*x])/(2*x) - c*f*Cos[e]*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*SinIntegral[f*x] - c*f*

Sec[e + f*x]*Sin[2*e]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*SinIntegral[2*f*x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{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) (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 {c \cos (e+f x) (1+\sin (e+f x))}{x^2} \, dx\\ &=\left (c \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) (1+\sin (e+f x))}{x^2} \, dx\\ &=\left (c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \left (\frac {\cos (e+f x)}{x^2}+\frac {\sin (2 e+2 f x)}{2 x^2}\right ) \, dx\\ &=\frac {1}{2} \left (c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \frac {\sin (2 e+2 f x)}{x^2} \, dx+\left (c \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 {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x}-\frac {c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \sin (2 e+2 f x)}{2 x}+\left (c f \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \frac {\cos (2 e+2 f x)}{x} \, dx-\left (c 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 {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x}-\frac {c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \sin (2 e+2 f x)}{2 x}-\left (c 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 (c f \cos (2 e) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \frac {\cos (2 f x)}{x} \, dx-\left (c 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-\left (c f \sec (e+f x) \sin (2 e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \frac {\sin (2 f x)}{x} \, dx\\ &=-\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x}+c f \cos (2 e) \text {Ci}(2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}-c 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)}-\frac {c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \sin (2 e+2 f x)}{2 x}-c 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)-c f \sec (e+f x) \sin (2 e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(2 f x)\\ \end {align*}

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Mathematica [C] time = 1.44, size = 231, normalized size = 0.85 \[ \frac {c e^{-i (e+f x)} \sqrt {-i c e^{-i (e+f x)} \left (e^{i (e+f x)}+i\right )^2} \left (-2 i f x e^{i (e+2 f x)} \text {Ei}(-i f x)+2 i f x e^{3 i e+2 i f x} \text {Ei}(i f x)+2 f x e^{2 i (2 e+f x)} \text {Ei}(2 i f x)-2 e^{i (e+f x)}-2 e^{3 i (e+f x)}+i e^{4 i (e+f x)}+2 f x e^{2 i f x} \text {Ei}(-2 i f x)-i\right ) \sqrt {a-a \sin (e+f x)}}{2 \sqrt {2} x \left (1+e^{2 i (e+f x)}\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(c*Sqrt[((-I)*c*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x))]*(-I - 2*E^(I*(e + f*x)) - 2*E^((3*I)*(e + f*x)) + I*

E^((4*I)*(e + f*x)) - (2*I)*E^(I*(e + 2*f*x))*f*x*ExpIntegralEi[(-I)*f*x] + (2*I)*E^((3*I)*e + (2*I)*f*x)*f*x*

ExpIntegralEi[I*f*x] + 2*E^((2*I)*f*x)*f*x*ExpIntegralEi[(-2*I)*f*x] + 2*E^((2*I)*(2*e + f*x))*f*x*ExpIntegral

Ei[(2*I)*f*x])*Sqrt[a - a*Sin[e + f*x]])/(2*Sqrt[2]*E^(I*(e + f*x))*(1 + E^((2*I)*(e + 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((c+c*sin(f*x+e))^(3/2)*(a-a*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((c+c*sin(f*x+e))^(3/2)*(a-a*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/