3.148 \(\int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx\)

Optimal. Leaf size=110 \[ -\frac {1}{2} d \sin \left (\frac {c}{2}\right ) \text {Ci}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}-\frac {1}{2} d \cos \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}-\frac {\sqrt {a \cos (c+d x)+a}}{x} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3319, 3297, 3303, 3299, 3302} \[ -\frac {1}{2} d \sin \left (\frac {c}{2}\right ) \text {CosIntegral}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}-\frac {1}{2} d \cos \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}-\frac {\sqrt {a \cos (c+d x)+a}}{x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]/x^2,x]

[Out]

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

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 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx &=\left (\sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \frac {\sin \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )}{x^2} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{x}-\frac {1}{2} \left (d \sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{x}-\frac {1}{2} \left (d \cos \left (\frac {c}{2}\right ) \sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right )\right ) \int \frac {\sin \left (\frac {d x}{2}\right )}{x} \, dx-\frac {1}{2} \left (d \sqrt {a+a \cos (c+d x)} \csc \left (\frac {1}{2} \left (c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )\right ) \int \frac {\cos \left (\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{x}-\frac {1}{2} d \sqrt {a+a \cos (c+d x)} \text {Ci}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )-\frac {1}{2} d \cos \left (\frac {c}{2}\right ) \sqrt {a+a \cos (c+d x)} \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Si}\left (\frac {d x}{2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 75, normalized size = 0.68 \[ -\frac {\sqrt {a (\cos (c+d x)+1)} \left (d x \sin \left (\frac {c}{2}\right ) \text {Ci}\left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+d x \cos \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+2\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]/x^2,x]

[Out]

-1/2*(Sqrt[a*(1 + Cos[c + d*x])]*(2 + d*x*CosIntegral[(d*x)/2]*Sec[(c + d*x)/2]*Sin[c/2] + d*x*Cos[c/2]*Sec[(c
+ d*x)/2]*SinIntegral[(d*x)/2]))/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*cos(d*x+c))^(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 [C]  time = 1.08, size = 560, normalized size = 5.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)/x^2,x, algorithm="giac")

[Out]

1/4*sqrt(2)*(d*x*imag_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 - d*x*
imag_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 + 2*d*x*sgn(cos(1/2*d*
x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*d*x)^2*tan(1/4*c)^2 - 2*d*x*real_part(cos_integral(1/2*d*x))*sgn(cos
(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 2*d*x*real_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c
))*tan(1/4*d*x)^2*tan(1/4*c) - d*x*imag_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 +
d*x*imag_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 - 2*d*x*sgn(cos(1/2*d*x + 1/2*
c))*sin_integral(1/2*d*x)*tan(1/4*d*x)^2 + d*x*imag_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(
1/4*c)^2 - d*x*imag_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c)^2 + 2*d*x*sgn(cos(1/2*d*
x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*c)^2 - 2*d*x*real_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1/2*
c))*tan(1/4*c) - 2*d*x*real_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c) - 4*sgn(cos(1/2*
d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 - d*x*imag_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c)) + d
*x*imag_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c)) - 2*d*x*sgn(cos(1/2*d*x + 1/2*c))*sin_integral(
1/2*d*x) + 4*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 + 16*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)*tan(1/4*c) +
4*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c)^2 - 4*sgn(cos(1/2*d*x + 1/2*c)))*sqrt(a)/(x*tan(1/4*d*x)^2*tan(1/4*c)^
2 + x*tan(1/4*d*x)^2 + x*tan(1/4*c)^2 + x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +a \cos \left (d x +c \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(1/2)/x^2,x)

[Out]

int((a+a*cos(d*x+c))^(1/2)/x^2,x)

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maxima [C]  time = 0.98, size = 198, normalized size = 1.80 \[ -\frac {{\left (4 \, {\left (E_{2}\left (\frac {1}{2} i \, d x\right ) + E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right )^{3} + 4 \, {\left (E_{2}\left (\frac {1}{2} i \, d x\right ) + E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right ) \sin \left (\frac {1}{2} \, c\right )^{2} - {\left (4 i \, E_{2}\left (\frac {1}{2} i \, d x\right ) - 4 i \, E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \sin \left (\frac {1}{2} \, c\right )^{3} + 4 \, {\left (E_{2}\left (\frac {1}{2} i \, d x\right ) + E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right ) - {\left ({\left (4 i \, E_{2}\left (\frac {1}{2} i \, d x\right ) - 4 i \, E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right )^{2} + 4 i \, E_{2}\left (\frac {1}{2} i \, d x\right ) - 4 i \, E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \sin \left (\frac {1}{2} \, c\right )\right )} \sqrt {a} d}{8 \, {\left ({\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} {\left (d x + c\right )} - {\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)/x^2,x, algorithm="maxima")

[Out]

-1/8*(4*(exp_integral_e(2, 1/2*I*d*x) + exp_integral_e(2, -1/2*I*d*x))*cos(1/2*c)^3 + 4*(exp_integral_e(2, 1/2
*I*d*x) + exp_integral_e(2, -1/2*I*d*x))*cos(1/2*c)*sin(1/2*c)^2 - (4*I*exp_integral_e(2, 1/2*I*d*x) - 4*I*exp
_integral_e(2, -1/2*I*d*x))*sin(1/2*c)^3 + 4*(exp_integral_e(2, 1/2*I*d*x) + exp_integral_e(2, -1/2*I*d*x))*co
s(1/2*c) - ((4*I*exp_integral_e(2, 1/2*I*d*x) - 4*I*exp_integral_e(2, -1/2*I*d*x))*cos(1/2*c)^2 + 4*I*exp_inte
gral_e(2, 1/2*I*d*x) - 4*I*exp_integral_e(2, -1/2*I*d*x))*sin(1/2*c))*sqrt(a)*d/((sqrt(2)*cos(1/2*c)^2 + sqrt(
2)*sin(1/2*c)^2)*(d*x + c) - (sqrt(2)*cos(1/2*c)^2 + sqrt(2)*sin(1/2*c)^2)*c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^(1/2)/x^2,x)

[Out]

int((a + a*cos(c + d*x))^(1/2)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(1/2)/x**2,x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))/x**2, x)

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