3.149 \(\int \frac {\sqrt {a+a \cos (c+d x)}}{x^3} \, dx\)

Optimal. Leaf size=151 \[ -\frac {1}{8} d^2 \cos \left (\frac {c}{2}\right ) \text {Ci}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}+\frac {1}{8} d^2 \sin \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}-\frac {\sqrt {a \cos (c+d x)+a}}{2 x^2}+\frac {d \tan \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}}{4 x} \]

[Out]

-1/2*(a+a*cos(d*x+c))^(1/2)/x^2-1/8*d^2*Ci(1/2*d*x)*cos(1/2*c)*sec(1/2*d*x+1/2*c)*(a+a*cos(d*x+c))^(1/2)+1/8*d
^2*sec(1/2*d*x+1/2*c)*Si(1/2*d*x)*sin(1/2*c)*(a+a*cos(d*x+c))^(1/2)+1/4*d*(a+a*cos(d*x+c))^(1/2)*tan(1/2*d*x+1
/2*c)/x

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Rubi [A]  time = 0.16, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {1}{8} d^2 \cos \left (\frac {c}{2}\right ) \text {CosIntegral}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}+\frac {1}{8} d^2 \sin \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}-\frac {\sqrt {a \cos (c+d x)+a}}{2 x^2}+\frac {d \tan \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]/x^3,x]

[Out]

-Sqrt[a + a*Cos[c + d*x]]/(2*x^2) - (d^2*Cos[c/2]*Sqrt[a + a*Cos[c + d*x]]*CosIntegral[(d*x)/2]*Sec[c/2 + (d*x
)/2])/8 + (d^2*Sqrt[a + a*Cos[c + d*x]]*Sec[c/2 + (d*x)/2]*Sin[c/2]*SinIntegral[(d*x)/2])/8 + (d*Sqrt[a + a*Co
s[c + d*x]]*Tan[c/2 + (d*x)/2])/(4*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 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^3} \, 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^3} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{2 x^2}-\frac {1}{4} \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^2} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{2 x^2}+\frac {d \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{4 x}-\frac {1}{8} \left (d^2 \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 {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{2 x^2}+\frac {d \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{4 x}-\frac {1}{8} \left (d^2 \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 {\cos \left (\frac {d x}{2}\right )}{x} \, dx+\frac {1}{8} \left (d^2 \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 {\sin \left (\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+a \cos (c+d x)}}{2 x^2}-\frac {1}{8} d^2 \cos \left (\frac {c}{2}\right ) \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 )+\frac {1}{8} d^2 \sqrt {a+a \cos (c+d x)} \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right )+\frac {d \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{4 x}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 98, normalized size = 0.65 \[ \frac {\sqrt {a (\cos (c+d x)+1)} \left (-d^2 x^2 \cos \left (\frac {c}{2}\right ) \text {Ci}\left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+d^2 x^2 \sin \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+2 d x \tan \left (\frac {1}{2} (c+d x)\right )-4\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]/x^3,x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*(-4 - d^2*x^2*Cos[c/2]*CosIntegral[(d*x)/2]*Sec[(c + d*x)/2] + d^2*x^2*Sec[(c + d*
x)/2]*Sin[c/2]*SinIntegral[(d*x)/2] + 2*d*x*Tan[(c + d*x)/2]))/(8*x^2)

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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^3,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 = 1.23, size = 662, normalized size = 4.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)/x^3,x, algorithm="giac")

[Out]

1/16*sqrt(2)*(d^2*x^2*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^2*x^2*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 + 2*d^2*x^2*i
mag_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^2*x^2*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) + 4*d^2*x^2*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) - d^2*x^2*real_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1
/2*c))*tan(1/4*d*x)^2 - d^2*x^2*real_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 + d
^2*x^2*real_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c)^2 + d^2*x^2*real_part(cos_integra
l(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c)^2 + 2*d^2*x^2*imag_part(cos_integral(1/2*d*x))*sgn(cos(1/2*d
*x + 1/2*c))*tan(1/4*c) - 2*d^2*x^2*imag_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c) + 4
*d^2*x^2*sgn(cos(1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*c) - d^2*x^2*real_part(cos_integral(1/2*d*x))
*sgn(cos(1/2*d*x + 1/2*c)) - d^2*x^2*real_part(cos_integral(-1/2*d*x))*sgn(cos(1/2*d*x + 1/2*c)) - 8*d*x*sgn(c
os(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 8*d*x*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)*tan(1/4*c)^2 - 8
*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 + 8*d*x*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x) + 8*d*x*
sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c) + 8*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 + 32*sgn(cos(1/2*d*x + 1/2*c
))*tan(1/4*d*x)*tan(1/4*c) + 8*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c)^2 - 8*sgn(cos(1/2*d*x + 1/2*c)))*sqrt(a)/(
x^2*tan(1/4*d*x)^2*tan(1/4*c)^2 + x^2*tan(1/4*d*x)^2 + x^2*tan(1/4*c)^2 + x^2)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +a \cos \left (d x +c \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(1/2)/x^3,x)

[Out]

int((a+a*cos(d*x+c))^(1/2)/x^3,x)

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maxima [C]  time = 1.52, size = 232, normalized size = 1.54 \[ -\frac {{\left (4 \, {\left (E_{3}\left (\frac {1}{2} i \, d x\right ) + E_{3}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right )^{3} + 4 \, {\left (E_{3}\left (\frac {1}{2} i \, d x\right ) + E_{3}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right ) \sin \left (\frac {1}{2} \, c\right )^{2} - {\left (4 i \, E_{3}\left (\frac {1}{2} i \, d x\right ) - 4 i \, E_{3}\left (-\frac {1}{2} i \, d x\right )\right )} \sin \left (\frac {1}{2} \, c\right )^{3} + 4 \, {\left (E_{3}\left (\frac {1}{2} i \, d x\right ) + E_{3}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right ) - {\left ({\left (4 i \, E_{3}\left (\frac {1}{2} i \, d x\right ) - 4 i \, E_{3}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right )^{2} + 4 i \, E_{3}\left (\frac {1}{2} i \, d x\right ) - 4 i \, E_{3}\left (-\frac {1}{2} i \, d x\right )\right )} \sin \left (\frac {1}{2} \, c\right )\right )} \sqrt {a} d^{2}}{8 \, {\left ({\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} {\left (d x + c\right )}^{2} - 2 \, {\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} {\left (d x + c\right )} c + {\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^(1/2)/x^3,x)

[Out]

int((a + a*cos(c + d*x))^(1/2)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(1/2)/x**3,x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))/x**3, x)

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