3.127 \(\int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx\)
Optimal. Leaf size=174 \[ -\frac {1}{8} d^2 \sin \left (\frac {1}{4} (2 c+\pi )\right ) \text {Ci}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}-\frac {1}{8} d^2 \cos \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}-\frac {\sqrt {a \sin (c+d x)+a}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}}{4 x} \]
[Out]
-1/2*(a+a*sin(d*x+c))^(1/2)/x^2-1/4*d*cot(1/2*c+1/4*Pi+1/2*d*x)*(a+a*sin(d*x+c))^(1/2)/x-1/8*d^2*cos(1/2*c+1/4
*Pi)*csc(1/2*c+1/4*Pi+1/2*d*x)*Si(1/2*d*x)*(a+a*sin(d*x+c))^(1/2)-1/8*d^2*Ci(1/2*d*x)*csc(1/2*c+1/4*Pi+1/2*d*x
)*sin(1/2*c+1/4*Pi)*(a+a*sin(d*x+c))^(1/2)
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Rubi [A] time = 0.19, antiderivative size = 174, 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} \[ -\frac {1}{8} d^2 \sin \left (\frac {1}{4} (2 c+\pi )\right ) \text {CosIntegral}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}-\frac {1}{8} d^2 \cos \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}-\frac {\sqrt {a \sin (c+d x)+a}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}}{4 x} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + a*Sin[c + d*x]]/x^3,x]
[Out]
-Sqrt[a + a*Sin[c + d*x]]/(2*x^2) - (d*Cot[c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]])/(4*x) - (d^2*CosInt
egral[(d*x)/2]*Csc[c/2 + Pi/4 + (d*x)/2]*Sin[(2*c + Pi)/4]*Sqrt[a + a*Sin[c + d*x]])/8 - (d^2*Cos[(2*c + Pi)/4
]*Csc[c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]]*SinIntegral[(d*x)/2])/8
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 \sin (c+d x)}}{x^3} \, dx &=\left (\csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{x^3} \, dx\\ &=-\frac {\sqrt {a+a \sin (c+d x)}}{2 x^2}+\frac {1}{4} \left (d \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\cos \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{x^2} \, dx\\ &=-\frac {\sqrt {a+a \sin (c+d x)}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{4 x}-\frac {1}{8} \left (d^2 \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\cos \left (\frac {c}{2}-\frac {\pi }{4}+\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+a \sin (c+d x)}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{4 x}-\frac {1}{8} \left (d^2 \cos \left (\frac {1}{4} (2 c+\pi )\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sin \left (\frac {d x}{2}\right )}{x} \, dx-\frac {1}{8} \left (d^2 \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {1}{4} (2 c+\pi )\right ) \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\cos \left (\frac {d x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+a \sin (c+d x)}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{4 x}-\frac {1}{8} d^2 \text {Ci}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {1}{4} (2 c+\pi )\right ) \sqrt {a+a \sin (c+d x)}-\frac {1}{8} d^2 \cos \left (\frac {1}{4} (2 c+\pi )\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)} \text {Si}\left (\frac {d x}{2}\right )\\ \end {align*}
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Mathematica [A] time = 0.36, size = 153, normalized size = 0.88 \[ -\frac {\sqrt {a (\sin (c+d x)+1)} \left (d^2 x^2 \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \text {Ci}\left (\frac {d x}{2}\right )+d^2 x^2 \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \text {Si}\left (\frac {d x}{2}\right )-2 d x \sin \left (\frac {1}{2} (c+d x)\right )+4 \sin \left (\frac {1}{2} (c+d x)\right )+2 d x \cos \left (\frac {1}{2} (c+d x)\right )+4 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 x^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + a*Sin[c + d*x]]/x^3,x]
[Out]
-1/8*(Sqrt[a*(1 + Sin[c + d*x])]*(4*Cos[(c + d*x)/2] + 2*d*x*Cos[(c + d*x)/2] + d^2*x^2*CosIntegral[(d*x)/2]*(
Cos[c/2] + Sin[c/2]) + 4*Sin[(c + d*x)/2] - 2*d*x*Sin[(c + d*x)/2] + d^2*x^2*(Cos[c/2] - Sin[c/2])*SinIntegral
[(d*x)/2]))/(x^2*(Cos[(c + d*x)/2] + Sin[(c + d*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*sin(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 = 2.08, size = 1487, normalized size = 8.55 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))^(1/2)/x^3,x, algorithm="giac")
[Out]
1/16*sqrt(2)*(d^2*x^2*imag_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 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(-1/4*pi + 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(-1/4*pi + 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(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^
2 + 2*d^2*x^2*sgn(cos(-1/4*pi + 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^2*x^
2*imag_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 2*d^2*x^2*i
mag_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 2*d^2*x^2*rea
l_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 2*d^2*x^2*real_p
art(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) + 4*d^2*x^2*sgn(cos(
-1/4*pi + 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*imag_part(cos_integral(1
/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 + d^2*x^2*imag_part(cos_integral(-1/2*d*x))*sgn(co
s(-1/4*pi + 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(-1/4*pi + 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(-1/4*pi + 1/2*d*x + 1/2*c))*t
an(1/4*d*x)^2 - 2*d^2*x^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*d*x)^2 + d^2*x^2*i
mag_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^2 - d^2*x^2*imag_part(cos_integ
ral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^2 + d^2*x^2*real_part(cos_integral(1/2*d*x))*sgn
(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^2 + d^2*x^2*real_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/
2*d*x + 1/2*c))*tan(1/4*c)^2 + 2*d^2*x^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*c)^
2 + 2*d^2*x^2*imag_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 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(-1/4*pi + 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(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) - 2*d^2*x^2*real_part(cos_integral(-1/2*d*x))*sgn(c
os(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) + 4*d^2*x^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x
)*tan(1/4*c) - 4*d*x*sgn(cos(-1/4*pi + 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_i
ntegral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + d^2*x^2*imag_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c)) - d^2*x^2*real_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - d^2*x
^2*real_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 2*d^2*x^2*sgn(cos(-1/4*pi + 1/2*d*x
+ 1/2*c))*sin_integral(1/2*d*x) - 8*d*x*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 8*d*x
*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)*tan(1/4*c)^2 + 4*d*x*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan
(1/4*d*x)^2 + 16*d*x*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)*tan(1/4*c) + 4*d*x*sgn(cos(-1/4*pi + 1/2
*d*x + 1/2*c))*tan(1/4*c)^2 - 8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 + 8*d*x*sgn(co
s(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x) + 8*d*x*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) + 16*sgn(cos
(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) + 16*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)*t
an(1/4*c)^2 - 4*d*x*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2
+ 32*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)*tan(1/4*c) + 8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(
1/4*c)^2 - 16*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x) - 16*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/
4*c) - 8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))*sqrt(a)/(sqrt(2)*x^2*tan(1/4*d*x)^2*tan(1/4*c)^2 + sqrt(2)*x^2*t
an(1/4*d*x)^2 + sqrt(2)*x^2*tan(1/4*c)^2 + sqrt(2)*x^2)
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maple [F] time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +a \sin \left (d x +c \right )}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(d*x+c))^(1/2)/x^3,x)
[Out]
int((a+a*sin(d*x+c))^(1/2)/x^3,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (d x + c\right ) + a}}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))^(1/2)/x^3,x, algorithm="maxima")
[Out]
integrate(sqrt(a*sin(d*x + c) + a)/x^3, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{x^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(c + d*x))^(1/2)/x^3,x)
[Out]
int((a + a*sin(c + d*x))^(1/2)/x^3, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))**(1/2)/x**3,x)
[Out]
Integral(sqrt(a*(sin(c + d*x) + 1))/x**3, x)
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