3.125 \(\int \frac {\sqrt {a+a \sin (c+d x)}}{x} \, dx\)

Optimal. Leaf size=101 \[ \sin \left (\frac {1}{4} (2 c+\pi )\right ) \text {Ci}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}+\cos \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \]

[Out]

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)+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)

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3319, 3303, 3299, 3302} \[ \sin \left (\frac {1}{4} (2 c+\pi )\right ) \text {CosIntegral}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a}+\cos \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[c + d*x]]/x,x]

[Out]

CosIntegral[(d*x)/2]*Csc[c/2 + Pi/4 + (d*x)/2]*Sin[(2*c + Pi)/4]*Sqrt[a + a*Sin[c + d*x]] + Cos[(2*c + Pi)/4]*
Csc[c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]]*SinIntegral[(d*x)/2]

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} \, 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} \, dx\\ &=\left (\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+\left (\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\\ &=\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)}+\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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 83, normalized size = 0.82 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \text {Ci}\left (\frac {d x}{2}\right )+\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[c + d*x]]/x,x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*(CosIntegral[(d*x)/2]*(Cos[c/2] + Sin[c/2]) + (Cos[c/2] - Sin[c/2])*SinIntegral[(d
*x)/2]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])

________________________________________________________________________________________

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,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

________________________________________________________________________________________

giac [C]  time = 2.25, size = 383, normalized size = 3.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(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 - imag_part(co
s_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c)^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 + 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*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*c)^2 + 2*imag_part(c
os_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) - 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*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*real_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*c) + 4*sgn
(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*c) - imag_part(cos_integral(1/2*d*x))*sgn(cos(-
1/4*pi + 1/2*d*x + 1/2*c)) + imag_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - real_part
(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - real_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*p
i + 1/2*d*x + 1/2*c)) - 2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x))*sqrt(a)/(sqrt(2)*tan(1/4*
c)^2 + sqrt(2))

________________________________________________________________________________________

maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +a \sin \left (d x +c \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (d x + c\right ) + a}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sin(d*x + c) + a)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(1/2)/x,x)

[Out]

int((a + a*sin(c + d*x))^(1/2)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________