\(\int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx\) [148]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=-\frac {\sqrt {a+a \cos (c+d x)}}{x}-\frac {1}{2} d \sqrt {a+a \cos (c+d x)} \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {c}{2}\right )-\frac {1}{2} d \cos \left (\frac {c}{2}\right ) \sqrt {a+a \cos (c+d x)} \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Si}\left (\frac {d x}{2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=-\frac {\sqrt {a (1+\cos (c+d x))} \left (2+d x \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}\right )+d x \cos \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \text {Si}\left (\frac {d x}{2}\right )\right )}{2 x} \] Input:

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

Output:

-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
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.74, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3800, 3042, 3778, 25, 3042, 3784, 3042, 3780, 3783}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}{x^2}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \int \frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \int \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{2}\right )}{x^2}dx\)

\(\Big \downarrow \) 3778

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {1}{2} d \int -\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}dx-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (-\frac {1}{2} d \int \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}dx-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (-\frac {1}{2} d \int \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}dx-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

\(\Big \downarrow \) 3784

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (-\frac {1}{2} d \left (\sin \left (\frac {c}{2}\right ) \int \frac {\cos \left (\frac {d x}{2}\right )}{x}dx+\cos \left (\frac {c}{2}\right ) \int \frac {\sin \left (\frac {d x}{2}\right )}{x}dx\right )-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (-\frac {1}{2} d \left (\sin \left (\frac {c}{2}\right ) \int \frac {\sin \left (\frac {d x}{2}+\frac {\pi }{2}\right )}{x}dx+\cos \left (\frac {c}{2}\right ) \int \frac {\sin \left (\frac {d x}{2}\right )}{x}dx\right )-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

\(\Big \downarrow \) 3780

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (-\frac {1}{2} d \left (\sin \left (\frac {c}{2}\right ) \int \frac {\sin \left (\frac {d x}{2}+\frac {\pi }{2}\right )}{x}dx+\cos \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right )\right )-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

\(\Big \downarrow \) 3783

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (-\frac {1}{2} d \left (\sin \left (\frac {c}{2}\right ) \operatorname {CosIntegral}\left (\frac {d x}{2}\right )+\cos \left (\frac {c}{2}\right ) \text {Si}\left (\frac {d x}{2}\right )\right )-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{x}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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] - Simp[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 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(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])
 

Maple [F]

\[\int \frac {\sqrt {a +a \cos \left (d x +c \right )}}{x^{2}}d x\]

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

\[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=\int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}{x^{2}}\, dx \] Input:

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=-\frac {{\left ({\left (E_{2}\left (\frac {1}{2} i \, d x\right ) + E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right )^{3} + {\left (E_{2}\left (\frac {1}{2} i \, d x\right ) + E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right ) \sin \left (\frac {1}{2} \, c\right )^{2} + {\left (-i \, E_{2}\left (\frac {1}{2} i \, d x\right ) + i \, E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \sin \left (\frac {1}{2} \, c\right )^{3} + {\left (E_{2}\left (\frac {1}{2} i \, d x\right ) + E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right ) + {\left ({\left (-i \, E_{2}\left (\frac {1}{2} i \, d x\right ) + i \, E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \cos \left (\frac {1}{2} \, c\right )^{2} - i \, E_{2}\left (\frac {1}{2} i \, d x\right ) + i \, E_{2}\left (-\frac {1}{2} i \, d x\right )\right )} \sin \left (\frac {1}{2} \, c\right )\right )} \sqrt {a} d}{2 \, {\left ({\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} {\left (d x + c\right )} - {\left (\sqrt {2} \cos \left (\frac {1}{2} \, c\right )^{2} + \sqrt {2} \sin \left (\frac {1}{2} \, c\right )^{2}\right )} c\right )}} \] Input:

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

Output:

-1/2*((exp_integral_e(2, 1/2*I*d*x) + exp_integral_e(2, -1/2*I*d*x))*cos(1 
/2*c)^3 + (exp_integral_e(2, 1/2*I*d*x) + exp_integral_e(2, -1/2*I*d*x))*c 
os(1/2*c)*sin(1/2*c)^2 + (-I*exp_integral_e(2, 1/2*I*d*x) + I*exp_integral 
_e(2, -1/2*I*d*x))*sin(1/2*c)^3 + (exp_integral_e(2, 1/2*I*d*x) + exp_inte 
gral_e(2, -1/2*I*d*x))*cos(1/2*c) + ((-I*exp_integral_e(2, 1/2*I*d*x) + I* 
exp_integral_e(2, -1/2*I*d*x))*cos(1/2*c)^2 - I*exp_integral_e(2, 1/2*I*d* 
x) + 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)
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.38 (sec) , antiderivative size = 560, normalized size of antiderivative = 5.09 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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_p 
art(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))*t 
an(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))*si 
n_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))*t 
an(1/4*d*x)*tan(1/4*c) + 4*sgn(cos(1/2*d*x + 1/2*c))*tan(1/4*c)^2 - 4*s...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=\int \frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}}{x^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+a \cos (c+d x)}}{x^2} \, dx=\frac {\sqrt {a}\, \left (-2 \sqrt {\cos \left (d x +c \right )+1}-\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sin \left (d x +c \right )}{\cos \left (d x +c \right ) x +x}d x \right ) d x \right )}{2 x} \] Input:

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

Output:

(sqrt(a)*( - 2*sqrt(cos(c + d*x) + 1) - int((sqrt(cos(c + d*x) + 1)*sin(c 
+ d*x))/(cos(c + d*x)*x + x),x)*d*x))/(2*x)