Integrand size = 8, antiderivative size = 48 \[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-\frac {2 \operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}+\frac {\text {Si}(2 a+2 b x)}{b} \] Output:
(b*x+a)*Ci(b*x+a)^2/b-2*Ci(b*x+a)*sin(b*x+a)/b+Si(2*b*x+2*a)/b
Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2-2 \operatorname {CosIntegral}(a+b x) \sin (a+b x)+\text {Si}(2 (a+b x))}{b} \] Input:
Integrate[CosIntegral[a + b*x]^2,x]
Output:
((a + b*x)*CosIntegral[a + b*x]^2 - 2*CosIntegral[a + b*x]*Sin[a + b*x] + SinIntegral[2*(a + b*x)])/b
Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {7060, 7066, 4906, 27, 3042, 3780}
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.
\(\displaystyle \int \operatorname {CosIntegral}(a+b x)^2 \, dx\) |
\(\Big \downarrow \) 7060 |
\(\displaystyle \frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-2 \int \cos (a+b x) \operatorname {CosIntegral}(a+b x)dx\) |
\(\Big \downarrow \) 7066 |
\(\displaystyle \frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-2 \left (\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\int \frac {\cos (a+b x) \sin (a+b x)}{a+b x}dx\right )\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-2 \left (\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\int \frac {\sin (2 a+2 b x)}{2 (a+b x)}dx\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-2 \left (\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{a+b x}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-2 \left (\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{a+b x}dx\right )\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {(a+b x) \operatorname {CosIntegral}(a+b x)^2}{b}-2 \left (\frac {\operatorname {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\text {Si}(2 a+2 b x)}{2 b}\right )\) |
Input:
Int[CosIntegral[a + b*x]^2,x]
Output:
((a + b*x)*CosIntegral[a + b*x]^2)/b - 2*((CosIntegral[a + b*x]*Sin[a + b* x])/b - SinIntegral[2*a + 2*b*x]/(2*b))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[CosIntegral[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[(a + b*x)*(CosInte gral[a + b*x]^2/b), x] - Simp[2 Int[Cos[a + b*x]*CosIntegral[a + b*x], x] , x] /; FreeQ[{a, b}, x]
Int[Cos[(a_.) + (b_.)*(x_)]*CosIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[Sin[a + b*x]*(CosIntegral[c + d*x]/b), x] - Simp[d/b Int[Sin[a + b*x] *(Cos[c + d*x]/(c + d*x)), x], x] /; FreeQ[{a, b, c, d}, x]
Time = 11.92 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\operatorname {Ci}\left (b x +a \right )^{2} \left (b x +a \right )-2 \,\operatorname {Ci}\left (b x +a \right ) \sin \left (b x +a \right )+\operatorname {Si}\left (2 b x +2 a \right )}{b}\) | \(43\) |
default | \(\frac {\operatorname {Ci}\left (b x +a \right )^{2} \left (b x +a \right )-2 \,\operatorname {Ci}\left (b x +a \right ) \sin \left (b x +a \right )+\operatorname {Si}\left (2 b x +2 a \right )}{b}\) | \(43\) |
Input:
int(Ci(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(Ci(b*x+a)^2*(b*x+a)-2*Ci(b*x+a)*sin(b*x+a)+Si(2*b*x+2*a))
Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.83 \[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\frac {2 \, {\left (\pi b^{2} x + \pi a b\right )} \operatorname {C}\left (b x + a\right )^{2} - 4 \, b \operatorname {C}\left (b x + a\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{2 \, \pi b^{2}} \] Input:
integrate(fresnel_cos(b*x+a)^2,x, algorithm="fricas")
Output:
1/2*(2*(pi*b^2*x + pi*a*b)*fresnel_cos(b*x + a)^2 - 4*b*fresnel_cos(b*x + a)*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) + sqrt(2)*sqrt(b^2)*fresnel _sin(sqrt(2)*sqrt(b^2)*(b*x + a)/b))/(pi*b^2)
\[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\int \operatorname {Ci}^{2}{\left (a + b x \right )}\, dx \] Input:
integrate(Ci(b*x+a)**2,x)
Output:
Integral(Ci(a + b*x)**2, x)
\[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\int { \operatorname {C}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(fresnel_cos(b*x+a)^2,x, algorithm="maxima")
Output:
integrate(fresnel_cos(b*x + a)^2, x)
\[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\int { \operatorname {C}\left (b x + a\right )^{2} \,d x } \] Input:
integrate(fresnel_cos(b*x+a)^2,x, algorithm="giac")
Output:
integrate(fresnel_cos(b*x + a)^2, x)
Timed out. \[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\int {\mathrm {cosint}\left (a+b\,x\right )}^2 \,d x \] Input:
int(cosint(a + b*x)^2,x)
Output:
int(cosint(a + b*x)^2, x)
\[ \int \operatorname {CosIntegral}(a+b x)^2 \, dx=\int \mathit {ci} \left (b x +a \right )^{2}d x \] Input:
int(Ci(b*x+a)^2,x)
Output:
int(ci(a + b*x)**2,x)