Integrand size = 10, antiderivative size = 184 \[ \int x^3 \text {Si}(a+b x) \, dx=\frac {a \cos (a+b x)}{2 b^4}-\frac {a^3 \cos (a+b x)}{4 b^4}-\frac {3 x \cos (a+b x)}{2 b^3}+\frac {a^2 x \cos (a+b x)}{4 b^3}-\frac {a x^2 \cos (a+b x)}{4 b^2}+\frac {x^3 \cos (a+b x)}{4 b}+\frac {3 \sin (a+b x)}{2 b^4}-\frac {a^2 \sin (a+b x)}{4 b^4}+\frac {a x \sin (a+b x)}{2 b^3}-\frac {3 x^2 \sin (a+b x)}{4 b^2}-\frac {a^4 \text {Si}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {Si}(a+b x) \]
1/2*a*cos(b*x+a)/b^4-1/4*a^3*cos(b*x+a)/b^4-3/2*x*cos(b*x+a)/b^3+1/4*a^2*x *cos(b*x+a)/b^3-1/4*a*x^2*cos(b*x+a)/b^2+1/4*x^3*cos(b*x+a)/b-1/4*a^4*Si(b *x+a)/b^4+1/4*x^4*Si(b*x+a)+3/2*sin(b*x+a)/b^4-1/4*a^2*sin(b*x+a)/b^4+1/2* a*x*sin(b*x+a)/b^3-3/4*x^2*sin(b*x+a)/b^2
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.52 \[ \int x^3 \text {Si}(a+b x) \, dx=\frac {\left (2 a-a^3-6 b x+a^2 b x-a b^2 x^2+b^3 x^3\right ) \cos (a+b x)-\left (-6+a^2-2 a b x+3 b^2 x^2\right ) \sin (a+b x)+\left (-a^4+b^4 x^4\right ) \text {Si}(a+b x)}{4 b^4} \]
((2*a - a^3 - 6*b*x + a^2*b*x - a*b^2*x^2 + b^3*x^3)*Cos[a + b*x] - (-6 + a^2 - 2*a*b*x + 3*b^2*x^2)*Sin[a + b*x] + (-a^4 + b^4*x^4)*SinIntegral[a + b*x])/(4*b^4)
Time = 0.64 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7057, 7293, 2009}
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.
\(\displaystyle \int x^3 \text {Si}(a+b x) \, dx\) |
\(\Big \downarrow \) 7057 |
\(\displaystyle \frac {1}{4} x^4 \text {Si}(a+b x)-\frac {1}{4} b \int \frac {x^4 \sin (a+b x)}{a+b x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} x^4 \text {Si}(a+b x)-\frac {1}{4} b \int \left (\frac {\sin (a+b x) a^4}{b^4 (a+b x)}-\frac {\sin (a+b x) a^3}{b^4}+\frac {x \sin (a+b x) a^2}{b^3}-\frac {x^2 \sin (a+b x) a}{b^2}+\frac {x^3 \sin (a+b x)}{b}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} x^4 \text {Si}(a+b x)-\frac {1}{4} b \left (\frac {a^4 \text {Si}(a+b x)}{b^5}+\frac {a^3 \cos (a+b x)}{b^5}+\frac {a^2 \sin (a+b x)}{b^5}-\frac {a^2 x \cos (a+b x)}{b^4}-\frac {6 \sin (a+b x)}{b^5}-\frac {2 a \cos (a+b x)}{b^5}-\frac {2 a x \sin (a+b x)}{b^4}+\frac {6 x \cos (a+b x)}{b^4}+\frac {3 x^2 \sin (a+b x)}{b^3}+\frac {a x^2 \cos (a+b x)}{b^3}-\frac {x^3 \cos (a+b x)}{b^2}\right )\) |
(x^4*SinIntegral[a + b*x])/4 - (b*((-2*a*Cos[a + b*x])/b^5 + (a^3*Cos[a + b*x])/b^5 + (6*x*Cos[a + b*x])/b^4 - (a^2*x*Cos[a + b*x])/b^4 + (a*x^2*Cos [a + b*x])/b^3 - (x^3*Cos[a + b*x])/b^2 - (6*Sin[a + b*x])/b^5 + (a^2*Sin[ a + b*x])/b^5 - (2*a*x*Sin[a + b*x])/b^4 + (3*x^2*Sin[a + b*x])/b^3 + (a^4 *SinIntegral[a + b*x])/b^5))/4
3.1.18.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] : > Simp[(c + d*x)^(m + 1)*(SinIntegral[a + b*x]/(d*(m + 1))), x] - Simp[b/(d *(m + 1)) Int[(c + d*x)^(m + 1)*(Sin[a + b*x]/(a + b*x)), x], x] /; FreeQ [{a, b, c, d, m}, x] && NeQ[m, -1]
Time = 0.45 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.85
method | result | size |
parts | \(\frac {x^{4} \operatorname {Si}\left (b x +a \right )}{4}-\frac {a^{4} \operatorname {Si}\left (b x +a \right )+4 a^{3} \cos \left (b x +a \right )+6 a^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )-4 a \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )-\left (b x +a \right )^{3} \cos \left (b x +a \right )+3 \left (b x +a \right )^{2} \sin \left (b x +a \right )-6 \sin \left (b x +a \right )+6 \left (b x +a \right ) \cos \left (b x +a \right )}{4 b^{4}}\) | \(156\) |
derivativedivides | \(\frac {\frac {\operatorname {Si}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {Si}\left (b x +a \right )}{4}-a^{3} \cos \left (b x +a \right )-\frac {3 a^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{2}+a \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {\left (b x +a \right )^{3} \cos \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right )^{2} \sin \left (b x +a \right )}{4}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \left (b x +a \right ) \cos \left (b x +a \right )}{2}}{b^{4}}\) | \(157\) |
default | \(\frac {\frac {\operatorname {Si}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {Si}\left (b x +a \right )}{4}-a^{3} \cos \left (b x +a \right )-\frac {3 a^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{2}+a \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )+\frac {\left (b x +a \right )^{3} \cos \left (b x +a \right )}{4}-\frac {3 \left (b x +a \right )^{2} \sin \left (b x +a \right )}{4}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \left (b x +a \right ) \cos \left (b x +a \right )}{2}}{b^{4}}\) | \(157\) |
1/4*x^4*Si(b*x+a)-1/4/b^4*(a^4*Si(b*x+a)+4*a^3*cos(b*x+a)+6*a^2*(sin(b*x+a )-(b*x+a)*cos(b*x+a))-4*a*(-(b*x+a)^2*cos(b*x+a)+2*cos(b*x+a)+2*(b*x+a)*si n(b*x+a))-(b*x+a)^3*cos(b*x+a)+3*(b*x+a)^2*sin(b*x+a)-6*sin(b*x+a)+6*(b*x+ a)*cos(b*x+a))
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.50 \[ \int x^3 \text {Si}(a+b x) \, dx=\frac {{\left (b^{3} x^{3} - a b^{2} x^{2} - a^{3} + {\left (a^{2} - 6\right )} b x + 2 \, a\right )} \cos \left (b x + a\right ) - {\left (3 \, b^{2} x^{2} - 2 \, a b x + a^{2} - 6\right )} \sin \left (b x + a\right ) + {\left (b^{4} x^{4} - a^{4}\right )} \operatorname {Si}\left (b x + a\right )}{4 \, b^{4}} \]
1/4*((b^3*x^3 - a*b^2*x^2 - a^3 + (a^2 - 6)*b*x + 2*a)*cos(b*x + a) - (3*b ^2*x^2 - 2*a*b*x + a^2 - 6)*sin(b*x + a) + (b^4*x^4 - a^4)*sin_integral(b* x + a))/b^4
\[ \int x^3 \text {Si}(a+b x) \, dx=\int x^{3} \operatorname {Si}{\left (a + b x \right )}\, dx \]
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.67 \[ \int x^3 \text {Si}(a+b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {Si}\left (b x + a\right ) - \frac {a^{4} {\left (-i \, {\rm Ei}\left (i \, b x + i \, a\right ) + i \, {\rm Ei}\left (-i \, b x - i \, a\right )\right )} - 2 \, {\left ({\left (b x + a\right )}^{3} - 4 \, {\left (b x + a\right )}^{2} a - 4 \, a^{3} + 6 \, {\left (a^{2} - 1\right )} {\left (b x + a\right )} + 8 \, a\right )} \cos \left (b x + a\right ) + 2 \, {\left (3 \, {\left (b x + a\right )}^{2} - 8 \, {\left (b x + a\right )} a + 6 \, a^{2} - 6\right )} \sin \left (b x + a\right )}{8 \, b^{4}} \]
1/4*x^4*sin_integral(b*x + a) - 1/8*(a^4*(-I*Ei(I*b*x + I*a) + I*Ei(-I*b*x - I*a)) - 2*((b*x + a)^3 - 4*(b*x + a)^2*a - 4*a^3 + 6*(a^2 - 1)*(b*x + a ) + 8*a)*cos(b*x + a) + 2*(3*(b*x + a)^2 - 8*(b*x + a)*a + 6*a^2 - 6)*sin( b*x + a))/b^4
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.28 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.84 \[ \int x^3 \text {Si}(a+b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {Si}\left (b x + a\right ) - \frac {{\left (2 \, b^{3} x^{3} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a b^{2} x^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + a^{4} \Im \left ( \operatorname {Ci}\left (b x + a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - a^{4} \Im \left ( \operatorname {Ci}\left (-b x - a\right ) \right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a^{4} \operatorname {Si}\left (b x + a\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, b^{3} x^{3} + 2 \, a^{2} b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a b^{2} x^{2} + a^{4} \Im \left ( \operatorname {Ci}\left (b x + a\right ) \right ) - a^{4} \Im \left ( \operatorname {Ci}\left (-b x - a\right ) \right ) + 2 \, a^{4} \operatorname {Si}\left (b x + a\right ) + 12 \, b^{2} x^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 2 \, a^{3} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 2 \, a^{2} b x - 8 \, a b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 12 \, b x \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, a^{3} + 4 \, a^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + 4 \, a \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 12 \, b x - 4 \, a - 24 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )} b}{8 \, {\left (b^{5} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b^{5}\right )}} \]
1/4*x^4*sin_integral(b*x + a) - 1/8*(2*b^3*x^3*tan(1/2*b*x + 1/2*a)^2 - 2* a*b^2*x^2*tan(1/2*b*x + 1/2*a)^2 + a^4*imag_part(cos_integral(b*x + a))*ta n(1/2*b*x + 1/2*a)^2 - a^4*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x + 1/2*a)^2 + 2*a^4*sin_integral(b*x + a)*tan(1/2*b*x + 1/2*a)^2 - 2*b^3*x^3 + 2*a^2*b*x*tan(1/2*b*x + 1/2*a)^2 + 2*a*b^2*x^2 + a^4*imag_part(cos_inte gral(b*x + a)) - a^4*imag_part(cos_integral(-b*x - a)) + 2*a^4*sin_integra l(b*x + a) + 12*b^2*x^2*tan(1/2*b*x + 1/2*a) - 2*a^3*tan(1/2*b*x + 1/2*a)^ 2 - 2*a^2*b*x - 8*a*b*x*tan(1/2*b*x + 1/2*a) - 12*b*x*tan(1/2*b*x + 1/2*a) ^2 + 2*a^3 + 4*a^2*tan(1/2*b*x + 1/2*a) + 4*a*tan(1/2*b*x + 1/2*a)^2 + 12* b*x - 4*a - 24*tan(1/2*b*x + 1/2*a))*b/(b^5*tan(1/2*b*x + 1/2*a)^2 + b^5)
Timed out. \[ \int x^3 \text {Si}(a+b x) \, dx=\int x^3\,\mathrm {sinint}\left (a+b\,x\right ) \,d x \]