Integrand size = 10, antiderivative size = 227 \[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\frac {2 a \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2}-\frac {a^4 \operatorname {FresnelC}(a+b x)}{4 b^4}+\frac {3 \operatorname {FresnelC}(a+b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \operatorname {FresnelC}(a+b x)+\frac {3 a^2 \operatorname {FresnelS}(a+b x)}{2 b^4 \pi }+\frac {a^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {3 a^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }+\frac {a (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }-\frac {(a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi } \]
2*a*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi^2-3/4*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)/b^4/P i^2-1/4*a^4*FresnelC(b*x+a)/b^4+3/4*FresnelC(b*x+a)/b^4/Pi^2+1/4*x^4*Fresn elC(b*x+a)+3/2*a^2*FresnelS(b*x+a)/b^4/Pi+a^3*sin(1/2*Pi*(b*x+a)^2)/b^4/Pi -3/2*a^2*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)/b^4/Pi+a*(b*x+a)^2*sin(1/2*Pi*(b*x+ a)^2)/b^4/Pi-1/4*(b*x+a)^3*sin(1/2*Pi*(b*x+a)^2)/b^4/Pi
Time = 0.22 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.73 \[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\frac {5 a \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-3 b x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+\left (3-a^4 \pi ^2+b^4 \pi ^2 x^4\right ) \operatorname {FresnelC}(a+b x)+6 a^2 \pi \operatorname {FresnelS}(a+b x)+a^3 \pi \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-a^2 b \pi x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+a b^2 \pi x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-b^3 \pi x^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \]
(5*a*Cos[(Pi*(a + b*x)^2)/2] - 3*b*x*Cos[(Pi*(a + b*x)^2)/2] + (3 - a^4*Pi ^2 + b^4*Pi^2*x^4)*FresnelC[a + b*x] + 6*a^2*Pi*FresnelS[a + b*x] + a^3*Pi *Sin[(Pi*(a + b*x)^2)/2] - a^2*b*Pi*x*Sin[(Pi*(a + b*x)^2)/2] + a*b^2*Pi*x ^2*Sin[(Pi*(a + b*x)^2)/2] - b^3*Pi*x^3*Sin[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi ^2)
Time = 0.39 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6983, 3915, 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.
\(\displaystyle \int x^3 \operatorname {FresnelC}(a+b x) \, dx\) |
\(\Big \downarrow \) 6983 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {FresnelC}(a+b x)-\frac {1}{4} b \int x^4 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )dx\) |
\(\Big \downarrow \) 3915 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {FresnelC}(a+b x)-\frac {\int \left (\cos \left (\frac {1}{2} \pi (a+b x)^2\right ) a^4-4 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) a^3+6 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) a^2-4 (a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) a+(a+b x)^4 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )\right )d(a+b x)}{4 b^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {FresnelC}(a+b x)-\frac {a^4 \operatorname {FresnelC}(a+b x)-\frac {4 a^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {6 a^2 \operatorname {FresnelS}(a+b x)}{\pi }+\frac {6 a^2 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {3 \operatorname {FresnelC}(a+b x)}{\pi ^2}-\frac {4 a (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }+\frac {(a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {8 a \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2}+\frac {3 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2}}{4 b^4}\) |
(x^4*FresnelC[a + b*x])/4 - ((-8*a*Cos[(Pi*(a + b*x)^2)/2])/Pi^2 + (3*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/Pi^2 + a^4*FresnelC[a + b*x] - (3*FresnelC[ a + b*x])/Pi^2 - (6*a^2*FresnelS[a + b*x])/Pi - (4*a^3*Sin[(Pi*(a + b*x)^2 )/2])/Pi + (6*a^2*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/Pi - (4*a*(a + b*x)^2 *Sin[(Pi*(a + b*x)^2)/2])/Pi + ((a + b*x)^3*Sin[(Pi*(a + b*x)^2)/2])/Pi)/( 4*b^4)
3.2.34.3.1 Defintions of rubi rules used
Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_ .) + (h_.)*(x_))^(m_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat or[n], 1]}, Simp[k/f^(m + 1) Subst[Int[ExpandIntegrand[(a + b*Cos[c + d*x ^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x ]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
Int[FresnelC[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> S imp[(c + d*x)^(m + 1)*(FresnelC[a + b*x]/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[(c + d*x)^(m + 1)*Cos[(Pi/2)*(a + b*x)^2], x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0]
Time = 0.57 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {FresnelC}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {FresnelC}\left (b x +a \right )}{4}+\frac {a^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {3 a^{2} \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }+\frac {3 a^{2} \operatorname {FresnelS}\left (b x +a \right )}{2 \pi }+\frac {a \left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 a \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {\left (b x +a \right )^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }+\frac {-\frac {3 \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }+\frac {3 \,\operatorname {FresnelC}\left (b x +a \right )}{4 \pi }}{\pi }}{b^{4}}\) | \(187\) |
default | \(\frac {\frac {\operatorname {FresnelC}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {FresnelC}\left (b x +a \right )}{4}+\frac {a^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {3 a^{2} \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }+\frac {3 a^{2} \operatorname {FresnelS}\left (b x +a \right )}{2 \pi }+\frac {a \left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 a \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {\left (b x +a \right )^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }+\frac {-\frac {3 \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }+\frac {3 \,\operatorname {FresnelC}\left (b x +a \right )}{4 \pi }}{\pi }}{b^{4}}\) | \(187\) |
parts | \(\frac {x^{4} \operatorname {FresnelC}\left (b x +a \right )}{4}-\frac {b \left (\frac {x^{3} \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (\frac {x^{2} \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (\frac {x \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (\frac {\sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}-\frac {\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b}-\frac {2 \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b^{2} \pi }\right )}{b}-\frac {3 \left (-\frac {x \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}+\frac {\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \pi }\right )}{4}\) | \(484\) |
1/b^4*(1/4*FresnelC(b*x+a)*b^4*x^4-1/4*a^4*FresnelC(b*x+a)+a^3/Pi*sin(1/2* Pi*(b*x+a)^2)-3/2*a^2/Pi*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)+3/2*a^2/Pi*FresnelS (b*x+a)+a/Pi*(b*x+a)^2*sin(1/2*Pi*(b*x+a)^2)+2*a/Pi^2*cos(1/2*Pi*(b*x+a)^2 )-1/4/Pi*(b*x+a)^3*sin(1/2*Pi*(b*x+a)^2)+3/4/Pi*(-1/Pi*(b*x+a)*cos(1/2*Pi* (b*x+a)^2)+1/Pi*FresnelC(b*x+a)))
Time = 0.26 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.78 \[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\frac {\pi ^{2} b^{5} x^{4} \operatorname {C}\left (b x + a\right ) + 6 \, \pi a^{2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi ^{2} a^{4} - 3\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (3 \, b^{2} x - 5 \, a b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (\pi b^{4} x^{3} - \pi a b^{3} x^{2} + \pi a^{2} b^{2} x - \pi a^{3} b\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{4 \, \pi ^{2} b^{5}} \]
1/4*(pi^2*b^5*x^4*fresnel_cos(b*x + a) + 6*pi*a^2*sqrt(b^2)*fresnel_sin(sq rt(b^2)*(b*x + a)/b) - (pi^2*a^4 - 3)*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b) - (3*b^2*x - 5*a*b)*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - (pi*b^4*x^3 - pi*a*b^3*x^2 + pi*a^2*b^2*x - pi*a^3*b)*sin(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2))/(pi^2*b^5)
\[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\int x^{3} C\left (a + b x\right )\, dx \]
Result contains complex when optimal does not.
Time = 1.10 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.21 \[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\frac {1}{4} \, x^{4} \operatorname {C}\left (b x + a\right ) + \frac {{\left (16 \, {\left (-i \, \pi ^{2} e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + i \, \pi ^{2} e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{4} + 32 \, {\left (\pi \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + \pi \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a^{2} + 16 \, {\left ({\left (-i \, \pi ^{2} e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + i \, \pi ^{2} e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{3} + 2 \, {\left (\pi \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + \pi \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a\right )} b x + {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \pi ^{\frac {5}{2}} {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \pi ^{\frac {5}{2}} {\left (\operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )}\right )} a^{4} + 12 \, {\left (-\left (i + 1\right ) \, \sqrt {2} \pi \Gamma \left (\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \pi \Gamma \left (\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a^{2} + \left (4 i - 4\right ) \, \sqrt {2} \Gamma \left (\frac {5}{2}, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) - \left (4 i + 4\right ) \, \sqrt {2} \Gamma \left (\frac {5}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} \sqrt {2 \, \pi b^{2} x^{2} + 4 \, \pi a b x + 2 \, \pi a^{2}}\right )} b}{32 \, {\left (\pi ^{3} b^{6} x + \pi ^{3} a b^{5}\right )}} \]
1/4*x^4*fresnel_cos(b*x + a) + 1/32*(16*(-I*pi^2*e^(1/2*I*pi*b^2*x^2 + I*p i*a*b*x + 1/2*I*pi*a^2) + I*pi^2*e^(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I *pi*a^2))*a^4 + 32*(pi*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a ^2) + pi*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 + 16 *((-I*pi^2*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + I*pi^2*e^(-1 /2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^3 + 2*(pi*gamma(2, 1/2*I*p i*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi*gamma(2, -1/2*I*pi*b^2*x^2 - I *pi*a*b*x - 1/2*I*pi*a^2))*a)*b*x + (((I - 1)*sqrt(2)*pi^(5/2)*(erf(sqrt(1 /2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2)) - 1) - (I + 1)*sqrt(2)*pi^(5 /2)*(erf(sqrt(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - 1))*a^4 + 12*(-(I + 1)*sqrt(2)*pi*gamma(3/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*p i*a^2) + (I - 1)*sqrt(2)*pi*gamma(3/2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/ 2*I*pi*a^2))*a^2 + (4*I - 4)*sqrt(2)*gamma(5/2, 1/2*I*pi*b^2*x^2 + I*pi*a* b*x + 1/2*I*pi*a^2) - (4*I + 4)*sqrt(2)*gamma(5/2, -1/2*I*pi*b^2*x^2 - I*p i*a*b*x - 1/2*I*pi*a^2))*sqrt(2*pi*b^2*x^2 + 4*pi*a*b*x + 2*pi*a^2))*b/(pi ^3*b^6*x + pi^3*a*b^5)
\[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\int { x^{3} \operatorname {C}\left (b x + a\right ) \,d x } \]
Timed out. \[ \int x^3 \operatorname {FresnelC}(a+b x) \, dx=\int x^3\,\mathrm {FresnelC}\left (a+b\,x\right ) \,d x \]