\(\int x \cos ((a+b x)^2) \, dx\) [86]

Optimal result

Integrand size = 10, antiderivative size = 47 \[ \int x \cos \left ((a+b x)^2\right ) \, dx=-\frac {a \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} (a+b x)\right )}{b^2}+\frac {\sin \left ((a+b x)^2\right )}{2 b^2} \]

[Out]

1/2*sin((b*x+a)^2)/b^2-1/2*a*FresnelC((b*x+a)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/b^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3515, 3433, 3461, 2717} \[ \int x \cos \left ((a+b x)^2\right ) \, dx=\frac {\sin \left ((a+b x)^2\right )}{2 b^2}-\frac {\sqrt {\frac {\pi }{2}} a \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} (a+b x)\right )}{b^2} \]

[In]

Int[x*Cos[(a + b*x)^2],x]

[Out]

-((a*Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*(a + b*x)])/b^2) + Sin[(a + b*x)^2]/(2*b^2)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3515

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (-a \cos \left (x^2\right )+x \cos \left (x^2\right )\right ) \, dx,x,a+b x\right )}{b^2} \\ & = \frac {\text {Subst}\left (\int x \cos \left (x^2\right ) \, dx,x,a+b x\right )}{b^2}-\frac {a \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,a+b x\right )}{b^2} \\ & = -\frac {a \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} (a+b x)\right )}{b^2}+\frac {\text {Subst}\left (\int \cos (x) \, dx,x,(a+b x)^2\right )}{2 b^2} \\ & = -\frac {a \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} (a+b x)\right )}{b^2}+\frac {\sin \left ((a+b x)^2\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int x \cos \left ((a+b x)^2\right ) \, dx=\frac {-a \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} (a+b x)\right )+\sin \left ((a+b x)^2\right )}{2 b^2} \]

[In]

Integrate[x*Cos[(a + b*x)^2],x]

[Out]

(-(a*Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*(a + b*x)]) + Sin[(a + b*x)^2])/(2*b^2)

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.34

method	result	size
default	\(\frac {\sin \left (x^{2} b^{2}+2 a b x +a^{2}\right )}{2 b^{2}}-\frac {a \sqrt {2}\, \sqrt {\pi }\, \operatorname {C}\left (\frac {\sqrt {2}\, \left (b^{2} x +a b \right )}{\sqrt {\pi }\, \sqrt {b^{2}}}\right )}{2 b \sqrt {b^{2}}}\)	\(63\)
risch	\(\frac {\left (-1\right )^{\frac {3}{4}} a \sqrt {\pi }\, \operatorname {erf}\left (b \left (-1\right )^{\frac {1}{4}} x +\left (-1\right )^{\frac {1}{4}} a \right )}{4 b^{2}}+\frac {a \sqrt {\pi }\, \operatorname {erf}\left (-b \sqrt {-i}\, x +\frac {i a}{\sqrt {-i}}\right )}{4 b^{2} \sqrt {-i}}+\frac {\sin \left (\left (b x +a \right )^{2}\right )}{2 b^{2}}\)	\(71\)
parts	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {C}\left (\frac {\sqrt {2}\, \left (b^{2} x +a b \right )}{\sqrt {\pi }\, \sqrt {b^{2}}}\right ) x}{2 \sqrt {b^{2}}}-\frac {\pi \left (\operatorname {C}\left (\frac {\sqrt {2}\, b^{2} x}{\sqrt {\pi }\, \sqrt {b^{2}}}+\frac {\sqrt {2}\, a b}{\sqrt {\pi }\, \sqrt {b^{2}}}\right ) \left (\frac {\sqrt {2}\, b^{2} x}{\sqrt {\pi }\, \sqrt {b^{2}}}+\frac {\sqrt {2}\, a b}{\sqrt {\pi }\, \sqrt {b^{2}}}\right )-\frac {\sin \left (\frac {\pi \left (\frac {\sqrt {2}\, b^{2} x}{\sqrt {\pi }\, \sqrt {b^{2}}}+\frac {\sqrt {2}\, a b}{\sqrt {\pi }\, \sqrt {b^{2}}}\right )^{2}}{2}\right )}{\pi }\right )}{2 b^{2}}\)	\(151\)

[In]

int(x*cos((b*x+a)^2),x,method=_RETURNVERBOSE)

[Out]

1/2/b^2*sin(b^2*x^2+2*a*b*x+a^2)-1/2*a/b*2^(1/2)*Pi^(1/2)/(b^2)^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)/(b^2)^(1/2)*(b
^2*x+a*b))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.34 \[ \int x \cos \left ((a+b x)^2\right ) \, dx=-\frac {\sqrt {2} \pi a \sqrt {\frac {b^{2}}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (b x + a\right )} \sqrt {\frac {b^{2}}{\pi }}}{b}\right ) - b \sin \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{2 \, b^{3}} \]

[In]

integrate(x*cos((b*x+a)^2),x, algorithm="fricas")

[Out]

-1/2*(sqrt(2)*pi*a*sqrt(b^2/pi)*fresnel_cos(sqrt(2)*(b*x + a)*sqrt(b^2/pi)/b) - b*sin(b^2*x^2 + 2*a*b*x + a^2)
)/b^3

Sympy [F]

\[ \int x \cos \left ((a+b x)^2\right ) \, dx=\int x \cos {\left (a^{2} + 2 a b x + b^{2} x^{2} \right )}\, dx \]

[In]

integrate(x*cos((b*x+a)**2),x)

[Out]

Integral(x*cos(a**2 + 2*a*b*x + b**2*x**2), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 199, normalized size of antiderivative = 4.23 \[ \int x \cos \left ((a+b x)^2\right ) \, dx=\frac {2 \, b x {\left (-i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )} + i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}\right )} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (-\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}}\right ) - 1\right )} + \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}}\right ) - 1\right )}\right )} a + 2 \, a {\left (-i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )} + i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}\right )}}{8 \, {\left (b^{3} x + a b^{2}\right )}} \]

[In]

integrate(x*cos((b*x+a)^2),x, algorithm="maxima")

[Out]

1/8*(2*b*x*(-I*e^(I*b^2*x^2 + 2*I*a*b*x + I*a^2) + I*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)) - sqrt(b^2*x^2 + 2*a*
b*x + a^2)*(-(I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b^2*x^2 + 2*I*a*b*x + I*a^2)) - 1) + (I + 1)*sqrt(2)*sqrt(pi
)*(erf(sqrt(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)) - 1))*a + 2*a*(-I*e^(I*b^2*x^2 + 2*I*a*b*x + I*a^2) + I*e^(-I*b^2
*x^2 - 2*I*a*b*x - I*a^2)))/(b^3*x + a*b^2)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.53 \[ \int x \cos \left ((a+b x)^2\right ) \, dx=-\frac {-\frac {\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } a \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} {\left (x + \frac {a}{b}\right )} {\left | b \right |}\right )}{{\left | b \right |}} + \frac {2 i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )}}{b}}{8 \, b} - \frac {\frac {\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } a \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} {\left (x + \frac {a}{b}\right )} {\left | b \right |}\right )}{{\left | b \right |}} - \frac {2 i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}}{b}}{8 \, b} \]

[In]

integrate(x*cos((b*x+a)^2),x, algorithm="giac")

[Out]

-1/8*(-(I + 1)*sqrt(2)*sqrt(pi)*a*erf((1/2*I - 1/2)*sqrt(2)*(x + a/b)*abs(b))/abs(b) + 2*I*e^(I*b^2*x^2 + 2*I*
a*b*x + I*a^2)/b)/b - 1/8*((I - 1)*sqrt(2)*sqrt(pi)*a*erf(-(1/2*I + 1/2)*sqrt(2)*(x + a/b)*abs(b))/abs(b) - 2*
I*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)/b)/b

Mupad [B] (verification not implemented)

Time = 13.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int x \cos \left ((a+b x)^2\right ) \, dx=\frac {\sin \left ({\left (a+b\,x\right )}^2\right )}{2\,b^2}-\frac {\sqrt {2}\,a\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\left (a+b\,x\right )}{\sqrt {\pi }}\right )}{2\,b^2} \]

[In]

int(x*cos((a + b*x)^2),x)

[Out]

sin((a + b*x)^2)/(2*b^2) - (2^(1/2)*a*pi^(1/2)*fresnelc((2^(1/2)*(a + b*x))/pi^(1/2)))/(2*b^2)