\(\int \frac {\cos (a+b x^2)}{\sqrt {x}} \, dx\) [26]

Optimal result

Integrand size = 14, antiderivative size = 81 \[ \int \frac {\cos \left (a+b x^2\right )}{\sqrt {x}} \, dx=-\frac {e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{4 \sqrt [4]{-i b x^2}}-\frac {e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{4 \sqrt [4]{i b x^2}} \]

[Out]

-1/4*exp(I*a)*GAMMA(1/4,-I*b*x^2)*x^(1/2)/(-I*b*x^2)^(1/4)-1/4*GAMMA(1/4,I*b*x^2)*x^(1/2)/exp(I*a)/(I*b*x^2)^(
1/4)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3471, 2250} \[ \int \frac {\cos \left (a+b x^2\right )}{\sqrt {x}} \, dx=-\frac {e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{4 \sqrt [4]{-i b x^2}}-\frac {e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{4 \sqrt [4]{i b x^2}} \]

[In]

Int[Cos[a + b*x^2]/Sqrt[x],x]

[Out]

-1/4*(E^(I*a)*Sqrt[x]*Gamma[1/4, (-I)*b*x^2])/((-I)*b*x^2)^(1/4) - (Sqrt[x]*Gamma[1/4, I*b*x^2])/(4*E^(I*a)*(I
*b*x^2)^(1/4))

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 3471

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^((-c)*I - d*I*x^n),
x], x] + Dist[1/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-i a-i b x^2}}{\sqrt {x}} \, dx+\frac {1}{2} \int \frac {e^{i a+i b x^2}}{\sqrt {x}} \, dx \\ & = -\frac {e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{4 \sqrt [4]{-i b x^2}}-\frac {e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{4 \sqrt [4]{i b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \frac {\cos \left (a+b x^2\right )}{\sqrt {x}} \, dx=-\frac {\sqrt {x} \left (\sqrt [4]{-i b x^2} \Gamma \left (\frac {1}{4},i b x^2\right ) (\cos (a)-i \sin (a))+\sqrt [4]{i b x^2} \Gamma \left (\frac {1}{4},-i b x^2\right ) (\cos (a)+i \sin (a))\right )}{4 \sqrt [4]{b^2 x^4}} \]

[In]

Integrate[Cos[a + b*x^2]/Sqrt[x],x]

[Out]

-1/4*(Sqrt[x]*(((-I)*b*x^2)^(1/4)*Gamma[1/4, I*b*x^2]*(Cos[a] - I*Sin[a]) + (I*b*x^2)^(1/4)*Gamma[1/4, (-I)*b*
x^2]*(Cos[a] + I*Sin[a])))/(b^2*x^4)^(1/4)

Maple [C] (verified)

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

Time = 0.39 (sec) , antiderivative size = 338, normalized size of antiderivative = 4.17

method	result	size
meijerg	\(\frac {\cos \left (a \right ) \sqrt {\pi }\, 2^{\frac {1}{4}} \left (\frac {6 \,2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {1}{8}} \left (\frac {8 x^{4} b^{2}}{27}+\frac {2}{3}\right ) \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, x^{\frac {3}{2}} b}+\frac {4 \,2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {1}{8}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{\sqrt {\pi }\, x^{\frac {3}{2}} b}-\frac {16 x^{\frac {9}{2}} \left (b^{2}\right )^{\frac {1}{8}} b^{2} 2^{\frac {3}{4}} \sin \left (b \,x^{2}\right ) s_{\frac {7}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{9 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}-\frac {4 x^{\frac {9}{2}} \left (b^{2}\right )^{\frac {1}{8}} b^{2} 2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {3}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{\sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{4 \left (b^{2}\right )^{\frac {1}{8}}}-\frac {\sin \left (a \right ) \sqrt {\pi }\, 2^{\frac {1}{4}} \left (\frac {4 \sqrt {x}\, 2^{\frac {3}{4}} b^{\frac {1}{4}} \sin \left (b \,x^{2}\right )}{5 \sqrt {\pi }}-\frac {16 \sqrt {x}\, 2^{\frac {3}{4}} b^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{5 \sqrt {\pi }}-\frac {4 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \sin \left (b \,x^{2}\right ) s_{\frac {3}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{5 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}+\frac {16 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {7}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{5 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{4 b^{\frac {1}{4}}}\)	\(338\)

[In]

int(cos(b*x^2+a)/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4*cos(a)*Pi^(1/2)*2^(1/4)/(b^2)^(1/8)*(6/Pi^(1/2)/x^(3/2)*2^(3/4)*(b^2)^(1/8)*(8/27*x^4*b^2+2/3)/b*sin(b*x^2
)+4/Pi^(1/2)/x^(3/2)*2^(3/4)*(b^2)^(1/8)/b*(cos(b*x^2)*x^2*b-sin(b*x^2))-16/9/Pi^(1/2)*x^(9/2)*(b^2)^(1/8)*b^2
*2^(3/4)/(b*x^2)^(7/4)*sin(b*x^2)*LommelS1(7/4,3/2,b*x^2)-4/Pi^(1/2)*x^(9/2)*(b^2)^(1/8)*b^2*2^(3/4)/(b*x^2)^(
11/4)*(cos(b*x^2)*x^2*b-sin(b*x^2))*LommelS1(3/4,1/2,b*x^2))-1/4*sin(a)*Pi^(1/2)*2^(1/4)/b^(1/4)*(4/5/Pi^(1/2)
*x^(1/2)*2^(3/4)*b^(1/4)*sin(b*x^2)-16/5/Pi^(1/2)*x^(1/2)*2^(3/4)*b^(1/4)*(cos(b*x^2)*x^2*b-sin(b*x^2))-4/5/Pi
^(1/2)*x^(9/2)*b^(9/4)*2^(3/4)/(b*x^2)^(7/4)*sin(b*x^2)*LommelS1(3/4,3/2,b*x^2)+16/5/Pi^(1/2)*x^(9/2)*b^(9/4)*
2^(3/4)/(b*x^2)^(11/4)*(cos(b*x^2)*x^2*b-sin(b*x^2))*LommelS1(7/4,1/2,b*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.59 \[ \int \frac {\cos \left (a+b x^2\right )}{\sqrt {x}} \, dx=\frac {\left (i \, b\right )^{\frac {3}{4}} {\left (i \, \cos \left (a\right ) + \sin \left (a\right )\right )} \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + \left (-i \, b\right )^{\frac {3}{4}} {\left (-i \, \cos \left (a\right ) + \sin \left (a\right )\right )} \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )}{4 \, b} \]

[In]

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

[Out]

1/4*((I*b)^(3/4)*(I*cos(a) + sin(a))*gamma(1/4, I*b*x^2) + (-I*b)^(3/4)*(-I*cos(a) + sin(a))*gamma(1/4, -I*b*x
^2))/b

Sympy [F]

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

[In]

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

[Out]

Integral(cos(a + b*x**2)/sqrt(x), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos \left (a+b x^2\right )}{\sqrt {x}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> Encountered operator mismatch in maxima-to-sr translation

Giac [F]

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

[In]

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

[Out]

integrate(cos(b*x^2 + a)/sqrt(x), x)

Mupad [F(-1)]

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

[In]

int(cos(a + b*x^2)/x^(1/2),x)

[Out]

int(cos(a + b*x^2)/x^(1/2), x)