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

Optimal. Leaf size=81 \[ -\frac {e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{4 \sqrt [4]{-i b x^2}}-\frac {e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{4 \sqrt [4]{i b x^2}} \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3390, 2218} \[ -\frac {e^{i a} \sqrt {x} \text {Gamma}\left (\frac {1}{4},-i b x^2\right )}{4 \sqrt [4]{-i b x^2}}-\frac {e^{-i a} \sqrt {x} \text {Gamma}\left (\frac {1}{4},i b x^2\right )}{4 \sqrt [4]{i b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]/Sqrt[x],x]

[Out]

-(E^(I*a)*Sqrt[x]*Gamma[1/4, (-I)*b*x^2])/(4*((-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 2218

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

Rule 3390

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

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Mathematica [A]  time = 0.07, size = 89, normalized size = 1.10 \[ -\frac {\sqrt {x} \left (\sqrt [4]{-i b x^2} (\cos (a)-i \sin (a)) \Gamma \left (\frac {1}{4},i b x^2\right )+\sqrt [4]{i b x^2} (\cos (a)+i \sin (a)) \Gamma \left (\frac {1}{4},-i b x^2\right )\right )}{4 \sqrt [4]{b^2 x^4}} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.95, size = 44, normalized size = 0.54 \[ \frac {i \, \left (i \, b\right )^{\frac {3}{4}} e^{\left (-i \, a\right )} \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) - i \, \left (-i \, b\right )^{\frac {3}{4}} e^{\left (i \, a\right )} \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x^{2} + a\right )}{\sqrt {x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x^2 + a)/sqrt(x), x)

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maple [C]  time = 0.08, size = 338, normalized size = 4.17 \[ \frac {\cos \relax (a ) \sqrt {\pi }\, 2^{\frac {1}{4}} \left (\frac {6 \,2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {1}{8}} \left (\frac {8 x^{4} b^{2}}{27}+\frac {2}{3}\right ) \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, x^{\frac {3}{2}} b}+\frac {4 \,2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {1}{8}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{\sqrt {\pi }\, x^{\frac {3}{2}} b}-\frac {16 x^{\frac {9}{2}} \left (b^{2}\right )^{\frac {1}{8}} b^{2} 2^{\frac {3}{4}} \sin \left (b \,x^{2}\right ) \LommelS 1 \left (\frac {7}{4}, \frac {3}{2}, b \,x^{2}\right )}{9 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}-\frac {4 x^{\frac {9}{2}} \left (b^{2}\right )^{\frac {1}{8}} b^{2} 2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) \LommelS 1 \left (\frac {3}{4}, \frac {1}{2}, b \,x^{2}\right )}{\sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{4 \left (b^{2}\right )^{\frac {1}{8}}}-\frac {\sin \relax (a ) \sqrt {\pi }\, 2^{\frac {1}{4}} \left (\frac {4 \sqrt {x}\, 2^{\frac {3}{4}} b^{\frac {1}{4}} \sin \left (b \,x^{2}\right )}{5 \sqrt {\pi }}-\frac {16 \sqrt {x}\, 2^{\frac {3}{4}} b^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{5 \sqrt {\pi }}-\frac {4 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \sin \left (b \,x^{2}\right ) \LommelS 1 \left (\frac {3}{4}, \frac {3}{2}, b \,x^{2}\right )}{5 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}+\frac {16 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) \LommelS 1 \left (\frac {7}{4}, \frac {1}{2}, b \,x^{2}\right )}{5 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{4 b^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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maxima [B]  time = 2.84, size = 135, normalized size = 1.67 \[ -\frac {{\left ({\left (\sqrt {\sqrt {2} + 2} {\left (\Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )} - \sqrt {-\sqrt {2} + 2} {\left (i \, \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )}\right )} \cos \relax (a) - {\left (\sqrt {-\sqrt {2} + 2} {\left (\Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )} + \sqrt {\sqrt {2} + 2} {\left (i \, \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )}\right )} \sin \relax (a)\right )} \sqrt {x}}{8 \, \left (b x^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((sqrt(sqrt(2) + 2)*(gamma(1/4, I*b*x^2) + gamma(1/4, -I*b*x^2)) - sqrt(-sqrt(2) + 2)*(I*gamma(1/4, I*b*x
^2) - I*gamma(1/4, -I*b*x^2)))*cos(a) - (sqrt(-sqrt(2) + 2)*(gamma(1/4, I*b*x^2) + gamma(1/4, -I*b*x^2)) + sqr
t(sqrt(2) + 2)*(I*gamma(1/4, I*b*x^2) - I*gamma(1/4, -I*b*x^2)))*sin(a))*sqrt(x)/(b*x^2)^(1/4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (b\,x^2+a\right )}{\sqrt {x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^2)/x^(1/2),x)

[Out]

int(cos(a + b*x^2)/x^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (a + b x^{2} \right )}}{\sqrt {x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x**2)/sqrt(x), x)

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