\(\int \sqrt {x} \cos (a+b x^2) \, dx\) [25]

Optimal result

Integrand size = 14, antiderivative size = 81 \[ \int \sqrt {x} \cos \left (a+b x^2\right ) \, dx=-\frac {e^{i a} x^{3/2} \Gamma \left (\frac {3}{4},-i b x^2\right )}{4 \left (-i b x^2\right )^{3/4}}-\frac {e^{-i a} x^{3/2} \Gamma \left (\frac {3}{4},i b x^2\right )}{4 \left (i b x^2\right )^{3/4}} \]

[Out]

-1/4*exp(I*a)*x^(3/2)*GAMMA(3/4,-I*b*x^2)/(-I*b*x^2)^(3/4)-1/4*x^(3/2)*GAMMA(3/4,I*b*x^2)/exp(I*a)/(I*b*x^2)^(
3/4)

Rubi [A] (verified)

Time = 0.07 (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} \[ \int \sqrt {x} \cos \left (a+b x^2\right ) \, dx=-\frac {e^{i a} x^{3/2} \Gamma \left (\frac {3}{4},-i b x^2\right )}{4 \left (-i b x^2\right )^{3/4}}-\frac {e^{-i a} x^{3/2} \Gamma \left (\frac {3}{4},i b x^2\right )}{4 \left (i b x^2\right )^{3/4}} \]

[In]

Int[Sqrt[x]*Cos[a + b*x^2],x]

[Out]

-1/4*(E^(I*a)*x^(3/2)*Gamma[3/4, (-I)*b*x^2])/((-I)*b*x^2)^(3/4) - (x^(3/2)*Gamma[3/4, I*b*x^2])/(4*E^(I*a)*(I
*b*x^2)^(3/4))

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 e^{-i a-i b x^2} \sqrt {x} \, dx+\frac {1}{2} \int e^{i a+i b x^2} \sqrt {x} \, dx \\ & = -\frac {e^{i a} x^{3/2} \Gamma \left (\frac {3}{4},-i b x^2\right )}{4 \left (-i b x^2\right )^{3/4}}-\frac {e^{-i a} x^{3/2} \Gamma \left (\frac {3}{4},i b x^2\right )}{4 \left (i b x^2\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \sqrt {x} \cos \left (a+b x^2\right ) \, dx=-\frac {x^{3/2} \left (\left (-i b x^2\right )^{3/4} \Gamma \left (\frac {3}{4},i b x^2\right ) (\cos (a)-i \sin (a))+\left (i b x^2\right )^{3/4} \Gamma \left (\frac {3}{4},-i b x^2\right ) (\cos (a)+i \sin (a))\right )}{4 \left (b^2 x^4\right )^{3/4}} \]

[In]

Integrate[Sqrt[x]*Cos[a + b*x^2],x]

[Out]

-1/4*(x^(3/2)*(((-I)*b*x^2)^(3/4)*Gamma[3/4, I*b*x^2]*(Cos[a] - I*Sin[a]) + (I*b*x^2)^(3/4)*Gamma[3/4, (-I)*b*
x^2]*(Cos[a] + I*Sin[a])))/(b^2*x^4)^(3/4)

Maple [C] (verified)

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

Time = 0.43 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.58

method	result	size
meijerg	\(\frac {2^{\frac {3}{4}} \cos \left (a \right ) \sqrt {\pi }\, \left (\frac {4 \,2^{\frac {1}{4}} \left (b^{2}\right )^{\frac {3}{8}} \sin \left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \sqrt {x}\, b}+\frac {4 \,2^{\frac {1}{4}} \left (b^{2}\right )^{\frac {3}{8}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{3 \sqrt {\pi }\, \sqrt {x}\, b}-\frac {x^{\frac {7}{2}} \left (b^{2}\right )^{\frac {3}{8}} 2^{\frac {1}{4}} b \sin \left (b \,x^{2}\right ) s_{\frac {1}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {5}{4}}}-\frac {4 x^{\frac {7}{2}} \left (b^{2}\right )^{\frac {3}{8}} 2^{\frac {1}{4}} b \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {5}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {9}{4}}}\right )}{4 \left (b^{2}\right )^{\frac {3}{8}}}-\frac {2^{\frac {3}{4}} \sin \left (a \right ) \sqrt {\pi }\, \left (\frac {4 x^{\frac {3}{2}} 2^{\frac {1}{4}} b^{\frac {3}{4}} \sin \left (b \,x^{2}\right )}{7 \sqrt {\pi }}-\frac {4 x^{\frac {7}{2}} b^{\frac {7}{4}} 2^{\frac {1}{4}} \sin \left (b \,x^{2}\right ) s_{\frac {5}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{7 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {5}{4}}}-\frac {x^{\frac {7}{2}} b^{\frac {7}{4}} 2^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {1}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{\sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {9}{4}}}\right )}{4 b^{\frac {3}{4}}}\)	\(290\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(x)*cos(a + b*x**2), x)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(x)*cos(b*x^2 + a), x)

Mupad [F(-1)]

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

[In]

int(x^(1/2)*cos(a + b*x^2),x)

[Out]

int(x^(1/2)*cos(a + b*x^2), x)