3.24 \(\int x^{3/2} \cos (a+b x^2) \, dx\)

Optimal. Leaf size=111 \[ \frac {\sqrt {x} \sin \left (a+b x^2\right )}{2 b}-\frac {i e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{16 b \sqrt [4]{-i b x^2}}+\frac {i e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{16 b \sqrt [4]{i b x^2}} \]

[Out]

-1/16*I*exp(I*a)*GAMMA(1/4,-I*b*x^2)*x^(1/2)/b/(-I*b*x^2)^(1/4)+1/16*I*GAMMA(1/4,I*b*x^2)*x^(1/2)/b/exp(I*a)/(
I*b*x^2)^(1/4)+1/2*sin(b*x^2+a)*x^(1/2)/b

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Rubi [A]  time = 0.08, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3386, 3389, 2218} \[ -\frac {i e^{i a} \sqrt {x} \text {Gamma}\left (\frac {1}{4},-i b x^2\right )}{16 b \sqrt [4]{-i b x^2}}+\frac {i e^{-i a} \sqrt {x} \text {Gamma}\left (\frac {1}{4},i b x^2\right )}{16 b \sqrt [4]{i b x^2}}+\frac {\sqrt {x} \sin \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*Cos[a + b*x^2],x]

[Out]

((-I/16)*E^(I*a)*Sqrt[x]*Gamma[1/4, (-I)*b*x^2])/(b*((-I)*b*x^2)^(1/4)) + ((I/16)*Sqrt[x]*Gamma[1/4, I*b*x^2])
/(b*E^(I*a)*(I*b*x^2)^(1/4)) + (Sqrt[x]*Sin[a + b*x^2])/(2*b)

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 3386

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

Rule 3389

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] - Dist[I/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 x^{3/2} \cos \left (a+b x^2\right ) \, dx &=\frac {\sqrt {x} \sin \left (a+b x^2\right )}{2 b}-\frac {\int \frac {\sin \left (a+b x^2\right )}{\sqrt {x}} \, dx}{4 b}\\ &=\frac {\sqrt {x} \sin \left (a+b x^2\right )}{2 b}-\frac {i \int \frac {e^{-i a-i b x^2}}{\sqrt {x}} \, dx}{8 b}+\frac {i \int \frac {e^{i a+i b x^2}}{\sqrt {x}} \, dx}{8 b}\\ &=-\frac {i e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{16 b \sqrt [4]{-i b x^2}}+\frac {i e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{16 b \sqrt [4]{i b x^2}}+\frac {\sqrt {x} \sin \left (a+b x^2\right )}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 111, normalized size = 1.00 \[ \frac {b x^{9/2} \left (8 \sqrt [4]{b^2 x^4} \sin \left (a+b x^2\right )+\sqrt [4]{i b x^2} (\sin (a)-i \cos (a)) \Gamma \left (\frac {1}{4},-i b x^2\right )+\sqrt [4]{-i b x^2} (\sin (a)+i \cos (a)) \Gamma \left (\frac {1}{4},i b x^2\right )\right )}{16 \left (b^2 x^4\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*Cos[a + b*x^2],x]

[Out]

(b*x^(9/2)*((I*b*x^2)^(1/4)*Gamma[1/4, (-I)*b*x^2]*((-I)*Cos[a] + Sin[a]) + ((-I)*b*x^2)^(1/4)*Gamma[1/4, I*b*
x^2]*(I*Cos[a] + Sin[a]) + 8*(b^2*x^4)^(1/4)*Sin[a + b*x^2]))/(16*(b^2*x^4)^(5/4))

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fricas [A]  time = 0.60, size = 56, normalized size = 0.50 \[ \frac {\left (i \, b\right )^{\frac {3}{4}} e^{\left (-i \, a\right )} \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + \left (-i \, b\right )^{\frac {3}{4}} e^{\left (i \, a\right )} \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right ) + 8 \, b \sqrt {x} \sin \left (b x^{2} + a\right )}{16 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*cos(b*x^2+a),x, algorithm="fricas")

[Out]

1/16*((I*b)^(3/4)*e^(-I*a)*gamma(1/4, I*b*x^2) + (-I*b)^(3/4)*e^(I*a)*gamma(1/4, -I*b*x^2) + 8*b*sqrt(x)*sin(b
*x^2 + a))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \cos \left (b x^{2} + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*cos(b*x^2+a),x, algorithm="giac")

[Out]

integrate(x^(3/2)*cos(b*x^2 + a), x)

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maple [C]  time = 0.08, size = 290, normalized size = 2.61 \[ \frac {2^{\frac {1}{4}} \cos \relax (a ) \sqrt {\pi }\, \left (\frac {2 \sqrt {x}\, 2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {5}{8}} \sin \left (b \,x^{2}\right )}{5 \sqrt {\pi }\, b}+\frac {2 \sqrt {x}\, 2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {5}{8}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{5 \sqrt {\pi }\, b}+\frac {x^{\frac {9}{2}} \left (b^{2}\right )^{\frac {5}{8}} 2^{\frac {3}{4}} b \sin \left (b \,x^{2}\right ) \LommelS 1 \left (\frac {3}{4}, \frac {3}{2}, b \,x^{2}\right )}{10 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}-\frac {2 x^{\frac {9}{2}} \left (b^{2}\right )^{\frac {5}{8}} 2^{\frac {3}{4}} b \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) \LommelS 1 \left (\frac {7}{4}, \frac {1}{2}, b \,x^{2}\right )}{5 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{2 \left (b^{2}\right )^{\frac {5}{8}}}-\frac {2^{\frac {1}{4}} \sin \relax (a ) \sqrt {\pi }\, \left (\frac {2 x^{\frac {5}{2}} 2^{\frac {3}{4}} b^{\frac {5}{4}} \sin \left (b \,x^{2}\right )}{9 \sqrt {\pi }}-\frac {2 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \sin \left (b \,x^{2}\right ) \LommelS 1 \left (\frac {7}{4}, \frac {3}{2}, b \,x^{2}\right )}{9 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}-\frac {x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) \LommelS 1 \left (\frac {3}{4}, \frac {1}{2}, b \,x^{2}\right )}{2 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{2 b^{\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*cos(b*x^2+a),x)

[Out]

1/2*2^(1/4)/(b^2)^(5/8)*cos(a)*Pi^(1/2)*(2/5/Pi^(1/2)*x^(1/2)*2^(3/4)*(b^2)^(5/8)/b*sin(b*x^2)+2/5/Pi^(1/2)*x^
(1/2)*2^(3/4)*(b^2)^(5/8)/b*(cos(b*x^2)*x^2*b-sin(b*x^2))+1/10/Pi^(1/2)*x^(9/2)*(b^2)^(5/8)*2^(3/4)*b/(b*x^2)^
(7/4)*sin(b*x^2)*LommelS1(3/4,3/2,b*x^2)-2/5/Pi^(1/2)*x^(9/2)*(b^2)^(5/8)*2^(3/4)*b/(b*x^2)^(11/4)*(cos(b*x^2)
*x^2*b-sin(b*x^2))*LommelS1(7/4,1/2,b*x^2))-1/2*2^(1/4)/b^(5/4)*sin(a)*Pi^(1/2)*(2/9/Pi^(1/2)*x^(5/2)*2^(3/4)*
b^(5/4)*sin(b*x^2)-2/9/Pi^(1/2)*x^(9/2)*b^(9/4)*2^(3/4)/(b*x^2)^(7/4)*sin(b*x^2)*LommelS1(7/4,3/2,b*x^2)-1/2/P
i^(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(3/4,1/2,b*x^2))

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maxima [B]  time = 1.34, size = 158, normalized size = 1.42 \[ \frac {16 \, \left (b x^{2}\right )^{\frac {1}{4}} \sqrt {x} \sin \left (b x^{2} + a\right ) + {\left ({\left (\sqrt {-\sqrt {2} + 2} {\left (\Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )} + \sqrt {\sqrt {2} + 2} {\left (i \, \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )}\right )} \cos \relax (a) + {\left (\sqrt {\sqrt {2} + 2} {\left (\Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )} + \sqrt {-\sqrt {2} + 2} {\left (-i \, \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + i \, \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right )\right )}\right )} \sin \relax (a)\right )} \sqrt {x}}{32 \, \left (b x^{2}\right )^{\frac {1}{4}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*cos(b*x^2+a),x, algorithm="maxima")

[Out]

1/32*(16*(b*x^2)^(1/4)*sqrt(x)*sin(b*x^2 + a) + ((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)) + sqrt(-sqrt(2) + 2)*(-I*gamma(1/4, I*b*x^2) + I*gamma(1/4, -I*b*x^2)))*s
in(a))*sqrt(x))/((b*x^2)^(1/4)*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{3/2}\,\cos \left (b\,x^2+a\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*cos(a + b*x^2),x)

[Out]

int(x^(3/2)*cos(a + b*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {3}{2}} \cos {\left (a + b x^{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*cos(b*x**2+a),x)

[Out]

Integral(x**(3/2)*cos(a + b*x**2), x)

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