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

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

[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)

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Rubi [A]  time = 0.0786466, antiderivative size = 111, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.157765, size = 111, normalized size = 1. \[ \frac{b x^{9/2} \left (\sqrt [4]{i b x^2} (\sin (a)-i \cos (a)) \text{Gamma}\left (\frac{1}{4},-i b x^2\right )+\sqrt [4]{-i b x^2} (\sin (a)+i \cos (a)) \text{Gamma}\left (\frac{1}{4},i b x^2\right )+8 \sqrt [4]{b^2 x^4} \sin \left (a+b x^2\right )\right )}{16 \left (b^2 x^4\right )^{5/4}} \]

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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Maple [C]  time = 0.077, size = 290, normalized size = 2.6 \begin{align*}{\frac{\sqrt [4]{2}\cos \left ( a \right ) \sqrt{\pi }}{2} \left ({\frac{2\,{2}^{3/4}\sin \left ( b{x}^{2} \right ) }{5\,\sqrt{\pi }b}\sqrt{x} \left ({b}^{2} \right ) ^{{\frac{5}{8}}}}+{\frac{2\,{2}^{3/4} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{5\,\sqrt{\pi }b}\sqrt{x} \left ({b}^{2} \right ) ^{{\frac{5}{8}}}}+{\frac{{2}^{{\frac{3}{4}}}b\sin \left ( b{x}^{2} \right ) }{10\,\sqrt{\pi }}{x}^{{\frac{9}{2}}} \left ({b}^{2} \right ) ^{{\frac{5}{8}}}{\it LommelS1} \left ({\frac{3}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{7}{4}}}}-{\frac{2\,{2}^{3/4}b \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{5\,\sqrt{\pi }}{x}^{{\frac{9}{2}}} \left ({b}^{2} \right ) ^{{\frac{5}{8}}}{\it LommelS1} \left ({\frac{7}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{11}{4}}}} \right ) \left ({b}^{2} \right ) ^{-{\frac{5}{8}}}}-{\frac{\sqrt [4]{2}\sin \left ( a \right ) \sqrt{\pi }}{2} \left ({\frac{2\,{2}^{3/4}\sin \left ( b{x}^{2} \right ) }{9\,\sqrt{\pi }}{x}^{{\frac{5}{2}}}{b}^{{\frac{5}{4}}}}-{\frac{2\,{2}^{3/4}\sin \left ( b{x}^{2} \right ) }{9\,\sqrt{\pi }}{x}^{{\frac{9}{2}}}{b}^{{\frac{9}{4}}}{\it LommelS1} \left ({\frac{7}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{7}{4}}}}-{\frac{{2}^{{\frac{3}{4}}} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{2\,\sqrt{\pi }}{x}^{{\frac{9}{2}}}{b}^{{\frac{9}{4}}}{\it LommelS1} \left ({\frac{3}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{11}{4}}}} \right ){b}^{-{\frac{5}{4}}}} \end{align*}

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

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Maxima [B]  time = 1.52483, size = 390, normalized size = 3.51 \begin{align*} \frac{16 \, \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}} \sqrt{x} \sin \left (b x^{2} + a\right ) +{\left ({\left ({\left (i \, \Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) +{\left ({\left (\Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) + i \, \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )\right )} \sqrt{x}}{32 \, \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}} b} \end{align*}

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*(x^2*abs(b))^(1/4)*sqrt(x)*sin(b*x^2 + a) + (((I*gamma(1/4, I*b*x^2) - I*gamma(1/4, -I*b*x^2))*cos(1/
8*pi + 1/4*arctan2(0, b)) + (I*gamma(1/4, I*b*x^2) - I*gamma(1/4, -I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b))
+ (gamma(1/4, I*b*x^2) + gamma(1/4, -I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) - (gamma(1/4, I*b*x^2) + gamma(
1/4, -I*b*x^2))*sin(-1/8*pi + 1/4*arctan2(0, b)))*cos(a) + ((gamma(1/4, I*b*x^2) + gamma(1/4, -I*b*x^2))*cos(1
/8*pi + 1/4*arctan2(0, b)) + (gamma(1/4, I*b*x^2) + gamma(1/4, -I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) + (
-I*gamma(1/4, I*b*x^2) + I*gamma(1/4, -I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) + (I*gamma(1/4, I*b*x^2) - I*
gamma(1/4, -I*b*x^2))*sin(-1/8*pi + 1/4*arctan2(0, b)))*sin(a))*sqrt(x))/((x^2*abs(b))^(1/4)*b)

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Fricas [A]  time = 1.75403, size = 170, normalized size = 1.53 \begin{align*} \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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{3}{2}} \cos{\left (a + b x^{2} \right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{3}{2}} \cos \left (b x^{2} + a\right )\,{d x} \end{align*}

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)