3.23 \(\int x^{5/2} \cos (a+b x^2) \, dx\)

Optimal. Leaf size=111 \[ -\frac{3 i e^{i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},-i b x^2\right )}{16 b \left (-i b x^2\right )^{3/4}}+\frac{3 i e^{-i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},i b x^2\right )}{16 b \left (i b x^2\right )^{3/4}}+\frac{x^{3/2} \sin \left (a+b x^2\right )}{2 b} \]

[Out]

(((-3*I)/16)*E^(I*a)*x^(3/2)*Gamma[3/4, (-I)*b*x^2])/(b*((-I)*b*x^2)^(3/4)) + (((3*I)/16)*x^(3/2)*Gamma[3/4, I
*b*x^2])/(b*E^(I*a)*(I*b*x^2)^(3/4)) + (x^(3/2)*Sin[a + b*x^2])/(2*b)

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Rubi [A]  time = 0.0811548, 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} \[ -\frac{3 i e^{i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},-i b x^2\right )}{16 b \left (-i b x^2\right )^{3/4}}+\frac{3 i e^{-i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},i b x^2\right )}{16 b \left (i b x^2\right )^{3/4}}+\frac{x^{3/2} \sin \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*Cos[a + b*x^2],x]

[Out]

(((-3*I)/16)*E^(I*a)*x^(3/2)*Gamma[3/4, (-I)*b*x^2])/(b*((-I)*b*x^2)^(3/4)) + (((3*I)/16)*x^(3/2)*Gamma[3/4, I
*b*x^2])/(b*E^(I*a)*(I*b*x^2)^(3/4)) + (x^(3/2)*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^{5/2} \cos \left (a+b x^2\right ) \, dx &=\frac{x^{3/2} \sin \left (a+b x^2\right )}{2 b}-\frac{3 \int \sqrt{x} \sin \left (a+b x^2\right ) \, dx}{4 b}\\ &=\frac{x^{3/2} \sin \left (a+b x^2\right )}{2 b}-\frac{(3 i) \int e^{-i a-i b x^2} \sqrt{x} \, dx}{8 b}+\frac{(3 i) \int e^{i a+i b x^2} \sqrt{x} \, dx}{8 b}\\ &=-\frac{3 i e^{i a} x^{3/2} \Gamma \left (\frac{3}{4},-i b x^2\right )}{16 b \left (-i b x^2\right )^{3/4}}+\frac{3 i e^{-i a} x^{3/2} \Gamma \left (\frac{3}{4},i b x^2\right )}{16 b \left (i b x^2\right )^{3/4}}+\frac{x^{3/2} \sin \left (a+b x^2\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.201149, size = 113, normalized size = 1.02 \[ \frac{b x^{11/2} \left (3 \left (i b x^2\right )^{3/4} (\sin (a)-i \cos (a)) \text{Gamma}\left (\frac{3}{4},-i b x^2\right )+3 \left (-i b x^2\right )^{3/4} (\sin (a)+i \cos (a)) \text{Gamma}\left (\frac{3}{4},i b x^2\right )+8 \left (b^2 x^4\right )^{3/4} \sin \left (a+b x^2\right )\right )}{16 \left (b^2 x^4\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*Cos[a + b*x^2],x]

[Out]

(b*x^(11/2)*(3*(I*b*x^2)^(3/4)*Gamma[3/4, (-I)*b*x^2]*((-I)*Cos[a] + Sin[a]) + 3*((-I)*b*x^2)^(3/4)*Gamma[3/4,
I*b*x^2]*(I*Cos[a] + Sin[a]) + 8*(b^2*x^4)^(3/4)*Sin[a + b*x^2]))/(16*(b^2*x^4)^(7/4))

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Maple [C]  time = 0.142, size = 229, normalized size = 2.1 \begin{align*}{\frac{{2}^{{\frac{3}{4}}}\cos \left ( a \right ) \sqrt{\pi }}{2} \left ({\frac{2\,\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{7\,\sqrt{\pi }b}{x}^{{\frac{3}{2}}} \left ({b}^{2} \right ) ^{{\frac{7}{8}}}}+{\frac{3\,\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{14\,\sqrt{\pi }}{x}^{{\frac{7}{2}}} \left ({b}^{2} \right ) ^{{\frac{7}{8}}}{\it LommelS1} \left ({\frac{5}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{5}{4}}}}+{\frac{3\,\sqrt [4]{2} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{8\,\sqrt{\pi }}{x}^{{\frac{7}{2}}} \left ({b}^{2} \right ) ^{{\frac{7}{8}}}{\it LommelS1} \left ({\frac{1}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{9}{4}}}} \right ) \left ({b}^{2} \right ) ^{-{\frac{7}{8}}}}-{\frac{{2}^{{\frac{3}{4}}}\sin \left ( a \right ) \sqrt{\pi }}{2} \left ( -{\frac{\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{8\,\sqrt{\pi }}{x}^{{\frac{7}{2}}}{b}^{{\frac{7}{4}}}{\it LommelS1} \left ({\frac{1}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{5}{4}}}}-{\frac{\sqrt [4]{2} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{2\,\sqrt{\pi }}{x}^{{\frac{7}{2}}}{b}^{{\frac{7}{4}}}{\it LommelS1} \left ({\frac{5}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{9}{4}}}} \right ){b}^{-{\frac{7}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*cos(b*x^2+a),x)

[Out]

1/2*2^(3/4)/(b^2)^(7/8)*cos(a)*Pi^(1/2)*(2/7/Pi^(1/2)*x^(3/2)*2^(1/4)*(b^2)^(7/8)*sin(b*x^2)/b+3/14/Pi^(1/2)*x
^(7/2)*(b^2)^(7/8)*2^(1/4)/(b*x^2)^(5/4)*sin(b*x^2)*LommelS1(5/4,3/2,b*x^2)+3/8/Pi^(1/2)*x^(7/2)*(b^2)^(7/8)*2
^(1/4)/(b*x^2)^(9/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))*LommelS1(1/4,1/2,b*x^2))-1/2*2^(3/4)/b^(7/4)*sin(a)*Pi^(1/2
)*(-1/8/Pi^(1/2)*x^(7/2)*b^(7/4)*2^(1/4)/(b*x^2)^(5/4)*sin(b*x^2)*LommelS1(1/4,3/2,b*x^2)-1/2/Pi^(1/2)*x^(7/2)
*b^(7/4)*2^(1/4)/(b*x^2)^(9/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))*LommelS1(5/4,1/2,b*x^2))

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Maxima [B]  time = 1.47386, size = 397, normalized size = 3.58 \begin{align*} \frac{16 \, x^{2}{\left | b \right |} \sin \left (b x^{2} + a\right ) - \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}{\left ({\left ({\left (-3 i \, \Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + 3 i \, \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (-3 i \, \Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + 3 i \, \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) - 3 \,{\left (\Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) + 3 \,{\left (\Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) -{\left (3 \,{\left (\Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) + 3 \,{\left (\Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (3 i \, \Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) - 3 i \, \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (-3 i \, \Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + 3 i \, \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )\right )}}{32 \, b \sqrt{x}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*cos(b*x^2+a),x, algorithm="maxima")

[Out]

1/32*(16*x^2*abs(b)*sin(b*x^2 + a) - (x^2*abs(b))^(1/4)*(((-3*I*gamma(3/4, I*b*x^2) + 3*I*gamma(3/4, -I*b*x^2)
)*cos(3/8*pi + 3/4*arctan2(0, b)) + (-3*I*gamma(3/4, I*b*x^2) + 3*I*gamma(3/4, -I*b*x^2))*cos(-3/8*pi + 3/4*ar
ctan2(0, b)) - 3*(gamma(3/4, I*b*x^2) + gamma(3/4, -I*b*x^2))*sin(3/8*pi + 3/4*arctan2(0, b)) + 3*(gamma(3/4,
I*b*x^2) + gamma(3/4, -I*b*x^2))*sin(-3/8*pi + 3/4*arctan2(0, b)))*cos(a) - (3*(gamma(3/4, I*b*x^2) + gamma(3/
4, -I*b*x^2))*cos(3/8*pi + 3/4*arctan2(0, b)) + 3*(gamma(3/4, I*b*x^2) + gamma(3/4, -I*b*x^2))*cos(-3/8*pi + 3
/4*arctan2(0, b)) - (3*I*gamma(3/4, I*b*x^2) - 3*I*gamma(3/4, -I*b*x^2))*sin(3/8*pi + 3/4*arctan2(0, b)) - (-3
*I*gamma(3/4, I*b*x^2) + 3*I*gamma(3/4, -I*b*x^2))*sin(-3/8*pi + 3/4*arctan2(0, b)))*sin(a)))/(b*sqrt(x)*abs(b
))

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Fricas [A]  time = 1.70188, size = 176, normalized size = 1.59 \begin{align*} \frac{8 \, b x^{\frac{3}{2}} \sin \left (b x^{2} + a\right ) + 3 \, \left (i \, b\right )^{\frac{1}{4}} e^{\left (-i \, a\right )} \Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + 3 \, \left (-i \, b\right )^{\frac{1}{4}} e^{\left (i \, a\right )} \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right )}{16 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*cos(b*x^2+a),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*cos(b*x**2+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{5}{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^(5/2)*cos(b*x^2+a),x, algorithm="giac")

[Out]

integrate(x^(5/2)*cos(b*x^2 + a), x)