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

Optimal. Leaf size=132 \[ -\frac{i e^{2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{64 \sqrt [4]{2} b \sqrt [4]{-i b x^2}}+\frac{i e^{-2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{64 \sqrt [4]{2} b \sqrt [4]{i b x^2}}+\frac{\sqrt{x} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}+\frac{x^{5/2}}{5} \]

[Out]

x^(5/2)/5 - ((I/64)*E^((2*I)*a)*Sqrt[x]*Gamma[1/4, (-2*I)*b*x^2])/(2^(1/4)*b*((-I)*b*x^2)^(1/4)) + ((I/64)*Sqr
t[x]*Gamma[1/4, (2*I)*b*x^2])/(2^(1/4)*b*E^((2*I)*a)*(I*b*x^2)^(1/4)) + (Sqrt[x]*Sin[2*(a + b*x^2)])/(8*b)

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Rubi [A]  time = 0.123891, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3402, 3404, 3386, 3355, 2208} \[ -\frac{i e^{2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{64 \sqrt [4]{2} b \sqrt [4]{-i b x^2}}+\frac{i e^{-2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{64 \sqrt [4]{2} b \sqrt [4]{i b x^2}}+\frac{\sqrt{x} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}+\frac{x^{5/2}}{5} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*Cos[a + b*x^2]^2,x]

[Out]

x^(5/2)/5 - ((I/64)*E^((2*I)*a)*Sqrt[x]*Gamma[1/4, (-2*I)*b*x^2])/(2^(1/4)*b*((-I)*b*x^2)^(1/4)) + ((I/64)*Sqr
t[x]*Gamma[1/4, (2*I)*b*x^2])/(2^(1/4)*b*E^((2*I)*a)*(I*b*x^2)^(1/4)) + (Sqrt[x]*Sin[2*(a + b*x^2)])/(8*b)

Rule 3402

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_.)*(x_))^(m_), x_Symbol] :> With[{k = Denominator[m
]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*Cos[c + (d*x^(k*n))/e^n])^p, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e}, x] && IntegerQ[p] && IGtQ[n, 0] && FractionQ[m]

Rule 3404

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 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 3355

Int[Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[I/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] - Dist[I/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f}, x] && IGtQ[n, 2]

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rubi steps

\begin{align*} \int x^{3/2} \cos ^2\left (a+b x^2\right ) \, dx &=2 \operatorname{Subst}\left (\int x^4 \cos ^2\left (a+b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{x^4}{2}+\frac{1}{2} x^4 \cos \left (2 a+2 b x^4\right )\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{5/2}}{5}+\operatorname{Subst}\left (\int x^4 \cos \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{5/2}}{5}+\frac{\sqrt{x} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}-\frac{\operatorname{Subst}\left (\int \sin \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )}{8 b}\\ &=\frac{x^{5/2}}{5}+\frac{\sqrt{x} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}-\frac{i \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} \, dx,x,\sqrt{x}\right )}{16 b}+\frac{i \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} \, dx,x,\sqrt{x}\right )}{16 b}\\ &=\frac{x^{5/2}}{5}-\frac{i e^{2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},-2 i b x^2\right )}{64 \sqrt [4]{2} b \sqrt [4]{-i b x^2}}+\frac{i e^{-2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},2 i b x^2\right )}{64 \sqrt [4]{2} b \sqrt [4]{i b x^2}}+\frac{\sqrt{x} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.344793, size = 142, normalized size = 1.08 \[ \frac{b x^{9/2} \left (5\ 2^{3/4} \sqrt [4]{i b x^2} (\sin (2 a)-i \cos (2 a)) \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )+5\ 2^{3/4} \sqrt [4]{-i b x^2} (\sin (2 a)+i \cos (2 a)) \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )+16 \sqrt [4]{b^2 x^4} \left (5 \sin \left (2 \left (a+b x^2\right )\right )+8 b x^2\right )\right )}{640 \left (b^2 x^4\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*Cos[a + b*x^2]^2,x]

[Out]

(b*x^(9/2)*(5*2^(3/4)*(I*b*x^2)^(1/4)*Gamma[1/4, (-2*I)*b*x^2]*((-I)*Cos[2*a] + Sin[2*a]) + 5*2^(3/4)*((-I)*b*
x^2)^(1/4)*Gamma[1/4, (2*I)*b*x^2]*(I*Cos[2*a] + Sin[2*a]) + 16*(b^2*x^4)^(1/4)*(8*b*x^2 + 5*Sin[2*(a + b*x^2)
])))/(640*(b^2*x^4)^(5/4))

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{2}}} \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.41165, size = 424, normalized size = 3.21 \begin{align*} \frac{2^{\frac{3}{4}}{\left (16 \cdot 2^{\frac{1}{4}}{\left (8 \, b x^{\frac{5}{2}} + 5 \, \sqrt{x} \sin \left (2 \, b x^{2} + 2 \, a\right )\right )} \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}} +{\left ({\left ({\left (5 i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - 5 i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (5 i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - 5 i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) + 5 \,{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) - 5 \,{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) +{\left (5 \,{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) + 5 \,{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (-5 i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + 5 i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (5 i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - 5 i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \sqrt{x}\right )}}{1280 \, \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)^2,x, algorithm="maxima")

[Out]

1/1280*2^(3/4)*(16*2^(1/4)*(8*b*x^(5/2) + 5*sqrt(x)*sin(2*b*x^2 + 2*a))*(x^2*abs(b))^(1/4) + (((5*I*gamma(1/4,
2*I*b*x^2) - 5*I*gamma(1/4, -2*I*b*x^2))*cos(1/8*pi + 1/4*arctan2(0, b)) + (5*I*gamma(1/4, 2*I*b*x^2) - 5*I*g
amma(1/4, -2*I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) + 5*(gamma(1/4, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*s
in(1/8*pi + 1/4*arctan2(0, b)) - 5*(gamma(1/4, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*sin(-1/8*pi + 1/4*arctan2(
0, b)))*cos(2*a) + (5*(gamma(1/4, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*cos(1/8*pi + 1/4*arctan2(0, b)) + 5*(ga
mma(1/4, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) + (-5*I*gamma(1/4, 2*I*b*x^2) +
5*I*gamma(1/4, -2*I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) + (5*I*gamma(1/4, 2*I*b*x^2) - 5*I*gamma(1/4, -2*
I*b*x^2))*sin(-1/8*pi + 1/4*arctan2(0, b)))*sin(2*a))*sqrt(x))/((x^2*abs(b))^(1/4)*b)

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Fricas [A]  time = 1.82749, size = 236, normalized size = 1.79 \begin{align*} \frac{5 \, \left (2 i \, b\right )^{\frac{3}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + 5 \, \left (-2 i \, b\right )^{\frac{3}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right ) + 32 \,{\left (4 \, b^{2} x^{2} + 5 \, b \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right )\right )} \sqrt{x}}{640 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*(5*(2*I*b)^(3/4)*e^(-2*I*a)*gamma(1/4, 2*I*b*x^2) + 5*(-2*I*b)^(3/4)*e^(2*I*a)*gamma(1/4, -2*I*b*x^2) +
32*(4*b^2*x^2 + 5*b*cos(b*x^2 + a)*sin(b*x^2 + a))*sqrt(x))/b^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{3}{2}} \cos ^{2}{\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)**2,x)

[Out]

Integral(x**(3/2)*cos(a + b*x**2)**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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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