3.63 \(\int \frac{a+b \sin (c+d x^3)}{x^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{b e^{i c} d x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{2 \left (-i d x^3\right )^{2/3}}-\frac{b e^{-i c} d x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{2 \left (i d x^3\right )^{2/3}}-\frac{a}{x}-\frac{b \sin \left (c+d x^3\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0780452, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {14, 3387, 3390, 2218} \[ -\frac{b e^{i c} d x^2 \text{Gamma}\left (\frac{2}{3},-i d x^3\right )}{2 \left (-i d x^3\right )^{2/3}}-\frac{b e^{-i c} d x^2 \text{Gamma}\left (\frac{2}{3},i d x^3\right )}{2 \left (i d x^3\right )^{2/3}}-\frac{a}{x}-\frac{b \sin \left (c+d x^3\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])/x^2,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3387

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

Rule 3390

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]

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

Mathematica [A]  time = 0.210745, size = 120, normalized size = 1.19 \[ \frac{-i b \left (-i d x^3\right )^{5/3} (\cos (c)-i \sin (c)) \text{Gamma}\left (\frac{2}{3},i d x^3\right )+i b \left (i d x^3\right )^{5/3} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{2}{3},-i d x^3\right )-2 \left (d^2 x^6\right )^{2/3} \left (a+b \sin \left (c+d x^3\right )\right )}{2 x \left (d^2 x^6\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])/x^2,x]

[Out]

((-I)*b*((-I)*d*x^3)^(5/3)*Gamma[2/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + I*b*(I*d*x^3)^(5/3)*Gamma[2/3, (-I)*d*x^3
]*(Cos[c] + I*Sin[c]) - 2*(d^2*x^6)^(2/3)*(a + b*Sin[c + d*x^3]))/(2*x*(d^2*x^6)^(2/3))

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\sin \left ( d{x}^{3}+c \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))/x^2,x)

[Out]

int((a+b*sin(d*x^3+c))/x^2,x)

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Maxima [B]  time = 1.15554, size = 366, normalized size = 3.62 \begin{align*} -\frac{\left (x^{3}{\left | d \right |}\right )^{\frac{1}{3}}{\left ({\left ({\left (i \, \Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) -{\left (\Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) +{\left ({\left (\Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (-\frac{1}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (-\frac{1}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{6} \, \pi + \frac{1}{3} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} b}{12 \, x} - \frac{a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(x^3*abs(d))^(1/3)*(((I*gamma(-1/3, I*d*x^3) - I*gamma(-1/3, -I*d*x^3))*cos(1/6*pi + 1/3*arctan2(0, d))
+ (I*gamma(-1/3, I*d*x^3) - I*gamma(-1/3, -I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) - (gamma(-1/3, I*d*x^3)
+ gamma(-1/3, -I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) + (gamma(-1/3, I*d*x^3) + gamma(-1/3, -I*d*x^3))*sin(
-1/6*pi + 1/3*arctan2(0, d)))*cos(c) + ((gamma(-1/3, I*d*x^3) + gamma(-1/3, -I*d*x^3))*cos(1/6*pi + 1/3*arctan
2(0, d)) + (gamma(-1/3, I*d*x^3) + gamma(-1/3, -I*d*x^3))*cos(-1/6*pi + 1/3*arctan2(0, d)) + (I*gamma(-1/3, I*
d*x^3) - I*gamma(-1/3, -I*d*x^3))*sin(1/6*pi + 1/3*arctan2(0, d)) + (-I*gamma(-1/3, I*d*x^3) + I*gamma(-1/3, -
I*d*x^3))*sin(-1/6*pi + 1/3*arctan2(0, d)))*sin(c))*b/x - a/x

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Fricas [A]  time = 1.7596, size = 180, normalized size = 1.78 \begin{align*} \frac{i \, b \left (i \, d\right )^{\frac{1}{3}} x e^{\left (-i \, c\right )} \Gamma \left (\frac{2}{3}, i \, d x^{3}\right ) - i \, b \left (-i \, d\right )^{\frac{1}{3}} x e^{\left (i \, c\right )} \Gamma \left (\frac{2}{3}, -i \, d x^{3}\right ) - 2 \, b \sin \left (d x^{3} + c\right ) - 2 \, a}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*b*(I*d)^(1/3)*x*e^(-I*c)*gamma(2/3, I*d*x^3) - I*b*(-I*d)^(1/3)*x*e^(I*c)*gamma(2/3, -I*d*x^3) - 2*b*si
n(d*x^3 + c) - 2*a)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))/x**2,x)

[Out]

Integral((a + b*sin(c + d*x**3))/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(d*x^3 + c) + a)/x^2, x)