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

Optimal. Leaf size=101 \[ -\frac{b e^{i c} d x \text{Gamma}\left (\frac{1}{3},-i d x^3\right )}{4 \sqrt [3]{-i d x^3}}-\frac{b e^{-i c} d x \text{Gamma}\left (\frac{1}{3},i d x^3\right )}{4 \sqrt [3]{i d x^3}}-\frac{a}{2 x^2}-\frac{b \sin \left (c+d x^3\right )}{2 x^2} \]

[Out]

-a/(2*x^2) - (b*d*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/(4*((-I)*d*x^3)^(1/3)) - (b*d*x*Gamma[1/3, I*d*x^3])/(4*E^
(I*c)*(I*d*x^3)^(1/3)) - (b*Sin[c + d*x^3])/(2*x^2)

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Rubi [A]  time = 0.0549399, 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, 3356, 2208} \[ -\frac{b e^{i c} d x \text{Gamma}\left (\frac{1}{3},-i d x^3\right )}{4 \sqrt [3]{-i d x^3}}-\frac{b e^{-i c} d x \text{Gamma}\left (\frac{1}{3},i d x^3\right )}{4 \sqrt [3]{i d x^3}}-\frac{a}{2 x^2}-\frac{b \sin \left (c+d x^3\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])/x^3,x]

[Out]

-a/(2*x^2) - (b*d*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/(4*((-I)*d*x^3)^(1/3)) - (b*d*x*Gamma[1/3, I*d*x^3])/(4*E^
(I*c)*(I*d*x^3)^(1/3)) - (b*Sin[c + d*x^3])/(2*x^2)

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 3356

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

Mathematica [A]  time = 0.192598, size = 120, normalized size = 1.19 \[ \frac{-i b \left (-i d x^3\right )^{4/3} (\cos (c)-i \sin (c)) \text{Gamma}\left (\frac{1}{3},i d x^3\right )+i b \left (i d x^3\right )^{4/3} (\cos (c)+i \sin (c)) \text{Gamma}\left (\frac{1}{3},-i d x^3\right )-2 \sqrt [3]{d^2 x^6} \left (a+b \sin \left (c+d x^3\right )\right )}{4 x^2 \sqrt [3]{d^2 x^6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*b*((-I)*d*x^3)^(4/3)*Gamma[1/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + I*b*(I*d*x^3)^(4/3)*Gamma[1/3, (-I)*d*x^3
]*(Cos[c] + I*Sin[c]) - 2*(d^2*x^6)^(1/3)*(a + b*Sin[c + d*x^3]))/(4*x^2*(d^2*x^6)^(1/3))

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\sin \left ( d{x}^{3}+c \right ) }{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.17775, size = 366, normalized size = 3.62 \begin{align*} -\frac{\left (x^{3}{\left | d \right |}\right )^{\frac{2}{3}}{\left ({\left ({\left (i \, \Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) -{\left (\Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) +{\left ({\left (\Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) + \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \cos \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) - i \, \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (-\frac{2}{3}, i \, d x^{3}\right ) + i \, \Gamma \left (-\frac{2}{3}, -i \, d x^{3}\right )\right )} \sin \left (-\frac{1}{3} \, \pi + \frac{2}{3} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} b}{12 \, x^{2}} - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.73536, size = 188, normalized size = 1.86 \begin{align*} \frac{i \, b \left (i \, d\right )^{\frac{2}{3}} x^{2} e^{\left (-i \, c\right )} \Gamma \left (\frac{1}{3}, i \, d x^{3}\right ) - i \, b \left (-i \, d\right )^{\frac{2}{3}} x^{2} e^{\left (i \, c\right )} \Gamma \left (\frac{1}{3}, -i \, d x^{3}\right ) - 2 \, b \sin \left (d x^{3} + c\right ) - 2 \, a}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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