3.1.68 \(\int \frac {a+b \sin (c+d x^3)}{x^6} \, dx\) [68]

Optimal. Leaf size=126 \[ -\frac {a}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2}-\frac {3 i b d^2 e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{20 \sqrt [3]{-i d x^3}}+\frac {3 i b d^2 e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{20 \sqrt [3]{i d x^3}}-\frac {b \sin \left (c+d x^3\right )}{5 x^5} \]

[Out]

-1/5*a/x^5-3/10*b*d*cos(d*x^3+c)/x^2-3/20*I*b*d^2*exp(I*c)*x*GAMMA(1/3,-I*d*x^3)/(-I*d*x^3)^(1/3)+3/20*I*b*d^2
*x*GAMMA(1/3,I*d*x^3)/exp(I*c)/(I*d*x^3)^(1/3)-1/5*b*sin(d*x^3+c)/x^5

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Rubi [A]
time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, 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 = {14, 3468, 3469, 3436, 2239} \begin {gather*} -\frac {3 i b e^{i c} d^2 x \text {Gamma}\left (\frac {1}{3},-i d x^3\right )}{20 \sqrt [3]{-i d x^3}}+\frac {3 i b e^{-i c} d^2 x \text {Gamma}\left (\frac {1}{3},i d x^3\right )}{20 \sqrt [3]{i d x^3}}-\frac {a}{5 x^5}-\frac {b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])/x^6,x]

[Out]

-1/5*a/x^5 - (3*b*d*Cos[c + d*x^3])/(10*x^2) - (((3*I)/20)*b*d^2*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3])/((-I)*d*x^3
)^(1/3) + (((3*I)/20)*b*d^2*x*Gamma[1/3, I*d*x^3])/(E^(I*c)*(I*d*x^3)^(1/3)) - (b*Sin[c + d*x^3])/(5*x^5)

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 2239

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]

Rule 3436

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 3468

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 3469

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

Rubi steps

\begin {align*} \int \frac {a+b \sin \left (c+d x^3\right )}{x^6} \, dx &=\int \left (\frac {a}{x^6}+\frac {b \sin \left (c+d x^3\right )}{x^6}\right ) \, dx\\ &=-\frac {a}{5 x^5}+b \int \frac {\sin \left (c+d x^3\right )}{x^6} \, dx\\ &=-\frac {a}{5 x^5}-\frac {b \sin \left (c+d x^3\right )}{5 x^5}+\frac {1}{5} (3 b d) \int \frac {\cos \left (c+d x^3\right )}{x^3} \, dx\\ &=-\frac {a}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2}-\frac {b \sin \left (c+d x^3\right )}{5 x^5}-\frac {1}{10} \left (9 b d^2\right ) \int \sin \left (c+d x^3\right ) \, dx\\ &=-\frac {a}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2}-\frac {b \sin \left (c+d x^3\right )}{5 x^5}-\frac {1}{20} \left (9 i b d^2\right ) \int e^{-i c-i d x^3} \, dx+\frac {1}{20} \left (9 i b d^2\right ) \int e^{i c+i d x^3} \, dx\\ &=-\frac {a}{5 x^5}-\frac {3 b d \cos \left (c+d x^3\right )}{10 x^2}-\frac {3 i b d^2 e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{20 \sqrt [3]{-i d x^3}}+\frac {3 i b d^2 e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{20 \sqrt [3]{i d x^3}}-\frac {b \sin \left (c+d x^3\right )}{5 x^5}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 146, normalized size = 1.16 \begin {gather*} \frac {3 b d^2 x^6 \sqrt [3]{i d x^3} \Gamma \left (\frac {1}{3},-i d x^3\right ) (-i \cos (c)+\sin (c))+3 b d^2 x^6 \sqrt [3]{-i d x^3} \Gamma \left (\frac {1}{3},i d x^3\right ) (i \cos (c)+\sin (c))-2 \sqrt [3]{d^2 x^6} \left (2 a+3 b d x^3 \cos \left (c+d x^3\right )+2 b \sin \left (c+d x^3\right )\right )}{20 x^5 \sqrt [3]{d^2 x^6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])/x^6,x]

[Out]

(3*b*d^2*x^6*(I*d*x^3)^(1/3)*Gamma[1/3, (-I)*d*x^3]*((-I)*Cos[c] + Sin[c]) + 3*b*d^2*x^6*((-I)*d*x^3)^(1/3)*Ga
mma[1/3, I*d*x^3]*(I*Cos[c] + Sin[c]) - 2*(d^2*x^6)^(1/3)*(2*a + 3*b*d*x^3*Cos[c + d*x^3] + 2*b*Sin[c + d*x^3]
))/(20*x^5*(d^2*x^6)^(1/3))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {a +b \sin \left (d \,x^{3}+c \right )}{x^{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))/x^6,x)

[Out]

int((a+b*sin(d*x^3+c))/x^6,x)

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Maxima [A]
time = 0.32, size = 91, normalized size = 0.72 \begin {gather*} -\frac {\left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} b d}{12 \, x^{2}} - \frac {a}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))/x^6,x, algorithm="maxima")

[Out]

-1/12*(d*x^3)^(2/3)*(((-I*sqrt(3) - 1)*gamma(-5/3, I*d*x^3) + (I*sqrt(3) - 1)*gamma(-5/3, -I*d*x^3))*cos(c) -
((sqrt(3) - I)*gamma(-5/3, I*d*x^3) + (sqrt(3) + I)*gamma(-5/3, -I*d*x^3))*sin(c))*b*d/x^2 - 1/5*a/x^5

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Fricas [A]
time = 0.11, size = 83, normalized size = 0.66 \begin {gather*} \frac {3 \, b \left (i \, d\right )^{\frac {2}{3}} d x^{5} e^{\left (-i \, c\right )} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) + 3 \, b \left (-i \, d\right )^{\frac {2}{3}} d x^{5} e^{\left (i \, c\right )} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) - 6 \, b d x^{3} \cos \left (d x^{3} + c\right ) - 4 \, b \sin \left (d x^{3} + c\right ) - 4 \, a}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))/x^6,x, algorithm="fricas")

[Out]

1/20*(3*b*(I*d)^(2/3)*d*x^5*e^(-I*c)*gamma(1/3, I*d*x^3) + 3*b*(-I*d)^(2/3)*d*x^5*e^(I*c)*gamma(1/3, -I*d*x^3)
- 6*b*d*x^3*cos(d*x^3 + c) - 4*b*sin(d*x^3 + c) - 4*a)/x^5

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \sin {\left (c + d x^{3} \right )}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))/x**6,x)

[Out]

Integral((a + b*sin(c + d*x**3))/x**6, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))/x^6,x, algorithm="giac")

[Out]

integrate((b*sin(d*x^3 + c) + a)/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\sin \left (d\,x^3+c\right )}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))/x^6,x)

[Out]

int((a + b*sin(c + d*x^3))/x^6, x)

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