3.1.60 \(\int \frac {a+b \sin (c+d x^3)}{x^4} \, dx\) [60]

Optimal. Leaf size=53 \[ -\frac {a}{3 x^3}+\frac {1}{3} b d \cos (c) \text {Ci}\left (d x^3\right )-\frac {b \sin \left (c+d x^3\right )}{3 x^3}-\frac {1}{3} b d \sin (c) \text {Si}\left (d x^3\right ) \]

[Out]

-1/3*a/x^3+1/3*b*d*Ci(d*x^3)*cos(c)-1/3*b*d*Si(d*x^3)*sin(c)-1/3*b*sin(d*x^3+c)/x^3

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Rubi [A]
time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {14, 3460, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {a}{3 x^3}+\frac {1}{3} b d \cos (c) \text {CosIntegral}\left (d x^3\right )-\frac {1}{3} b d \sin (c) \text {Si}\left (d x^3\right )-\frac {b \sin \left (c+d x^3\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])/x^4,x]

[Out]

-1/3*a/x^3 + (b*d*Cos[c]*CosIntegral[d*x^3])/3 - (b*Sin[c + d*x^3])/(3*x^3) - (b*d*Sin[c]*SinIntegral[d*x^3])/
3

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 3378

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

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3460

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

\begin {align*} \int \frac {a+b \sin \left (c+d x^3\right )}{x^4} \, dx &=\int \left (\frac {a}{x^4}+\frac {b \sin \left (c+d x^3\right )}{x^4}\right ) \, dx\\ &=-\frac {a}{3 x^3}+b \int \frac {\sin \left (c+d x^3\right )}{x^4} \, dx\\ &=-\frac {a}{3 x^3}+\frac {1}{3} b \text {Subst}\left (\int \frac {\sin (c+d x)}{x^2} \, dx,x,x^3\right )\\ &=-\frac {a}{3 x^3}-\frac {b \sin \left (c+d x^3\right )}{3 x^3}+\frac {1}{3} (b d) \text {Subst}\left (\int \frac {\cos (c+d x)}{x} \, dx,x,x^3\right )\\ &=-\frac {a}{3 x^3}-\frac {b \sin \left (c+d x^3\right )}{3 x^3}+\frac {1}{3} (b d \cos (c)) \text {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,x^3\right )-\frac {1}{3} (b d \sin (c)) \text {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,x^3\right )\\ &=-\frac {a}{3 x^3}+\frac {1}{3} b d \cos (c) \text {Ci}\left (d x^3\right )-\frac {b \sin \left (c+d x^3\right )}{3 x^3}-\frac {1}{3} b d \sin (c) \text {Si}\left (d x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 48, normalized size = 0.91 \begin {gather*} -\frac {a-b d x^3 \cos (c) \text {Ci}\left (d x^3\right )+b \sin \left (c+d x^3\right )+b d x^3 \sin (c) \text {Si}\left (d x^3\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])/x^4,x]

[Out]

-1/3*(a - b*d*x^3*Cos[c]*CosIntegral[d*x^3] + b*Sin[c + d*x^3] + b*d*x^3*Sin[c]*SinIntegral[d*x^3])/x^3

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \sin \left (d \,x^{3}+c \right )}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))/x^4,x)

[Out]

int((a+b*sin(d*x^3+c))/x^4,x)

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Maxima [C] Result contains complex when optimal does not.
time = 0.34, size = 57, normalized size = 1.08 \begin {gather*} \frac {1}{6} \, {\left ({\left (\Gamma \left (-1, i \, d x^{3}\right ) + \Gamma \left (-1, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left (i \, \Gamma \left (-1, i \, d x^{3}\right ) - i \, \Gamma \left (-1, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} b d - \frac {a}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 65, normalized size = 1.23 \begin {gather*} -\frac {2 \, b d x^{3} \sin \left (c\right ) \operatorname {Si}\left (d x^{3}\right ) - {\left (b d x^{3} \operatorname {Ci}\left (d x^{3}\right ) + b d x^{3} \operatorname {Ci}\left (-d x^{3}\right )\right )} \cos \left (c\right ) + 2 \, b \sin \left (d x^{3} + c\right ) + 2 \, a}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))/x^4,x, algorithm="fricas")

[Out]

-1/6*(2*b*d*x^3*sin(c)*sin_integral(d*x^3) - (b*d*x^3*cos_integral(d*x^3) + b*d*x^3*cos_integral(-d*x^3))*cos(
c) + 2*b*sin(d*x^3 + c) + 2*a)/x^3

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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^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))/x**4,x)

[Out]

Integral((a + b*sin(c + d*x**3))/x**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (45) = 90\).
time = 7.44, size = 99, normalized size = 1.87 \begin {gather*} \frac {{\left (d x^{3} + c\right )} b d^{2} \cos \left (c\right ) \operatorname {Ci}\left (d x^{3}\right ) - b c d^{2} \cos \left (c\right ) \operatorname {Ci}\left (d x^{3}\right ) - {\left (d x^{3} + c\right )} b d^{2} \sin \left (c\right ) \operatorname {Si}\left (d x^{3}\right ) + b c d^{2} \sin \left (c\right ) \operatorname {Si}\left (d x^{3}\right ) - b d^{2} \sin \left (d x^{3} + c\right ) - a d^{2}}{3 \, d^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))/x^4,x, algorithm="giac")

[Out]

1/3*((d*x^3 + c)*b*d^2*cos(c)*cos_integral(d*x^3) - b*c*d^2*cos(c)*cos_integral(d*x^3) - (d*x^3 + c)*b*d^2*sin
(c)*sin_integral(d*x^3) + b*c*d^2*sin(c)*sin_integral(d*x^3) - b*d^2*sin(d*x^3 + c) - a*d^2)/(d^2*x^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a+b\,\sin \left (d\,x^3+c\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))/x^4,x)

[Out]

int((a + b*sin(c + d*x^3))/x^4, x)

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