3.6 \(\int \frac {a+b \sin (c+d x^2)}{x^5} \, dx\)

Optimal. Leaf size=74 \[ -\frac {a}{4 x^4}-\frac {1}{4} b d^2 \sin (c) \text {Ci}\left (d x^2\right )-\frac {1}{4} b d^2 \cos (c) \text {Si}\left (d x^2\right )-\frac {b d \cos \left (c+d x^2\right )}{4 x^2}-\frac {b \sin \left (c+d x^2\right )}{4 x^4} \]

[Out]

-1/4*a/x^4-1/4*b*d*cos(d*x^2+c)/x^2-1/4*b*d^2*cos(c)*Si(d*x^2)-1/4*b*d^2*Ci(d*x^2)*sin(c)-1/4*b*sin(d*x^2+c)/x
^4

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Rubi [A]  time = 0.13, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {14, 3379, 3297, 3303, 3299, 3302} \[ -\frac {a}{4 x^4}-\frac {1}{4} b d^2 \sin (c) \text {CosIntegral}\left (d x^2\right )-\frac {1}{4} b d^2 \cos (c) \text {Si}\left (d x^2\right )-\frac {b \sin \left (c+d x^2\right )}{4 x^4}-\frac {b d \cos \left (c+d x^2\right )}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])/x^5,x]

[Out]

-a/(4*x^4) - (b*d*Cos[c + d*x^2])/(4*x^2) - (b*d^2*CosIntegral[d*x^2]*Sin[c])/4 - (b*Sin[c + d*x^2])/(4*x^4) -
(b*d^2*Cos[c]*SinIntegral[d*x^2])/4

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 3297

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 3299

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 3302

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 3303

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 3379

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^2\right )}{x^5} \, dx &=\int \left (\frac {a}{x^5}+\frac {b \sin \left (c+d x^2\right )}{x^5}\right ) \, dx\\ &=-\frac {a}{4 x^4}+b \int \frac {\sin \left (c+d x^2\right )}{x^5} \, dx\\ &=-\frac {a}{4 x^4}+\frac {1}{2} b \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac {a}{4 x^4}-\frac {b \sin \left (c+d x^2\right )}{4 x^4}+\frac {1}{4} (b d) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {a}{4 x^4}-\frac {b d \cos \left (c+d x^2\right )}{4 x^2}-\frac {b \sin \left (c+d x^2\right )}{4 x^4}-\frac {1}{4} \left (b d^2\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x} \, dx,x,x^2\right )\\ &=-\frac {a}{4 x^4}-\frac {b d \cos \left (c+d x^2\right )}{4 x^2}-\frac {b \sin \left (c+d x^2\right )}{4 x^4}-\frac {1}{4} \left (b d^2 \cos (c)\right ) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,x^2\right )-\frac {1}{4} \left (b d^2 \sin (c)\right ) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,x^2\right )\\ &=-\frac {a}{4 x^4}-\frac {b d \cos \left (c+d x^2\right )}{4 x^2}-\frac {1}{4} b d^2 \text {Ci}\left (d x^2\right ) \sin (c)-\frac {b \sin \left (c+d x^2\right )}{4 x^4}-\frac {1}{4} b d^2 \cos (c) \text {Si}\left (d x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 86, normalized size = 1.16 \[ -\frac {a}{4 x^4}-\frac {1}{4} b d^2 \left (\sin (c) \text {Ci}\left (d x^2\right )+\cos (c) \text {Si}\left (d x^2\right )\right )-\frac {b \cos \left (d x^2\right ) \left (d x^2 \cos (c)+\sin (c)\right )}{4 x^4}+\frac {b \sin \left (d x^2\right ) \left (d x^2 \sin (c)-\cos (c)\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^2])/x^5,x]

[Out]

-1/4*a/x^4 - (b*Cos[d*x^2]*(d*x^2*Cos[c] + Sin[c]))/(4*x^4) + (b*(-Cos[c] + d*x^2*Sin[c])*Sin[d*x^2])/(4*x^4)
- (b*d^2*(CosIntegral[d*x^2]*Sin[c] + Cos[c]*SinIntegral[d*x^2]))/4

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fricas [A]  time = 0.83, size = 85, normalized size = 1.15 \[ -\frac {2 \, b d^{2} x^{4} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) + 2 \, b d x^{2} \cos \left (d x^{2} + c\right ) + 2 \, b \sin \left (d x^{2} + c\right ) + {\left (b d^{2} x^{4} \operatorname {Ci}\left (d x^{2}\right ) + b d^{2} x^{4} \operatorname {Ci}\left (-d x^{2}\right )\right )} \sin \relax (c) + 2 \, a}{8 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^2+c))/x^5,x, algorithm="fricas")

[Out]

-1/8*(2*b*d^2*x^4*cos(c)*sin_integral(d*x^2) + 2*b*d*x^2*cos(d*x^2 + c) + 2*b*sin(d*x^2 + c) + (b*d^2*x^4*cos_
integral(d*x^2) + b*d^2*x^4*cos_integral(-d*x^2))*sin(c) + 2*a)/x^4

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giac [B]  time = 0.52, size = 204, normalized size = 2.76 \[ -\frac {{\left (d x^{2} + c\right )}^{2} b d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \relax (c) - 2 \, {\left (d x^{2} + c\right )} b c d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \relax (c) + b c^{2} d^{3} \operatorname {Ci}\left (d x^{2}\right ) \sin \relax (c) + {\left (d x^{2} + c\right )}^{2} b d^{3} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) - 2 \, {\left (d x^{2} + c\right )} b c d^{3} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) + b c^{2} d^{3} \cos \relax (c) \operatorname {Si}\left (d x^{2}\right ) + {\left (d x^{2} + c\right )} b d^{3} \cos \left (d x^{2} + c\right ) - b c d^{3} \cos \left (d x^{2} + c\right ) + b d^{3} \sin \left (d x^{2} + c\right ) + a d^{3}}{4 \, {\left ({\left (d x^{2} + c\right )}^{2} - 2 \, {\left (d x^{2} + c\right )} c + c^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^2+c))/x^5,x, algorithm="giac")

[Out]

-1/4*((d*x^2 + c)^2*b*d^3*cos_integral(d*x^2)*sin(c) - 2*(d*x^2 + c)*b*c*d^3*cos_integral(d*x^2)*sin(c) + b*c^
2*d^3*cos_integral(d*x^2)*sin(c) + (d*x^2 + c)^2*b*d^3*cos(c)*sin_integral(d*x^2) - 2*(d*x^2 + c)*b*c*d^3*cos(
c)*sin_integral(d*x^2) + b*c^2*d^3*cos(c)*sin_integral(d*x^2) + (d*x^2 + c)*b*d^3*cos(d*x^2 + c) - b*c*d^3*cos
(d*x^2 + c) + b*d^3*sin(d*x^2 + c) + a*d^3)/(((d*x^2 + c)^2 - 2*(d*x^2 + c)*c + c^2)*d)

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maple [A]  time = 0.02, size = 65, normalized size = 0.88 \[ -\frac {a}{4 x^{4}}+b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{4 x^{4}}+\frac {d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{2 x^{2}}-d \left (\frac {\cos \relax (c ) \Si \left (d \,x^{2}\right )}{2}+\frac {\sin \relax (c ) \Ci \left (d \,x^{2}\right )}{2}\right )\right )}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^2+c))/x^5,x)

[Out]

-1/4*a/x^4+b*(-1/4/x^4*sin(d*x^2+c)+1/2*d*(-1/2/x^2*cos(d*x^2+c)-d*(1/2*cos(c)*Si(d*x^2)+1/2*sin(c)*Ci(d*x^2))
))

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maxima [C]  time = 1.00, size = 58, normalized size = 0.78 \[ \frac {1}{4} \, {\left ({\left (i \, \Gamma \left (-2, i \, d x^{2}\right ) - i \, \Gamma \left (-2, -i \, d x^{2}\right )\right )} \cos \relax (c) + {\left (\Gamma \left (-2, i \, d x^{2}\right ) + \Gamma \left (-2, -i \, d x^{2}\right )\right )} \sin \relax (c)\right )} b d^{2} - \frac {a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^2+c))/x^5,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\sin \left (d\,x^2+c\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^2))/x^5,x)

[Out]

int((a + b*sin(c + d*x^2))/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \sin {\left (c + d x^{2} \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**2+c))/x**5,x)

[Out]

Integral((a + b*sin(c + d*x**2))/x**5, x)

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