3.72 \(\int \frac {(a+b \sin (c+d x^3))^2}{x^4} \, dx\)

Optimal. Leaf size=122 \[ -\frac {2 a^2+b^2}{6 x^3}+\frac {2}{3} a b d \cos (c) \text {Ci}\left (d x^3\right )-\frac {2}{3} a b d \sin (c) \text {Si}\left (d x^3\right )-\frac {2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac {1}{3} b^2 d \sin (2 c) \text {Ci}\left (2 d x^3\right )+\frac {1}{3} b^2 d \cos (2 c) \text {Si}\left (2 d x^3\right )+\frac {b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3403, 6, 3380, 3297, 3303, 3299, 3302, 3379} \[ -\frac {2 a^2+b^2}{6 x^3}+\frac {2}{3} a b d \cos (c) \text {CosIntegral}\left (d x^3\right )-\frac {2}{3} a b d \sin (c) \text {Si}\left (d x^3\right )-\frac {2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac {1}{3} b^2 d \sin (2 c) \text {CosIntegral}\left (2 d x^3\right )+\frac {1}{3} b^2 d \cos (2 c) \text {Si}\left (2 d x^3\right )+\frac {b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2/x^4,x]

[Out]

-(2*a^2 + b^2)/(6*x^3) + (b^2*Cos[2*(c + d*x^3)])/(6*x^3) + (2*a*b*d*Cos[c]*CosIntegral[d*x^3])/3 + (b^2*d*Cos
Integral[2*d*x^3]*Sin[2*c])/3 - (2*a*b*Sin[c + d*x^3])/(3*x^3) - (2*a*b*d*Sin[c]*SinIntegral[d*x^3])/3 + (b^2*
d*Cos[2*c]*SinIntegral[2*d*x^3])/3

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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]))

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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]))

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^4} \, dx &=\int \left (\frac {a^2}{x^4}+\frac {b^2}{2 x^4}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^4}+\frac {2 a b \sin \left (c+d x^3\right )}{x^4}\right ) \, dx\\ &=\int \left (\frac {a^2+\frac {b^2}{2}}{x^4}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^4}+\frac {2 a b \sin \left (c+d x^3\right )}{x^4}\right ) \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}+(2 a b) \int \frac {\sin \left (c+d x^3\right )}{x^4} \, dx-\frac {1}{2} b^2 \int \frac {\cos \left (2 c+2 d x^3\right )}{x^4} \, dx\\ &=-\frac {2 a^2+b^2}{6 x^3}+\frac {1}{3} (2 a b) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^2} \, dx,x,x^3\right )-\frac {1}{6} b^2 \operatorname {Subst}\left (\int \frac {\cos (2 c+2 d x)}{x^2} \, dx,x,x^3\right )\\ &=-\frac {2 a^2+b^2}{6 x^3}+\frac {b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3}-\frac {2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac {1}{3} (2 a b d) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x} \, dx,x,x^3\right )+\frac {1}{3} \left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {\sin (2 c+2 d x)}{x} \, dx,x,x^3\right )\\ &=-\frac {2 a^2+b^2}{6 x^3}+\frac {b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3}-\frac {2 a b \sin \left (c+d x^3\right )}{3 x^3}+\frac {1}{3} (2 a b d \cos (c)) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,x^3\right )+\frac {1}{3} \left (b^2 d \cos (2 c)\right ) \operatorname {Subst}\left (\int \frac {\sin (2 d x)}{x} \, dx,x,x^3\right )-\frac {1}{3} (2 a b d \sin (c)) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,x^3\right )+\frac {1}{3} \left (b^2 d \sin (2 c)\right ) \operatorname {Subst}\left (\int \frac {\cos (2 d x)}{x} \, dx,x,x^3\right )\\ &=-\frac {2 a^2+b^2}{6 x^3}+\frac {b^2 \cos \left (2 \left (c+d x^3\right )\right )}{6 x^3}+\frac {2}{3} a b d \cos (c) \text {Ci}\left (d x^3\right )+\frac {1}{3} b^2 d \text {Ci}\left (2 d x^3\right ) \sin (2 c)-\frac {2 a b \sin \left (c+d x^3\right )}{3 x^3}-\frac {2}{3} a b d \sin (c) \text {Si}\left (d x^3\right )+\frac {1}{3} b^2 d \cos (2 c) \text {Si}\left (2 d x^3\right )\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 116, normalized size = 0.95 \[ \frac {-2 a^2+4 a b d x^3 \cos (c) \text {Ci}\left (d x^3\right )-4 a b d x^3 \sin (c) \text {Si}\left (d x^3\right )-4 a b \sin \left (c+d x^3\right )+2 b^2 d x^3 \sin (2 c) \text {Ci}\left (2 d x^3\right )+2 b^2 d x^3 \cos (2 c) \text {Si}\left (2 d x^3\right )+b^2 \cos \left (2 \left (c+d x^3\right )\right )-b^2}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x^3])^2/x^4,x]

[Out]

(-2*a^2 - b^2 + b^2*Cos[2*(c + d*x^3)] + 4*a*b*d*x^3*Cos[c]*CosIntegral[d*x^3] + 2*b^2*d*x^3*CosIntegral[2*d*x
^3]*Sin[2*c] - 4*a*b*Sin[c + d*x^3] - 4*a*b*d*x^3*Sin[c]*SinIntegral[d*x^3] + 2*b^2*d*x^3*Cos[2*c]*SinIntegral
[2*d*x^3])/(6*x^3)

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fricas [A]  time = 0.67, size = 147, normalized size = 1.20 \[ \frac {2 \, b^{2} d x^{3} \cos \left (2 \, c\right ) \operatorname {Si}\left (2 \, d x^{3}\right ) - 4 \, a b d x^{3} \sin \relax (c) \operatorname {Si}\left (d x^{3}\right ) + 2 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 4 \, a b \sin \left (d x^{3} + c\right ) - 2 \, a^{2} - 2 \, b^{2} + 2 \, {\left (a b d x^{3} \operatorname {Ci}\left (d x^{3}\right ) + a b d x^{3} \operatorname {Ci}\left (-d x^{3}\right )\right )} \cos \relax (c) + {\left (b^{2} d x^{3} \operatorname {Ci}\left (2 \, d x^{3}\right ) + b^{2} d x^{3} \operatorname {Ci}\left (-2 \, d x^{3}\right )\right )} \sin \left (2 \, c\right )}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^4,x, algorithm="fricas")

[Out]

1/6*(2*b^2*d*x^3*cos(2*c)*sin_integral(2*d*x^3) - 4*a*b*d*x^3*sin(c)*sin_integral(d*x^3) + 2*b^2*cos(d*x^3 + c
)^2 - 4*a*b*sin(d*x^3 + c) - 2*a^2 - 2*b^2 + 2*(a*b*d*x^3*cos_integral(d*x^3) + a*b*d*x^3*cos_integral(-d*x^3)
)*cos(c) + (b^2*d*x^3*cos_integral(2*d*x^3) + b^2*d*x^3*cos_integral(-2*d*x^3))*sin(2*c))/x^3

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giac [B]  time = 1.24, size = 226, normalized size = 1.85 \[ \frac {4 \, {\left (d x^{3} + c\right )} a b d^{2} \cos \relax (c) \operatorname {Ci}\left (d x^{3}\right ) - 4 \, a b c d^{2} \cos \relax (c) \operatorname {Ci}\left (d x^{3}\right ) + 2 \, {\left (d x^{3} + c\right )} b^{2} d^{2} \operatorname {Ci}\left (2 \, d x^{3}\right ) \sin \left (2 \, c\right ) - 2 \, b^{2} c d^{2} \operatorname {Ci}\left (2 \, d x^{3}\right ) \sin \left (2 \, c\right ) - 4 \, {\left (d x^{3} + c\right )} a b d^{2} \sin \relax (c) \operatorname {Si}\left (d x^{3}\right ) + 4 \, a b c d^{2} \sin \relax (c) \operatorname {Si}\left (d x^{3}\right ) - 2 \, {\left (d x^{3} + c\right )} b^{2} d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{3}\right ) + 2 \, b^{2} c d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{3}\right ) + b^{2} d^{2} \cos \left (2 \, d x^{3} + 2 \, c\right ) - 4 \, a b d^{2} \sin \left (d x^{3} + c\right ) - 2 \, a^{2} d^{2} - b^{2} d^{2}}{6 \, d^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^4,x, algorithm="giac")

[Out]

1/6*(4*(d*x^3 + c)*a*b*d^2*cos(c)*cos_integral(d*x^3) - 4*a*b*c*d^2*cos(c)*cos_integral(d*x^3) + 2*(d*x^3 + c)
*b^2*d^2*cos_integral(2*d*x^3)*sin(2*c) - 2*b^2*c*d^2*cos_integral(2*d*x^3)*sin(2*c) - 4*(d*x^3 + c)*a*b*d^2*s
in(c)*sin_integral(d*x^3) + 4*a*b*c*d^2*sin(c)*sin_integral(d*x^3) - 2*(d*x^3 + c)*b^2*d^2*cos(2*c)*sin_integr
al(-2*d*x^3) + 2*b^2*c*d^2*cos(2*c)*sin_integral(-2*d*x^3) + b^2*d^2*cos(2*d*x^3 + 2*c) - 4*a*b*d^2*sin(d*x^3
+ c) - 2*a^2*d^2 - b^2*d^2)/(d^2*x^3)

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maple [F]  time = 0.67, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x^3+c))^2/x^4,x)

[Out]

int((a+b*sin(d*x^3+c))^2/x^4,x)

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maxima [C]  time = 0.48, size = 124, normalized size = 1.02 \[ \frac {1}{3} \, {\left ({\left (\Gamma \left (-1, i \, d x^{3}\right ) + \Gamma \left (-1, -i \, d x^{3}\right )\right )} \cos \relax (c) - {\left (i \, \Gamma \left (-1, i \, d x^{3}\right ) - i \, \Gamma \left (-1, -i \, d x^{3}\right )\right )} \sin \relax (c)\right )} a b d + \frac {{\left ({\left ({\left (i \, \Gamma \left (-1, 2 i \, d x^{3}\right ) - i \, \Gamma \left (-1, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left (\Gamma \left (-1, 2 i \, d x^{3}\right ) + \Gamma \left (-1, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} - 1\right )} b^{2}}{6 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x^3+c))^2/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))^2/x^4,x)

[Out]

int((a + b*sin(c + d*x^3))^2/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x**3+c))**2/x**4,x)

[Out]

Integral((a + b*sin(c + d*x**3))**2/x**4, x)

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