3.1.16 \(\int \frac {(a+b \sin (c+d x^2))^2}{x^3} \, dx\) [16]

Optimal. Leaf size=115 \[ -\frac {2 a^2+b^2}{4 x^2}+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2}+a b d \cos (c) \text {Ci}\left (d x^2\right )+\frac {1}{2} b^2 d \text {Ci}\left (2 d x^2\right ) \sin (2 c)-\frac {a b \sin \left (c+d x^2\right )}{x^2}-a b d \sin (c) \text {Si}\left (d x^2\right )+\frac {1}{2} b^2 d \cos (2 c) \text {Si}\left (2 d x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 115, 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 = {3484, 6, 3461, 3378, 3384, 3380, 3383, 3460} \begin {gather*} -\frac {2 a^2+b^2}{4 x^2}+a b d \cos (c) \text {CosIntegral}\left (d x^2\right )-a b d \sin (c) \text {Si}\left (d x^2\right )-\frac {a b \sin \left (c+d x^2\right )}{x^2}+\frac {1}{2} b^2 d \sin (2 c) \text {CosIntegral}\left (2 d x^2\right )+\frac {1}{2} b^2 d \cos (2 c) \text {Si}\left (2 d x^2\right )+\frac {b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2/x^3,x]

[Out]

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

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 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]))

Rule 3461

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 3484

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

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Mathematica [A]
time = 0.15, size = 116, normalized size = 1.01 \begin {gather*} \frac {-2 a^2-b^2+b^2 \cos \left (2 \left (c+d x^2\right )\right )+4 a b d x^2 \cos (c) \text {Ci}\left (d x^2\right )+2 b^2 d x^2 \text {Ci}\left (2 d x^2\right ) \sin (2 c)-4 a b \sin \left (c+d x^2\right )-4 a b d x^2 \sin (c) \text {Si}\left (d x^2\right )+2 b^2 d x^2 \cos (2 c) \text {Si}\left (2 d x^2\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.28, size = 203, normalized size = 1.77

method	result	size
risch	\(-\frac {\mathrm {csgn}\left (d \,x^{2}\right ) {\mathrm e}^{-2 i c} \pi \,b^{2} d}{4}+\frac {\sinIntegral \left (2 d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d}{2}-\frac {i \expIntegral \left (1, -2 i d \,x^{2}\right ) {\mathrm e}^{-2 i c} b^{2} d}{4}+\frac {i b^{2} d \expIntegral \left (1, -2 i d \,x^{2}\right ) {\mathrm e}^{2 i c}}{4}-\frac {a b d \expIntegral \left (1, -i d \,x^{2}\right ) {\mathrm e}^{i c}}{2}+\frac {i \mathrm {csgn}\left (d \,x^{2}\right ) \pi \,{\mathrm e}^{-i c} a b d}{2}-i \sinIntegral \left (d \,x^{2}\right ) {\mathrm e}^{-i c} a b d -\frac {\expIntegral \left (1, -i d \,x^{2}\right ) {\mathrm e}^{-i c} a b d}{2}-\frac {a^{2}}{2 x^{2}}-\frac {b^{2}}{4 x^{2}}-\frac {a b \sin \left (d \,x^{2}+c \right )}{x^{2}}+\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{4 x^{2}}\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*csgn(d*x^2)*exp(-2*I*c)*Pi*b^2*d+1/2*Si(2*d*x^2)*exp(-2*I*c)*b^2*d-1/4*I*Ei(1,-2*I*d*x^2)*exp(-2*I*c)*b^2
*d+1/4*I*b^2*d*Ei(1,-2*I*d*x^2)*exp(2*I*c)-1/2*a*b*d*Ei(1,-I*d*x^2)*exp(I*c)+1/2*I*csgn(d*x^2)*Pi*exp(-I*c)*a*
b*d-I*Si(d*x^2)*exp(-I*c)*a*b*d-1/2*Ei(1,-I*d*x^2)*exp(-I*c)*a*b*d-1/2*a^2/x^2-1/4*b^2/x^2-a*b*sin(d*x^2+c)/x^
2+1/4*b^2*cos(2*d*x^2+2*c)/x^2

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Maxima [C] Result contains complex when optimal does not.
time = 0.41, size = 124, normalized size = 1.08 \begin {gather*} \frac {1}{2} \, {\left ({\left (\Gamma \left (-1, i \, d x^{2}\right ) + \Gamma \left (-1, -i \, d x^{2}\right )\right )} \cos \left (c\right ) - {\left (i \, \Gamma \left (-1, i \, d x^{2}\right ) - i \, \Gamma \left (-1, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b d + \frac {{\left ({\left ({\left (i \, \Gamma \left (-1, 2 i \, d x^{2}\right ) - i \, \Gamma \left (-1, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + {\left (\Gamma \left (-1, 2 i \, d x^{2}\right ) + \Gamma \left (-1, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{2} - 1\right )} b^{2}}{4 \, x^{2}} - \frac {a^{2}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 147, normalized size = 1.28 \begin {gather*} \frac {2 \, b^{2} d x^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (2 \, d x^{2}\right ) - 4 \, a b d x^{2} \sin \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) + 2 \, b^{2} \cos \left (d x^{2} + c\right )^{2} - 4 \, a b \sin \left (d x^{2} + c\right ) - 2 \, a^{2} - 2 \, b^{2} + 2 \, {\left (a b d x^{2} \operatorname {Ci}\left (d x^{2}\right ) + a b d x^{2} \operatorname {Ci}\left (-d x^{2}\right )\right )} \cos \left (c\right ) + {\left (b^{2} d x^{2} \operatorname {Ci}\left (2 \, d x^{2}\right ) + b^{2} d x^{2} \operatorname {Ci}\left (-2 \, d x^{2}\right )\right )} \sin \left (2 \, c\right )}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (108) = 216\).
time = 3.63, size = 226, normalized size = 1.97 \begin {gather*} \frac {4 \, {\left (d x^{2} + c\right )} a b d^{2} \cos \left (c\right ) \operatorname {Ci}\left (d x^{2}\right ) - 4 \, a b c d^{2} \cos \left (c\right ) \operatorname {Ci}\left (d x^{2}\right ) + 2 \, {\left (d x^{2} + c\right )} b^{2} d^{2} \operatorname {Ci}\left (2 \, d x^{2}\right ) \sin \left (2 \, c\right ) - 2 \, b^{2} c d^{2} \operatorname {Ci}\left (2 \, d x^{2}\right ) \sin \left (2 \, c\right ) - 4 \, {\left (d x^{2} + c\right )} a b d^{2} \sin \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) + 4 \, a b c d^{2} \sin \left (c\right ) \operatorname {Si}\left (d x^{2}\right ) - 2 \, {\left (d x^{2} + c\right )} b^{2} d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) + 2 \, b^{2} c d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{2}\right ) + b^{2} d^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 4 \, a b d^{2} \sin \left (d x^{2} + c\right ) - 2 \, a^{2} d^{2} - b^{2} d^{2}}{4 \, d^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^2))^2/x^3,x)

[Out]

int((a + b*sin(c + d*x^2))^2/x^3, x)

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