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

Optimal. Leaf size=80 \[ \frac {1}{2} \left (2 a^2+b^2\right ) \log (x)+\frac {2}{3} a b \sin (c) \text {Ci}\left (d x^3\right )+\frac {2}{3} a b \cos (c) \text {Si}\left (d x^3\right )-\frac {1}{6} b^2 \cos (2 c) \text {Ci}\left (2 d x^3\right )+\frac {1}{6} b^2 \sin (2 c) \text {Si}\left (2 d x^3\right ) \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3403, 6, 3378, 3376, 3375, 3377} \[ \frac {1}{2} \left (2 a^2+b^2\right ) \log (x)+\frac {2}{3} a b \sin (c) \text {CosIntegral}\left (d x^3\right )+\frac {2}{3} a b \cos (c) \text {Si}\left (d x^3\right )-\frac {1}{6} b^2 \cos (2 c) \text {CosIntegral}\left (2 d x^3\right )+\frac {1}{6} b^2 \sin (2 c) \text {Si}\left (2 d x^3\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^3])^2/x,x]

[Out]

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

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 3375

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3376

Int[Cos[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CosIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3377

Int[Sin[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Sin[c], Int[Cos[d*x^n]/x, x], x] + Dist[Cos[c], Int[Si
n[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

Rule 3378

Int[Cos[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Cos[c], Int[Cos[d*x^n]/x, x], x] - Dist[Sin[c], Int[Si
n[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

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

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Mathematica [A]  time = 0.17, size = 71, normalized size = 0.89 \[ \frac {1}{2} \left (2 a^2+b^2\right ) \log (x)-\frac {1}{6} b \left (-4 a \sin (c) \text {Ci}\left (d x^3\right )-4 a \cos (c) \text {Si}\left (d x^3\right )+b \cos (2 c) \text {Ci}\left (2 d x^3\right )-b \sin (2 c) \text {Si}\left (2 d x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 95, normalized size = 1.19 \[ \frac {1}{6} \, b^{2} \sin \left (2 \, c\right ) \operatorname {Si}\left (2 \, d x^{3}\right ) + \frac {2}{3} \, a b \cos \relax (c) \operatorname {Si}\left (d x^{3}\right ) - \frac {1}{12} \, {\left (b^{2} \operatorname {Ci}\left (2 \, d x^{3}\right ) + b^{2} \operatorname {Ci}\left (-2 \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + \frac {1}{2} \, {\left (2 \, a^{2} + b^{2}\right )} \log \relax (x) + \frac {1}{3} \, {\left (a b \operatorname {Ci}\left (d x^{3}\right ) + a b \operatorname {Ci}\left (-d x^{3}\right )\right )} \sin \relax (c) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*sin(2*c)*sin_integral(2*d*x^3) + 2/3*a*b*cos(c)*sin_integral(d*x^3) - 1/12*(b^2*cos_integral(2*d*x^3)
+ b^2*cos_integral(-2*d*x^3))*cos(2*c) + 1/2*(2*a^2 + b^2)*log(x) + 1/3*(a*b*cos_integral(d*x^3) + a*b*cos_int
egral(-d*x^3))*sin(c)

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giac [A]  time = 0.75, size = 79, normalized size = 0.99 \[ -\frac {1}{6} \, b^{2} \cos \left (2 \, c\right ) \operatorname {Ci}\left (2 \, d x^{3}\right ) + \frac {2}{3} \, a b \operatorname {Ci}\left (d x^{3}\right ) \sin \relax (c) + \frac {2}{3} \, a b \cos \relax (c) \operatorname {Si}\left (d x^{3}\right ) - \frac {1}{6} \, b^{2} \sin \left (2 \, c\right ) \operatorname {Si}\left (-2 \, d x^{3}\right ) + \frac {1}{3} \, a^{2} \log \left (d x^{3}\right ) + \frac {1}{6} \, b^{2} \log \left (d x^{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.59, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \sin \left (d \,x^{3}+c \right )\right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.48, size = 108, normalized size = 1.35 \[ -\frac {1}{3} \, {\left ({\left (i \, {\rm Ei}\left (i \, d x^{3}\right ) - i \, {\rm Ei}\left (-i \, d x^{3}\right )\right )} \cos \relax (c) - {\left ({\rm Ei}\left (i \, d x^{3}\right ) + {\rm Ei}\left (-i \, d x^{3}\right )\right )} \sin \relax (c)\right )} a b - \frac {1}{12} \, {\left ({\left ({\rm Ei}\left (2 i \, d x^{3}\right ) + {\rm Ei}\left (-2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left (-i \, {\rm Ei}\left (2 i \, d x^{3}\right ) + i \, {\rm Ei}\left (-2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right ) - 6 \, \log \relax (x)\right )} b^{2} + a^{2} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((I*Ei(I*d*x^3) - I*Ei(-I*d*x^3))*cos(c) - (Ei(I*d*x^3) + Ei(-I*d*x^3))*sin(c))*a*b - 1/12*((Ei(2*I*d*x^3
) + Ei(-2*I*d*x^3))*cos(2*c) - (-I*Ei(2*I*d*x^3) + I*Ei(-2*I*d*x^3))*sin(2*c) - 6*log(x))*b^2 + a^2*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^3))^2/x,x)

[Out]

int((a + b*sin(c + d*x^3))^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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