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

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

[Out]

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

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Rubi [A]  time = 0.105475, antiderivative size = 74, normalized size of antiderivative = 1., 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} \[ \frac{1}{2} \left (2 a^2+b^2\right ) \log (x)+a b \sin (c) \text{CosIntegral}\left (d x^2\right )+a b \cos (c) \text{Si}\left (d x^2\right )-\frac{1}{4} b^2 \cos (2 c) \text{CosIntegral}\left (2 d x^2\right )+\frac{1}{4} b^2 \sin (2 c) \text{Si}\left (2 d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2/x,x]

[Out]

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

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]

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[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 3376

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

Rule 3375

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[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]

Rubi steps

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

Mathematica [A]  time = 0.1655, size = 71, normalized size = 0.96 \[ \frac{1}{2} \left (2 a^2+b^2\right ) \log (x)-\frac{1}{4} b \left (-4 a \sin (c) \text{CosIntegral}\left (d x^2\right )-4 a \cos (c) \text{Si}\left (d x^2\right )+b \cos (2 c) \text{CosIntegral}\left (2 d x^2\right )-b \sin (2 c) \text{Si}\left (2 d x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 69, normalized size = 0.9 \begin{align*} \ln \left ( x \right ){a}^{2}+{\frac{\ln \left ( x \right ){b}^{2}}{2}}+{\frac{{b}^{2}{\it Si} \left ( 2\,d{x}^{2} \right ) \sin \left ( 2\,c \right ) }{4}}-{\frac{{b}^{2}{\it Ci} \left ( 2\,d{x}^{2} \right ) \cos \left ( 2\,c \right ) }{4}}+ab\cos \left ( c \right ){\it Si} \left ( d{x}^{2} \right ) +ab{\it Ci} \left ( d{x}^{2} \right ) \sin \left ( c \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.21379, size = 146, normalized size = 1.97 \begin{align*} -\frac{1}{2} \,{\left ({\left (i \,{\rm Ei}\left (i \, d x^{2}\right ) - i \,{\rm Ei}\left (-i \, d x^{2}\right )\right )} \cos \left (c\right ) -{\left ({\rm Ei}\left (i \, d x^{2}\right ) +{\rm Ei}\left (-i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b - \frac{1}{8} \,{\left ({\left ({\rm Ei}\left (2 i \, d x^{2}\right ) +{\rm Ei}\left (-2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) -{\left (-i \,{\rm Ei}\left (2 i \, d x^{2}\right ) + i \,{\rm Ei}\left (-2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right ) - 4 \, \log \left (x\right )\right )} b^{2} + a^{2} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((I*Ei(I*d*x^2) - I*Ei(-I*d*x^2))*cos(c) - (Ei(I*d*x^2) + Ei(-I*d*x^2))*sin(c))*a*b - 1/8*((Ei(2*I*d*x^2)
+ Ei(-2*I*d*x^2))*cos(2*c) - (-I*Ei(2*I*d*x^2) + I*Ei(-2*I*d*x^2))*sin(2*c) - 4*log(x))*b^2 + a^2*log(x)

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Fricas [A]  time = 2.04536, size = 321, normalized size = 4.34 \begin{align*} \frac{1}{4} \, b^{2} \sin \left (2 \, c\right ) \operatorname{Si}\left (2 \, d x^{2}\right ) + a b \cos \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) - \frac{1}{8} \,{\left (b^{2} \operatorname{Ci}\left (2 \, d x^{2}\right ) + b^{2} \operatorname{Ci}\left (-2 \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + \frac{1}{2} \,{\left (2 \, a^{2} + b^{2}\right )} \log \left (x\right ) + \frac{1}{2} \,{\left (a b \operatorname{Ci}\left (d x^{2}\right ) + a b \operatorname{Ci}\left (-d x^{2}\right )\right )} \sin \left (c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15248, size = 104, normalized size = 1.41 \begin{align*} -\frac{1}{4} \, b^{2} \cos \left (2 \, c\right ) \operatorname{Ci}\left (2 \, d x^{2}\right ) + a b \operatorname{Ci}\left (d x^{2}\right ) \sin \left (c\right ) + a b \cos \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) - \frac{1}{4} \, b^{2} \sin \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{2}\right ) + \frac{1}{2} \, a^{2} \log \left (d x^{2}\right ) + \frac{1}{4} \, b^{2} \log \left (d x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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