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

Optimal. Leaf size=88 \[ -\frac{a}{x}+\sqrt{2 \pi } b \sqrt{d} \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\sqrt{2 \pi } b \sqrt{d} \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{b \sin \left (c+d x^2\right )}{x} \]

[Out]

-(a/x) + b*Sqrt[d]*Sqrt[2*Pi]*Cos[c]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x] - b*Sqrt[d]*Sqrt[2*Pi]*FresnelS[Sqrt[d]*Sq
rt[2/Pi]*x]*Sin[c] - (b*Sin[c + d*x^2])/x

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Rubi [A]  time = 0.0744913, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {14, 3387, 3354, 3352, 3351} \[ -\frac{a}{x}+\sqrt{2 \pi } b \sqrt{d} \cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\sqrt{2 \pi } b \sqrt{d} \sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{b \sin \left (c+d x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])/x^2,x]

[Out]

-(a/x) + b*Sqrt[d]*Sqrt[2*Pi]*Cos[c]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x] - b*Sqrt[d]*Sqrt[2*Pi]*FresnelS[Sqrt[d]*Sq
rt[2/Pi]*x]*Sin[c] - (b*Sin[c + d*x^2])/x

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 3387

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[((e*x)^(m + 1)*Sin[c + d*x^n])/(e*(m + 1
)), x] - Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \frac{a+b \sin \left (c+d x^2\right )}{x^2} \, dx &=\int \left (\frac{a}{x^2}+\frac{b \sin \left (c+d x^2\right )}{x^2}\right ) \, dx\\ &=-\frac{a}{x}+b \int \frac{\sin \left (c+d x^2\right )}{x^2} \, dx\\ &=-\frac{a}{x}-\frac{b \sin \left (c+d x^2\right )}{x}+(2 b d) \int \cos \left (c+d x^2\right ) \, dx\\ &=-\frac{a}{x}-\frac{b \sin \left (c+d x^2\right )}{x}+(2 b d \cos (c)) \int \cos \left (d x^2\right ) \, dx-(2 b d \sin (c)) \int \sin \left (d x^2\right ) \, dx\\ &=-\frac{a}{x}+b \sqrt{d} \sqrt{2 \pi } \cos (c) C\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-b \sqrt{d} \sqrt{2 \pi } S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right ) \sin (c)-\frac{b \sin \left (c+d x^2\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.182091, size = 91, normalized size = 1.03 \[ -\frac{a}{x}+\sqrt{2 \pi } b \sqrt{d} \left (\cos (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\sin (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )\right )-\frac{b \sin (c) \cos \left (d x^2\right )}{x}-\frac{b \cos (c) \sin \left (d x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/x) - (b*Cos[d*x^2]*Sin[c])/x + b*Sqrt[d]*Sqrt[2*Pi]*(Cos[c]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x] - FresnelS[Sqrt
[d]*Sqrt[2/Pi]*x]*Sin[c]) - (b*Cos[c]*Sin[d*x^2])/x

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Maple [A]  time = 0.008, size = 66, normalized size = 0.8 \begin{align*} -{\frac{a}{x}}+b \left ( -{\frac{\sin \left ( d{x}^{2}+c \right ) }{x}}+\sqrt{d}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( c \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{d}} \right ) -\sin \left ( c \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{d}} \right ) \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/x+b*(-1/x*sin(d*x^2+c)+d^(1/2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2))-sin(c)*Fresnel
S(x*d^(1/2)*2^(1/2)/Pi^(1/2))))

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Maxima [C]  time = 1.16251, size = 366, normalized size = 4.16 \begin{align*} -\frac{\sqrt{x^{2}{\left | d \right |}}{\left ({\left ({\left (i \, \Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) - i \, \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) - i \, \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) -{\left (\Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) +{\left ({\left (\Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) - i \, \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (-\frac{1}{2}, i \, d x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -i \, d x^{2}\right )\right )} \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} b}{8 \, x} - \frac{a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(x^2*abs(d))*(((I*gamma(-1/2, I*d*x^2) - I*gamma(-1/2, -I*d*x^2))*cos(1/4*pi + 1/2*arctan2(0, d)) + (
I*gamma(-1/2, I*d*x^2) - I*gamma(-1/2, -I*d*x^2))*cos(-1/4*pi + 1/2*arctan2(0, d)) - (gamma(-1/2, I*d*x^2) + g
amma(-1/2, -I*d*x^2))*sin(1/4*pi + 1/2*arctan2(0, d)) + (gamma(-1/2, I*d*x^2) + gamma(-1/2, -I*d*x^2))*sin(-1/
4*pi + 1/2*arctan2(0, d)))*cos(c) + ((gamma(-1/2, I*d*x^2) + gamma(-1/2, -I*d*x^2))*cos(1/4*pi + 1/2*arctan2(0
, d)) + (gamma(-1/2, I*d*x^2) + gamma(-1/2, -I*d*x^2))*cos(-1/4*pi + 1/2*arctan2(0, d)) + (I*gamma(-1/2, I*d*x
^2) - I*gamma(-1/2, -I*d*x^2))*sin(1/4*pi + 1/2*arctan2(0, d)) + (-I*gamma(-1/2, I*d*x^2) + I*gamma(-1/2, -I*d
*x^2))*sin(-1/4*pi + 1/2*arctan2(0, d)))*sin(c))*b/x - a/x

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Fricas [A]  time = 1.93645, size = 221, normalized size = 2.51 \begin{align*} \frac{\sqrt{2} \pi b x \sqrt{\frac{d}{\pi }} \cos \left (c\right ) \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) - \sqrt{2} \pi b x \sqrt{\frac{d}{\pi }} \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) \sin \left (c\right ) - b \sin \left (d x^{2} + c\right ) - a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*pi*b*x*sqrt(d/pi)*cos(c)*fresnel_cos(sqrt(2)*x*sqrt(d/pi)) - sqrt(2)*pi*b*x*sqrt(d/pi)*fresnel_sin(sq
rt(2)*x*sqrt(d/pi))*sin(c) - b*sin(d*x^2 + c) - a)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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