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

Optimal. Leaf size=114 \[ -\frac{a}{3 x^3}-\frac{2}{3} \sqrt{2 \pi } b d^{3/2} \sin (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\frac{2}{3} \sqrt{2 \pi } b d^{3/2} \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{b \sin \left (c+d x^2\right )}{3 x^3}-\frac{2 b d \cos \left (c+d x^2\right )}{3 x} \]

[Out]

-a/(3*x^3) - (2*b*d*Cos[c + d*x^2])/(3*x) - (2*b*d^(3/2)*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x])/3 -
(2*b*d^(3/2)*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/3 - (b*Sin[c + d*x^2])/(3*x^3)

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Rubi [A]  time = 0.0907686, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {14, 3387, 3388, 3353, 3352, 3351} \[ -\frac{a}{3 x^3}-\frac{2}{3} \sqrt{2 \pi } b d^{3/2} \sin (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\frac{2}{3} \sqrt{2 \pi } b d^{3/2} \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{b \sin \left (c+d x^2\right )}{3 x^3}-\frac{2 b d \cos \left (c+d x^2\right )}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])/x^4,x]

[Out]

-a/(3*x^3) - (2*b*d*Cos[c + d*x^2])/(3*x) - (2*b*d^(3/2)*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x])/3 -
(2*b*d^(3/2)*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/3 - (b*Sin[c + d*x^2])/(3*x^3)

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 3388

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

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[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^4} \, dx &=\int \left (\frac{a}{x^4}+\frac{b \sin \left (c+d x^2\right )}{x^4}\right ) \, dx\\ &=-\frac{a}{3 x^3}+b \int \frac{\sin \left (c+d x^2\right )}{x^4} \, dx\\ &=-\frac{a}{3 x^3}-\frac{b \sin \left (c+d x^2\right )}{3 x^3}+\frac{1}{3} (2 b d) \int \frac{\cos \left (c+d x^2\right )}{x^2} \, dx\\ &=-\frac{a}{3 x^3}-\frac{2 b d \cos \left (c+d x^2\right )}{3 x}-\frac{b \sin \left (c+d x^2\right )}{3 x^3}-\frac{1}{3} \left (4 b d^2\right ) \int \sin \left (c+d x^2\right ) \, dx\\ &=-\frac{a}{3 x^3}-\frac{2 b d \cos \left (c+d x^2\right )}{3 x}-\frac{b \sin \left (c+d x^2\right )}{3 x^3}-\frac{1}{3} \left (4 b d^2 \cos (c)\right ) \int \sin \left (d x^2\right ) \, dx-\frac{1}{3} \left (4 b d^2 \sin (c)\right ) \int \cos \left (d x^2\right ) \, dx\\ &=-\frac{a}{3 x^3}-\frac{2 b d \cos \left (c+d x^2\right )}{3 x}-\frac{2}{3} b d^{3/2} \sqrt{2 \pi } \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{2}{3} b d^{3/2} \sqrt{2 \pi } C\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right ) \sin (c)-\frac{b \sin \left (c+d x^2\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.215891, size = 119, normalized size = 1.04 \[ -\frac{a}{3 x^3}-\frac{2}{3} \sqrt{2 \pi } b d^{3/2} \left (\sin (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )+\cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )\right )-\frac{b \cos \left (d x^2\right ) \left (2 d x^2 \cos (c)+\sin (c)\right )}{3 x^3}+\frac{b \sin \left (d x^2\right ) \left (2 d x^2 \sin (c)-\cos (c)\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(3*x^3) - (b*Cos[d*x^2]*(2*d*x^2*Cos[c] + Sin[c]))/(3*x^3) - (2*b*d^(3/2)*Sqrt[2*Pi]*(Cos[c]*FresnelS[Sqrt[
d]*Sqrt[2/Pi]*x] + FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c]))/3 + (b*(-Cos[c] + 2*d*x^2*Sin[c])*Sin[d*x^2])/(3*x^
3)

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Maple [A]  time = 0.009, size = 83, normalized size = 0.7 \begin{align*} -{\frac{a}{3\,{x}^{3}}}+b \left ( -{\frac{\sin \left ( d{x}^{2}+c \right ) }{3\,{x}^{3}}}+{\frac{2\,d}{3} \left ( -{\frac{\cos \left ( d{x}^{2}+c \right ) }{x}}-\sqrt{d}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( c \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{d}} \right ) +\sin \left ( c \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{d}} \right ) \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^4,x)

[Out]

-1/3*a/x^3+b*(-1/3/x^3*sin(d*x^2+c)+2/3*d*(-1/x*cos(d*x^2+c)-d^(1/2)*2^(1/2)*Pi^(1/2)*(cos(c)*FresnelS(x*d^(1/
2)*2^(1/2)/Pi^(1/2))+sin(c)*FresnelC(x*d^(1/2)*2^(1/2)/Pi^(1/2)))))

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Maxima [C]  time = 1.17759, size = 369, normalized size = 3.24 \begin{align*} -\frac{\sqrt{x^{2}{\left | d \right |}}{\left ({\left ({\left (i \, \Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) - i \, \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \cos \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) - i \, \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \cos \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right ) -{\left (\Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \sin \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \sin \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right )\right )} \cos \left (c\right ) +{\left ({\left (\Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \cos \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right ) +{\left (\Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) + \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \cos \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right ) +{\left (i \, \Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) - i \, \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \sin \left (\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right ) +{\left (-i \, \Gamma \left (-\frac{3}{2}, i \, d x^{2}\right ) + i \, \Gamma \left (-\frac{3}{2}, -i \, d x^{2}\right )\right )} \sin \left (-\frac{3}{4} \, \pi + \frac{3}{2} \, \arctan \left (0, d\right )\right )\right )} \sin \left (c\right )\right )} b{\left | d \right |}}{8 \, x} - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(x^2*abs(d))*(((I*gamma(-3/2, I*d*x^2) - I*gamma(-3/2, -I*d*x^2))*cos(3/4*pi + 3/2*arctan2(0, d)) + (
I*gamma(-3/2, I*d*x^2) - I*gamma(-3/2, -I*d*x^2))*cos(-3/4*pi + 3/2*arctan2(0, d)) - (gamma(-3/2, I*d*x^2) + g
amma(-3/2, -I*d*x^2))*sin(3/4*pi + 3/2*arctan2(0, d)) + (gamma(-3/2, I*d*x^2) + gamma(-3/2, -I*d*x^2))*sin(-3/
4*pi + 3/2*arctan2(0, d)))*cos(c) + ((gamma(-3/2, I*d*x^2) + gamma(-3/2, -I*d*x^2))*cos(3/4*pi + 3/2*arctan2(0
, d)) + (gamma(-3/2, I*d*x^2) + gamma(-3/2, -I*d*x^2))*cos(-3/4*pi + 3/2*arctan2(0, d)) + (I*gamma(-3/2, I*d*x
^2) - I*gamma(-3/2, -I*d*x^2))*sin(3/4*pi + 3/2*arctan2(0, d)) + (-I*gamma(-3/2, I*d*x^2) + I*gamma(-3/2, -I*d
*x^2))*sin(-3/4*pi + 3/2*arctan2(0, d)))*sin(c))*b*abs(d)/x - 1/3*a/x^3

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Fricas [A]  time = 2.06092, size = 284, normalized size = 2.49 \begin{align*} -\frac{2 \, \sqrt{2} \pi b d x^{3} \sqrt{\frac{d}{\pi }} \cos \left (c\right ) \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) + 2 \, \sqrt{2} \pi b d x^{3} \sqrt{\frac{d}{\pi }} \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) \sin \left (c\right ) + 2 \, b d x^{2} \cos \left (d x^{2} + c\right ) + b \sin \left (d x^{2} + c\right ) + a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*sqrt(2)*pi*b*d*x^3*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt(d/pi)) + 2*sqrt(2)*pi*b*d*x^3*sqrt(d/p
i)*fresnel_cos(sqrt(2)*x*sqrt(d/pi))*sin(c) + 2*b*d*x^2*cos(d*x^2 + c) + b*sin(d*x^2 + c) + a)/x^3

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sin(c + d*x**2))/x**4, 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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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