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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.23, size = 119, normalized size = 1.04 \[ -\frac {a}{3 x^3}-\frac {2}{3} \sqrt {2 \pi } b d^{3/2} \left (\sin (c) C\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )+\cos (c) S\left (\sqrt {d} \sqrt {\frac {2}{\pi }} x\right )\right )-\frac {b \cos \left (d x^2\right ) \left (2 d x^2 \cos (c)+\sin (c)\right )}{3 x^3}+\frac {b \sin \left (d x^2\right ) \left (2 d x^2 \sin (c)-\cos (c)\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a/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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fricas [A]  time = 0.65, size = 98, normalized size = 0.86 \[ -\frac {2 \, \sqrt {2} \pi b d x^{3} \sqrt {\frac {d}{\pi }} \cos \relax (c) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) + 2 \, \sqrt {2} \pi b d x^{3} \sqrt {\frac {d}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {d}{\pi }}\right ) \sin \relax (c) + 2 \, b d x^{2} \cos \left (d x^{2} + c\right ) + b \sin \left (d x^{2} + c\right ) + a}{3 \, x^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \sin \left (d x^{2} + c\right ) + a}{x^{4}}\,{d x} \]

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)

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maple [A]  time = 0.03, size = 83, normalized size = 0.73 \[ -\frac {a}{3 x^{3}}+b \left (-\frac {\sin \left (d \,x^{2}+c \right )}{3 x^{3}}+\frac {2 d \left (-\frac {\cos \left (d \,x^{2}+c \right )}{x}-\sqrt {d}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (c ) \mathrm {S}\left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \relax (c ) \FresnelC \left (\frac {x \sqrt {d}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )\right )}{3}\right ) \]

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 = 1.04, size = 82, normalized size = 0.72 \[ -\frac {\sqrt {d x^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, i \, d x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -i \, d x^{2}\right )\right )} \cos \relax (c) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, i \, d x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -i \, d x^{2}\right )\right )} \sin \relax (c)\right )} b d}{8 \, x} - \frac {a}{3 \, x^{3}} \]

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(d*x^2)*((-(I + 1)*sqrt(2)*gamma(-3/2, I*d*x^2) + (I - 1)*sqrt(2)*gamma(-3/2, -I*d*x^2))*cos(c) + ((I
- 1)*sqrt(2)*gamma(-3/2, I*d*x^2) - (I + 1)*sqrt(2)*gamma(-3/2, -I*d*x^2))*sin(c))*b*d/x - 1/3*a/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\sin \left (d\,x^2+c\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x^2))/x^4,x)

[Out]

int((a + b*sin(c + d*x^2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \sin {\left (c + d x^{2} \right )}}{x^{4}}\, dx \]

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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