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

Optimal. Leaf size=239 \[ -\frac{2 a^2+b^2}{6 x^3}-\frac{4}{3} \sqrt{2 \pi } a b d^{3/2} \sin (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\frac{4}{3} \sqrt{2 \pi } a b d^{3/2} \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac{4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac{4}{3} \sqrt{\pi } b^2 d^{3/2} \cos (2 c) \text{FresnelC}\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )-\frac{4}{3} \sqrt{\pi } b^2 d^{3/2} \sin (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )-\frac{2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}+\frac{b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3} \]

[Out]

-(2*a^2 + b^2)/(6*x^3) - (4*a*b*d*Cos[c + d*x^2])/(3*x) + (b^2*Cos[2*c + 2*d*x^2])/(6*x^3) + (4*b^2*d^(3/2)*Sq
rt[Pi]*Cos[2*c]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2
/Pi]*x])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/3 - (4*b^2*d^(3/2)*Sqrt[Pi]*Fres
nelS[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/3 - (2*a*b*Sin[c + d*x^2])/(3*x^3) - (2*b^2*d*Sin[2*c + 2*d*x^2])/(3*x)

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Rubi [A]  time = 0.19728, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3403, 6, 3388, 3387, 3354, 3352, 3351, 3353} \[ -\frac{2 a^2+b^2}{6 x^3}-\frac{4}{3} \sqrt{2 \pi } a b d^{3/2} \sin (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )-\frac{4}{3} \sqrt{2 \pi } a b d^{3/2} \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac{4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac{4}{3} \sqrt{\pi } b^2 d^{3/2} \cos (2 c) \text{FresnelC}\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )-\frac{4}{3} \sqrt{\pi } b^2 d^{3/2} \sin (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )-\frac{2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}+\frac{b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2/x^4,x]

[Out]

-(2*a^2 + b^2)/(6*x^3) - (4*a*b*d*Cos[c + d*x^2])/(3*x) + (b^2*Cos[2*c + 2*d*x^2])/(6*x^3) + (4*b^2*d^(3/2)*Sq
rt[Pi]*Cos[2*c]*FresnelC[(2*Sqrt[d]*x)/Sqrt[Pi]])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2
/Pi]*x])/3 - (4*a*b*d^(3/2)*Sqrt[2*Pi]*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c])/3 - (4*b^2*d^(3/2)*Sqrt[Pi]*Fres
nelS[(2*Sqrt[d]*x)/Sqrt[Pi]]*Sin[2*c])/3 - (2*a*b*Sin[c + d*x^2])/(3*x^3) - (2*b^2*d*Sin[2*c + 2*d*x^2])/(3*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]

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 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 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]

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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^4} \, dx &=\int \left (\frac{a^2}{x^4}+\frac{b^2}{2 x^4}-\frac{b^2 \cos \left (2 c+2 d x^2\right )}{2 x^4}+\frac{2 a b \sin \left (c+d x^2\right )}{x^4}\right ) \, dx\\ &=\int \left (\frac{a^2+\frac{b^2}{2}}{x^4}-\frac{b^2 \cos \left (2 c+2 d x^2\right )}{2 x^4}+\frac{2 a b \sin \left (c+d x^2\right )}{x^4}\right ) \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}+(2 a b) \int \frac{\sin \left (c+d x^2\right )}{x^4} \, dx-\frac{1}{2} b^2 \int \frac{\cos \left (2 c+2 d x^2\right )}{x^4} \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}+\frac{b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}-\frac{2 a b \sin \left (c+d x^2\right )}{3 x^3}+\frac{1}{3} (4 a b d) \int \frac{\cos \left (c+d x^2\right )}{x^2} \, dx+\frac{1}{3} \left (2 b^2 d\right ) \int \frac{\sin \left (2 c+2 d x^2\right )}{x^2} \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}-\frac{4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac{b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}-\frac{2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac{2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}-\frac{1}{3} \left (8 a b d^2\right ) \int \sin \left (c+d x^2\right ) \, dx+\frac{1}{3} \left (8 b^2 d^2\right ) \int \cos \left (2 c+2 d x^2\right ) \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}-\frac{4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac{b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}-\frac{2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac{2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}-\frac{1}{3} \left (8 a b d^2 \cos (c)\right ) \int \sin \left (d x^2\right ) \, dx+\frac{1}{3} \left (8 b^2 d^2 \cos (2 c)\right ) \int \cos \left (2 d x^2\right ) \, dx-\frac{1}{3} \left (8 a b d^2 \sin (c)\right ) \int \cos \left (d x^2\right ) \, dx-\frac{1}{3} \left (8 b^2 d^2 \sin (2 c)\right ) \int \sin \left (2 d x^2\right ) \, dx\\ &=-\frac{2 a^2+b^2}{6 x^3}-\frac{4 a b d \cos \left (c+d x^2\right )}{3 x}+\frac{b^2 \cos \left (2 c+2 d x^2\right )}{6 x^3}+\frac{4}{3} b^2 d^{3/2} \sqrt{\pi } \cos (2 c) C\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )-\frac{4}{3} a b d^{3/2} \sqrt{2 \pi } \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )-\frac{4}{3} a b d^{3/2} \sqrt{2 \pi } C\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right ) \sin (c)-\frac{4}{3} b^2 d^{3/2} \sqrt{\pi } S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right ) \sin (2 c)-\frac{2 a b \sin \left (c+d x^2\right )}{3 x^3}-\frac{2 b^2 d \sin \left (2 c+2 d x^2\right )}{3 x}\\ \end{align*}

Mathematica [A]  time = 0.663367, size = 226, normalized size = 0.95 \[ -\frac{2 a^2+8 \sqrt{2 \pi } a b d^{3/2} x^3 \sin (c) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{d} x\right )+8 \sqrt{2 \pi } a b d^{3/2} x^3 \cos (c) S\left (\sqrt{d} \sqrt{\frac{2}{\pi }} x\right )+4 a b \sin \left (c+d x^2\right )+8 a b d x^2 \cos \left (c+d x^2\right )-8 \sqrt{\pi } b^2 d^{3/2} x^3 \cos (2 c) \text{FresnelC}\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )+8 \sqrt{\pi } b^2 d^{3/2} x^3 \sin (2 c) S\left (\frac{2 \sqrt{d} x}{\sqrt{\pi }}\right )+4 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )+b^2}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*a^2 + b^2 + 8*a*b*d*x^2*Cos[c + d*x^2] - b^2*Cos[2*(c + d*x^2)] - 8*b^2*d^(3/2)*Sqrt[Pi]*x^3*Cos[2*c]*Fres
nelC[(2*Sqrt[d]*x)/Sqrt[Pi]] + 8*a*b*d^(3/2)*Sqrt[2*Pi]*x^3*Cos[c]*FresnelS[Sqrt[d]*Sqrt[2/Pi]*x] + 8*a*b*d^(3
/2)*Sqrt[2*Pi]*x^3*FresnelC[Sqrt[d]*Sqrt[2/Pi]*x]*Sin[c] + 8*b^2*d^(3/2)*Sqrt[Pi]*x^3*FresnelS[(2*Sqrt[d]*x)/S
qrt[Pi]]*Sin[2*c] + 4*a*b*Sin[c + d*x^2] + 4*b^2*d*x^2*Sin[2*(c + d*x^2)])/(6*x^3)

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Maple [A]  time = 0.014, size = 175, normalized size = 0.7 \begin{align*} -{\frac{1}{3\,{x}^{3}} \left ({a}^{2}+{\frac{{b}^{2}}{2}} \right ) }-{\frac{{b}^{2}}{2} \left ( -{\frac{\cos \left ( 2\,d{x}^{2}+2\,c \right ) }{3\,{x}^{3}}}-{\frac{4\,d}{3} \left ( -{\frac{\sin \left ( 2\,d{x}^{2}+2\,c \right ) }{x}}+2\,\sqrt{d}\sqrt{\pi } \left ( \cos \left ( 2\,c \right ){\it FresnelC} \left ( 2\,{\frac{x\sqrt{d}}{\sqrt{\pi }}} \right ) -\sin \left ( 2\,c \right ){\it FresnelS} \left ( 2\,{\frac{x\sqrt{d}}{\sqrt{\pi }}} \right ) \right ) \right ) } \right ) }+2\,ab \left ( -1/3\,{\frac{\sin \left ( d{x}^{2}+c \right ) }{{x}^{3}}}+2/3\,d \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{d}\sqrt{2}}{\sqrt{\pi }}} \right ) +\sin \left ( c \right ){\it FresnelC} \left ({\frac{x\sqrt{d}\sqrt{2}}{\sqrt{\pi }}} \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))^2/x^4,x)

[Out]

-1/3*(a^2+1/2*b^2)/x^3-1/2*b^2*(-1/3/x^3*cos(2*d*x^2+2*c)-4/3*d*(-1/x*sin(2*d*x^2+2*c)+2*d^(1/2)*Pi^(1/2)*(cos
(2*c)*FresnelC(2*x*d^(1/2)/Pi^(1/2))-sin(2*c)*FresnelS(2*x*d^(1/2)/Pi^(1/2)))))+2*a*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 = 1.25199, size = 756, normalized size = 3.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*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))*a*b*abs(d)/x + 1/24*(sqrt(2)*sqrt(x^2*abs(d))*((3*(gamma(-3/2
, 2*I*d*x^2) + gamma(-3/2, -2*I*d*x^2))*cos(3/4*pi + 3/2*arctan2(0, d)) + 3*(gamma(-3/2, 2*I*d*x^2) + gamma(-3
/2, -2*I*d*x^2))*cos(-3/4*pi + 3/2*arctan2(0, d)) + (3*I*gamma(-3/2, 2*I*d*x^2) - 3*I*gamma(-3/2, -2*I*d*x^2))
*sin(3/4*pi + 3/2*arctan2(0, d)) + (-3*I*gamma(-3/2, 2*I*d*x^2) + 3*I*gamma(-3/2, -2*I*d*x^2))*sin(-3/4*pi + 3
/2*arctan2(0, d)))*cos(2*c) + ((-3*I*gamma(-3/2, 2*I*d*x^2) + 3*I*gamma(-3/2, -2*I*d*x^2))*cos(3/4*pi + 3/2*ar
ctan2(0, d)) + (-3*I*gamma(-3/2, 2*I*d*x^2) + 3*I*gamma(-3/2, -2*I*d*x^2))*cos(-3/4*pi + 3/2*arctan2(0, d)) +
3*(gamma(-3/2, 2*I*d*x^2) + gamma(-3/2, -2*I*d*x^2))*sin(3/4*pi + 3/2*arctan2(0, d)) - 3*(gamma(-3/2, 2*I*d*x^
2) + gamma(-3/2, -2*I*d*x^2))*sin(-3/4*pi + 3/2*arctan2(0, d)))*sin(2*c))*x^2*abs(d) - 4)*b^2/x^3 - 1/3*a^2/x^
3

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Fricas [A]  time = 2.44602, size = 556, normalized size = 2.33 \begin{align*} -\frac{4 \, \sqrt{2} \pi a b d x^{3} \sqrt{\frac{d}{\pi }} \cos \left (c\right ) \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) + 4 \, \sqrt{2} \pi a b d x^{3} \sqrt{\frac{d}{\pi }} \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{d}{\pi }}\right ) \sin \left (c\right ) - 4 \, \pi b^{2} d x^{3} \sqrt{\frac{d}{\pi }} \cos \left (2 \, c\right ) \operatorname{C}\left (2 \, x \sqrt{\frac{d}{\pi }}\right ) + 4 \, \pi b^{2} d x^{3} \sqrt{\frac{d}{\pi }} \operatorname{S}\left (2 \, x \sqrt{\frac{d}{\pi }}\right ) \sin \left (2 \, c\right ) + 4 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} + a^{2} + b^{2} + 2 \,{\left (2 \, b^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b\right )} \sin \left (d x^{2} + c\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(4*sqrt(2)*pi*a*b*d*x^3*sqrt(d/pi)*cos(c)*fresnel_sin(sqrt(2)*x*sqrt(d/pi)) + 4*sqrt(2)*pi*a*b*d*x^3*sqrt
(d/pi)*fresnel_cos(sqrt(2)*x*sqrt(d/pi))*sin(c) - 4*pi*b^2*d*x^3*sqrt(d/pi)*cos(2*c)*fresnel_cos(2*x*sqrt(d/pi
)) + 4*pi*b^2*d*x^3*sqrt(d/pi)*fresnel_sin(2*x*sqrt(d/pi))*sin(2*c) + 4*a*b*d*x^2*cos(d*x^2 + c) - b^2*cos(d*x
^2 + c)^2 + a^2 + b^2 + 2*(2*b^2*d*x^2*cos(d*x^2 + c) + a*b)*sin(d*x^2 + c))/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x^{2} + c\right ) + a\right )}^{2}}{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))^2/x^4,x, algorithm="giac")

[Out]

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