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

Optimal. Leaf size=198 \[ -\frac{b f \cos (a) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2}-\frac{\sqrt{2 \pi } \sqrt{b} \cos (a) (d e-c f) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d^2}+\frac{\sqrt{2 \pi } \sqrt{b} \sin (a) (d e-c f) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^2}+\frac{(c+d x) (d e-c f) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^2}+\frac{b f \sin (a) \text{Si}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2}+\frac{f (c+d x)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.259283, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {3433, 3359, 3387, 3354, 3352, 3351, 3379, 3297, 3303, 3299, 3302} \[ -\frac{b f \cos (a) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2}-\frac{\sqrt{2 \pi } \sqrt{b} \cos (a) (d e-c f) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d^2}+\frac{\sqrt{2 \pi } \sqrt{b} \sin (a) (d e-c f) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^2}+\frac{(c+d x) (d e-c f) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^2}+\frac{b f \sin (a) \text{Si}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2}+\frac{f (c+d x)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3433

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

Rule 3359

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(
a + b*Sin[c + d/x^n])^p/x^2, x], x, 1/(e + f*x)], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[n,
0] && EqQ[n, -2]

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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3297

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

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int (e+f x) \sin \left (a+\frac{b}{(c+d x)^2}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (d e \left (1-\frac{c f}{d e}\right ) \sin \left (a+\frac{b}{x^2}\right )+f x \sin \left (a+\frac{b}{x^2}\right )\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac{f \operatorname{Subst}\left (\int x \sin \left (a+\frac{b}{x^2}\right ) \, dx,x,c+d x\right )}{d^2}+\frac{(d e-c f) \operatorname{Subst}\left (\int \sin \left (a+\frac{b}{x^2}\right ) \, dx,x,c+d x\right )}{d^2}\\ &=-\frac{f \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^2} \, dx,x,\frac{1}{(c+d x)^2}\right )}{2 d^2}-\frac{(d e-c f) \operatorname{Subst}\left (\int \frac{\sin \left (a+b x^2\right )}{x^2} \, dx,x,\frac{1}{c+d x}\right )}{d^2}\\ &=\frac{(d e-c f) (c+d x) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^2}+\frac{f (c+d x)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{2 d^2}-\frac{(b f) \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x} \, dx,x,\frac{1}{(c+d x)^2}\right )}{2 d^2}-\frac{(2 b (d e-c f)) \operatorname{Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\frac{1}{c+d x}\right )}{d^2}\\ &=\frac{(d e-c f) (c+d x) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^2}+\frac{f (c+d x)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{2 d^2}-\frac{(b f \cos (a)) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,\frac{1}{(c+d x)^2}\right )}{2 d^2}-\frac{(2 b (d e-c f) \cos (a)) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{c+d x}\right )}{d^2}+\frac{(b f \sin (a)) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x} \, dx,x,\frac{1}{(c+d x)^2}\right )}{2 d^2}+\frac{(2 b (d e-c f) \sin (a)) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac{1}{c+d x}\right )}{d^2}\\ &=-\frac{b f \cos (a) \text{Ci}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2}-\frac{\sqrt{b} (d e-c f) \sqrt{2 \pi } \cos (a) C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^2}+\frac{\sqrt{b} (d e-c f) \sqrt{2 \pi } S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right ) \sin (a)}{d^2}+\frac{(d e-c f) (c+d x) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^2}+\frac{f (c+d x)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{2 d^2}+\frac{b f \sin (a) \text{Si}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.771596, size = 242, normalized size = 1.22 \[ \frac{c^2 (-f) \sin \left (a+\frac{b}{(c+d x)^2}\right )-b f \cos (a) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )+2 d^2 e x \sin \left (a+\frac{b}{(c+d x)^2}\right )+d^2 f x^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )-2 \sqrt{2 \pi } \sqrt{b} \cos (a) (d e-c f) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )+2 \sqrt{2 \pi } \sqrt{b} d e \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )+2 c d e \sin \left (a+\frac{b}{(c+d x)^2}\right )-2 \sqrt{2 \pi } \sqrt{b} c f \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )+b f \sin (a) \text{Si}\left (\frac{b}{(c+d x)^2}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*f*Cos[a]*CosIntegral[b/(c + d*x)^2]) - 2*Sqrt[b]*(d*e - c*f)*Sqrt[2*Pi]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/P
i])/(c + d*x)] + 2*Sqrt[b]*d*e*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]*Sin[a] - 2*Sqrt[b]*c*f*Sqrt
[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]*Sin[a] + 2*c*d*e*Sin[a + b/(c + d*x)^2] - c^2*f*Sin[a + b/(c +
d*x)^2] + 2*d^2*e*x*Sin[a + b/(c + d*x)^2] + d^2*f*x^2*Sin[a + b/(c + d*x)^2] + b*f*Sin[a]*SinIntegral[b/(c +
d*x)^2])/(2*d^2)

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Maple [A]  time = 0.013, size = 150, normalized size = 0.8 \begin{align*}{\frac{1}{{d}^{2}} \left ( - \left ( cf-de \right ) \left ( dx+c \right ) \sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) + \left ( cf-de \right ) \sqrt{b}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) -\sin \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) \right ) +{\frac{f \left ( dx+c \right ) ^{2}}{2}\sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }-fb \left ({\frac{\cos \left ( a \right ) }{2}{\it Ci} \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }-{\frac{\sin \left ( a \right ) }{2}{\it Si} \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2*(-(c*f-d*e)*(d*x+c)*sin(a+b/(d*x+c)^2)+(c*f-d*e)*b^(1/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelC(b^(1/2)*2^(1/
2)/Pi^(1/2)/(d*x+c))-sin(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)))+1/2*f*(d*x+c)^2*sin(a+b/(d*x+c)^2)-f*b
*(1/2*cos(a)*Ci(b/(d*x+c)^2)-1/2*sin(a)*Si(b/(d*x+c)^2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (f x^{2} + 2 \, e x\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \int \frac{{\left (b d f x^{2} + 2 \, b d e x\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}}\,{d x} + \int \frac{{\left (b d f x^{2} + 2 \, b d e x\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left ({\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} +{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(f*x^2 + 2*e*x)*sin((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) + integrate(1/2*(b*d*f*
x^2 + 2*b*d*e*x)*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3
*c^2*d*x + c^3), x) + integrate(1/2*(b*d*f*x^2 + 2*b*d*e*x)*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 +
2*c*d*x + c^2))/((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 +
2*c*d*x + c^2))^2 + (d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*sin((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^
2 + 2*c*d*x + c^2))^2), x)

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Fricas [A]  time = 1.98199, size = 670, normalized size = 3.38 \begin{align*} -\frac{4 \, \sqrt{2} \pi{\left (d^{2} e - c d f\right )} \sqrt{\frac{b}{\pi d^{2}}} \cos \left (a\right ) \operatorname{C}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) - 4 \, \sqrt{2} \pi{\left (d^{2} e - c d f\right )} \sqrt{\frac{b}{\pi d^{2}}} \operatorname{S}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) \sin \left (a\right ) - 2 \, b f \sin \left (a\right ) \operatorname{Si}\left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) +{\left (b f \operatorname{Ci}\left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + b f \operatorname{Ci}\left (-\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )\right )} \cos \left (a\right ) - 2 \,{\left (d^{2} f x^{2} + 2 \, d^{2} e x + 2 \, c d e - c^{2} f\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{4 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*pi*(d^2*e - c*d*f)*sqrt(b/(pi*d^2))*cos(a)*fresnel_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) -
4*sqrt(2)*pi*(d^2*e - c*d*f)*sqrt(b/(pi*d^2))*fresnel_sin(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c))*sin(a) - 2*b*
f*sin(a)*sin_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) + (b*f*cos_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) + b*f*cos_
integral(-b/(d^2*x^2 + 2*c*d*x + c^2)))*cos(a) - 2*(d^2*f*x^2 + 2*d^2*e*x + 2*c*d*e - c^2*f)*sin((a*d^2*x^2 +
2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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