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

Optimal. Leaf size=233 \[ \frac{2 \sqrt{2 \pi } b^{3/2} f^2 S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{3 d^3}-\frac{b f (d e-c f) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )}{d^3}-\frac{\sqrt{2 \pi } \sqrt{b} (d e-c f)^2 \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d^3}+\frac{f (c+d x)^2 (d e-c f) \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{(c+d x) (d e-c f)^2 \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}+\frac{2 b f^2 (c+d x) \cos \left (\frac{b}{(c+d x)^2}\right )}{3 d^3} \]

[Out]

(2*b*f^2*(c + d*x)*Cos[b/(c + d*x)^2])/(3*d^3) - (b*f*(d*e - c*f)*CosIntegral[b/(c + d*x)^2])/d^3 - (Sqrt[b]*(
d*e - c*f)^2*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d^3 + (2*b^(3/2)*f^2*Sqrt[2*Pi]*FresnelS[(Sq
rt[b]*Sqrt[2/Pi])/(c + d*x)])/(3*d^3) + ((d*e - c*f)^2*(c + d*x)*Sin[b/(c + d*x)^2])/d^3 + (f*(d*e - c*f)*(c +
d*x)^2*Sin[b/(c + d*x)^2])/d^3 + (f^2*(c + d*x)^3*Sin[b/(c + d*x)^2])/(3*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.249208, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {3433, 3359, 3387, 3352, 3379, 3297, 3302, 3409, 3388, 3351} \[ \frac{2 \sqrt{2 \pi } b^{3/2} f^2 S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{3 d^3}-\frac{b f (d e-c f) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )}{d^3}-\frac{\sqrt{2 \pi } \sqrt{b} (d e-c f)^2 \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d^3}+\frac{f (c+d x)^2 (d e-c f) \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{(c+d x) (d e-c f)^2 \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}+\frac{2 b f^2 (c+d x) \cos \left (\frac{b}{(c+d x)^2}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*Sin[b/(c + d*x)^2],x]

[Out]

(2*b*f^2*(c + d*x)*Cos[b/(c + d*x)^2])/(3*d^3) - (b*f*(d*e - c*f)*CosIntegral[b/(c + d*x)^2])/d^3 - (Sqrt[b]*(
d*e - c*f)^2*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d^3 + (2*b^(3/2)*f^2*Sqrt[2*Pi]*FresnelS[(Sq
rt[b]*Sqrt[2/Pi])/(c + d*x)])/(3*d^3) + ((d*e - c*f)^2*(c + d*x)*Sin[b/(c + d*x)^2])/d^3 + (f*(d*e - c*f)*(c +
d*x)^2*Sin[b/(c + d*x)^2])/d^3 + (f^2*(c + d*x)^3*Sin[b/(c + d*x)^2])/(3*d^3)

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 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 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 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]

Rule 3409

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

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 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 (e+f x)^2 \sin \left (\frac{b}{(c+d x)^2}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (d^2 e^2 \left (1+\frac{c f (-2 d e+c f)}{d^2 e^2}\right ) \sin \left (\frac{b}{x^2}\right )+2 d e f \left (1-\frac{c f}{d e}\right ) x \sin \left (\frac{b}{x^2}\right )+f^2 x^2 \sin \left (\frac{b}{x^2}\right )\right ) \, dx,x,c+d x\right )}{d^3}\\ &=\frac{f^2 \operatorname{Subst}\left (\int x^2 \sin \left (\frac{b}{x^2}\right ) \, dx,x,c+d x\right )}{d^3}+\frac{(2 f (d e-c f)) \operatorname{Subst}\left (\int x \sin \left (\frac{b}{x^2}\right ) \, dx,x,c+d x\right )}{d^3}+\frac{(d e-c f)^2 \operatorname{Subst}\left (\int \sin \left (\frac{b}{x^2}\right ) \, dx,x,c+d x\right )}{d^3}\\ &=-\frac{f^2 \operatorname{Subst}\left (\int \frac{\sin \left (b x^2\right )}{x^4} \, dx,x,\frac{1}{c+d x}\right )}{d^3}-\frac{(f (d e-c f)) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x^2} \, dx,x,\frac{1}{(c+d x)^2}\right )}{d^3}-\frac{(d e-c f)^2 \operatorname{Subst}\left (\int \frac{\sin \left (b x^2\right )}{x^2} \, dx,x,\frac{1}{c+d x}\right )}{d^3}\\ &=\frac{(d e-c f)^2 (c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f (d e-c f) (c+d x)^2 \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}-\frac{\left (2 b f^2\right ) \operatorname{Subst}\left (\int \frac{\cos \left (b x^2\right )}{x^2} \, dx,x,\frac{1}{c+d x}\right )}{3 d^3}-\frac{(b f (d e-c f)) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,\frac{1}{(c+d x)^2}\right )}{d^3}-\frac{\left (2 b (d e-c f)^2\right ) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{c+d x}\right )}{d^3}\\ &=\frac{2 b f^2 (c+d x) \cos \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}-\frac{b f (d e-c f) \text{Ci}\left (\frac{b}{(c+d x)^2}\right )}{d^3}-\frac{\sqrt{b} (d e-c f)^2 \sqrt{2 \pi } C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^3}+\frac{(d e-c f)^2 (c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f (d e-c f) (c+d x)^2 \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}+\frac{\left (4 b^2 f^2\right ) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac{1}{c+d x}\right )}{3 d^3}\\ &=\frac{2 b f^2 (c+d x) \cos \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}-\frac{b f (d e-c f) \text{Ci}\left (\frac{b}{(c+d x)^2}\right )}{d^3}-\frac{\sqrt{b} (d e-c f)^2 \sqrt{2 \pi } C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^3}+\frac{2 b^{3/2} f^2 \sqrt{2 \pi } S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{3 d^3}+\frac{(d e-c f)^2 (c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f (d e-c f) (c+d x)^2 \sin \left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (\frac{b}{(c+d x)^2}\right )}{3 d^3}\\ \end{align*}

Mathematica [A]  time = 0.492441, size = 265, normalized size = 1.14 \[ \frac{2 \sqrt{2 \pi } b^{3/2} f^2 S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )-3 c^2 d e f \sin \left (\frac{b}{(c+d x)^2}\right )+c^3 f^2 \sin \left (\frac{b}{(c+d x)^2}\right )+3 b f (c f-d e) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )+3 c d^2 e^2 \sin \left (\frac{b}{(c+d x)^2}\right )+3 d^3 e^2 x \sin \left (\frac{b}{(c+d x)^2}\right )+3 d^3 e f x^2 \sin \left (\frac{b}{(c+d x)^2}\right )+d^3 f^2 x^3 \sin \left (\frac{b}{(c+d x)^2}\right )-3 \sqrt{2 \pi } \sqrt{b} (d e-c f)^2 \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )+2 b c f^2 \cos \left (\frac{b}{(c+d x)^2}\right )+2 b d f^2 x \cos \left (\frac{b}{(c+d x)^2}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*c*f^2*Cos[b/(c + d*x)^2] + 2*b*d*f^2*x*Cos[b/(c + d*x)^2] + 3*b*f*(-(d*e) + c*f)*CosIntegral[b/(c + d*x)^
2] - 3*Sqrt[b]*(d*e - c*f)^2*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + 2*b^(3/2)*f^2*Sqrt[2*Pi]*Fr
esnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + 3*c*d^2*e^2*Sin[b/(c + d*x)^2] - 3*c^2*d*e*f*Sin[b/(c + d*x)^2] + c^3
*f^2*Sin[b/(c + d*x)^2] + 3*d^3*e^2*x*Sin[b/(c + d*x)^2] + 3*d^3*e*f*x^2*Sin[b/(c + d*x)^2] + d^3*f^2*x^3*Sin[
b/(c + d*x)^2])/(3*d^3)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 225, normalized size = 1. \begin{align*} -{\frac{1}{{d}^{3}} \left ( - \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2} \right ) \left ( dx+c \right ) \sin \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) + \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2} \right ) \sqrt{b}\sqrt{2}\sqrt{\pi }{\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) -{\frac{ \left ( -2\,c{f}^{2}+2\,def \right ) \left ( dx+c \right ) ^{2}}{2}\sin \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }+{\frac{ \left ( -2\,c{f}^{2}+2\,def \right ) b}{2}{\it Ci} \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }-{\frac{{f}^{2} \left ( dx+c \right ) ^{3}}{3}\sin \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }+{\frac{2\,{f}^{2}b}{3} \left ( - \left ( dx+c \right ) \cos \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) -\sqrt{b}\sqrt{2}\sqrt{\pi }{\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, b f^{2} x \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) +{\left (\frac{c^{3} f^{2} \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d^{3}} - 2 \, \int \frac{2 \, b^{2} f^{2} x \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left ({\left (b d^{2} e f - b c d f^{2}\right )} x^{2} +{\left (b d^{2} e^{2} - b c^{2} f^{2}\right )} x\right )} \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}}\,{d x} - 2 \, \int \frac{2 \, b^{2} f^{2} x \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left ({\left (b d^{2} e f - b c d f^{2}\right )} x^{2} +{\left (b d^{2} e^{2} - b c^{2} f^{2}\right )} x\right )} \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left ({\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )} \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} +{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )} \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2}\right )}}\,{d x}\right )} d^{2} +{\left (d^{2} f^{2} x^{3} + 3 \, d^{2} e f x^{2} + 3 \, d^{2} e^{2} x\right )} \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{3 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*b*f^2*x*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) - 3*d^2*integrate(1/3*(2*b^2*d*f^2*x*sin(b/(d^2*x^2 + 2*c*d*x
+ c^2)) + (b*c^3*f^2 - 3*(b*d^3*e*f - b*c*d^2*f^2)*x^2 - 3*(b*d^3*e^2 - b*c^2*d*f^2)*x)*cos(b/(d^2*x^2 + 2*c*d
*x + c^2)))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2), x) - 3*d^2*integrate(1/3*(2*b^2*d*f^2*x*sin(b/(d^
2*x^2 + 2*c*d*x + c^2)) + (b*c^3*f^2 - 3*(b*d^3*e*f - b*c*d^2*f^2)*x^2 - 3*(b*d^3*e^2 - b*c^2*d*f^2)*x)*cos(b/
(d^2*x^2 + 2*c*d*x + c^2)))/((d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2)*cos(b/(d^2*x^2 + 2*c*d*x + c^2))^
2 + (d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2)*sin(b/(d^2*x^2 + 2*c*d*x + c^2))^2), x) + (d^2*f^2*x^3 + 3
*d^2*e*f*x^2 + 3*d^2*e^2*x)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^2

________________________________________________________________________________________

Fricas [A]  time = 1.85872, size = 713, normalized size = 3.06 \begin{align*} \frac{4 \, \sqrt{2} \pi b d f^{2} \sqrt{\frac{b}{\pi d^{2}}} \operatorname{S}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) - 6 \, \sqrt{2} \pi{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \sqrt{\frac{b}{\pi d^{2}}} \operatorname{C}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) + 4 \,{\left (b d f^{2} x + b c f^{2}\right )} \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left (b d e f - b c f^{2}\right )} \operatorname{Ci}\left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left (b d e f - b c f^{2}\right )} \operatorname{Ci}\left (-\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \,{\left (d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 3 \, c d^{2} e^{2} - 3 \, c^{2} d e f + c^{3} f^{2}\right )} \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{6 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*sqrt(2)*pi*b*d*f^2*sqrt(b/(pi*d^2))*fresnel_sin(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) - 6*sqrt(2)*pi*(d
^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*sqrt(b/(pi*d^2))*fresnel_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + 4*(b*d*
f^2*x + b*c*f^2)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) - 3*(b*d*e*f - b*c*f^2)*cos_integral(b/(d^2*x^2 + 2*c*d*x +
c^2)) - 3*(b*d*e*f - b*c*f^2)*cos_integral(-b/(d^2*x^2 + 2*c*d*x + c^2)) + 2*(d^3*f^2*x^3 + 3*d^3*e*f*x^2 + 3*
d^3*e^2*x + 3*c*d^2*e^2 - 3*c^2*d*e*f + c^3*f^2)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^3

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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