3.154 \(\int (e+f x)^2 \sin (b (c+d x)^2) \, dx\)

Optimal. Leaf size=150 \[ \frac{\sqrt{\frac{\pi }{2}} f^2 \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{2 b^{3/2} d^3}+\frac{\sqrt{\frac{\pi }{2}} (d e-c f)^2 S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d^3}-\frac{f (d e-c f) \cos \left (b (c+d x)^2\right )}{b d^3}-\frac{f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3} \]

[Out]

-((f*(d*e - c*f)*Cos[b*(c + d*x)^2])/(b*d^3)) - (f^2*(c + d*x)*Cos[b*(c + d*x)^2])/(2*b*d^3) + (f^2*Sqrt[Pi/2]
*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(2*b^(3/2)*d^3) + ((d*e - c*f)^2*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/P
i]*(c + d*x)])/(Sqrt[b]*d^3)

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Rubi [A]  time = 0.163992, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3433, 3351, 3379, 2638, 3385, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} f^2 \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{2 b^{3/2} d^3}+\frac{\sqrt{\frac{\pi }{2}} (d e-c f)^2 S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d^3}-\frac{f (d e-c f) \cos \left (b (c+d x)^2\right )}{b d^3}-\frac{f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*Sin[b*(c + d*x)^2],x]

[Out]

-((f*(d*e - c*f)*Cos[b*(c + d*x)^2])/(b*d^3)) - (f^2*(c + d*x)*Cos[b*(c + d*x)^2])/(2*b*d^3) + (f^2*Sqrt[Pi/2]
*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(2*b^(3/2)*d^3) + ((d*e - c*f)^2*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/P
i]*(c + d*x)])/(Sqrt[b]*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 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3385

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

Rubi steps

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

Mathematica [A]  time = 0.618085, size = 117, normalized size = 0.78 \[ \frac{2 \sqrt{2 \pi } b (d e-c f)^2 S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )-2 \sqrt{b} f \cos \left (b (c+d x)^2\right ) (-c f+2 d e+d f x)+\sqrt{2 \pi } f^2 \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{4 b^{3/2} d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b]*f*(2*d*e - c*f + d*f*x)*Cos[b*(c + d*x)^2] + f^2*Sqrt[2*Pi]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)]
+ 2*b*(d*e - c*f)^2*Sqrt[2*Pi]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(4*b^(3/2)*d^3)

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Maple [B]  time = 0.005, size = 291, normalized size = 1.9 \begin{align*} -{\frac{{f}^{2}x\cos \left ({d}^{2}{x}^{2}b+2\,cdxb+{c}^{2}b \right ) }{2\,{d}^{2}b}}-{\frac{{f}^{2}c}{d} \left ( -{\frac{\cos \left ({d}^{2}{x}^{2}b+2\,cdxb+{c}^{2}b \right ) }{2\,{d}^{2}b}}-{\frac{c\sqrt{2}\sqrt{\pi }}{2\,d}{\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}} \right ) }+{\frac{{f}^{2}\sqrt{2}\sqrt{\pi }}{4\,{d}^{2}b}{\it FresnelC} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}}-{\frac{ef\cos \left ({d}^{2}{x}^{2}b+2\,cdxb+{c}^{2}b \right ) }{{d}^{2}b}}-{\frac{efc\sqrt{2}\sqrt{\pi }}{d}{\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}}+{\frac{\sqrt{2}\sqrt{\pi }{e}^{2}}{2}{\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*f^2/d^2/b*x*cos(b*d^2*x^2+2*b*c*d*x+b*c^2)-f^2*c/d*(-1/2/d^2/b*cos(b*d^2*x^2+2*b*c*d*x+b*c^2)-1/2*c/d*2^(
1/2)*Pi^(1/2)/(d^2*b)^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d)))+1/4*f^2/d^2/b*2^(1/2)*Pi
^(1/2)/(d^2*b)^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-e*f/d^2/b*cos(b*d^2*x^2+2*b*c*d*
x+b*c^2)-e*f*c/d*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))+1/2*2
^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*e^2*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))

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Maxima [C]  time = 3.69617, size = 2245, normalized size = 14.97 \begin{align*} \text{result too large to display} \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]

1/8*sqrt(pi)*((I*cos(1/4*pi + 1/2*arctan2(0, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arcta
n2(0, b)) - sin(-1/4*pi + 1/2*arctan2(0, b)))*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (I*cos(1/4*pi + 1/2*arctan2(0
, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) - sin(1/4*pi + 1/2*arctan2(0, b)) + sin(-1/4*pi + 1/2*arctan2(0, b)
))*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))*e^2/(d*sqrt(abs(b))) - 1/2*((e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + e
^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2) - (((-2*I*sqrt(pi)*(erf(sqrt(I
*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + 2*I*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1)
)*b*c*d*x + (-2*I*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + 2*I*sqrt(pi)*(erf(sqrt(-I*b*
d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2)*cos(1/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)) - 2*((sq
rt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x -
I*b*c^2)) - 1))*b*c*d*x + (sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + sqrt(pi)*(erf(sqrt(
-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2)*sin(1/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)))*sq
rt(abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2)))*e*f/(b*d^2*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2)) + 1/32*(16*(b*d
^2*x^2 + 2*b*c*d*x + b*c^2)^2*c*(e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b
*c^2)) - (((-I*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + I*sqrt(pi)*(erf(sqrt(-I*b*d^2*x
^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2*d*x + (-I*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) -
1) + I*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^3)*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*
b*c^2)*cos(1/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)) - ((sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x
+ I*b*c^2)) - 1) + sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2*d*x + (sqrt(pi)*(erf
(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) -
1))*b*c^3)*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2)*sin(1/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)) + (b
*d^3*x^3*(4*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 4*I*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*
b*c^2)) + b*c*d^2*x^2*(12*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 12*I*gamma(3/2, -I*b*d^2*x^2 - 2
*I*b*c*d*x - I*b*c^2)) + b*c^2*d*x*(12*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 12*I*gamma(3/2, -I*
b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) + b*c^3*(4*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 4*I*gamma(3
/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)))*cos(3/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)) + 4*(b*d^3*
x^3*(gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) + 3*b
*c*d^2*x^2*(gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)
) + 3*b*c^2*d*x*(gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b
*c^2)) + b*c^3*(gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*
c^2)))*sin(3/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)))*sqrt(abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2)))*f
^2/((b*d^2*x^2 + 2*b*c*d*x + b*c^2)^2*b*d^3)

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Fricas [A]  time = 1.61034, size = 392, normalized size = 2.61 \begin{align*} \frac{\sqrt{2} \pi \sqrt{\frac{b d^{2}}{\pi }} f^{2} \operatorname{C}\left (\frac{\sqrt{2} \sqrt{\frac{b d^{2}}{\pi }}{\left (d x + c\right )}}{d}\right ) + 2 \, \sqrt{2} \pi{\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \sqrt{\frac{b d^{2}}{\pi }} \operatorname{S}\left (\frac{\sqrt{2} \sqrt{\frac{b d^{2}}{\pi }}{\left (d x + c\right )}}{d}\right ) - 2 \,{\left (b d^{2} f^{2} x + 2 \, b d^{2} e f - b c d f^{2}\right )} \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )}{4 \, b^{2} d^{4}} \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]

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

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

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Giac [C]  time = 1.19738, size = 903, normalized size = 6.02 \begin{align*} \text{result too large to display} \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]

-1/4*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))*e^2/(sqrt(b*d^2)*(
I*b*d^2/sqrt(b^2*d^4) + 1)) + 1/4*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)
*(x + c/d))*e^2/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) - 1/2*(-I*sqrt(2)*sqrt(pi)*c*f*erf(-1/2*sqrt(2)*sqr
t(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))*e/(sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)) + f*e^(-I*b*d^2*x^
2 - 2*I*b*c*d*x - I*b*c^2 + 1)/(b*d))/d - 1/2*(I*sqrt(2)*sqrt(pi)*c*f*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(-I*b*d^2/s
qrt(b^2*d^4) + 1)*(x + c/d))*e/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) + f*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I
*b*c^2 + 1)/(b*d))/d - 1/8*(I*sqrt(2)*sqrt(pi)*(2*b*c^2*f^2 - I*f^2)*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(I*b*d^2/sqr
t(b^2*d^4) + 1)*(x + c/d))/(sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*b) + 2*I*(d*f^2*(-I*x - I*c/d) + 2*I*c*f^2
)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)/(b*d))/d^2 - 1/8*(-I*sqrt(2)*sqrt(pi)*(2*b*c^2*f^2 + I*f^2)*erf(-1/
2*sqrt(2)*sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)*b) + 2
*I*(d*f^2*(-I*x - I*c/d) + 2*I*c*f^2)*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)/(b*d))/d^2