3.55 \(\int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=142 \[ \frac {\sqrt {\pi } \sqrt {d} \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}-\frac {\sqrt {\pi } \sqrt {d} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{4 b} \]

[Out]

1/8*cos(2*a-2*b*c/d)*FresnelC(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*d^(1/2)*Pi^(1/2)/b^(3/2)-1/8*FresnelS(
2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*sin(2*a-2*b*c/d)*d^(1/2)*Pi^(1/2)/b^(3/2)-1/4*cos(2*b*x+2*a)*(d*x+c)
^(1/2)/b

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Rubi [A]  time = 0.22, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4406, 12, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac {\sqrt {\pi } \sqrt {d} \cos \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{8 b^{3/2}}-\frac {\sqrt {\pi } \sqrt {d} \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3304

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

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 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 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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 134, normalized size = 0.94 \[ \frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )-2 \sqrt {\frac {b}{d}} \sqrt {c+d x} \cos (2 (a+b x))}{8 b \sqrt {\frac {b}{d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(-2*Sqrt[b/d]*Sqrt[c + d*x]*Cos[2*(a + b*x)] + Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*
x])/Sqrt[Pi]] - Sqrt[Pi]*FresnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d])/(8*b*Sqrt[b/d])

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fricas [A]  time = 0.47, size = 125, normalized size = 0.88 \[ \frac {\pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - \pi d \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (2 \, b \cos \left (b x + a\right )^{2} - b\right )} \sqrt {d x + c}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

1/8*(pi*d*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - pi*d*sqrt(b/(pi*d
))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 2*(2*b*cos(b*x + a)^2 - b)*sqrt(d*x + c
))/b^2

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giac [C]  time = 0.40, size = 402, normalized size = 2.83 \[ -\frac {4 \, {\left (\frac {i \, \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} - \frac {i \, \sqrt {\pi } d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}\right )} c - \frac {i \, \sqrt {\pi } {\left (4 \, b c + i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {2 i \, b c - 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {i \, \sqrt {\pi } {\left (4 \, b c - i \, d\right )} d \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (\frac {-2 i \, b c + 2 i \, a d}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} + \frac {2 \, \sqrt {d x + c} d e^{\left (\frac {2 i \, {\left (d x + c\right )} b - 2 i \, b c + 2 i \, a d}{d}\right )}}{b} + \frac {2 \, \sqrt {d x + c} d e^{\left (\frac {-2 i \, {\left (d x + c\right )} b + 2 i \, b c - 2 i \, a d}{d}\right )}}{b}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

-1/16*(4*(I*sqrt(pi)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sq
rt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) - I*sqrt(pi)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e
^((-2*I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)))*c - I*sqrt(pi)*(4*b*c + I*d)*d*erf(-sqrt(b*d
)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b)
+ I*sqrt(pi)*(4*b*c - I*d)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*b*c + 2*I*a*
d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) + 2*sqrt(d*x + c)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)
/b + 2*sqrt(d*x + c)*d*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b)/d

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maple [A]  time = 0.00, size = 142, normalized size = 1.00 \[ \frac {-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a),x)

[Out]

2/d*(-1/8/b*d*(d*x+c)^(1/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+1/16/b*d*Pi^(1/2)/(b/d)^(1/2)*(cos(2*(a*d-b*c)/d)
*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)-sin(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)
^(1/2)*b/d)))

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maxima [C]  time = 0.60, size = 209, normalized size = 1.47 \[ -\frac {\sqrt {2} {\left (8 \, \sqrt {2} \sqrt {d x + c} b \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + {\left (\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) + {\left (-\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{64 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

-1/64*sqrt(2)*(8*sqrt(2)*sqrt(d*x + c)*b*cos(2*((d*x + c)*b - b*c + a*d)/d) + ((I - 1)*4^(1/4)*sqrt(pi)*d*(b^2
/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I + 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt
(d*x + c)*sqrt(2*I*b/d)) + (-(I + 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (I - 1)*4^(1/4
)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*sin(a + b*x)*(c + d*x)^(1/2),x)

[Out]

int(cos(a + b*x)*sin(a + b*x)*(c + d*x)^(1/2), x)

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sympy [B]  time = 6.21, size = 389, normalized size = 2.74 \[ - \frac {b^{\frac {3}{2}} \sqrt {\frac {d}{b}} \left (c + d x\right )^{\frac {5}{2}} \cos {\left (2 a - \frac {2 b c}{d} \right )} \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {9}{4} \end {matrix}\middle | {- \frac {b^{2} \left (c + d x\right )^{2}}{d^{2}}} \right )}}{4 d^{\frac {5}{2}} \Gamma \left (\frac {7}{4}\right ) \Gamma \left (\frac {9}{4}\right )} - \frac {\sqrt {b} \sqrt {\frac {d}{b}} \left (c + d x\right )^{\frac {3}{2}} \sin {\left (2 a - \frac {2 b c}{d} \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {1}{2}, \frac {5}{4}, \frac {7}{4} \end {matrix}\middle | {- \frac {b^{2} \left (c + d x\right )^{2}}{d^{2}}} \right )}}{8 d^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {\pi } c \sqrt {\frac {d}{b}} \sin {\left (2 a - \frac {2 b c}{d} \right )} C\left (\frac {2 b \sqrt {c + d x}}{\sqrt {\pi } d \sqrt {\frac {b}{d}}}\right )}{2 d} + \frac {\sqrt {\pi } c \sqrt {\frac {d}{b}} \cos {\left (2 a - \frac {2 b c}{d} \right )} S\left (\frac {2 b \sqrt {c + d x}}{\sqrt {\pi } d \sqrt {\frac {b}{d}}}\right )}{2 d} + \frac {\sqrt {\pi } x \sqrt {\frac {d}{b}} \sin {\left (2 a - \frac {2 b c}{d} \right )} C\left (\frac {2 b \sqrt {c + d x}}{\sqrt {\pi } d \sqrt {\frac {b}{d}}}\right )}{2} + \frac {\sqrt {\pi } x \sqrt {\frac {d}{b}} \cos {\left (2 a - \frac {2 b c}{d} \right )} S\left (\frac {2 b \sqrt {c + d x}}{\sqrt {\pi } d \sqrt {\frac {b}{d}}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(3/2)*sqrt(d/b)*(c + d*x)**(5/2)*cos(2*a - 2*b*c/d)*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4
), -b**2*(c + d*x)**2/d**2)/(4*d**(5/2)*gamma(7/4)*gamma(9/4)) - sqrt(b)*sqrt(d/b)*(c + d*x)**(3/2)*sin(2*a -
2*b*c/d)*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2, 5/4, 7/4), -b**2*(c + d*x)**2/d**2)/(8*d**(3/2)*gamma(5
/4)*gamma(7/4)) + sqrt(pi)*c*sqrt(d/b)*sin(2*a - 2*b*c/d)*fresnelc(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/(
2*d) + sqrt(pi)*c*sqrt(d/b)*cos(2*a - 2*b*c/d)*fresnels(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/(2*d) + sqrt
(pi)*x*sqrt(d/b)*sin(2*a - 2*b*c/d)*fresnelc(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/2 + sqrt(pi)*x*sqrt(d/b
)*cos(2*a - 2*b*c/d)*fresnels(2*b*sqrt(c + d*x)/(sqrt(pi)*d*sqrt(b/d)))/2

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