3.52 \(\int (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=196 \[ -\frac{15 \sqrt{\pi } d^{5/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{128 b^{7/2}}+\frac{15 \sqrt{\pi } d^{5/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{128 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \cos (2 a+2 b x)}{64 b^3}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b} \]

[Out]

(15*d^2*Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(64*b^3) - ((c + d*x)^(5/2)*Cos[2*a + 2*b*x])/(4*b) - (15*d^(5/2)*Sqrt
[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(128*b^(7/2)) + (15*d^(5/2)*
Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(128*b^(7/2)) + (5*d*(c
+ d*x)^(3/2)*Sin[2*a + 2*b*x])/(16*b^2)

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Rubi [A]  time = 0.447523, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{15 \sqrt{\pi } d^{5/2} \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{\pi } \sqrt{d}}\right )}{128 b^{7/2}}+\frac{15 \sqrt{\pi } d^{5/2} \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{c+d x}}{\sqrt{d} \sqrt{\pi }}\right )}{128 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \cos (2 a+2 b x)}{64 b^3}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(15*d^2*Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(64*b^3) - ((c + d*x)^(5/2)*Cos[2*a + 2*b*x])/(4*b) - (15*d^(5/2)*Sqrt
[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(128*b^(7/2)) + (15*d^(5/2)*
Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(128*b^(7/2)) + (5*d*(c
+ d*x)^(3/2)*Sin[2*a + 2*b*x])/(16*b^2)

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]

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 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 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 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 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 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 (c+d x)^{5/2} \cos (a+b x) \sin (a+b x) \, dx &=\int \frac{1}{2} (c+d x)^{5/2} \sin (2 a+2 b x) \, dx\\ &=\frac{1}{2} \int (c+d x)^{5/2} \sin (2 a+2 b x) \, dx\\ &=-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac{(5 d) \int (c+d x)^{3/2} \cos (2 a+2 b x) \, dx}{8 b}\\ &=-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (15 d^2\right ) \int \sqrt{c+d x} \sin (2 a+2 b x) \, dx}{32 b^2}\\ &=\frac{15 d^2 \sqrt{c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (15 d^3\right ) \int \frac{\cos (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{128 b^3}\\ &=\frac{15 d^2 \sqrt{c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (15 d^3 \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}{128 b^3}+\frac{\left (15 d^3 \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}{128 b^3}\\ &=\frac{15 d^2 \sqrt{c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}-\frac{\left (15 d^2 \cos \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{64 b^3}+\frac{\left (15 d^2 \sin \left (2 a-\frac{2 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{64 b^3}\\ &=\frac{15 d^2 \sqrt{c+d x} \cos (2 a+2 b x)}{64 b^3}-\frac{(c+d x)^{5/2} \cos (2 a+2 b x)}{4 b}-\frac{15 d^{5/2} \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 )}{128 b^{7/2}}+\frac{15 d^{5/2} \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 )}{128 b^{7/2}}+\frac{5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{16 b^2}\\ \end{align*}

Mathematica [A]  time = 2.36958, size = 179, normalized size = 0.91 \[ \frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x} \left (20 b d (c+d x) \sin (2 (a+b x))-\cos (2 (a+b x)) \left (16 b^2 (c+d x)^2-15 d^2\right )\right )-15 \sqrt{\pi } d^2 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )+15 \sqrt{\pi } d^2 \sin \left (2 a-\frac{2 b c}{d}\right ) S\left (\frac{2 \sqrt{\frac{b}{d}} \sqrt{c+d x}}{\sqrt{\pi }}\right )}{128 b^3 \sqrt{\frac{b}{d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)*Cos[a + b*x]*Sin[a + b*x],x]

[Out]

(-15*d^2*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + 15*d^2*Sqrt[Pi]*Fresne
lS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] + 2*Sqrt[b/d]*Sqrt[c + d*x]*(-((-15*d^2 + 16*b^2
*(c + d*x)^2)*Cos[2*(a + b*x)]) + 20*b*d*(c + d*x)*Sin[2*(a + b*x)]))/(128*b^3*Sqrt[b/d])

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Maple [A]  time = 0.037, size = 234, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{d} \left ( -1/8\,{\frac{d \left ( dx+c \right ) ^{5/2}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }+5/8\,{\frac{d}{b} \left ( 1/4\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\sin \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }-3/4\,{\frac{d}{b} \left ( -1/4\,{\frac{d\sqrt{dx+c}}{b}\cos \left ( 2\,{\frac{ \left ( dx+c \right ) b}{d}}+2\,{\frac{ad-bc}{d}} \right ) }+1/8\,{\frac{d\sqrt{\pi }}{b} \left ( \cos \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) -\sin \left ( 2\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{dx+c}b}{d\sqrt{\pi }}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a),x)

[Out]

2/d*(-1/8/b*d*(d*x+c)^(5/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+5/8/b*d*(1/4/b*d*(d*x+c)^(3/2)*sin(2/d*(d*x+c)*b+
2*(a*d-b*c)/d)-3/4/b*d*(-1/4/b*d*(d*x+c)^(1/2)*cos(2/d*(d*x+c)*b+2*(a*d-b*c)/d)+1/8/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 = 1.95552, size = 910, normalized size = 4.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a),x, algorithm="maxima")

[Out]

1/1024*sqrt(2)*(160*sqrt(2)*(d*x + c)^(3/2)*b*d^2*sqrt(abs(b)/abs(d))*sin(2*((d*x + c)*b - b*c + a*d)/d) - 8*(
16*sqrt(2)*(d*x + c)^(5/2)*b^2*d*sqrt(abs(b)/abs(d)) - 15*sqrt(2)*sqrt(d*x + c)*d^3*sqrt(abs(b)/abs(d)))*cos(2
*((d*x + c)*b - b*c + a*d)/d) - ((15*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) +
15*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*sin(1/4*pi + 1/2*ar
ctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/
sqrt(d^2))))*d^3*cos(-2*(b*c - a*d)/d) - (15*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt
(d^2))) + 15*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*p
i + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arcta
n2(0, d/sqrt(d^2))))*d^3*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - ((15*sqrt(pi)*cos(1/4*pi +
1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0
, d/sqrt(d^2))) + 15*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*
sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*cos(-2*(b*c - a*d)/d) - (-15*I*sqrt(pi)*co
s(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) +
1/2*arctan2(0, d/sqrt(d^2))) + 15*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 15*
sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*sin(-2*(b*c - a*d)/d))*erf(sqrt(d
*x + c)*sqrt(-2*I*b/d)))/(b^3*d*sqrt(abs(b)/abs(d)))

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Fricas [A]  time = 0.550609, size = 540, normalized size = 2.76 \begin{align*} -\frac{15 \, \pi d^{3} \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 ) - 15 \, \pi d^{3} \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 (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 15 \, b d^{2} - 2 \,{\left (16 \, b^{3} d^{2} x^{2} + 32 \, b^{3} c d x + 16 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \cos \left (b x + a\right )^{2} + 40 \,{\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt{d x + c}}{128 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a),x, algorithm="fricas")

[Out]

-1/128*(15*pi*d^3*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - 15*pi*d^3
*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) - 2*(16*b^3*d^2*x^2 + 32*b^3
*c*d*x + 16*b^3*c^2 - 15*b*d^2 - 2*(16*b^3*d^2*x^2 + 32*b^3*c*d*x + 16*b^3*c^2 - 15*b*d^2)*cos(b*x + a)^2 + 40
*(b^2*d^2*x + b^2*c*d)*cos(b*x + a)*sin(b*x + a))*sqrt(d*x + c))/b^4

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)*cos(b*x+a)*sin(b*x+a),x)

[Out]

Timed out

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Giac [C]  time = 1.31783, size = 1327, normalized size = 6.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a),x, algorithm="giac")

[Out]

-1/256*(16*(sqrt(pi)*d^2*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) + sqrt(pi)*d^2*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)*c^2 + d^2*((I*s
qrt(pi)*(-16*I*b^2*c^2*d + 24*b*c*d^2 + 15*I*d^3)*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^3) - 2*I*(16*I*(d*x + c)^(5/2)*b^2*d - 32*I*(
d*x + c)^(3/2)*b^2*c*d + 16*I*sqrt(d*x + c)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d^2 - 24*sqrt(d*x + c)*b*c*d^2 -
15*I*sqrt(d*x + c)*d^3)*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b^3)/d^2 + (I*sqrt(pi)*(-16*I*b^2*c^2*d -
24*b*c*d^2 + 15*I*d^3)*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^3) - 2*I*(16*I*(d*x + c)^(5/2)*b^2*d - 32*I*(d*x + c)^(3/2)*b^2*c*d
+ 16*I*sqrt(d*x + c)*b^2*c^2*d - 20*(d*x + c)^(3/2)*b*d^2 + 24*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)
*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b^3)/d^2) + 8*c*(I*sqrt(pi)*(4*I*b*c*d - 3*d^2)*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)
+ I*sqrt(pi)*(4*I*b*c*d + 3*d^2)*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) - 2*I*(4*I*(d*x + c)^(3/2)*b*d - 4*I*sqrt(d*x + c)*b*c*
d - 3*sqrt(d*x + c)*d^2)*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b^2 - 2*I*(4*I*(d*x + c)^(3/2)*b*d - 4*I*
sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*d)/d)/b^2))/d