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

Optimal. Leaf size=228 \[ -\frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{4096 b^{7/2}}-\frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{4096 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^{7/2}}{28 d} \]

[Out]

(c + d*x)^(7/2)/(28*d) - (5*d*(c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(256*b^2) - (15*d^(5/2)*Sqrt[Pi/2]*Cos[4*a - (
4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4096*b^(7/2)) - (15*d^(5/2)*Sqrt[Pi/2]*Fres
nelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(4096*b^(7/2)) + (15*d^2*Sqrt[c + d*x
]*Sin[4*a + 4*b*x])/(2048*b^3) - ((c + d*x)^(5/2)*Sin[4*a + 4*b*x])/(32*b)

________________________________________________________________________________________

Rubi [A]  time = 0.377683, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {4406, 3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{4096 b^{7/2}}-\frac{15 \sqrt{\frac{\pi }{2}} d^{5/2} \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{4096 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac{(c+d x)^{7/2}}{28 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^(7/2)/(28*d) - (5*d*(c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(256*b^2) - (15*d^(5/2)*Sqrt[Pi/2]*Cos[4*a - (
4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4096*b^(7/2)) - (15*d^(5/2)*Sqrt[Pi/2]*Fres
nelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(4096*b^(7/2)) + (15*d^2*Sqrt[c + d*x
]*Sin[4*a + 4*b*x])/(2048*b^3) - ((c + d*x)^(5/2)*Sin[4*a + 4*b*x])/(32*b)

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 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 ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{8} (c+d x)^{5/2}-\frac{1}{8} (c+d x)^{5/2} \cos (4 a+4 b x)\right ) \, dx\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{1}{8} \int (c+d x)^{5/2} \cos (4 a+4 b x) \, dx\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac{(5 d) \int (c+d x)^{3/2} \sin (4 a+4 b x) \, dx}{64 b}\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}+\frac{\left (15 d^2\right ) \int \sqrt{c+d x} \cos (4 a+4 b x) \, dx}{512 b^2}\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}+\frac{15 d^2 \sqrt{c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}-\frac{\left (15 d^3\right ) \int \frac{\sin (4 a+4 b x)}{\sqrt{c+d x}} \, dx}{4096 b^3}\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}+\frac{15 d^2 \sqrt{c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}-\frac{\left (15 d^3 \cos \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{4 b c}{d}+4 b x\right )}{\sqrt{c+d x}} \, dx}{4096 b^3}-\frac{\left (15 d^3 \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{4 b c}{d}+4 b x\right )}{\sqrt{c+d x}} \, dx}{4096 b^3}\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}+\frac{15 d^2 \sqrt{c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}-\frac{\left (15 d^2 \cos \left (4 a-\frac{4 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{4 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{2048 b^3}-\frac{\left (15 d^2 \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{4 b x^2}{d}\right ) \, dx,x,\sqrt{c+d x}\right )}{2048 b^3}\\ &=\frac{(c+d x)^{7/2}}{28 d}-\frac{5 d (c+d x)^{3/2} \cos (4 a+4 b x)}{256 b^2}-\frac{15 d^{5/2} \sqrt{\frac{\pi }{2}} \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right )}{4096 b^{7/2}}-\frac{15 d^{5/2} \sqrt{\frac{\pi }{2}} C\left (\frac{2 \sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}}{\sqrt{d}}\right ) \sin \left (4 a-\frac{4 b c}{d}\right )}{4096 b^{7/2}}+\frac{15 d^2 \sqrt{c+d x} \sin (4 a+4 b x)}{2048 b^3}-\frac{(c+d x)^{5/2} \sin (4 a+4 b x)}{32 b}\\ \end{align*}

Mathematica [A]  time = 2.7336, size = 206, normalized size = 0.9 \[ \frac{\sqrt{\frac{b}{d}} \left (4 \sqrt{\frac{b}{d}} \sqrt{c+d x} \left (-7 d \sin (4 (a+b x)) \left (64 b^2 (c+d x)^2-15 d^2\right )-280 b d^2 (c+d x) \cos (4 (a+b x))+512 b^3 (c+d x)^3\right )-105 \sqrt{2 \pi } d^3 \sin \left (4 a-\frac{4 b c}{d}\right ) \text{FresnelC}\left (2 \sqrt{\frac{2}{\pi }} \sqrt{\frac{b}{d}} \sqrt{c+d x}\right )-105 \sqrt{2 \pi } d^3 \cos \left (4 a-\frac{4 b c}{d}\right ) S\left (2 \sqrt{\frac{b}{d}} \sqrt{\frac{2}{\pi }} \sqrt{c+d x}\right )\right )}{57344 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b/d]*(-105*d^3*Sqrt[2*Pi]*Cos[4*a - (4*b*c)/d]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] - 105*d^3*
Sqrt[2*Pi]*FresnelC[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]*Sin[4*a - (4*b*c)/d] + 4*Sqrt[b/d]*Sqrt[c + d*x]*(51
2*b^3*(c + d*x)^3 - 280*b*d^2*(c + d*x)*Cos[4*(a + b*x)] - 7*d*(-15*d^2 + 64*b^2*(c + d*x)^2)*Sin[4*(a + b*x)]
)))/(57344*b^4)

________________________________________________________________________________________

Maple [A]  time = 0.034, size = 251, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{7/2}}{56}}-{\frac{d \left ( dx+c \right ) ^{5/2}}{64\,b}\sin \left ( 4\,{\frac{ \left ( dx+c \right ) b}{d}}+4\,{\frac{ad-bc}{d}} \right ) }+{\frac{5\,d}{64\,b} \left ( -1/8\,{\frac{d \left ( dx+c \right ) ^{3/2}}{b}\cos \left ( 4\,{\frac{ \left ( dx+c \right ) b}{d}}+4\,{\frac{ad-bc}{d}} \right ) }+3/8\,{\frac{d}{b} \left ( 1/8\,{\frac{d\sqrt{dx+c}}{b}\sin \left ( 4\,{\frac{ \left ( dx+c \right ) b}{d}}+4\,{\frac{ad-bc}{d}} \right ) }-1/32\,{\frac{d\sqrt{2}\sqrt{\pi }}{b} \left ( \cos \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\frac{1}{\sqrt{{\frac{b}{d}}}}}} \right ) +\sin \left ( 4\,{\frac{ad-bc}{d}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{2}\sqrt{dx+c}b}{\sqrt{\pi }d}{\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)^2*sin(b*x+a)^2,x)

[Out]

2/d*(1/56*(d*x+c)^(7/2)-1/64/b*d*(d*x+c)^(5/2)*sin(4/d*(d*x+c)*b+4*(a*d-b*c)/d)+5/64/b*d*(-1/8/b*d*(d*x+c)^(3/
2)*cos(4/d*(d*x+c)*b+4*(a*d-b*c)/d)+3/8/b*d*(1/8/b*d*(d*x+c)^(1/2)*sin(4/d*(d*x+c)*b+4*(a*d-b*c)/d)-1/32/b*d*2
^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos(4*(a*d-b*c)/d)*FresnelS(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)+sin
(4*(a*d-b*c)/d)*FresnelC(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*(d*x+c)^(1/2)*b/d)))))

________________________________________________________________________________________

Maxima [C]  time = 2.13834, size = 922, normalized size = 4.04 \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)^2*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/229376*(8192*(d*x + c)^(7/2)*b^3*sqrt(abs(b)/abs(d)) - 4480*(d*x + c)^(3/2)*b*d^2*sqrt(abs(b)/abs(d))*cos(4*
((d*x + c)*b - b*c + a*d)/d) - ((105*I*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2)))
+ 105*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*sqrt(pi)*sin(1/4*pi + 1/
2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 105*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0,
d/sqrt(d^2))))*d^3*cos(-4*(b*c - a*d)/d) + (105*sqrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sq
rt(d^2))) + 105*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 105*I*sqrt(pi)*sin(1
/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/
2*arctan2(0, d/sqrt(d^2))))*d^3*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - ((-105*I*sqrt(pi)*co
s(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 105*I*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2(0, b) +
1/2*arctan2(0, d/sqrt(d^2))) + 105*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) - 1
05*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*cos(-4*(b*c - a*d)/d) + (105*s
qrt(pi)*cos(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*sqrt(pi)*cos(-1/4*pi + 1/2*arctan2
(0, b) + 1/2*arctan2(0, d/sqrt(d^2))) + 105*I*sqrt(pi)*sin(1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(
d^2))) - 105*I*sqrt(pi)*sin(-1/4*pi + 1/2*arctan2(0, b) + 1/2*arctan2(0, d/sqrt(d^2))))*d^3*sin(-4*(b*c - a*d)
/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)) - 112*(64*(d*x + c)^(5/2)*b^2*d*sqrt(abs(b)/abs(d)) - 15*sqrt(d*x + c)*
d^3*sqrt(abs(b)/abs(d)))*sin(4*((d*x + c)*b - b*c + a*d)/d))/(b^3*d*sqrt(abs(b)/abs(d)))

________________________________________________________________________________________

Fricas [A]  time = 0.648786, size = 840, normalized size = 3.68 \begin{align*} -\frac{105 \, \sqrt{2} \pi d^{4} \sqrt{\frac{b}{\pi d}} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{S}\left (2 \, \sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) + 105 \, \sqrt{2} \pi d^{4} \sqrt{\frac{b}{\pi d}} \operatorname{C}\left (2 \, \sqrt{2} \sqrt{d x + c} \sqrt{\frac{b}{\pi d}}\right ) \sin \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) - 16 \,{\left (128 \, b^{4} d^{3} x^{3} + 384 \, b^{4} c d^{2} x^{2} + 128 \, b^{4} c^{3} - 70 \, b^{2} c d^{2} - 560 \,{\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{4} + 560 \,{\left (b^{2} d^{3} x + b^{2} c d^{2}\right )} \cos \left (b x + a\right )^{2} + 2 \,{\left (192 \, b^{4} c^{2} d - 35 \, b^{2} d^{3}\right )} x - 7 \,{\left (2 \,{\left (64 \, b^{3} d^{3} x^{2} + 128 \, b^{3} c d^{2} x + 64 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right )^{3} -{\left (64 \, b^{3} d^{3} x^{2} + 128 \, b^{3} c d^{2} x + 64 \, b^{3} c^{2} d - 15 \, b d^{3}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt{d x + c}}{57344 \, b^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/57344*(105*sqrt(2)*pi*d^4*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(
pi*d))) + 105*sqrt(2)*pi*d^4*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c -
a*d)/d) - 16*(128*b^4*d^3*x^3 + 384*b^4*c*d^2*x^2 + 128*b^4*c^3 - 70*b^2*c*d^2 - 560*(b^2*d^3*x + b^2*c*d^2)*c
os(b*x + a)^4 + 560*(b^2*d^3*x + b^2*c*d^2)*cos(b*x + a)^2 + 2*(192*b^4*c^2*d - 35*b^2*d^3)*x - 7*(2*(64*b^3*d
^3*x^2 + 128*b^3*c*d^2*x + 64*b^3*c^2*d - 15*b*d^3)*cos(b*x + a)^3 - (64*b^3*d^3*x^2 + 128*b^3*c*d^2*x + 64*b^
3*c^2*d - 15*b*d^3)*cos(b*x + a))*sin(b*x + a))*sqrt(d*x + c))/(b^4*d)

________________________________________________________________________________________

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)**2*sin(b*x+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [C]  time = 1.63911, size = 1467, normalized size = 6.43 \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)^2*sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/1720320*(2240*(3*I*sqrt(2)*sqrt(pi)*d^2*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e
^((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) - 3*I*sqrt(2)*sqrt(pi)*d^2*erf(-sqrt(2)*sqrt(
b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) +
1)*b) - 64*(d*x + c)^(3/2) - 12*I*sqrt(d*x + c)*d*e^((4*I*(d*x + c)*b - 4*I*b*c + 4*I*a*d)/d)/b + 12*I*sqrt(d
*x + c)*d*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b)*c^2 - d^2*(4096*(15*(d*x + c)^(7/2) - 42*(d*x + c)^(
5/2)*c + 35*(d*x + c)^(3/2)*c^2)/d^2 + 105*(sqrt(2)*sqrt(pi)*(-64*I*b^2*c^2*d + 48*b*c*d^2 + 15*I*d^3)*d*erf(-
sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(
b^2*d^2) + 1)*b^3) - 4*(64*I*(d*x + c)^(5/2)*b^2*d - 128*I*(d*x + c)^(3/2)*b^2*c*d + 64*I*sqrt(d*x + c)*b^2*c^
2*d + 40*(d*x + c)^(3/2)*b*d^2 - 48*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^((-4*I*(d*x + c)*b + 4*I
*b*c - 4*I*a*d)/d)/b^3)/d^2 + 105*(sqrt(2)*sqrt(pi)*(64*I*b^2*c^2*d + 48*b*c*d^2 - 15*I*d^3)*d*erf(-sqrt(2)*sq
rt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2
) + 1)*b^3) - 4*(-64*I*(d*x + c)^(5/2)*b^2*d + 128*I*(d*x + c)^(3/2)*b^2*c*d - 64*I*sqrt(d*x + c)*b^2*c^2*d +
40*(d*x + c)^(3/2)*b*d^2 - 48*sqrt(d*x + c)*b*c*d^2 + 15*I*sqrt(d*x + c)*d^3)*e^((4*I*(d*x + c)*b - 4*I*b*c +
4*I*a*d)/d)/b^3)/d^2) - 112*(1536*(d*x + c)^(5/2) - 2560*(d*x + c)^(3/2)*c + 15*sqrt(2)*sqrt(pi)*(8*I*b*c*d -
3*d^2)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d
)*(I*b*d/sqrt(b^2*d^2) + 1)*b^2) + 15*sqrt(2)*sqrt(pi)*(-8*I*b*c*d - 3*d^2)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x
+ c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 60*
(-8*I*(d*x + c)^(3/2)*b*d + 8*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((4*I*(d*x + c)*b - 4*I*b*c + 4*I
*a*d)/d)/b^2 - 60*(8*I*(d*x + c)^(3/2)*b*d - 8*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^((-4*I*(d*x + c)
*b + 4*I*b*c - 4*I*a*d)/d)/b^2)*c)/d