3.153 \(\int (e+f x)^3 \sin (b (c+d x)^2) \, dx\)

Optimal. Leaf size=223 \[ \frac{3 \sqrt{\frac{\pi }{2}} f^2 (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{2 b^{3/2} d^4}+\frac{f^3 \sin \left (b (c+d x)^2\right )}{2 b^2 d^4}-\frac{3 f^2 (c+d x) (d e-c f) \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac{\sqrt{\frac{\pi }{2}} (d e-c f)^3 S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d^4}-\frac{3 f (d e-c f)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac{f^3 (c+d x)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4} \]

[Out]

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

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Rubi [A]  time = 0.311852, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3433, 3351, 3379, 2638, 3385, 3352, 3296, 2637} \[ \frac{3 \sqrt{\frac{\pi }{2}} f^2 (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{2 b^{3/2} d^4}+\frac{f^3 \sin \left (b (c+d x)^2\right )}{2 b^2 d^4}-\frac{3 f^2 (c+d x) (d e-c f) \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac{\sqrt{\frac{\pi }{2}} (d e-c f)^3 S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d^4}-\frac{3 f (d e-c f)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac{f^3 (c+d x)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^3*Sin[b*(c + d*x)^2],x]

[Out]

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

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]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 1.02722, size = 173, normalized size = 0.78 \[ \frac{4 \sqrt{2 \pi } b^{3/2} (d e-c f)^3 S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )-4 b f \cos \left (b (c+d x)^2\right ) \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right )-6 \sqrt{2 \pi } \sqrt{b} f^2 (c f-d e) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )+4 f^3 \sin \left (b (c+d x)^2\right )}{8 b^2 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^3*Sin[b*(c + d*x)^2],x]

[Out]

(-4*b*f*(c^2*f^2 - c*d*f*(3*e + f*x) + d^2*(3*e^2 + 3*e*f*x + f^2*x^2))*Cos[b*(c + d*x)^2] - 6*Sqrt[b]*f^2*(-(
d*e) + c*f)*Sqrt[2*Pi]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)] + 4*b^(3/2)*(d*e - c*f)^3*Sqrt[2*Pi]*FresnelS[Sq
rt[b]*Sqrt[2/Pi]*(c + d*x)] + 4*f^3*Sin[b*(c + d*x)^2])/(8*b^2*d^4)

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Maple [B]  time = 0.014, size = 586, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sin((d*x+c)^2*b),x)

[Out]

-1/2*f^3/d^2/b*x^2*cos(b*d^2*x^2+2*b*c*d*x+b*c^2)-f^3*c/d*(-1/2/d^2/b*x*cos(b*d^2*x^2+2*b*c*d*x+b*c^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/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)))+f^3/d^2/b*(1/2/d^2/b*sin(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)*Fre
snelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d)))-3/2*e*f^2/d^2/b*x*cos(b*d^2*x^2+2*b*c*d*x+b*c^2)-3*e*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)))+3/4*e*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))-3/2*e^2*f/d^2/b*cos(b*d^2*x^2+2*b*c*d*x+b*c^2)-3/2*e^2*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^3*Fr
esnelS(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 = 4.5003, size = 3780, normalized size = 16.95 \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)^3*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^3/(d*sqrt(abs(b))) - 3/4*((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^2*f/(b*d^2*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2)) + 3/32*(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)*(e
rf(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)))
*e*f^2/((b*d^2*x^2 + 2*b*c*d*x + b*c^2)^2*b*d^3) - 1/64*(16*(b*d^2*x^2 + 2*b*c*d*x + b*c^2)^2*(3*b*c^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)) - I*gamma(2, I*b*d^2*x^2 + 2*I*b
*c*d*x + I*b*c^2) + I*gamma(2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*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^2*c^3*
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^2*c^4)*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)) - 2*((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^2*c^3*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^2*c^4)*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^2*c*d^3*x^3*(24*I*gamma(3/2,
I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 24*I*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) + b^2*c^2*d^2*x
^2*(72*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 72*I*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^
2)) + b^2*c^3*d*x*(72*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 72*I*gamma(3/2, -I*b*d^2*x^2 - 2*I*b
*c*d*x - I*b*c^2)) + b^2*c^4*(24*I*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 24*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)) + 24*(b^2*c*d^3*x^3*(gam
ma(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^2*c^2*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^2*c^3*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^2*c^4*(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
^3/((b*d^2*x^2 + 2*b*c*d*x + b*c^2)^2*b^2*d^4)

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Fricas [A]  time = 1.67144, size = 582, normalized size = 2.61 \begin{align*} \frac{2 \, d f^{3} \sin \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right ) + 3 \, \sqrt{2} \pi{\left (d e f^{2} - c f^{3}\right )} \sqrt{\frac{b d^{2}}{\pi }} \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^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\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^{3} f^{3} x^{2} + 3 \, b d^{3} e^{2} f - 3 \, b c d^{2} e f^{2} + b c^{2} d f^{3} +{\left (3 \, b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x\right )} \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )}{4 \, b^{2} d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(b*(d*x+c)^2),x, algorithm="fricas")

[Out]

1/4*(2*d*f^3*sin(b*d^2*x^2 + 2*b*c*d*x + b*c^2) + 3*sqrt(2)*pi*(d*e*f^2 - c*f^3)*sqrt(b*d^2/pi)*fresnel_cos(sq
rt(2)*sqrt(b*d^2/pi)*(d*x + c)/d) + 2*sqrt(2)*pi*(b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*s
qrt(b*d^2/pi)*fresnel_sin(sqrt(2)*sqrt(b*d^2/pi)*(d*x + c)/d) - 2*(b*d^3*f^3*x^2 + 3*b*d^3*e^2*f - 3*b*c*d^2*e
*f^2 + b*c^2*d*f^3 + (3*b*d^3*e*f^2 - b*c*d^2*f^3)*x)*cos(b*d^2*x^2 + 2*b*c*d*x + b*c^2))/(b^2*d^5)

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

[Out]

Integral((e + f*x)**3*sin(b*c**2 + 2*b*c*d*x + b*d**2*x**2), x)

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Giac [C]  time = 1.24218, size = 1381, normalized size = 6.19 \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)^3*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^3/(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^3/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) - 1/4*(-3*I*sqrt(2)*sqrt(pi)*c*f*erf(-1/2*sqrt(2)*s
qrt(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)) + 3*f*e^(-I*b*
d^2*x^2 - 2*I*b*c*d*x - I*b*c^2 + 2)/(b*d))/d - 1/4*(3*I*sqrt(2)*sqrt(pi)*c*f*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)) + 3*f*e^(I*b*d^2*x^2 + 2*I
*b*c*d*x + I*b*c^2 + 2)/(b*d))/d - 1/8*(I*sqrt(2)*sqrt(pi)*(6*b*c^2*f^2 - 3*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))*e/(sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*b) + 2*I*(d*f^2*(-3*I*x -
3*I*c/d) + 6*I*c*f^2)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2 + 1)/(b*d))/d^2 - 1/8*(-I*sqrt(2)*sqrt(pi)*(6*b*
c^2*f^2 + 3*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))*e/(sqrt(b*d^2)*(-I*b*d
^2/sqrt(b^2*d^4) + 1)*b) + 2*I*(d*f^2*(-3*I*x - 3*I*c/d) + 6*I*c*f^2)*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2 +
1)/(b*d))/d^2 + 1/8*(sqrt(2)*sqrt(pi)*(2*I*b*c^3*f^3 + 3*c*f^3)*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*(b*d^2*f^3*(x + c/d)^2 - 3*b*c*d*f^3*(x
+ c/d) + 3*b*c^2*f^3 - I*f^3)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)/(b^2*d))/d^3 + 1/8*(sqrt(2)*sqrt(pi)*(
-2*I*b*c^3*f^3 + 3*c*f^3)*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*(b*d^2*f^3*(x + c/d)^2 - 3*b*c*d*f^3*(x + c/d) + 3*b*c^2*f^3 + I*f^3)*e^(I*b
*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)/(b^2*d))/d^3