3.177 \(\int (e+f x)^2 \sin (a+\frac{b}{(c+d x)^2}) \, dx\)

Optimal. Leaf size=371 \[ \frac{2 \sqrt{2 \pi } b^{3/2} f^2 \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{3 d^3}+\frac{2 \sqrt{2 \pi } b^{3/2} f^2 \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{3 d^3}-\frac{b f \cos (a) (d e-c f) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )}{d^3}-\frac{\sqrt{2 \pi } \sqrt{b} \cos (a) (d e-c f)^2 \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d^3}+\frac{\sqrt{2 \pi } \sqrt{b} \sin (a) (d e-c f)^2 S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^3}+\frac{b f \sin (a) (d e-c f) \text{Si}\left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f (c+d x)^2 (d e-c f) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{(c+d x) (d e-c f)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{3 d^3}+\frac{2 b f^2 (c+d x) \cos \left (a+\frac{b}{(c+d x)^2}\right )}{3 d^3} \]

[Out]

(2*b*f^2*(c + d*x)*Cos[a + b/(c + d*x)^2])/(3*d^3) - (b*f*(d*e - c*f)*Cos[a]*CosIntegral[b/(c + d*x)^2])/d^3 -
(Sqrt[b]*(d*e - c*f)^2*Sqrt[2*Pi]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d^3 + (2*b^(3/2)*f^2*Sqrt[
2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/(3*d^3) + (2*b^(3/2)*f^2*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*S
qrt[2/Pi])/(c + d*x)]*Sin[a])/(3*d^3) + (Sqrt[b]*(d*e - c*f)^2*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d
*x)]*Sin[a])/d^3 + ((d*e - c*f)^2*(c + d*x)*Sin[a + b/(c + d*x)^2])/d^3 + (f*(d*e - c*f)*(c + d*x)^2*Sin[a + b
/(c + d*x)^2])/d^3 + (f^2*(c + d*x)^3*Sin[a + b/(c + d*x)^2])/(3*d^3) + (b*f*(d*e - c*f)*Sin[a]*SinIntegral[b/
(c + d*x)^2])/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.479071, antiderivative size = 371, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 14, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {3433, 3359, 3387, 3354, 3352, 3351, 3379, 3297, 3303, 3299, 3302, 3409, 3388, 3353} \[ \frac{2 \sqrt{2 \pi } b^{3/2} f^2 \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{3 d^3}+\frac{2 \sqrt{2 \pi } b^{3/2} f^2 \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{3 d^3}-\frac{b f \cos (a) (d e-c f) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )}{d^3}-\frac{\sqrt{2 \pi } \sqrt{b} \cos (a) (d e-c f)^2 \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d^3}+\frac{\sqrt{2 \pi } \sqrt{b} \sin (a) (d e-c f)^2 S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d^3}+\frac{b f \sin (a) (d e-c f) \text{Si}\left (\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f (c+d x)^2 (d e-c f) \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{(c+d x) (d e-c f)^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{d^3}+\frac{f^2 (c+d x)^3 \sin \left (a+\frac{b}{(c+d x)^2}\right )}{3 d^3}+\frac{2 b f^2 (c+d x) \cos \left (a+\frac{b}{(c+d x)^2}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*f^2*(c + d*x)*Cos[a + b/(c + d*x)^2])/(3*d^3) - (b*f*(d*e - c*f)*Cos[a]*CosIntegral[b/(c + d*x)^2])/d^3 -
(Sqrt[b]*(d*e - c*f)^2*Sqrt[2*Pi]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d^3 + (2*b^(3/2)*f^2*Sqrt[
2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/(3*d^3) + (2*b^(3/2)*f^2*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*S
qrt[2/Pi])/(c + d*x)]*Sin[a])/(3*d^3) + (Sqrt[b]*(d*e - c*f)^2*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d
*x)]*Sin[a])/d^3 + ((d*e - c*f)^2*(c + d*x)*Sin[a + b/(c + d*x)^2])/d^3 + (f*(d*e - c*f)*(c + d*x)^2*Sin[a + b
/(c + d*x)^2])/d^3 + (f^2*(c + d*x)^3*Sin[a + b/(c + d*x)^2])/(3*d^3) + (b*f*(d*e - c*f)*Sin[a]*SinIntegral[b/
(c + d*x)^2])/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 3359

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(
a + b*Sin[c + d/x^n])^p/x^2, x], x, 1/(e + f*x)], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[n,
0] && EqQ[n, -2]

Rule 3387

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

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, 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 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 3297

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

Rule 3303

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

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

Rule 3409

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sin[c + d/x^
n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2
]

Rule 3388

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

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rubi steps

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

Mathematica [A]  time = 1.46365, size = 467, normalized size = 1.26 \[ \frac{2 \sqrt{2 \pi } b^{3/2} f^2 \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )-3 c^2 d e f \sin \left (a+\frac{b}{(c+d x)^2}\right )+3 \sqrt{2 \pi } \sqrt{b} c^2 f^2 \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )+c^3 f^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )+3 b f \cos (a) (c f-d e) \text{CosIntegral}\left (\frac{b}{(c+d x)^2}\right )+3 \sqrt{2 \pi } \sqrt{b} d^2 e^2 \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )+3 c d^2 e^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )+3 d^3 e^2 x \sin \left (a+\frac{b}{(c+d x)^2}\right )+3 d^3 e f x^2 \sin \left (a+\frac{b}{(c+d x)^2}\right )+d^3 f^2 x^3 \sin \left (a+\frac{b}{(c+d x)^2}\right )+\sqrt{2 \pi } \sqrt{b} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right ) \left (2 b f^2 \sin (a)-3 \cos (a) (d e-c f)^2\right )-6 \sqrt{2 \pi } \sqrt{b} c d e f \sin (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )+3 b d e f \sin (a) \text{Si}\left (\frac{b}{(c+d x)^2}\right )-3 b c f^2 \sin (a) \text{Si}\left (\frac{b}{(c+d x)^2}\right )+2 b c f^2 \cos \left (a+\frac{b}{(c+d x)^2}\right )+2 b d f^2 x \cos \left (a+\frac{b}{(c+d x)^2}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*c*f^2*Cos[a + b/(c + d*x)^2] + 2*b*d*f^2*x*Cos[a + b/(c + d*x)^2] + 3*b*f*(-(d*e) + c*f)*Cos[a]*CosIntegr
al[b/(c + d*x)^2] + 2*b^(3/2)*f^2*Sqrt[2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + 3*Sqrt[b]*d^2*e
^2*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]*Sin[a] - 6*Sqrt[b]*c*d*e*f*Sqrt[2*Pi]*FresnelS[(Sqrt[b]
*Sqrt[2/Pi])/(c + d*x)]*Sin[a] + 3*Sqrt[b]*c^2*f^2*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]*Sin[a]
+ Sqrt[b]*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]*(-3*(d*e - c*f)^2*Cos[a] + 2*b*f^2*Sin[a]) + 3*c
*d^2*e^2*Sin[a + b/(c + d*x)^2] - 3*c^2*d*e*f*Sin[a + b/(c + d*x)^2] + c^3*f^2*Sin[a + b/(c + d*x)^2] + 3*d^3*
e^2*x*Sin[a + b/(c + d*x)^2] + 3*d^3*e*f*x^2*Sin[a + b/(c + d*x)^2] + d^3*f^2*x^3*Sin[a + b/(c + d*x)^2] + 3*b
*d*e*f*Sin[a]*SinIntegral[b/(c + d*x)^2] - 3*b*c*f^2*Sin[a]*SinIntegral[b/(c + d*x)^2])/(3*d^3)

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Maple [A]  time = 0.016, size = 302, normalized size = 0.8 \begin{align*} -{\frac{1}{{d}^{3}} \left ( - \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2} \right ) \left ( dx+c \right ) \sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) + \left ({c}^{2}{f}^{2}-2\,cdef+{d}^{2}{e}^{2} \right ) \sqrt{b}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) -\sin \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) \right ) -{\frac{ \left ( -2\,c{f}^{2}+2\,def \right ) \left ( dx+c \right ) ^{2}}{2}\sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }+ \left ( -2\,c{f}^{2}+2\,def \right ) b \left ({\frac{\cos \left ( a \right ) }{2}{\it Ci} \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }-{\frac{\sin \left ( a \right ) }{2}{\it Si} \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) } \right ) -{\frac{{f}^{2} \left ( dx+c \right ) ^{3}}{3}\sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) }+{\frac{2\,{f}^{2}b}{3} \left ( - \left ( dx+c \right ) \cos \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) -\sqrt{b}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^3*(-(c^2*f^2-2*c*d*e*f+d^2*e^2)*(d*x+c)*sin(a+b/(d*x+c)^2)+(c^2*f^2-2*c*d*e*f+d^2*e^2)*b^(1/2)*2^(1/2)*Pi
^(1/2)*(cos(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))-sin(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)))-1
/2*(-2*c*f^2+2*d*e*f)*(d*x+c)^2*sin(a+b/(d*x+c)^2)+(-2*c*f^2+2*d*e*f)*b*(1/2*cos(a)*Ci(b/(d*x+c)^2)-1/2*sin(a)
*Si(b/(d*x+c)^2))-1/3*f^2*(d*x+c)^3*sin(a+b/(d*x+c)^2)+2/3*f^2*b*(-(d*x+c)*cos(a+b/(d*x+c)^2)-b^(1/2)*2^(1/2)*
Pi^(1/2)*(cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))+sin(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)))
))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, b f^{2} x \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) +{\left (\frac{c^{3} f^{2} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d^{3}} - 2 \, \int \frac{2 \, b^{2} f^{2} x \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left ({\left (b d^{2} e f - b c d f^{2}\right )} x^{2} +{\left (b d^{2} e^{2} - b c^{2} f^{2}\right )} x\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}}\,{d x} - 2 \, \int \frac{2 \, b^{2} f^{2} x \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left ({\left (b d^{2} e f - b c d f^{2}\right )} x^{2} +{\left (b d^{2} e^{2} - b c^{2} f^{2}\right )} x\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \,{\left ({\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} +{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2}\right )}}\,{d x}\right )} d^{2} +{\left (d^{2} f^{2} x^{3} + 3 \, d^{2} e f x^{2} + 3 \, d^{2} e^{2} x\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{3 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*b*f^2*x*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) - 3*d^2*integrate(1/3*(2*b^2
*d*f^2*x*sin((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) + (b*c^3*f^2 - 3*(b*d^3*e*f - b*c*
d^2*f^2)*x^2 - 3*(b*d^3*e^2 - b*c^2*d*f^2)*x)*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2
)))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2), x) - 3*d^2*integrate(1/3*(2*b^2*d*f^2*x*sin((a*d^2*x^2 +
2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) + (b*c^3*f^2 - 3*(b*d^3*e*f - b*c*d^2*f^2)*x^2 - 3*(b*d^3*e^
2 - b*c^2*d*f^2)*x)*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/((d^5*x^3 + 3*c*d^4*x^
2 + 3*c^2*d^3*x + c^3*d^2)*cos((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2))^2 + (d^5*x^3 + 3
*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2)*sin((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2))^2), x)
+ (d^2*f^2*x^3 + 3*d^2*e*f*x^2 + 3*d^2*e^2*x)*sin((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2
)))/d^2

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Fricas [A]  time = 2.07514, size = 1040, normalized size = 2.8 \begin{align*} \frac{2 \, \sqrt{2}{\left (2 \, \pi b d f^{2} \sin \left (a\right ) - 3 \, \pi{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \cos \left (a\right )\right )} \sqrt{\frac{b}{\pi d^{2}}} \operatorname{C}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) + 2 \, \sqrt{2}{\left (2 \, \pi b d f^{2} \cos \left (a\right ) + 3 \, \pi{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \sin \left (a\right )\right )} \sqrt{\frac{b}{\pi d^{2}}} \operatorname{S}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) + 6 \,{\left (b d e f - b c f^{2}\right )} \sin \left (a\right ) \operatorname{Si}\left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \,{\left ({\left (b d e f - b c f^{2}\right )} \operatorname{Ci}\left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) +{\left (b d e f - b c f^{2}\right )} \operatorname{Ci}\left (-\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )\right )} \cos \left (a\right ) + 4 \,{\left (b d f^{2} x + b c f^{2}\right )} \cos \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \,{\left (d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 3 \, c d^{2} e^{2} - 3 \, c^{2} d e f + c^{3} f^{2}\right )} \sin \left (\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{6 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*sqrt(2)*(2*pi*b*d*f^2*sin(a) - 3*pi*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cos(a))*sqrt(b/(pi*d^2))*fresne
l_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + 2*sqrt(2)*(2*pi*b*d*f^2*cos(a) + 3*pi*(d^3*e^2 - 2*c*d^2*e*f + c
^2*d*f^2)*sin(a))*sqrt(b/(pi*d^2))*fresnel_sin(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + 6*(b*d*e*f - b*c*f^2)*s
in(a)*sin_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) - 3*((b*d*e*f - b*c*f^2)*cos_integral(b/(d^2*x^2 + 2*c*d*x + c
^2)) + (b*d*e*f - b*c*f^2)*cos_integral(-b/(d^2*x^2 + 2*c*d*x + c^2)))*cos(a) + 4*(b*d*f^2*x + b*c*f^2)*cos((a
*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) + 2*(d^3*f^2*x^3 + 3*d^3*e*f*x^2 + 3*d^3*e^2*x +
3*c*d^2*e^2 - 3*c^2*d*e*f + c^3*f^2)*sin((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/d^3

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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((f*x+e)**2*sin(a+b/(d*x+c)**2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^2*sin(a + b/(d*x + c)^2), x)