3.2.59 \(\int (e+f x)^3 \sin (\frac {b}{(c+d x)^2}) \, dx\) [159]

Optimal. Leaf size=337 \[ \frac {2 b f^2 (d e-c f) (c+d x) \cos \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {b f^3 (c+d x)^2 \cos \left (\frac {b}{(c+d x)^2}\right )}{4 d^4}-\frac {3 b f (d e-c f)^2 \text {Ci}\left (\frac {b}{(c+d x)^2}\right )}{2 d^4}-\frac {\sqrt {b} (d e-c f)^3 \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^4}+\frac {2 b^{3/2} f^2 (d e-c f) \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^4}+\frac {(d e-c f)^3 (c+d x) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {3 f (d e-c f)^2 (c+d x)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{2 d^4}+\frac {f^2 (d e-c f) (c+d x)^3 \sin \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {f^3 (c+d x)^4 \sin \left (\frac {b}{(c+d x)^2}\right )}{4 d^4}+\frac {b^2 f^3 \text {Si}\left (\frac {b}{(c+d x)^2}\right )}{4 d^4} \]

[Out]

-3/2*b*f*(-c*f+d*e)^2*Ci(b/(d*x+c)^2)/d^4+2*b*f^2*(-c*f+d*e)*(d*x+c)*cos(b/(d*x+c)^2)/d^4+1/4*b*f^3*(d*x+c)^2*
cos(b/(d*x+c)^2)/d^4+1/4*b^2*f^3*Si(b/(d*x+c)^2)/d^4+(-c*f+d*e)^3*(d*x+c)*sin(b/(d*x+c)^2)/d^4+3/2*f*(-c*f+d*e
)^2*(d*x+c)^2*sin(b/(d*x+c)^2)/d^4+f^2*(-c*f+d*e)*(d*x+c)^3*sin(b/(d*x+c)^2)/d^4+1/4*f^3*(d*x+c)^4*sin(b/(d*x+
c)^2)/d^4+2*b^(3/2)*f^2*(-c*f+d*e)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))*2^(1/2)*Pi^(1/2)/d^4-(-c*f+d*e)^
3*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))*b^(1/2)*2^(1/2)*Pi^(1/2)/d^4

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Rubi [A]
time = 0.29, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {3514, 3440, 3468, 3433, 3460, 3378, 3383, 3490, 3469, 3432, 3380} \begin {gather*} \frac {2 \sqrt {2 \pi } b^{3/2} f^2 (d e-c f) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^4}+\frac {b^2 f^3 \text {Si}\left (\frac {b}{(c+d x)^2}\right )}{4 d^4}-\frac {3 b f (d e-c f)^2 \text {CosIntegral}\left (\frac {b}{(c+d x)^2}\right )}{2 d^4}+\frac {f^2 (c+d x)^3 (d e-c f) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {2 b f^2 (c+d x) (d e-c f) \cos \left (\frac {b}{(c+d x)^2}\right )}{d^4}-\frac {\sqrt {2 \pi } \sqrt {b} (d e-c f)^3 \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{c+d x}\right )}{d^4}+\frac {3 f (c+d x)^2 (d e-c f)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{2 d^4}+\frac {(c+d x) (d e-c f)^3 \sin \left (\frac {b}{(c+d x)^2}\right )}{d^4}+\frac {f^3 (c+d x)^4 \sin \left (\frac {b}{(c+d x)^2}\right )}{4 d^4}+\frac {b f^3 (c+d x)^2 \cos \left (\frac {b}{(c+d x)^2}\right )}{4 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^3*Sin[b/(c + d*x)^2],x]

[Out]

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

Rule 3378

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 3380

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 3383

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 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3440

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 3460

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 3468

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 3469

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 3490

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 3514

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]

Rubi steps

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

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Mathematica [A]
time = 0.51, size = 440, normalized size = 1.31 \begin {gather*} \frac {8 b c d e f^2 \cos \left (\frac {b}{(c+d x)^2}\right )-7 b c^2 f^3 \cos \left (\frac {b}{(c+d x)^2}\right )+8 b d^2 e f^2 x \cos \left (\frac {b}{(c+d x)^2}\right )-6 b c d f^3 x \cos \left (\frac {b}{(c+d x)^2}\right )+b d^2 f^3 x^2 \cos \left (\frac {b}{(c+d x)^2}\right )-6 b f (d e-c f)^2 \text {Ci}\left (\frac {b}{(c+d x)^2}\right )-4 \sqrt {b} (d e-c f)^3 \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+8 b^{3/2} d e f^2 \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )-8 b^{3/2} c f^3 \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+4 c d^3 e^3 \sin \left (\frac {b}{(c+d x)^2}\right )-6 c^2 d^2 e^2 f \sin \left (\frac {b}{(c+d x)^2}\right )+4 c^3 d e f^2 \sin \left (\frac {b}{(c+d x)^2}\right )-c^4 f^3 \sin \left (\frac {b}{(c+d x)^2}\right )+4 d^4 e^3 x \sin \left (\frac {b}{(c+d x)^2}\right )+6 d^4 e^2 f x^2 \sin \left (\frac {b}{(c+d x)^2}\right )+4 d^4 e f^2 x^3 \sin \left (\frac {b}{(c+d x)^2}\right )+d^4 f^3 x^4 \sin \left (\frac {b}{(c+d x)^2}\right )+b^2 f^3 \text {Si}\left (\frac {b}{(c+d x)^2}\right )}{4 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*c*d*e*f^2*Cos[b/(c + d*x)^2] - 7*b*c^2*f^3*Cos[b/(c + d*x)^2] + 8*b*d^2*e*f^2*x*Cos[b/(c + d*x)^2] - 6*b*
c*d*f^3*x*Cos[b/(c + d*x)^2] + b*d^2*f^3*x^2*Cos[b/(c + d*x)^2] - 6*b*f*(d*e - c*f)^2*CosIntegral[b/(c + d*x)^
2] - 4*Sqrt[b]*(d*e - c*f)^3*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] + 8*b^(3/2)*d*e*f^2*Sqrt[2*Pi
]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)] - 8*b^(3/2)*c*f^3*Sqrt[2*Pi]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x
)] + 4*c*d^3*e^3*Sin[b/(c + d*x)^2] - 6*c^2*d^2*e^2*f*Sin[b/(c + d*x)^2] + 4*c^3*d*e*f^2*Sin[b/(c + d*x)^2] -
c^4*f^3*Sin[b/(c + d*x)^2] + 4*d^4*e^3*x*Sin[b/(c + d*x)^2] + 6*d^4*e^2*f*x^2*Sin[b/(c + d*x)^2] + 4*d^4*e*f^2
*x^3*Sin[b/(c + d*x)^2] + d^4*f^3*x^4*Sin[b/(c + d*x)^2] + b^2*f^3*SinIntegral[b/(c + d*x)^2])/(4*d^4)

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Maple [A]
time = 0.13, size = 365, normalized size = 1.08 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^4*(-(c^3*f^3-3*c^2*d*e*f^2+3*c*d^2*e^2*f-d^3*e^3)*(d*x+c)*sin(b/(d*x+c)^2)+(c^3*f^3-3*c^2*d*e*f^2+3*c*d^2*
e^2*f-d^3*e^3)*b^(1/2)*2^(1/2)*Pi^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))-1/2*(-3*c^2*f^3+6*c*d*e*f^2
-3*d^2*e^2*f)*(d*x+c)^2*sin(b/(d*x+c)^2)+1/2*(-3*c^2*f^3+6*c*d*e*f^2-3*d^2*e^2*f)*b*Ci(b/(d*x+c)^2)-1/3*(3*c*f
^3-3*d*e*f^2)*(d*x+c)^3*sin(b/(d*x+c)^2)+2/3*(3*c*f^3-3*d*e*f^2)*b*(-(d*x+c)*cos(b/(d*x+c)^2)-b^(1/2)*2^(1/2)*
Pi^(1/2)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)))+1/4*f^3*(d*x+c)^4*sin(b/(d*x+c)^2)-1/2*f^3*b*(-1/2*(d*x+c
)^2*cos(b/(d*x+c)^2)-1/2*b*Si(b/(d*x+c)^2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

-1/4*(4*d^3*integrate(-1/4*((3*b*c^4*f^3 - 4*b*c^3*d*f^2*e + 6*(b*c^2*d^2*f^3 - 2*b*c*d^3*f^2*e + b*d^4*f*e^2)
*x^2 + 4*(2*b*c^3*d*f^3 - 3*b*c^2*d^2*f^2*e + b*d^4*e^3)*x)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) - (b^2*d^2*f^3*x^
2 - 2*(3*b^2*c*d*f^3 - 4*b^2*d^2*f^2*e)*x)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^
4*x + c^3*d^3), x) + 4*d^3*integrate(-1/4*((3*b*c^4*f^3 - 4*b*c^3*d*f^2*e + 6*(b*c^2*d^2*f^3 - 2*b*c*d^3*f^2*e
+ b*d^4*f*e^2)*x^2 + 4*(2*b*c^3*d*f^3 - 3*b*c^2*d^2*f^2*e + b*d^4*e^3)*x)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) -
(b^2*d^2*f^3*x^2 - 2*(3*b^2*c*d*f^3 - 4*b^2*d^2*f^2*e)*x)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/((d^6*x^3 + 3*c*d^
5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cos(b/(d^2*x^2 + 2*c*d*x + c^2))^2 + (d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3
*d^3)*sin(b/(d^2*x^2 + 2*c*d*x + c^2))^2), x) - (b*d*f^3*x^2 - 2*(3*b*c*f^3 - 4*b*d*f^2*e)*x)*cos(b/(d^2*x^2 +
2*c*d*x + c^2)) - (d^3*f^3*x^4 + 4*d^3*f^2*x^3*e + 6*d^3*f*x^2*e^2 + 4*d^3*x*e^3)*sin(b/(d^2*x^2 + 2*c*d*x +
c^2)))/d^3

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Fricas [A]
time = 0.42, size = 448, normalized size = 1.33 \begin {gather*} \frac {b^{2} f^{3} \operatorname {Si}\left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, \sqrt {2} {\left (\pi c^{3} d f^{3} - 3 \, \pi c^{2} d^{2} f^{2} e + 3 \, \pi c d^{3} f e^{2} - \pi d^{4} e^{3}\right )} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {C}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) - 8 \, \sqrt {2} {\left (\pi b c d f^{3} - \pi b d^{2} f^{2} e\right )} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {S}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) + {\left (b d^{2} f^{3} x^{2} - 6 \, b c d f^{3} x - 7 \, b c^{2} f^{3} + 8 \, {\left (b d^{2} f^{2} x + b c d f^{2}\right )} e\right )} \cos \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \, {\left (b c^{2} f^{3} - 2 \, b c d f^{2} e + b d^{2} f e^{2}\right )} \operatorname {Ci}\left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 3 \, {\left (b c^{2} f^{3} - 2 \, b c d f^{2} e + b d^{2} f e^{2}\right )} \operatorname {Ci}\left (-\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + {\left (d^{4} f^{3} x^{4} - c^{4} f^{3} + 4 \, {\left (d^{4} x + c d^{3}\right )} e^{3} + 6 \, {\left (d^{4} f x^{2} - c^{2} d^{2} f\right )} e^{2} + 4 \, {\left (d^{4} f^{2} x^{3} + c^{3} d f^{2}\right )} e\right )} \sin \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{4 \, d^{4}} \end {gather*}

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*(b^2*f^3*sin_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) + 4*sqrt(2)*(pi*c^3*d*f^3 - 3*pi*c^2*d^2*f^2*e + 3*pi*c
*d^3*f*e^2 - pi*d^4*e^3)*sqrt(b/(pi*d^2))*fresnel_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) - 8*sqrt(2)*(pi*b*
c*d*f^3 - pi*b*d^2*f^2*e)*sqrt(b/(pi*d^2))*fresnel_sin(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + (b*d^2*f^3*x^2
- 6*b*c*d*f^3*x - 7*b*c^2*f^3 + 8*(b*d^2*f^2*x + b*c*d*f^2)*e)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) - 3*(b*c^2*f^3
- 2*b*c*d*f^2*e + b*d^2*f*e^2)*cos_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) - 3*(b*c^2*f^3 - 2*b*c*d*f^2*e + b*d
^2*f*e^2)*cos_integral(-b/(d^2*x^2 + 2*c*d*x + c^2)) + (d^4*f^3*x^4 - c^4*f^3 + 4*(d^4*x + c*d^3)*e^3 + 6*(d^4
*f*x^2 - c^2*d^2*f)*e^2 + 4*(d^4*f^2*x^3 + c^3*d*f^2)*e)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e + f x\right )^{3} \sin {\left (\frac {b}{c^{2} + 2 c d x + d^{2} x^{2}} \right )}\, dx \end {gather*}

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*c*d*x + d**2*x**2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \sin \left (\frac {b}{{\left (c+d\,x\right )}^2}\right )\,{\left (e+f\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b/(c + d*x)^2)*(e + f*x)^3,x)

[Out]

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

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