3.2.60 \(\int (e+f x)^2 \sin (\frac {b}{(c+d x)^2}) \, dx\) [160]

Optimal. Leaf size=233 \[ \frac {2 b f^2 (c+d x) \cos \left (\frac {b}{(c+d x)^2}\right )}{3 d^3}-\frac {b f (d e-c f) \text {Ci}\left (\frac {b}{(c+d x)^2}\right )}{d^3}-\frac {\sqrt {b} (d e-c f)^2 \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^3}+\frac {2 b^{3/2} f^2 \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{3 d^3}+\frac {(d e-c f)^2 (c+d x) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^3}+\frac {f (d e-c f) (c+d x)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{d^3}+\frac {f^2 (c+d x)^3 \sin \left (\frac {b}{(c+d x)^2}\right )}{3 d^3} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3514, 3440, 3468, 3433, 3460, 3378, 3383, 3490, 3469, 3432} \begin {gather*} \frac {2 \sqrt {2 \pi } b^{3/2} f^2 S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{3 d^3}-\frac {b f (d e-c f) \text {CosIntegral}\left (\frac {b}{(c+d x)^2}\right )}{d^3}-\frac {\sqrt {2 \pi } \sqrt {b} (d e-c f)^2 \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{c+d x}\right )}{d^3}+\frac {f (c+d x)^2 (d e-c f) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^3}+\frac {(c+d x) (d e-c f)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{d^3}+\frac {f^2 (c+d x)^3 \sin \left (\frac {b}{(c+d x)^2}\right )}{3 d^3}+\frac {2 b f^2 (c+d x) \cos \left (\frac {b}{(c+d x)^2}\right )}{3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*Sin[b/(c + d*x)^2],x]

[Out]

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

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

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Mathematica [A]
time = 0.30, size = 265, normalized size = 1.14 \begin {gather*} \frac {2 b c f^2 \cos \left (\frac {b}{(c+d x)^2}\right )+2 b d f^2 x \cos \left (\frac {b}{(c+d x)^2}\right )+3 b f (-d e+c f) \text {Ci}\left (\frac {b}{(c+d x)^2}\right )-3 \sqrt {b} (d e-c f)^2 \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+2 b^{3/2} f^2 \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+3 c d^2 e^2 \sin \left (\frac {b}{(c+d x)^2}\right )-3 c^2 d e f \sin \left (\frac {b}{(c+d x)^2}\right )+c^3 f^2 \sin \left (\frac {b}{(c+d x)^2}\right )+3 d^3 e^2 x \sin \left (\frac {b}{(c+d x)^2}\right )+3 d^3 e f x^2 \sin \left (\frac {b}{(c+d x)^2}\right )+d^3 f^2 x^3 \sin \left (\frac {b}{(c+d x)^2}\right )}{3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 225, normalized size = 0.97

method	result	size
derivativedivides	\(-\frac {-\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )-\frac {\left (-2 c \,f^{2}+2 d e f \right ) \left (d x +c \right )^{2} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}+\frac {\left (-2 c \,f^{2}+2 d e f \right ) b \cosineIntegral \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {f^{2} \left (d x +c \right )^{3} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{3}+\frac {2 f^{2} b \left (-\left (d x +c \right ) \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )}{3}}{d^{3}}\)	\(225\)
default	\(-\frac {-\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )-\frac {\left (-2 c \,f^{2}+2 d e f \right ) \left (d x +c \right )^{2} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}+\frac {\left (-2 c \,f^{2}+2 d e f \right ) b \cosineIntegral \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {f^{2} \left (d x +c \right )^{3} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{3}+\frac {2 f^{2} b \left (-\left (d x +c \right ) \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )}{3}}{d^{3}}\)	\(225\)
risch	\(-\frac {e^{2} b \sqrt {\pi }\, \erf \left (\frac {\sqrt {-i b}}{d x +c}\right )}{2 d \sqrt {-i b}}-\frac {i f^{2} b^{2} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-i b}}{d x +c}\right )}{3 d^{3} \sqrt {-i b}}-\frac {f^{2} c^{2} b \sqrt {\pi }\, \erf \left (\frac {\sqrt {-i b}}{d x +c}\right )}{2 d^{3} \sqrt {-i b}}-\frac {f^{2} c b \expIntegral \left (1, -\frac {i b}{\left (d x +c \right )^{2}}\right )}{2 d^{3}}+\frac {e f b \expIntegral \left (1, -\frac {i b}{\left (d x +c \right )^{2}}\right )}{2 d^{2}}+\frac {e f c b \sqrt {\pi }\, \erf \left (\frac {\sqrt {-i b}}{d x +c}\right )}{d^{2} \sqrt {-i b}}-\frac {e^{2} b \sqrt {\pi }\, \erf \left (\frac {\sqrt {i b}}{d x +c}\right )}{2 d \sqrt {i b}}+\frac {i f^{2} b^{2} \sqrt {\pi }\, \erf \left (\frac {\sqrt {i b}}{d x +c}\right )}{3 d^{3} \sqrt {i b}}-\frac {f^{2} c^{2} b \sqrt {\pi }\, \erf \left (\frac {\sqrt {i b}}{d x +c}\right )}{2 d^{3} \sqrt {i b}}-\frac {f^{2} c b \expIntegral \left (1, \frac {i b}{\left (d x +c \right )^{2}}\right )}{2 d^{3}}+\frac {e f b \expIntegral \left (1, \frac {i b}{\left (d x +c \right )^{2}}\right )}{2 d^{2}}+\frac {e f c b \sqrt {\pi }\, \erf \left (\frac {\sqrt {i b}}{d x +c}\right )}{d^{2} \sqrt {i b}}+\frac {2 b \,f^{2} \left (d x +c \right ) \cos \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{3 d^{3}}+\frac {\left (d^{3} f^{2} x^{3}+3 d^{3} e f \,x^{2}+3 d^{3} e^{2} x +c^{3} f^{2}-3 c^{2} d e f +3 c \,d^{2} e^{2}\right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{3 d^{3}}\)	\(457\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^3*(-(c^2*f^2-2*c*d*e*f+d^2*e^2)*(d*x+c)*sin(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)*FresnelC(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(b/(d*x+c)^2)+1/2*(-2*c*f^
2+2*d*e*f)*b*Ci(b/(d*x+c)^2)-1/3*f^2*(d*x+c)^3*sin(b/(d*x+c)^2)+2/3*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))))

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

[Out]

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

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Fricas [A]
time = 0.40, size = 303, normalized size = 1.30 \begin {gather*} \frac {4 \, \sqrt {2} \pi b d f^{2} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {S}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) - 6 \, \sqrt {2} {\left (\pi c^{2} d f^{2} - 2 \, \pi c d^{2} f e + \pi d^{3} e^{2}\right )} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {C}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) + 4 \, {\left (b d f^{2} x + b c f^{2}\right )} \cos \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 3 \, {\left (b c f^{2} - b d f e\right )} \operatorname {Ci}\left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 3 \, {\left (b c f^{2} - b d f e\right )} \operatorname {Ci}\left (-\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \, {\left (d^{3} f^{2} x^{3} + c^{3} f^{2} + 3 \, {\left (d^{3} x + c d^{2}\right )} e^{2} + 3 \, {\left (d^{3} f x^{2} - c^{2} d f\right )} e\right )} \sin \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{6 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*sqrt(2)*pi*b*d*f^2*sqrt(b/(pi*d^2))*fresnel_sin(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) - 6*sqrt(2)*(pi*c
^2*d*f^2 - 2*pi*c*d^2*f*e + pi*d^3*e^2)*sqrt(b/(pi*d^2))*fresnel_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + 4
*(b*d*f^2*x + b*c*f^2)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)) + 3*(b*c*f^2 - b*d*f*e)*cos_integral(b/(d^2*x^2 + 2*c*
d*x + c^2)) + 3*(b*c*f^2 - b*d*f*e)*cos_integral(-b/(d^2*x^2 + 2*c*d*x + c^2)) + 2*(d^3*f^2*x^3 + c^3*f^2 + 3*
(d^3*x + c*d^2)*e^2 + 3*(d^3*f*x^2 - c^2*d*f)*e)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^3

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

[Out]

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

[Out]

integrate((f*x + e)^2*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 )}^2 \,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)^2,x)

[Out]

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

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