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

Optimal. Leaf size=120 \[ -\frac {b f \text {Ci}\left (\frac {b}{(c+d x)^2}\right )}{2 d^2}-\frac {\sqrt {2 \pi } \sqrt {b} (d e-c f) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d^2}+\frac {(c+d x) (d e-c f) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{2 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3433, 3359, 3387, 3352, 3379, 3297, 3302} \[ -\frac {b f \text {CosIntegral}\left (\frac {b}{(c+d x)^2}\right )}{2 d^2}-\frac {\sqrt {2 \pi } \sqrt {b} (d e-c f) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{c+d x}\right )}{d^2}+\frac {(c+d x) (d e-c f) \sin \left (\frac {b}{(c+d x)^2}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (\frac {b}{(c+d x)^2}\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*Sin[b/(c + d*x)^2],x]

[Out]

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

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 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 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 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 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 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 95, normalized size = 0.79 \[ -\frac {b f \text {Ci}\left (\frac {b}{(c+d x)^2}\right )+2 \sqrt {2 \pi } \sqrt {b} (d e-c f) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+(c+d x) \sin \left (\frac {b}{(c+d x)^2}\right ) (c f-2 d e-d f x)}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*f*CosIntegral[b/(c + d*x)^2] + 2*Sqrt[b]*(d*e - c*f)*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x
)] + (c + d*x)*(-2*d*e + c*f - d*f*x)*Sin[b/(c + d*x)^2])/d^2

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 155, normalized size = 1.29 \[ -\frac {4 \, \sqrt {2} \pi {\left (d^{2} e - c d f\right )} \sqrt {\frac {b}{\pi d^{2}}} \operatorname {C}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) + b f \operatorname {Ci}\left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + b f \operatorname {Ci}\left (-\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 2 \, {\left (d^{2} f x^{2} + 2 \, d^{2} e x + 2 \, c d e - c^{2} f\right )} \sin \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{4 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*pi*(d^2*e - c*d*f)*sqrt(b/(pi*d^2))*fresnel_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) + b*f*co
s_integral(b/(d^2*x^2 + 2*c*d*x + c^2)) + b*f*cos_integral(-b/(d^2*x^2 + 2*c*d*x + c^2)) - 2*(d^2*f*x^2 + 2*d^
2*e*x + 2*c*d*e - c^2*f)*sin(b/(d^2*x^2 + 2*c*d*x + c^2)))/d^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 101, normalized size = 0.84 \[ \frac {-\left (c f -d e \right ) \left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\left (c f -d e \right ) \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )+\frac {f \left (d x +c \right )^{2} \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}-\frac {f b \Ci \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{2}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (f x^{2} + 2 \, e x\right )} \sin \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \int \frac {{\left (b d f x^{2} + 2 \, b d e x\right )} \cos \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )}}\,{d x} + \int \frac {{\left (b d f x^{2} + 2 \, b d e x\right )} \cos \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, {\left ({\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \cos \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} + {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \sin \left (\frac {b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (\frac {b}{{\left (c+d\,x\right )}^2}\right )\,\left (e+f\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b/(c + d*x)^2)*(e + f*x),x)

[Out]

int(sin(b/(c + d*x)^2)*(e + f*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________