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

Optimal. Leaf size=69 \[ \frac {\sqrt {\frac {\pi }{2}} (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^2}-\frac {f \cos \left (b (c+d x)^2\right )}{2 b d^2} \]

[Out]

-1/2*f*cos(b*(d*x+c)^2)/b/d^2+1/2*(-c*f+d*e)*FresnelS((d*x+c)*b^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/d^2/b
^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3433, 3351, 3379, 2638} \[ \frac {\sqrt {\frac {\pi }{2}} (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^2}-\frac {f \cos \left (b (c+d x)^2\right )}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*Sin[b*(c + d*x)^2],x]

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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 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 (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 (b x^2\right )+f x \sin \left (b x^2\right )\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {f \operatorname {Subst}\left (\int x \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^2}+\frac {(d e-c f) \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^2}\\ &=\frac {(d e-c f) \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^2}+\frac {f \operatorname {Subst}\left (\int \sin (b x) \, dx,x,(c+d x)^2\right )}{2 d^2}\\ &=-\frac {f \cos \left (b (c+d x)^2\right )}{2 b d^2}+\frac {(d e-c f) \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^2}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 66, normalized size = 0.96 \[ \frac {\sqrt {2 \pi } \sqrt {b} (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )-f \cos \left (b (c+d x)^2\right )}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 80, normalized size = 1.16 \[ \frac {\sqrt {2} \pi \sqrt {\frac {b d^{2}}{\pi }} {\left (d e - c f\right )} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) - d f \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )}{2 \, b d^{3}} \]

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]

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

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giac [C]  time = 0.93, size = 367, normalized size = 5.32 \[ -\frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right ) e}{4 \, \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} + \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right ) e}{4 \, \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } c f \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} + \frac {f e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}}{b d}}{4 \, d} - \frac {\frac {i \, \sqrt {2} \sqrt {\pi } c f \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} + \frac {f e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )}}{b d}}{4 \, d} \]

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]

-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/(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/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) - 1/4*(-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))/(sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)) + f*e^(-I*b*d^2*x^2 - 2*
I*b*c*d*x - I*b*c^2)/(b*d))/d - 1/4*(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))/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) + f*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)/(b*d
))/d

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maple [B]  time = 0.02, size = 120, normalized size = 1.74 \[ -\frac {f \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {f c \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 d \sqrt {d^{2} b}}+\frac {\sqrt {2}\, \sqrt {\pi }\, e \,\mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 \sqrt {d^{2} b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*f/d^2/b*cos(b*d^2*x^2+2*b*c*d*x+b*c^2)-1/2*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*FresnelS(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 = 0.86, size = 271, normalized size = 3.93 \[ \frac {\sqrt {2} \sqrt {\pi } e {\left (\left (i + 1\right ) \, \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {i \, b}}\right ) + \left (i - 1\right ) \, \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {-i \, b}}\right )\right )}}{8 \, \sqrt {b} d} - \frac {{\left (2 \, d x {\left (e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )} + e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}\right )} - \sqrt {b d^{2} x^{2} + 2 \, b c d x + b c^{2}} {\left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}}\right ) - 1\right )}\right )} c + 2 \, c {\left (e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )} + e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}\right )}\right )} f}{8 \, {\left (b d^{3} x + b c d^{2}\right )}} \]

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]

1/8*sqrt(2)*sqrt(pi)*e*((I + 1)*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (I - 1)*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))/
(sqrt(b)*d) - 1/8*(2*d*x*(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))
- sqrt(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*(-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2
)) - 1) + (I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*c + 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)))*f/(b*d^3*x + b*c*d^2)

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mupad [B]  time = 0.11, size = 58, normalized size = 0.84 \[ -\frac {f\,\cos \left (b\,{\left (c+d\,x\right )}^2\right )}{2\,b\,d^2}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )\,\left (c\,f-d\,e\right )}{2\,\sqrt {b}\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*(c + d*x)^2)*(e + f*x),x)

[Out]

- (f*cos(b*(c + d*x)^2))/(2*b*d^2) - (2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*b^(1/2)*(c + d*x))/pi^(1/2))*(c*f - d
*e))/(2*b^(1/2)*d^2)

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

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*b*c*d*x + b*d**2*x**2), x)

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