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]

-(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)

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Rubi [A]  time = 0.0726357, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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*}

Mathematica [A]  time = 0.174191, 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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Maple [B]  time = 0.006, size = 120, normalized size = 1.7 \begin{align*} -{\frac{f\cos \left ({d}^{2}{x}^{2}b+2\,cdxb+{c}^{2}b \right ) }{2\,{d}^{2}b}}-{\frac{cf\sqrt{2}\sqrt{\pi }}{2\,d}{\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}}+{\frac{\sqrt{2}\sqrt{\pi }e}{2}{\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}} \end{align*}

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 = 2.48203, size = 829, normalized size = 12.01 \begin{align*} \text{result too large to display} \end{align*}

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(pi)*((I*cos(1/4*pi + 1/2*arctan2(0, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arcta
n2(0, b)) - sin(-1/4*pi + 1/2*arctan2(0, b)))*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (I*cos(1/4*pi + 1/2*arctan2(0
, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) - sin(1/4*pi + 1/2*arctan2(0, b)) + sin(-1/4*pi + 1/2*arctan2(0, b)
))*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))*e/(d*sqrt(abs(b))) - 1/4*((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))*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2) - (((-2*I*sqrt(pi)*(erf(sqrt(I*b
*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + 2*I*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*
b*c*d*x + (-2*I*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + 2*I*sqrt(pi)*(erf(sqrt(-I*b*d^
2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2)*cos(1/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)) - 2*((sqrt
(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*
b*c^2)) - 1))*b*c*d*x + (sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + sqrt(pi)*(erf(sqrt(-I
*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2)*sin(1/2*arctan2(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2, 0)))*sqrt
(abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2)))*f/(b*d^2*abs(4*b*d^2*x^2 + 8*b*c*d*x + 4*b*c^2))

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Fricas [A]  time = 1.65385, size = 192, normalized size = 2.78 \begin{align*} \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}} \end{align*}

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

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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Giac [C]  time = 1.17714, size = 495, normalized size = 7.17 \begin{align*} -\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} \end{align*}

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