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

Optimal. Leaf size=122 \[ \frac{\sqrt{\frac{\pi }{2}} \sin (a) (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{\sqrt{b} d^2}+\frac{\sqrt{\frac{\pi }{2}} \cos (a) (d e-c f) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d^2}-\frac{f \cos \left (a+b (c+d x)^2\right )}{2 b d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17774, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3433, 3353, 3352, 3351, 3379, 2638} \[ \frac{\sqrt{\frac{\pi }{2}} \sin (a) (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )}{\sqrt{b} d^2}+\frac{\sqrt{\frac{\pi }{2}} \cos (a) (d e-c f) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d^2}-\frac{f \cos \left (a+b (c+d x)^2\right )}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(f*Cos[a + b*(c + d*x)^2])/(2*b*d^2) + ((d*e - c*f)*Sqrt[Pi/2]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])
/(Sqrt[b]*d^2) + ((d*e - c*f)*Sqrt[Pi/2]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)]*Sin[a])/(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 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

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

Mathematica [A]  time = 0.60075, size = 114, normalized size = 0.93 \[ \frac{\sqrt{2 \pi } \sqrt{b} \sin (a) (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} (c+d x)\right )+\sqrt{2 \pi } \sqrt{b} \cos (a) (d e-c f) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )-f \cos \left (a+b (c+d x)^2\right )}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 309, normalized size = 2.5 \begin{align*} -{\frac{f\cos \left ({d}^{2}{x}^{2}b+2\,cdxb+{c}^{2}b+a \right ) }{2\,{d}^{2}b}}-{\frac{cf\sqrt{2}\sqrt{\pi }}{2\,d} \left ( \cos \left ({\frac{{b}^{2}{c}^{2}{d}^{2}-{d}^{2}b \left ({c}^{2}b+a \right ) }{{d}^{2}b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ) -\sin \left ({\frac{{b}^{2}{c}^{2}{d}^{2}-{d}^{2}b \left ({c}^{2}b+a \right ) }{{d}^{2}b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ) \right ){\frac{1}{\sqrt{{d}^{2}b}}}}+{\frac{\sqrt{2}\sqrt{\pi }e}{2} \left ( \cos \left ({\frac{{b}^{2}{c}^{2}{d}^{2}-{d}^{2}b \left ({c}^{2}b+a \right ) }{{d}^{2}b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ) -\sin \left ({\frac{{b}^{2}{c}^{2}{d}^{2}-{d}^{2}b \left ({c}^{2}b+a \right ) }{{d}^{2}b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ) \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(a+(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+a)-1/2*f*c/d*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*(cos((b^2*c^2*d^2-d^2*b
*(b*c^2+a))/d^2/b)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-sin((b^2*c^2*d^2-d^2*b*(b*c^2+a))/
d^2/b)*FresnelC(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*(cos((b^
2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/b)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-sin((b^2*c^2*d^2-d^
2*b*(b*c^2+a))/d^2/b)*FresnelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d)))

________________________________________________________________________________________

Maxima [C]  time = 2.53045, size = 1435, normalized size = 11.76 \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(a+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*ar
ctan2(0, b)) + sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a) - (cos(1/4*pi + 1/2*arctan2(0, b)) + cos(-1/4*pi + 1/2
*arctan2(0, b)) - I*sin(1/4*pi + 1/2*arctan2(0, b)) + I*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*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)))*cos(a) + (cos(1/4*pi + 1/2*arctan2(0, b)) + cos(-1/4*
pi + 1/2*arctan2(0, b)) + I*sin(1/4*pi + 1/2*arctan2(0, b)) - I*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*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))*cos(a) + (I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - I*e^(-I*b*d^2*x^2 - 2*I
*b*c*d*x - I*b*c^2))*sin(a))*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))*cos(a) + 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))*sin(a))*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))*cos(a) + 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))*sin(a))*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))*cos(a) + (-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))*sin(a))*b*c*d*x + (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(sq
rt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*cos(a) + (-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))*sin(a))*b*c^2)*sin(1/2*ar
ctan2(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))

________________________________________________________________________________________

Fricas [A]  time = 1.82045, size = 335, normalized size = 2.75 \begin{align*} \frac{\sqrt{2} \pi \sqrt{\frac{b d^{2}}{\pi }}{\left (d e - c f\right )} \cos \left (a\right ) \operatorname{S}\left (\frac{\sqrt{2} \sqrt{\frac{b d^{2}}{\pi }}{\left (d x + c\right )}}{d}\right ) + \sqrt{2} \pi \sqrt{\frac{b d^{2}}{\pi }}{\left (d e - c f\right )} \operatorname{C}\left (\frac{\sqrt{2} \sqrt{\frac{b d^{2}}{\pi }}{\left (d x + c\right )}}{d}\right ) \sin \left (a\right ) - d f \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}{2 \, b d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [C]  time = 1.1967, size = 525, normalized size = 4.3 \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^{\left (i \, a + 1\right )}}{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^{\left (-i \, a + 1\right )}}{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 ) e^{\left (i \, a\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} + i \, a\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 ) e^{\left (-i \, a\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} - i \, a\right )}}{b d}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(a+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^(I*a + 1)/(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^(-I*a + 1)/(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))*e^(I*a)/(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 + I*a)/(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))*e^(-I*a)/(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 - I*a)/(b*d))/d