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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 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 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 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 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 (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*}

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Mathematica [A]  time = 0.60, size = 114, normalized size = 0.93 \[ \frac {\sqrt {2 \pi } \sqrt {b} \sin (a) (d e-c f) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )+\sqrt {2 \pi } \sqrt {b} \cos (a) (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )-f \cos \left (a+b (c+d x)^2\right )}{2 b d^2} \]

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)

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fricas [A]  time = 0.67, size = 131, normalized size = 1.07 \[ \frac {\sqrt {2} \pi \sqrt {\frac {b d^{2}}{\pi }} {\left (d e - c f\right )} \cos \relax (a) \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) + \sqrt {2} \pi \sqrt {\frac {b d^{2}}{\pi }} {\left (d e - c f\right )} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) \sin \relax (a) - d f \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}{2 \, b d^{3}} \]

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)

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giac [C]  time = 0.72, size = 389, normalized size = 3.19 \[ \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right ) e^{\left (i \, a + 1\right )}}{4 \, \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} - \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right ) e^{\left (-i \, a + 1\right )}}{4 \, \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} - \frac {\frac {i \, \sqrt {2} \sqrt {\pi } c f \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right ) e^{\left (i \, a\right )}}{\sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} + \frac {f e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2} + i \, a\right )}}{b d}}{4 \, d} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } c f \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right ) e^{\left (-i \, a\right )}}{\sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} + \frac {f e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2} - i \, a\right )}}{b d}}{4 \, d} \]

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

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maple [B]  time = 0.03, size = 309, normalized size = 2.53 \[ -\frac {f \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}+a \right )}{2 d^{2} b}-\frac {f c \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {b^{2} c^{2} d^{2}-d^{2} b \left (b \,c^{2}+a \right )}{d^{2} b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )-\sin \left (\frac {b^{2} c^{2} d^{2}-d^{2} b \left (b \,c^{2}+a \right )}{d^{2} b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )\right )}{2 d \sqrt {d^{2} b}}+\frac {\sqrt {2}\, \sqrt {\pi }\, e \left (\cos \left (\frac {b^{2} c^{2} d^{2}-d^{2} b \left (b \,c^{2}+a \right )}{d^{2} b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )-\sin \left (\frac {b^{2} c^{2} d^{2}-d^{2} b \left (b \,c^{2}+a \right )}{d^{2} b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )\right )}{2 \sqrt {d^{2} b}} \]

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)))

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maxima [C]  time = 1.22, size = 481, normalized size = 3.94 \[ -\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, \cos \relax (a) + \left (i - 1\right ) \, \sin \relax (a)\right )} \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {i \, b}}\right ) + {\left (-\left (i - 1\right ) \, \cos \relax (a) + \left (i + 1\right ) \, \sin \relax (a)\right )} \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {-i \, b}}\right )\right )} e}{8 \, \sqrt {b} d} - \frac {{\left ({\left (1024 \, {\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 )} \cos \relax (a) + {\left (1024 i \, e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )} - 1024 i \, e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}\right )} \sin \relax (a)\right )} d x + 2 \, \sqrt {b d^{2} x^{2} + 2 \, b c d x + b c^{2}} {\left ({\left (\left (256 i + 256\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 (256 i - 256\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 )} \cos \relax (a) + {\left (-\left (256 i - 256\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 (256 i + 256\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 )} \sin \relax (a)\right )} c + {\left (1024 \, {\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 )} \cos \relax (a) + {\left (1024 i \, e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )} - 1024 i \, e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}\right )} \sin \relax (a)\right )} c\right )} f}{4096 \, {\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(a+b*(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/8*sqrt(2)*sqrt(pi)*((-(I + 1)*cos(a) + (I - 1)*sin(a))*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (-(I - 1)*cos(a)
+ (I + 1)*sin(a))*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))*e/(sqrt(b)*d) - 1/4096*((1024*(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) + (1024*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*
c^2) - 1024*I*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*d*x + 2*sqrt(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*((
(256*I + 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) - (256*I - 256)*sqrt(2)*sq
rt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*cos(a) + (-(256*I - 256)*sqrt(2)*sqrt(pi)*(erf(s
qrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + (256*I + 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b
*c*d*x - I*b*c^2)) - 1))*sin(a))*c + (1024*(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) + (1024*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 1024*I*e^(-I*b*d^2*x^2 - 2*I*b*c*
d*x - I*b*c^2))*sin(a))*c)*f/(b*d^3*x + b*c*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(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)

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