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

Optimal. Leaf size=140 \[ \frac {\sqrt {\frac {\pi }{2}} \sin \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {e \cos \left (a+b x+c x^2\right )}{2 c} \]

[Out]

-1/2*e*cos(c*x^2+b*x+a)/c+1/4*(-b*e+2*c*d)*cos(a-1/4/c*b^2)*FresnelS(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*2
^(1/2)*Pi^(1/2)/c^(3/2)+1/4*(-b*e+2*c*d)*FresnelC(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a-1/4/c*b^2)*2^(
1/2)*Pi^(1/2)/c^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3461, 3447, 3351, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} \sin \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {e \cos \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Sin[a + b*x + c*x^2],x]

[Out]

-(e*Cos[a + b*x + c*x^2])/(2*c) + ((2*c*d - b*e)*Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*S
qrt[2*Pi])])/(2*c^(3/2)) + ((2*c*d - b*e)*Sqrt[Pi/2]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4
*c)])/(2*c^(3/2))

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 3447

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Cos[(b^2 - 4*a*c)/(4*c)], Int[Sin[(b + 2*c*x)^2/
(4*c)], x], x] - Dist[Sin[(b^2 - 4*a*c)/(4*c)], Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] &&
NeQ[b^2 - 4*a*c, 0]

Rule 3461

Int[((d_.) + (e_.)*(x_))*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> -Simp[(e*Cos[a + b*x + c*x^2])/(
2*c), x] + Dist[(2*c*d - b*e)/(2*c), Int[Sin[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*
d - b*e, 0]

Rubi steps

\begin {align*} \int (d+e x) \sin \left (a+b x+c x^2\right ) \, dx &=-\frac {e \cos \left (a+b x+c x^2\right )}{2 c}+\frac {(2 c d-b e) \int \sin \left (a+b x+c x^2\right ) \, dx}{2 c}\\ &=-\frac {e \cos \left (a+b x+c x^2\right )}{2 c}+\frac {\left ((2 c d-b e) \cos \left (a-\frac {b^2}{4 c}\right )\right ) \int \sin \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{2 c}+\frac {\left ((2 c d-b e) \sin \left (a-\frac {b^2}{4 c}\right )\right ) \int \cos \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{2 c}\\ &=-\frac {e \cos \left (a+b x+c x^2\right )}{2 c}+\frac {(2 c d-b e) \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {(2 c d-b e) \sqrt {\frac {\pi }{2}} C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )}{2 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.66, size = 128, normalized size = 0.91 \[ \frac {\sqrt {2 \pi } \sin \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+\sqrt {2 \pi } \cos \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-2 \sqrt {c} e \cos (a+x (b+c x))}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c]*e*Cos[a + x*(b + c*x)] + (2*c*d - b*e)*Sqrt[2*Pi]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]
*Sqrt[2*Pi])] + (2*c*d - b*e)*Sqrt[2*Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(4*c^(
3/2))

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fricas [A]  time = 0.42, size = 135, normalized size = 0.96 \[ \frac {\sqrt {2} \pi {\left (2 \, c d - b e\right )} \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + \sqrt {2} \pi {\left (2 \, c d - b e\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - 2 \, c e \cos \left (c x^{2} + b x + a\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*pi*(2*c*d - b*e)*sqrt(c/pi)*cos(-1/4*(b^2 - 4*a*c)/c)*fresnel_sin(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/
pi)/c) + sqrt(2)*pi*(2*c*d - b*e)*sqrt(c/pi)*fresnel_cos(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c)*sin(-1/4*(b^2 -
4*a*c)/c) - 2*c*e*cos(c*x^2 + b*x + a))/c^2

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giac [C]  time = 0.19, size = 327, normalized size = 2.34 \[ \frac {i \, \sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{4 \, {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - \frac {i \, \sqrt {2} \sqrt {\pi } d \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{4 \, {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - \frac {\frac {i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c - 4 \, c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + 2 \, e^{\left (i \, c x^{2} + i \, b x + i \, a + 1\right )}}{8 \, c} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c - 4 \, c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + 2 \, e^{\left (-i \, c x^{2} - i \, b x - i \, a + 1\right )}}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*sqrt(2)*sqrt(pi)*d*erf(-1/4*sqrt(2)*(2*x + b/c)*(-I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(I*b^2 - 4*I*a*c
)/c)/((-I*c/abs(c) + 1)*sqrt(abs(c))) - 1/4*I*sqrt(2)*sqrt(pi)*d*erf(-1/4*sqrt(2)*(2*x + b/c)*(I*c/abs(c) + 1)
*sqrt(abs(c)))*e^(-1/4*(-I*b^2 + 4*I*a*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) - 1/8*(I*sqrt(2)*sqrt(pi)*b*erf(-
1/4*sqrt(2)*(2*x + b/c)*(-I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(I*b^2 - 4*I*a*c - 4*c)/c)/((-I*c/abs(c) + 1)*
sqrt(abs(c))) + 2*e^(I*c*x^2 + I*b*x + I*a + 1))/c - 1/8*(-I*sqrt(2)*sqrt(pi)*b*erf(-1/4*sqrt(2)*(2*x + b/c)*(
I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(-I*b^2 + 4*I*a*c - 4*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) + 2*e^(-I*c*
x^2 - I*b*x - I*a + 1))/c

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maple [A]  time = 0.02, size = 184, normalized size = 1.31 \[ -\frac {e \cos \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {e b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}+\frac {\sqrt {2}\, \sqrt {\pi }\, d \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{2 \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e*cos(c*x^2+b*x+a)/c-1/4*e*b/c^(3/2)*2^(1/2)*Pi^(1/2)*(cos((1/4*b^2-c*a)/c)*FresnelS(2^(1/2)/Pi^(1/2)/c^(
1/2)*(c*x+1/2*b))-sin((1/4*b^2-c*a)/c)*FresnelC(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b)))+1/2*2^(1/2)*Pi^(1/2)/c^
(1/2)*d*(cos((1/4*b^2-c*a)/c)*FresnelS(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b))-sin((1/4*b^2-c*a)/c)*FresnelC(2^(
1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b)))

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maxima [C]  time = 0.90, size = 695, normalized size = 4.96 \[ -\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + \left (i - 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x + i \, b}{2 \, \sqrt {i \, c}}\right ) + {\left (-\left (i - 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + \left (i + 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x + i \, b}{2 \, \sqrt {-i \, c}}\right )\right )} d}{8 \, \sqrt {c}} + \frac {{\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left ({\left (-\left (2 i + 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} + \left (2 i - 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b c \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + {\left (\left (2 i - 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (2 i + 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b c \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} x - {\left (4 \, c {\left (e^{\left (\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )} + e^{\left (-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )}\right )} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - c {\left (-4 i \, e^{\left (\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )} + 4 i \, e^{\left (-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )}\right )} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}\right )} e}{16 \, c^{2} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(pi)*((-(I + 1)*cos(-1/4*(b^2 - 4*a*c)/c) + (I - 1)*sin(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*I*c
*x + I*b)/sqrt(I*c)) + (-(I - 1)*cos(-1/4*(b^2 - 4*a*c)/c) + (I + 1)*sin(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*I*c
*x + I*b)/sqrt(-I*c)))*d/sqrt(c) + 1/16*((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I
*b^2)/c)) - 1) + (I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*cos(-
1/4*(b^2 - 4*a*c)/c) + ((I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (I
+ 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*sin(-1/4*(b^2 - 4*a*c)/c
) + ((-(2*I + 2)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (2*I - 2)*sqrt(2)
*sqrt(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*cos(-1/4*(b^2 - 4*a*c)/c) + ((2*I - 2
)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (2*I + 2)*sqrt(2)*sqrt(pi)*(erf(
1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*sin(-1/4*(b^2 - 4*a*c)/c))*x - (4*c*(e^(1/4*(4*I*c^2
*x^2 + 4*I*b*c*x + I*b^2)/c) + e^(-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*cos(-1/4*(b^2 - 4*a*c)/c) - c*(-4
*I*e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + 4*I*e^(-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*sin(-1/4*(b
^2 - 4*a*c)/c))*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/c))*e/(c^2*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x + c*x^2)*(d + e*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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