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

Optimal. Leaf size=285 \[ \frac {\sqrt {\frac {\pi }{2}} \sin \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e)^2 C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}+\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e)^2 S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}+\frac {\sqrt {\frac {\pi }{2}} e^2 \cos \left (a-\frac {b^2}{4 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} e^2 \sin \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {e (2 c d-b e) \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac {e (d+e x) \cos \left (a+b x+c x^2\right )}{2 c} \]

[Out]

-1/4*e*(-b*e+2*c*d)*cos(c*x^2+b*x+a)/c^2-1/2*e*(e*x+d)*cos(c*x^2+b*x+a)/c+1/4*e^2*cos(a-1/4/c*b^2)*FresnelC(1/
2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/c^(3/2)+1/8*(-b*e+2*c*d)^2*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^(5/2)+1/8*(-b*e+2*c*d)^2*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^(5/2)-1/4*e^2*FresnelS(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 = 0.27, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3463, 3448, 3352, 3351, 3461, 3447} \[ \frac {\sqrt {\frac {\pi }{2}} \sin \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e)^2 \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{4 c^{5/2}}+\frac {\sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) (2 c d-b e)^2 S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{4 c^{5/2}}+\frac {\sqrt {\frac {\pi }{2}} e^2 \cos \left (a-\frac {b^2}{4 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{2 c^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} e^2 \sin \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {e (2 c d-b e) \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac {e (d+e x) \cos \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*(2*c*d - b*e)*Cos[a + b*x + c*x^2])/(4*c^2) - (e*(d + e*x)*Cos[a + b*x + c*x^2])/(2*c) + (e^2*Sqrt[Pi/2]*C
os[a - b^2/(4*c)]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(2*c^(3/2)) + ((2*c*d - b*e)^2*Sqrt[Pi/2]*Cos[a
- b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(4*c^(5/2)) + ((2*c*d - b*e)^2*Sqrt[Pi/2]*FresnelC[(b
+ 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(4*c^(5/2)) - (e^2*Sqrt[Pi/2]*FresnelS[(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 3448

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

Rule 3463

Int[((d_.) + (e_.)*(x_))^(m_)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> -Simp[(e*(d + e*x)^(m - 1)*
Cos[a + b*x + c*x^2])/(2*c), x] + (Dist[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Cos[a + b*x + c*x^2], x], x
] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Sin[a + b*x + c*x^2], x], x]) /; FreeQ[{a, b, c, d, e}, x]
&& NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rubi steps

\begin {align*} \int (d+e x)^2 \sin \left (a+b x+c x^2\right ) \, dx &=-\frac {e (d+e x) \cos \left (a+b x+c x^2\right )}{2 c}+\frac {e^2 \int \cos \left (a+b x+c x^2\right ) \, dx}{2 c}-\frac {(-2 c d+b e) \int (d+e x) \sin \left (a+b x+c x^2\right ) \, dx}{2 c}\\ &=-\frac {e (2 c d-b e) \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac {e (d+e x) \cos \left (a+b x+c x^2\right )}{2 c}+\frac {(2 c d-b e)^2 \int \sin \left (a+b x+c x^2\right ) \, dx}{4 c^2}+\frac {\left (e^2 \cos \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 {\left (e^2 \sin \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 {e (2 c d-b e) \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac {e (d+e x) \cos \left (a+b x+c x^2\right )}{2 c}+\frac {e^2 \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}-\frac {e^2 \sqrt {\frac {\pi }{2}} S\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}}+\frac {\left ((2 c d-b e)^2 \cos \left (a-\frac {b^2}{4 c}\right )\right ) \int \sin \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{4 c^2}+\frac {\left ((2 c d-b e)^2 \sin \left (a-\frac {b^2}{4 c}\right )\right ) \int \cos \left (\frac {(b+2 c x)^2}{4 c}\right ) \, dx}{4 c^2}\\ &=-\frac {e (2 c d-b e) \cos \left (a+b x+c x^2\right )}{4 c^2}-\frac {e (d+e x) \cos \left (a+b x+c x^2\right )}{2 c}+\frac {e^2 \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b^2}{4 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {(2 c d-b e)^2 \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 )}{4 c^{5/2}}+\frac {(2 c d-b e)^2 \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 )}{4 c^{5/2}}-\frac {e^2 \sqrt {\frac {\pi }{2}} S\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 = 1.33, size = 186, normalized size = 0.65 \[ \frac {\sqrt {2 \pi } C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (\sin \left (a-\frac {b^2}{4 c}\right ) (b e-2 c d)^2+2 c e^2 \cos \left (a-\frac {b^2}{4 c}\right )\right )+\sqrt {2 \pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \left (\cos \left (a-\frac {b^2}{4 c}\right ) (b e-2 c d)^2-2 c e^2 \sin \left (a-\frac {b^2}{4 c}\right )\right )+2 \sqrt {c} e \cos (a+x (b+c x)) (b e-2 c (2 d+e x))}{8 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*e*(b*e - 2*c*(2*d + e*x))*Cos[a + x*(b + c*x)] + Sqrt[2*Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi
])]*((-2*c*d + b*e)^2*Cos[a - b^2/(4*c)] - 2*c*e^2*Sin[a - b^2/(4*c)]) + Sqrt[2*Pi]*FresnelC[(b + 2*c*x)/(Sqrt
[c]*Sqrt[2*Pi])]*(2*c*e^2*Cos[a - b^2/(4*c)] + (-2*c*d + b*e)^2*Sin[a - b^2/(4*c)]))/(8*c^(5/2))

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fricas [A]  time = 0.43, size = 231, normalized size = 0.81 \[ \frac {\sqrt {2} {\left (2 \, \pi c e^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} {\left (2 \, \pi c e^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - 2 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \cos \left (c x^{2} + b x + a\right )}{8 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(2)*(2*pi*c*e^2*cos(-1/4*(b^2 - 4*a*c)/c) + pi*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*sin(-1/4*(b^2 - 4*a*
c)/c))*sqrt(c/pi)*fresnel_cos(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c) - sqrt(2)*(2*pi*c*e^2*sin(-1/4*(b^2 - 4*a*
c)/c) - pi*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*cos(-1/4*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_sin(1/2*sqrt(2)*(2*
c*x + b)*sqrt(c/pi)/c) - 2*(2*c^2*e^2*x + 4*c^2*d*e - b*c*e^2)*cos(c*x^2 + b*x + a))/c^3

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giac [C]  time = 0.39, size = 569, normalized size = 2.00 \[ \frac {i \, \sqrt {2} \sqrt {\pi } d^{2} \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^{2} \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 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}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + 2 \, d e^{\left (i \, c x^{2} + i \, b x + i \, a + 1\right )}}{4 \, c} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } b 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}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + 2 \, d e^{\left (-i \, c x^{2} - i \, b x - i \, a + 1\right )}}{4 \, c} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (b^{2} + 2 i \, c\right )} \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 - 8 \, c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 i \, {\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (i \, c x^{2} + i \, b x + i \, a + 2\right )}}{16 \, c^{2}} - \frac {\frac {i \, \sqrt {2} \sqrt {\pi } {\left (b^{2} - 2 i \, c\right )} \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 - 8 \, c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 i \, {\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (-i \, c x^{2} - i \, b x - i \, a + 2\right )}}{16 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*sqrt(2)*sqrt(pi)*d^2*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^2*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)*b*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 - 4*c)/c)/((-I*c/abs(c)
+ 1)*sqrt(abs(c))) + 2*d*e^(I*c*x^2 + I*b*x + I*a + 1))/c - 1/4*(-I*sqrt(2)*sqrt(pi)*b*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 - 4*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) +
2*d*e^(-I*c*x^2 - I*b*x - I*a + 1))/c - 1/16*(-I*sqrt(2)*sqrt(pi)*(b^2 + 2*I*c)*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 - 8*c)/c)/((-I*c/abs(c) + 1)*sqrt(abs(c))) - 2*I*(c*(2
*I*x + I*b/c) - 2*I*b)*e^(I*c*x^2 + I*b*x + I*a + 2))/c^2 - 1/16*(I*sqrt(2)*sqrt(pi)*(b^2 - 2*I*c)*erf(-1/4*sq
rt(2)*(2*x + b/c)*(I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(-I*b^2 + 4*I*a*c - 8*c)/c)/((I*c/abs(c) + 1)*sqrt(ab
s(c))) - 2*I*(c*(2*I*x + I*b/c) - 2*I*b)*e^(-I*c*x^2 - I*b*x - I*a + 2))/c^2

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maple [A]  time = 0.03, size = 399, normalized size = 1.40 \[ -\frac {e^{2} x \cos \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {e^{2} b \left (-\frac {\cos \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{2 c}+\frac {e^{2} \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}}-\frac {d e \cos \left (c \,x^{2}+b x +a \right )}{c}-\frac {d e b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {2}\, \sqrt {\pi }\, d^{2} \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{2 \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^2/c*x*cos(c*x^2+b*x+a)-1/2*e^2*b/c*(-1/2*cos(c*x^2+b*x+a)/c-1/4*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))))+1/4*e^2/c^(3/2)*2^(1/2)*Pi^(1/2)*(cos((1/4*b^2-c*a)/c)*FresnelC(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x
+1/2*b))+sin((1/4*b^2-c*a)/c)*FresnelS(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b)))-d*e/c*cos(c*x^2+b*x+a)-1/2*d*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^2*(cos((1/4*b^2-c*a)/c)*F
resnelS(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 = 2.24, size = 2259, normalized size = 7.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*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^2/sqrt(c) + 1/8*((-(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))*d*e/(c^2*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/c)) - 1/32*(((
(-(8*I + 8)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (8*I - 8)*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*c^3 + (-(32*I - 32)*sqrt(2)*gamma(3/2, 1/4
*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (32*I + 32)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/
c))*c^4)*cos(-1/4*(b^2 - 4*a*c)/c) + (((8*I - 8)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b
^2)/c)) - 1) - (8*I + 8)*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*c^3 +
(-(32*I + 32)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (32*I - 32)*sqrt(2)*gamma(3/2, -1
/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*c^4)*sin(-1/4*(b^2 - 4*a*c)/c))*x^3 + (((-(12*I + 12)*sqrt(2)*sqrt(pi
)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (12*I - 12)*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^3*c^2 + (-(48*I - 48)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c) + (48*I + 48)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*cos(-1/4*(b^2
- 4*a*c)/c) + (((12*I - 12)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (12*I
+ 12)*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^3*c^2 + (-(48*I + 48)*sqrt
(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (48*I - 48)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4
*I*b*c*x + I*b^2)/c))*b*c^3)*sin(-1/4*(b^2 - 4*a*c)/c))*x^2 - (8*b*c^2*(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) - b*c^2*(-8*I*e^(1/4*(4*I*c^2*
x^2 + 4*I*b*c*x + I*b^2)/c) + 8*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))*((4
*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2) + (((-(6*I + 6)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I
*b^2)/c)) - 1) + (6*I - 6)*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^4*c +
(-(24*I - 24)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (24*I + 24)*sqrt(2)*gamma(3/2, -1
/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*cos(-1/4*(b^2 - 4*a*c)/c) + (((6*I - 6)*sqrt(2)*sqrt(pi)*(er
f(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (6*I + 6)*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^4*c + (-(24*I + 24)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)
/c) + (24*I - 24)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*sin(-1/4*(b^2 - 4*a*c
)/c))*x + ((-(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^5 + (-(4*I - 4)*sqrt(2)*gamma(3/2, 1/
4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (4*I + 4)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c
))*b^3*c)*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^5 + (-(4*I
+ 4)*sqrt(2)*gamma(3/2, 1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (4*I - 4)*sqrt(2)*gamma(3/2, -1/4*(4*I*c^2
*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*sin(-1/4*(b^2 - 4*a*c)/c))*e^2/(c^4*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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