\(\int (d+e x)^2 \sin ^2(a+b x+c x^2) \, dx\) [33]

Optimal result

Integrand size = 21, antiderivative size = 291 \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {(d+e x)^3}{6 e}-\frac {(2 c d-b e)^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {e^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {e^2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}+\frac {(2 c d-b e)^2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{5/2}}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c} \]

[Out]

1/6*(e*x+d)^3/e-1/16*e*(-b*e+2*c*d)*sin(2*c*x^2+2*b*x+2*a)/c^2-1/8*e*(e*x+d)*sin(2*c*x^2+2*b*x+2*a)/c-1/16*(-b
*e+2*c*d)^2*cos(2*a-1/2*b^2/c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(1/2)/c^(5/2)+1/16*e^2*cos(2*a-1/2*b^2/
c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(1/2)/c^(3/2)+1/16*e^2*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a
-1/2*b^2/c)*Pi^(1/2)/c^(3/2)+1/16*(-b*e+2*c*d)^2*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a-1/2*b^2/c)*Pi^(1
/2)/c^(5/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3548, 3545, 3543, 3529, 3433, 3432, 3528} \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } e^2 \sin \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } e^2 \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {(d+e x)^3}{6 e} \]

[In]

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

[Out]

(d + e*x)^3/(6*e) - ((2*c*d - b*e)^2*Sqrt[Pi]*Cos[2*a - b^2/(2*c)]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(
16*c^(5/2)) + (e^2*Sqrt[Pi]*Cos[2*a - b^2/(2*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(16*c^(3/2)) + (e^2
*Sqrt[Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*Sin[2*a - b^2/(2*c)])/(16*c^(3/2)) + ((2*c*d - b*e)^2*Sqrt[
Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*Sin[2*a - b^2/(2*c)])/(16*c^(5/2)) - (e*(2*c*d - b*e)*Sin[2*a + 2
*b*x + 2*c*x^2])/(16*c^2) - (e*(d + e*x)*Sin[2*a + 2*b*x + 2*c*x^2])/(8*c)

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3528

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 3529

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 3543

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

Rule 3545

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*(S
in[a + b*x + c*x^2]/(2*c)), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Cos[a + b*x + c*x^2], x], x
] - Dist[e^2*((m - 1)/(2*c)), Int[(d + e*x)^(m - 2)*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]

Rule 3548

Int[((d_.) + (e_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]^(n_), x_Symbol] :> Int[ExpandTrigReduce[
(d + e*x)^m, Sin[a + b*x + c*x^2]^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} (d+e x)^2-\frac {1}{2} (d+e x)^2 \cos \left (2 a+2 b x+2 c x^2\right )\right ) \, dx \\ & = \frac {(d+e x)^3}{6 e}-\frac {1}{2} \int (d+e x)^2 \cos \left (2 a+2 b x+2 c x^2\right ) \, dx \\ & = \frac {(d+e x)^3}{6 e}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {e^2 \int \sin \left (2 a+2 b x+2 c x^2\right ) \, dx}{8 c}-\frac {(2 c d-b e) \int (d+e x) \cos \left (2 a+2 b x+2 c x^2\right ) \, dx}{4 c} \\ & = \frac {(d+e x)^3}{6 e}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}-\frac {(2 c d-b e)^2 \int \cos \left (2 a+2 b x+2 c x^2\right ) \, dx}{8 c^2}+\frac {\left (e^2 \cos \left (2 a-\frac {b^2}{2 c}\right )\right ) \int \sin \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c}+\frac {\left (e^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right ) \int \cos \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c} \\ & = \frac {(d+e x)^3}{6 e}+\frac {e^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {e^2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}-\frac {\left ((2 c d-b e)^2 \cos \left (2 a-\frac {b^2}{2 c}\right )\right ) \int \cos \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c^2}+\frac {\left ((2 c d-b e)^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right ) \int \sin \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx}{8 c^2} \\ & = \frac {(d+e x)^3}{6 e}-\frac {(2 c d-b e)^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {e^2 \sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {e^2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{3/2}}+\frac {(2 c d-b e)^2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{16 c^{5/2}}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.74 \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {-3 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left ((-2 c d+b e)^2 \cos \left (2 a-\frac {b^2}{2 c}\right )-c e^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right )+3 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left (c e^2 \cos \left (2 a-\frac {b^2}{2 c}\right )+(-2 c d+b e)^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right )+\sqrt {c} \left (8 c^2 x \left (3 d^2+3 d e x+e^2 x^2\right )-3 e (4 c d-b e+2 c e x) \sin (2 (a+x (b+c x)))\right )}{48 c^{5/2}} \]

[In]

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

[Out]

(-3*Sqrt[Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*((-2*c*d + b*e)^2*Cos[2*a - b^2/(2*c)] - c*e^2*Sin[2*a -
b^2/(2*c)]) + 3*Sqrt[Pi]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*(c*e^2*Cos[2*a - b^2/(2*c)] + (-2*c*d + b*e
)^2*Sin[2*a - b^2/(2*c)]) + Sqrt[c]*(8*c^2*x*(3*d^2 + 3*d*e*x + e^2*x^2) - 3*e*(4*c*d - b*e + 2*c*e*x)*Sin[2*(
a + x*(b + c*x))]))/(48*c^(5/2))

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {e^{2} x \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}+\frac {e^{2} b \left (\frac {\sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{4 c}+\frac {e^{2} \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )-\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{16 c^{\frac {3}{2}}}-\frac {d e \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}+\frac {d e b \sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}-\frac {\sqrt {\pi }\, d^{2} \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}}+\frac {d e \,x^{2}}{2}+\frac {d^{2} x}{2}+\frac {e^{2} x^{3}}{6}\)	\(378\)
risch	\(-\frac {\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, d^{2} {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {i c}}-\frac {e^{2} \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, b^{2} {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{64 \sqrt {i c}\, c^{2}}+\frac {i e^{2} \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{64 \sqrt {i c}\, c}+\frac {d e \,\operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right ) \sqrt {2}\, \sqrt {\pi }\, b \,{\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{16 \sqrt {i c}\, c}+\frac {\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, d^{2} {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}}+\frac {e^{2} \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, b^{2} {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{32 \sqrt {-2 i c}\, c^{2}}+\frac {i e^{2} \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{32 \sqrt {-2 i c}\, c}-\frac {d e \,\operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right ) \sqrt {\pi }\, b \,{\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}}}{8 \sqrt {-2 i c}\, c}+\frac {d^{2} x}{2}+\frac {e^{2} x^{3}}{6}+\frac {d e \,x^{2}}{2}+2 i \left (-\frac {e^{2} \left (-\frac {i x}{4 c}+\frac {i b}{8 c^{2}}\right )}{4}+\frac {i d e}{8 c}\right ) \sin \left (2 c \,x^{2}+2 b x +2 a \right )\)	\(544\)

[In]

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

[Out]

-1/8*e^2/c*x*sin(2*c*x^2+2*b*x+2*a)+1/4*e^2*b/c*(1/4*sin(2*c*x^2+2*b*x+2*a)/c-1/4*b/c^(3/2)*Pi^(1/2)*(cos(1/2*
(-4*a*c+b^2)/c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))+sin(1/2*(-4*a*c+b^2)/c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/
2))))+1/16*e^2/c^(3/2)*Pi^(1/2)*(cos(1/2*(-4*a*c+b^2)/c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2))-sin(1/2*(-4*a*c+
b^2)/c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2)))-1/4*d*e/c*sin(2*c*x^2+2*b*x+2*a)+1/4*d*e*b/c^(3/2)*Pi^(1/2)*(cos
(1/2*(-4*a*c+b^2)/c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))+sin(1/2*(-4*a*c+b^2)/c)*FresnelS((2*c*x+b)/c^(1/2)/P
i^(1/2)))-1/4*Pi^(1/2)/c^(1/2)*d^2*(cos(1/2*(-4*a*c+b^2)/c)*FresnelC((2*c*x+b)/c^(1/2)/Pi^(1/2))+sin(1/2*(-4*a
*c+b^2)/c)*FresnelS((2*c*x+b)/c^(1/2)/Pi^(1/2)))+1/2*d*e*x^2+1/2*d^2*x+1/6*e^2*x^3

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.88 \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {8 \, c^{3} e^{2} x^{3} + 24 \, c^{3} d e x^{2} + 24 \, c^{3} d^{2} x - 6 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \cos \left (c x^{2} + b x + a\right ) \sin \left (c x^{2} + b x + a\right ) + 3 \, {\left (\pi c e^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) - \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) + 3 \, {\left (\pi c e^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right )}{48 \, c^{3}} \]

[In]

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

[Out]

1/48*(8*c^3*e^2*x^3 + 24*c^3*d*e*x^2 + 24*c^3*d^2*x - 6*(2*c^2*e^2*x + 4*c^2*d*e - b*c*e^2)*cos(c*x^2 + b*x +
a)*sin(c*x^2 + b*x + a) + 3*(pi*c*e^2*sin(-1/2*(b^2 - 4*a*c)/c) - pi*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*cos(-1/
2*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_cos((2*c*x + b)*sqrt(c/pi)/c) + 3*(pi*c*e^2*cos(-1/2*(b^2 - 4*a*c)/c) +
pi*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*sin(-1/2*(b^2 - 4*a*c)/c))*sqrt(c/pi)*fresnel_sin((2*c*x + b)*sqrt(c/pi)
/c))/c^3

Sympy [F]

\[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\int \left (d + e x\right )^{2} \sin ^{2}{\left (a + b x + c x^{2} \right )}\, dx \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.26 (sec) , antiderivative size = 2361, normalized size of antiderivative = 8.11 \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I + 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2
*I*c*x + I*b)/sqrt(2*I*c)) + ((I + 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I - 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*
c*x + I*b)/sqrt(-2*I*c)))*c^(3/2) + 16*c^2*x)*d^2/c^2 + 1/32*sqrt(2)*((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*cos(-1/2*(b^2 - 4*a*c)/c) + (-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I
*b*c*x + I*b^2)/c)) - 1))*b^2*sin(-1/2*(b^2 - 4*a*c)/c) - 2*(((I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c
*x + I*b^2)/c)) - 1))*b*c*cos(-1/2*(b^2 - 4*a*c)/c) + ((I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*
b^2)/c)) - 1))*b*c*sin(-1/2*(b^2 - 4*a*c)/c))*x + 2*sqrt(2)*(4*c^2*x^2 - c*(-I*e^(1/2*(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c) + I*e^(-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*cos(-1/2*(b^2 - 4*a*c)/c) - c*(e^(1/2*(4*I*c^2*
x^2 + 4*I*b*c*x + I*b^2)/c) + e^(-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*sin(-1/2*(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)) - 1/384*sqrt(2)*(24*(((-(I - 1)*sq
rt(2)*sqrt(pi)*(erf(sqrt(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(
sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + 2*((I + 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c
^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I - 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*c^4)*co
s(-1/2*(b^2 - 4*a*c)/c) + ((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*c^3 + 2
*(-(I - 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*
c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*c^4)*sin(-1/2*(b^2 - 4*a*c)/c))*x^3 + 36*(((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(sq
rt(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(sqrt(1/2)*sqrt(-(4*I*c
^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + 2*((I + 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I
*b^2)/c) - (I - 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b*c^3)*cos(-1/2*(b^2 - 4*a*c)
/c) + ((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) + (I - 1)*sqr
t(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^3*c^2 + 2*(-(I - 1)*sqrt(2)*g
amma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x
+ I*b^2)/c))*b*c^3)*sin(-1/2*(b^2 - 4*a*c)/c))*x^2 - 4*sqrt(2)*(8*c^4*x^3 - 3*b*c^2*(I*e^(1/2*(4*I*c^2*x^2 + 4
*I*b*c*x + I*b^2)/c) - I*e^(-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*cos(-1/2*(b^2 - 4*a*c)/c) + 3*b*c^2*(e^
(1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + e^(-1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*sin(-1/2*(b^2 - 4*a*
c)/c))*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2) + 18*(((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b
^2)/c)) - 1))*b^4*c + 2*((I + 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I - 1)*sqrt(2)
*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*cos(-1/2*(b^2 - 4*a*c)/c) + ((-(I + 1)*sqrt(2)
*sqrt(pi)*(erf(sqrt(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(sqrt(
1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^4*c + 2*(-(I - 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c) + (I + 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^2*c^2)*sin(
-1/2*(b^2 - 4*a*c)/c))*x + 3*((-(I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^5 + 2*
((I + 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (I - 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^
2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*cos(-1/2*(b^2 - 4*a*c)/c) + 3*((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(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(sqrt(1/2)*sqrt(-(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c)) - 1))*b^5 + 2*(-(I - 1)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) +
(I + 1)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*b^3*c)*sin(-1/2*(b^2 - 4*a*c)/c))*e^2/(
c^4*((4*c^2*x^2 + 4*b*c*x + b^2)/c)^(3/2))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.01 \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {1}{6} \, e^{2} x^{3} + \frac {1}{2} \, d e x^{2} + \frac {1}{2} \, d^{2} x - \frac {-i \, {\left (c e^{2} {\left (2 \, x + \frac {b}{c}\right )} + 4 \, c d e - 2 \, b e^{2}\right )} e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )} + \frac {\sqrt {\pi } {\left (4 i \, c^{2} d^{2} - 4 i \, b c d e + i \, b^{2} e^{2} - c e^{2}\right )} \operatorname {erf}\left (-\frac {1}{2} i \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} - \frac {i \, {\left (c e^{2} {\left (2 \, x + \frac {b}{c}\right )} + 4 \, c d e - 2 \, b e^{2}\right )} e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )} + \frac {\sqrt {\pi } {\left (-4 i \, c^{2} d^{2} + 4 i \, b c d e - i \, b^{2} e^{2} - c e^{2}\right )} \operatorname {erf}\left (\frac {1}{2} i \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} \]

[In]

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

[Out]

1/6*e^2*x^3 + 1/2*d*e*x^2 + 1/2*d^2*x - 1/32*(-I*(c*e^2*(2*x + b/c) + 4*c*d*e - 2*b*e^2)*e^(2*I*c*x^2 + 2*I*b*
x + 2*I*a) + sqrt(pi)*(4*I*c^2*d^2 - 4*I*b*c*d*e + I*b^2*e^2 - c*e^2)*erf(-1/2*I*sqrt(c)*(2*x + b/c)*(I*c/abs(
c) + 1))*e^(-1/2*(I*b^2 - 4*I*a*c)/c)/(sqrt(c)*(I*c/abs(c) + 1)))/c^2 - 1/32*(I*(c*e^2*(2*x + b/c) + 4*c*d*e -
2*b*e^2)*e^(-2*I*c*x^2 - 2*I*b*x - 2*I*a) + sqrt(pi)*(-4*I*c^2*d^2 + 4*I*b*c*d*e - I*b^2*e^2 - c*e^2)*erf(1/2
*I*sqrt(c)*(2*x + b/c)*(-I*c/abs(c) + 1))*e^(-1/2*(-I*b^2 + 4*I*a*c)/c)/(sqrt(c)*(-I*c/abs(c) + 1)))/c^2

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx=\int {\sin \left (c\,x^2+b\,x+a\right )}^2\,{\left (d+e\,x\right )}^2 \,d x \]

[In]

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

[Out]

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