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

Optimal. Leaf size=291 \[ -\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } e^2 \sin \left (2 a-\frac {b^2}{2 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } e^2 \cos \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {(d+e x)^3}{6 e} \]

[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/c*b^2)*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/c*b^
2)*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/c*b^2)*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/c*b^2)*Pi^(1
/2)/c^(5/2)

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Rubi [A]  time = 0.34, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3467, 3464, 3447, 3351, 3352, 3462, 3448} \[ -\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) (2 c d-b e)^2 S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{5/2}}+\frac {\sqrt {\pi } e^2 \sin \left (2 a-\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{16 c^{3/2}}+\frac {\sqrt {\pi } e^2 \cos \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}-\frac {e (2 c d-b e) \sin \left (2 a+2 b x+2 c x^2\right )}{16 c^2}-\frac {e (d+e x) \sin \left (2 a+2 b x+2 c x^2\right )}{8 c}+\frac {(d+e x)^3}{6 e} \]

Antiderivative was successfully verified.

[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 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 3462

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 3464

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[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Sin[a + b*x + c*x^2], x], x
] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Cos[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 3467

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*} \int (d+e x)^2 \sin ^2\left (a+b x+c x^2\right ) \, dx &=\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 ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {e^2 \sqrt {\pi } C\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 ) C\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 ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{16 c^{3/2}}+\frac {e^2 \sqrt {\pi } C\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 } S\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*}

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Mathematica [A]  time = 1.22, size = 215, normalized size = 0.74 \[ \frac {-3 \sqrt {\pi } C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left (\cos \left (2 a-\frac {b^2}{2 c}\right ) (b e-2 c d)^2-c e^2 \sin \left (2 a-\frac {b^2}{2 c}\right )\right )+3 \sqrt {\pi } S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \left (\sin \left (2 a-\frac {b^2}{2 c}\right ) (b e-2 c d)^2+c e^2 \cos \left (2 a-\frac {b^2}{2 c}\right )\right )+\sqrt {c} \left (8 c^2 x \left (3 d^2+3 d e x+e^2 x^2\right )-3 e \sin (2 (a+x (b+c x))) (-b e+4 c d+2 c e x)\right )}{48 c^{5/2}} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.46, size = 257, normalized size = 0.88 \[ \frac {8 \, c^{3} e^{2} x^{3} + 24 \, c^{3} d e x^{2} + 24 \, c^{3} d^{2} x - 6 \, {\left (2 \, c^{2} e^{2} x + 4 \, c^{2} d e - b c e^{2}\right )} \cos \left (c x^{2} + b x + a\right ) \sin \left (c x^{2} + b x + a\right ) + 3 \, {\left (\pi c e^{2} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) - \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) + 3 \, {\left (\pi c e^{2} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \pi {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right )}{48 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [C]  time = 0.57, size = 538, normalized size = 1.85 \[ \frac {1}{6} \, x^{3} e^{2} + \frac {1}{2} \, d x^{2} e + \frac {1}{2} \, d^{2} x + \frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {1}{2} \, \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 )}}{8 \, \sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}} + \frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {1}{2} \, \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 )}}{8 \, \sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}} - \frac {\frac {\sqrt {\pi } b d \operatorname {erf}\left (-\frac {1}{2} \, \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}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}} - i \, d e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a + 1\right )}}{8 \, c} - \frac {\frac {\sqrt {\pi } b d \operatorname {erf}\left (-\frac {1}{2} \, \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}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}} + i \, d e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a + 1\right )}}{8 \, c} - \frac {{\left (c {\left (-2 i \, x - \frac {i \, b}{c}\right )} + 2 i \, b\right )} e^{\left (2 i \, c x^{2} + 2 i \, b x + 2 i \, a + 2\right )} - \frac {\sqrt {\pi } {\left (b^{2} + i \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \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 - 4 \, c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, c^{2}} - \frac {{\left (c {\left (2 i \, x + \frac {i \, b}{c}\right )} - 2 i \, b\right )} e^{\left (-2 i \, c x^{2} - 2 i \, b x - 2 i \, a + 2\right )} - \frac {\sqrt {\pi } {\left (b^{2} - i \, c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \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 - 4 \, c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}}}{32 \, 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)^2,x, algorithm="giac")

[Out]

1/6*x^3*e^2 + 1/2*d*x^2*e + 1/2*d^2*x + 1/8*sqrt(pi)*d^2*erf(-1/2*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)) + 1/8*sqrt(pi)*d^2*erf(-1/2*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)) - 1/8*(sqrt(pi)*b*d*erf(-1/2*sqrt(c)*(2*x + b
/c)*(-I*c/abs(c) + 1))*e^(-1/2*(I*b^2 - 4*I*a*c - 2*c)/c)/(sqrt(c)*(-I*c/abs(c) + 1)) - I*d*e^(2*I*c*x^2 + 2*I
*b*x + 2*I*a + 1))/c - 1/8*(sqrt(pi)*b*d*erf(-1/2*sqrt(c)*(2*x + b/c)*(I*c/abs(c) + 1))*e^(-1/2*(-I*b^2 + 4*I*
a*c - 2*c)/c)/(sqrt(c)*(I*c/abs(c) + 1)) + I*d*e^(-2*I*c*x^2 - 2*I*b*x - 2*I*a + 1))/c - 1/32*((c*(-2*I*x - I*
b/c) + 2*I*b)*e^(2*I*c*x^2 + 2*I*b*x + 2*I*a + 2) - sqrt(pi)*(b^2 + I*c)*erf(-1/2*sqrt(c)*(2*x + b/c)*(-I*c/ab
s(c) + 1))*e^(-1/2*(I*b^2 - 4*I*a*c - 4*c)/c)/(sqrt(c)*(-I*c/abs(c) + 1)))/c^2 - 1/32*((c*(2*I*x + I*b/c) - 2*
I*b)*e^(-2*I*c*x^2 - 2*I*b*x - 2*I*a + 2) - sqrt(pi)*(b^2 - I*c)*erf(-1/2*sqrt(c)*(2*x + b/c)*(I*c/abs(c) + 1)
)*e^(-1/2*(-I*b^2 + 4*I*a*c - 4*c)/c)/(sqrt(c)*(I*c/abs(c) + 1)))/c^2

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maple [A]  time = 0.07, size = 378, normalized size = 1.30 \[ -\frac {e^{2} x \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{8 c}+\frac {e^{2} b \left (\frac {\sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}-\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {-4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}\right )}{4 c}+\frac {e^{2} \sqrt {\pi }\, \left (\cos \left (\frac {-4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )-\sin \left (\frac {-4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{16 c^{\frac {3}{2}}}-\frac {d e \sin \left (2 c \,x^{2}+2 b x +2 a \right )}{4 c}+\frac {d e b \sqrt {\pi }\, \left (\cos \left (\frac {-4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 c^{\frac {3}{2}}}-\frac {\sqrt {\pi }\, d^{2} \left (\cos \left (\frac {-4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}}+\frac {x^{2} d e}{2}+\frac {d^{2} x}{2}+\frac {x^{3} e^{2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*x^2*d*e+1/2*d^2*x+1/6*x^3*e^2

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maxima [C]  time = 2.09, size = 2344, normalized size = 8.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*sin(c*x^2+b*x+a)^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 - 2)*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) + (2*I + 2)*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) + (-(2*I + 2)*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) + (2*I - 2)*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 + sqrt(2)*(8*c^2*x^2 + c*(2*I*e^(1/2*(4*I*c^2*x^2 + 4*
I*b*c*x + I*b^2)/c) - 2*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) - 2*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 - 24)*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) - (24*I + 24)*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 + (-(48*I + 48)*sqrt(2)*gam
ma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (48*I - 48)*sqrt(2)*gamma(3/2, -1/2*(4*I*c^2*x^2 + 4*I*b*c*
x + I*b^2)/c))*c^4)*cos(-1/2*(b^2 - 4*a*c)/c) + (((24*I + 24)*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) - (24*I - 24)*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 + ((48*I - 48)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (48*I +
48)*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 + ((((3
6*I - 36)*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) - (36*I + 36)*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 + (-(72*I + 72)*sqrt(2)*ga
mma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (72*I - 72)*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) + (((36*I + 36)*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) - (36*I - 36)*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 + ((72*I - 72)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (72*
I + 72)*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 +
2*sqrt(2)*(16*c^4*x^3 + b*c^2*(-6*I*e^(1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + 6*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) + 6*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 - 18)*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) - (18*I + 18)*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^4*c + (-(36*I + 36)*sqrt(2)*g
amma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (36*I - 36)*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) + (((18*I + 18)*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) - (18*I - 18)*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 + ((36*I - 36)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (36
*I + 36)*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 - 3)*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) - (3*I + 3)*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 + (-(6*I + 6)*sqrt(2)*gamma(3
/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + (6*I - 6)*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 + 3)*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) - (3*I - 3)*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 + ((6*I - 6)*sqrt(2)*gamma(3/2, 1/2*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) - (6*I + 6)*sqrt(2)*gam
ma(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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