3.2.54 \(\int (e+f x)^2 \sin (b (c+d x)^2) \, dx\) [154]

Optimal. Leaf size=150 \[ -\frac {f (d e-c f) \cos \left (b (c+d x)^2\right )}{b d^3}-\frac {f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3}+\frac {f^2 \sqrt {\frac {\pi }{2}} C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^3}+\frac {(d e-c f)^2 \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^3} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 150, 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 = {3514, 3432, 3460, 2718, 3466, 3433} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} f^2 \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} (c+d x)\right )}{2 b^{3/2} d^3}+\frac {\sqrt {\frac {\pi }{2}} (d e-c f)^2 S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^3}-\frac {f (d e-c f) \cos \left (b (c+d x)^2\right )}{b d^3}-\frac {f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*Sin[b*(c + d*x)^2],x]

[Out]

-((f*(d*e - c*f)*Cos[b*(c + d*x)^2])/(b*d^3)) - (f^2*(c + d*x)*Cos[b*(c + d*x)^2])/(2*b*d^3) + (f^2*Sqrt[Pi/2]
*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(2*b^(3/2)*d^3) + ((d*e - c*f)^2*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/P
i]*(c + d*x)])/(Sqrt[b]*d^3)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 3460

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 3466

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^(n - 1))*(e*x)^(m - n + 1)*(Cos[c +
d*x^n]/(d*n)), x] + Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3514

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)^2 \sin \left (b (c+d x)^2\right ) \, dx &=\frac {\text {Subst}\left (\int \left (d^2 e^2 \left (1+\frac {c f (-2 d e+c f)}{d^2 e^2}\right ) \sin \left (b x^2\right )+2 d e f \left (1-\frac {c f}{d e}\right ) x \sin \left (b x^2\right )+f^2 x^2 \sin \left (b x^2\right )\right ) \, dx,x,c+d x\right )}{d^3}\\ &=\frac {f^2 \text {Subst}\left (\int x^2 \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^3}+\frac {(2 f (d e-c f)) \text {Subst}\left (\int x \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^3}+\frac {(d e-c f)^2 \text {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^3}\\ &=-\frac {f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3}+\frac {(d e-c f)^2 \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^3}+\frac {f^2 \text {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,c+d x\right )}{2 b d^3}+\frac {(f (d e-c f)) \text {Subst}\left (\int \sin (b x) \, dx,x,(c+d x)^2\right )}{d^3}\\ &=-\frac {f (d e-c f) \cos \left (b (c+d x)^2\right )}{b d^3}-\frac {f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3}+\frac {f^2 \sqrt {\frac {\pi }{2}} C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^3}+\frac {(d e-c f)^2 \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^3}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 117, normalized size = 0.78 \begin {gather*} \frac {-2 \sqrt {b} f (2 d e-c f+d f x) \cos \left (b (c+d x)^2\right )+f^2 \sqrt {2 \pi } C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )+2 b (d e-c f)^2 \sqrt {2 \pi } S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{4 b^{3/2} d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(290\) vs. \(2(127)=254\).
time = 0.08, size = 291, normalized size = 1.94

method	result	size
default	\(-\frac {f^{2} x \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 b \,d^{2}}-\frac {f^{2} c \left (-\frac {\cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 b \,d^{2}}-\frac {c \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{2 d \sqrt {b \,d^{2}}}\right )}{d}+\frac {f^{2} \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{4 b \,d^{2} \sqrt {b \,d^{2}}}-\frac {e f \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{b \,d^{2}}-\frac {e f c \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{d \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, \sqrt {\pi }\, e^{2} \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{2 \sqrt {b \,d^{2}}}\)	\(291\)
risch	\(\frac {i e^{2} \sqrt {\pi }\, \erf \left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{4 d \sqrt {-i b}}+\frac {i f^{2} c^{2} \sqrt {\pi }\, \erf \left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{4 d^{3} \sqrt {-i b}}-\frac {f^{2} \sqrt {\pi }\, \erf \left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{8 b \,d^{3} \sqrt {-i b}}-\frac {i e f c \sqrt {\pi }\, \erf \left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{2 d^{2} \sqrt {-i b}}+\frac {i e^{2} \sqrt {\pi }\, \erf \left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{4 d \sqrt {i b}}+\frac {i f^{2} c^{2} \sqrt {\pi }\, \erf \left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{4 d^{3} \sqrt {i b}}+\frac {f^{2} \sqrt {\pi }\, \erf \left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{8 b \,d^{3} \sqrt {i b}}-\frac {i e f c \sqrt {\pi }\, \erf \left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{2 d^{2} \sqrt {i b}}+2 \left (\frac {i f^{2} \left (\frac {i x}{2 d^{2} b}-\frac {i c}{2 d^{3} b}\right )}{2}-\frac {e f}{2 b \,d^{2}}\right ) \cos \left (b \left (d x +c \right )^{2}\right )\)	\(386\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 1.08, size = 564, normalized size = 3.76 \begin {gather*} \frac {{\left (4 \, b c d x {\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 )} + 4 \, b c^{2} {\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 )} - \sqrt {b d^{2} x^{2} + 2 \, b c d x + b c^{2}} {\left ({\left (-\left (i + 1\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 (i - 1\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 )} b c^{2} - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right ) + \left (i + 1\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, -i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )\right )}\right )} f^{2}}{8 \, {\left (b^{2} d^{4} x + b^{2} c d^{3}\right )}} - \frac {{\left (2 \, d x {\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 )} - \sqrt {b d^{2} x^{2} + 2 \, b c d x + b c^{2}} {\left (-\left (i + 1\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 (i - 1\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 )} c + 2 \, c {\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 )}\right )} f e}{4 \, {\left (b d^{3} x + b c d^{2}\right )}} + \frac {\sqrt {2} \sqrt {\pi } {\left (\left (i + 1\right ) \, \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {i \, b}}\right ) + \left (i - 1\right ) \, \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {-i \, b}}\right )\right )} e^{2}}{8 \, \sqrt {b} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*b*c*d*x*(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)) + 4*b*c^2*
(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)) - sqrt(b*d^2*x^2 + 2*b*c*
d*x + b*c^2)*((-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + (I - 1)*sqrt(2
)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*b*c^2 - (I - 1)*sqrt(2)*gamma(3/2, I*b*d^2*x
^2 + 2*I*b*c*d*x + I*b*c^2) + (I + 1)*sqrt(2)*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)))*f^2/(b^2*d^4*
x + b^2*c*d^3) - 1/4*(2*d*x*(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
)) - sqrt(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*(-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*
c^2)) - 1) + (I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*c + 2*c*(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)))*f*e/(b*d^3*x + b*c*d^2) + 1/8*sqr
t(2)*sqrt(pi)*((I + 1)*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (I - 1)*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))*e^2/(sqrt
(b)*d)

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Fricas [A]
time = 0.48, size = 165, normalized size = 1.10 \begin {gather*} \frac {\sqrt {2} \pi \sqrt {\frac {b d^{2}}{\pi }} f^{2} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) + 2 \, \sqrt {2} {\left (\pi b c^{2} f^{2} - 2 \, \pi b c d f e + \pi b d^{2} e^{2}\right )} \sqrt {\frac {b d^{2}}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (b d^{2} f^{2} x - b c d f^{2} + 2 \, b d^{2} f e\right )} \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )}{4 \, b^{2} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 4.59, size = 669, normalized size = 4.46 \begin {gather*} -\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^{2}}{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^{2}}{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}{\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} + 1\right )}}{b d}}{2 \, 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}{\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} + 1\right )}}{b d}}{2 \, d} - \frac {\frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 \, b c^{2} f^{2} - i \, f^{2}\right )} \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 )}{\sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} b} + \frac {2 i \, {\left (d f^{2} {\left (-i \, x - \frac {i \, c}{d}\right )} + 2 i \, c f^{2}\right )} e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}}{b d}}{8 \, d^{2}} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 \, b c^{2} f^{2} + i \, f^{2}\right )} \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 )}{\sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} b} + \frac {2 i \, {\left (d f^{2} {\left (-i \, x - \frac {i \, c}{d}\right )} + 2 i \, c f^{2}\right )} e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )}}{b d}}{8 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sin(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^2/(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^2/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) - 1/2*(-I*sqrt(2)*sqrt(pi)*c*f*erf(-1/2*sqrt(2)*sqr
t(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))*e/(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 + 1)/(b*d))/d - 1/2*(I*sqrt(2)*sqrt(pi)*c*f*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(-I*b*d^2/s
qrt(b^2*d^4) + 1)*(x + c/d))*e/(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 + 1)/(b*d))/d - 1/8*(I*sqrt(2)*sqrt(pi)*(2*b*c^2*f^2 - I*f^2)*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(I*b*d^2/sqr
t(b^2*d^4) + 1)*(x + c/d))/(sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*b) + 2*I*(d*f^2*(-I*x - I*c/d) + 2*I*c*f^2
)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)/(b*d))/d^2 - 1/8*(-I*sqrt(2)*sqrt(pi)*(2*b*c^2*f^2 + I*f^2)*erf(-1/
2*sqrt(2)*sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)*b) + 2
*I*(d*f^2*(-I*x - I*c/d) + 2*I*c*f^2)*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)/(b*d))/d^2

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Mupad [B]
time = 0.19, size = 136, normalized size = 0.91 \begin {gather*} \frac {\cos \left (b\,{\left (c+d\,x\right )}^2\right )\,\left (c\,f^2-2\,d\,e\,f\right )}{2\,b\,d^3}-\frac {f^2\,x\,\cos \left (b\,{\left (c+d\,x\right )}^2\right )}{2\,b\,d^2}+\frac {\sqrt {2}\,f^2\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )}{4\,b^{3/2}\,d^3}+\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2\right )}{2\,\sqrt {b}\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*(c + d*x)^2)*(e + f*x)^2,x)

[Out]

(cos(b*(c + d*x)^2)*(c*f^2 - 2*d*e*f))/(2*b*d^3) - (f^2*x*cos(b*(c + d*x)^2))/(2*b*d^2) + (2^(1/2)*f^2*pi^(1/2
)*fresnelc((2^(1/2)*b^(1/2)*(c + d*x))/pi^(1/2)))/(4*b^(3/2)*d^3) + (2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*b^(1/2
)*(c + d*x))/pi^(1/2))*(c^2*f^2 + d^2*e^2 - 2*c*d*e*f))/(2*b^(1/2)*d^3)

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