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

3.2.54.1 Optimal result
3.2.54.2 Mathematica [A] (verified)
3.2.54.3 Rubi [A] (verified)
3.2.54.4 Maple [B] (verified)
3.2.54.5 Fricas [A] (verification not implemented)
3.2.54.6 Sympy [F]
3.2.54.7 Maxima [C] (verification not implemented)
3.2.54.8 Giac [C] (verification not implemented)
3.2.54.9 Mupad [B] (verification not implemented)

3.2.54.1 Optimal result

Integrand size = 18, antiderivative size = 150 \[ \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx=-\frac {f (d e-c f) \cos \left (b (c+d x)^2\right )}{b d^3}-\frac {f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^3}+\frac {f^2 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^3}+\frac {(d e-c f)^2 \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^3} \]

output
-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)
 
3.2.54.2 Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.78 \[ \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx=\frac {-2 \sqrt {b} f (2 d e-c f+d f x) \cos \left (b (c+d x)^2\right )+f^2 \sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )+2 b (d e-c f)^2 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{4 b^{3/2} d^3} \]

input
Integrate[(e + f*x)^2*Sin[b*(c + d*x)^2],x]
 
output
(-2*Sqrt[b]*f*(2*d*e - c*f + d*f*x)*Cos[b*(c + d*x)^2] + f^2*Sqrt[2*Pi]*Fr 
esnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)] + 2*b*(d*e - c*f)^2*Sqrt[2*Pi]*Fresne 
lS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(4*b^(3/2)*d^3)
 
3.2.54.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3914, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx\)

\(\Big \downarrow \) 3914

\(\displaystyle \frac {\int \left (\sin \left (b (c+d x)^2\right ) (d e-c f)^2+2 f (c+d x) \sin \left (b (c+d x)^2\right ) (d e-c f)+f^2 (c+d x)^2 \sin \left (b (c+d x)^2\right )\right )d(c+d x)}{d^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\sqrt {\frac {\pi }{2}} f^2 \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} (d e-c f)^2 \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b}}-\frac {f (d e-c f) \cos \left (b (c+d x)^2\right )}{b}-\frac {f^2 (c+d x) \cos \left (b (c+d x)^2\right )}{2 b}}{d^3}\)

input
Int[(e + f*x)^2*Sin[b*(c + d*x)^2],x]
 
output
(-((f*(d*e - c*f)*Cos[b*(c + d*x)^2])/b) - (f^2*(c + d*x)*Cos[b*(c + d*x)^ 
2])/(2*b) + (f^2*Sqrt[Pi/2]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(2*b^( 
3/2)) + ((d*e - c*f)^2*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/ 
Sqrt[b])/d^3
 

3.2.54.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3914
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat 
or[n], 1]}, Simp[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]
 
3.2.54.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(290\) vs. \(2(127)=254\).

Time = 0.62 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.94

method result size
default \(-\frac {f^{2} x \cos \left (d^{2} x^{2} b +2 c d x b +c^{2} b \right )}{2 b \,d^{2}}-\frac {f^{2} c \left (-\frac {\cos \left (d^{2} x^{2} b +2 c d x b +c^{2} b \right )}{2 b \,d^{2}}-\frac {c \sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{2 d \sqrt {b \,d^{2}}}\right )}{d}+\frac {f^{2} \sqrt {2}\, \sqrt {\pi }\, \operatorname {C}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{4 b \,d^{2} \sqrt {b \,d^{2}}}-\frac {e f \cos \left (d^{2} x^{2} b +2 c d x b +c^{2} b \right )}{b \,d^{2}}-\frac {e f c \sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{d \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, \sqrt {\pi }\, e^{2} \operatorname {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{2 \sqrt {b \,d^{2}}}\) \(291\)
risch \(\frac {i \operatorname {erf}\left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right ) \sqrt {\pi }\, e^{2}}{4 \sqrt {-i b}\, d}+\frac {i f^{2} c^{2} \sqrt {\pi }\, \operatorname {erf}\left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{4 d^{3} \sqrt {-i b}}-\frac {f^{2} \sqrt {\pi }\, \operatorname {erf}\left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{8 b \,d^{3} \sqrt {-i b}}-\frac {i e f c \sqrt {\pi }\, \operatorname {erf}\left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{2 d^{2} \sqrt {-i b}}+\frac {i e^{2} \sqrt {\pi }\, \operatorname {erf}\left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{4 d \sqrt {i b}}+\frac {i f^{2} c^{2} \sqrt {\pi }\, \operatorname {erf}\left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{4 d^{3} \sqrt {i b}}+\frac {f^{2} \sqrt {\pi }\, \operatorname {erf}\left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{8 b \,d^{3} \sqrt {i b}}-\frac {i e f c \sqrt {\pi }\, \operatorname {erf}\left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{2 d^{2} \sqrt {i b}}+2 \left (\frac {i f^{2} \left (\frac {i x}{2 d^{2} b}-\frac {i c}{2 b \,d^{3}}\right )}{2}-\frac {e f}{2 d^{2} b}\right ) \cos \left (b \left (d x +c \right )^{2}\right )\) \(386\)
parts \(\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right ) x^{2} f^{2}}{2 \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right ) f e x}{\sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, \sqrt {\pi }\, e^{2} \operatorname {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +c d b \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{2 \sqrt {b \,d^{2}}}-\frac {\pi f \left (\frac {\operatorname {S}\left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right ) \left (-2 b c d f \left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )+\frac {\sqrt {b \,d^{2}}\, \sqrt {2}\, \sqrt {\pi }\, f \left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )^{2}}{2}+2 e b \,d^{2} \left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )\right )}{b \,d^{2}}-\frac {-\frac {\sqrt {b \,d^{2}}\, \sqrt {2}\, f \left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right ) \cos \left (\frac {\pi \left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )^{2}}{2}\right )}{\sqrt {\pi }}+\frac {\sqrt {b \,d^{2}}\, \sqrt {2}\, f \,\operatorname {C}\left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{\sqrt {\pi }}-\frac {\left (-4 b c d f +4 e b \,d^{2}\right ) \cos \left (\frac {\pi \left (\frac {\sqrt {2}\, b \,d^{2} x}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}+\frac {\sqrt {2}\, c d b}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )^{2}}{2}\right )}{\pi }}{2 b \,d^{2}}\right )}{2 b \,d^{2}}\) \(555\)

input
int((f*x+e)^2*sin(b*(d*x+c)^2),x,method=_RETURNVERBOSE)
 
output
-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))
 
3.2.54.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.08 \[ \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx=\frac {\sqrt {2} \pi \sqrt {\frac {b d^{2}}{\pi }} f^{2} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) + 2 \, \sqrt {2} \pi {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \sqrt {\frac {b d^{2}}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (b d^{2} f^{2} x + 2 \, b d^{2} e f - b c d f^{2}\right )} \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )}{4 \, b^{2} d^{4}} \]

input
integrate((f*x+e)^2*sin(b*(d*x+c)^2),x, algorithm="fricas")
 
output
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*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*sqrt(b*d^2/p 
i)*fresnel_sin(sqrt(2)*sqrt(b*d^2/pi)*(d*x + c)/d) - 2*(b*d^2*f^2*x + 2*b* 
d^2*e*f - b*c*d*f^2)*cos(b*d^2*x^2 + 2*b*c*d*x + b*c^2))/(b^2*d^4)
 
3.2.54.6 Sympy [F]

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

input
integrate((f*x+e)**2*sin(b*(d*x+c)**2),x)
 
output
Integral((e + f*x)**2*sin(b*c**2 + 2*b*c*d*x + b*d**2*x**2), x)
 
3.2.54.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 564, normalized size of antiderivative = 3.76 \[ \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx=\frac {\sqrt {2} \sqrt {\pi } e^{2} {\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 )}}{8 \, \sqrt {b} d} - \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 )} e f}{4 \, {\left (b d^{3} x + b c d^{2}\right )}} + \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 )}} \]

input
integrate((f*x+e)^2*sin(b*(d*x+c)^2),x, algorithm="maxima")
 
output
1/8*sqrt(2)*sqrt(pi)*e^2*((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)))/(sqrt(b)*d) - 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)*(er 
f(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)))*e* 
f/(b*d^3*x + b*c*d^2) + 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)) - s 
qrt(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)
 
3.2.54.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.22 \[ \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx=-\frac {\frac {i \, \sqrt {2} \sqrt {\pi } {\left (2 i \, b d^{2} e^{2} - 4 i \, b c d e f + 2 i \, b c^{2} f^{2} - f^{2}\right )} \operatorname {erf}\left (-\frac {1}{2} i \, \sqrt {2} \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} b} + \frac {2 \, {\left (d f^{2} {\left (x + \frac {c}{d}\right )} + 2 \, d e f - 2 \, c f^{2}\right )} e^{\left (i \, b d^{2} x^{2} + 2 i \, b c d x + i \, b c^{2}\right )}}{b d}}{8 \, d^{2}} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } {\left (-2 i \, b d^{2} e^{2} + 4 i \, b c d e f - 2 i \, b c^{2} f^{2} - f^{2}\right )} \operatorname {erf}\left (\frac {1}{2} i \, \sqrt {2} \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right )}{\sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} b} + \frac {2 \, {\left (d f^{2} {\left (x + \frac {c}{d}\right )} + 2 \, d e f - 2 \, c f^{2}\right )} e^{\left (-i \, b d^{2} x^{2} - 2 i \, b c d x - i \, b c^{2}\right )}}{b d}}{8 \, d^{2}} \]

input
integrate((f*x+e)^2*sin(b*(d*x+c)^2),x, algorithm="giac")
 
output
-1/8*(I*sqrt(2)*sqrt(pi)*(2*I*b*d^2*e^2 - 4*I*b*c*d*e*f + 2*I*b*c^2*f^2 - 
f^2)*erf(-1/2*I*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*(d*f^2*(x + c/d) + 2*d*e* 
f - 2*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*I*b*d^2*e^2 + 4*I*b*c*d*e*f - 2*I*b*c^2*f^2 - f^2)*er 
f(1/2*I*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*(d*f^2*(x + c/d) + 2*d*e*f - 2* 
c*f^2)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)/(b*d))/d^2
 
3.2.54.9 Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91 \[ \int (e+f x)^2 \sin \left (b (c+d x)^2\right ) \, dx=\frac {\cos \left (b\,{\left (c+d\,x\right )}^2\right )\,\left (c\,f^2-2\,d\,e\,f\right )}{2\,b\,d^3}-\frac {f^2\,x\,\cos \left (b\,{\left (c+d\,x\right )}^2\right )}{2\,b\,d^2}+\frac {\sqrt {2}\,f^2\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )}{4\,b^{3/2}\,d^3}+\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )\,\left (c^2\,f^2-2\,c\,d\,e\,f+d^2\,e^2\right )}{2\,\sqrt {b}\,d^3} \]

input
int(sin(b*(c + d*x)^2)*(e + f*x)^2,x)
 
output
(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)