3.153 \(\int (e+f x)^3 \sin (b (c+d x)^2) \, dx\)

Optimal. Leaf size=223 \[ \frac {3 \sqrt {\frac {\pi }{2}} f^2 (d e-c f) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^4}+\frac {f^3 \sin \left (b (c+d x)^2\right )}{2 b^2 d^4}-\frac {3 f^2 (c+d x) (d e-c f) \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac {\sqrt {\frac {\pi }{2}} (d e-c f)^3 S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}-\frac {3 f (d e-c f)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3433, 3351, 3379, 2638, 3385, 3352, 3296, 2637} \[ \frac {3 \sqrt {\frac {\pi }{2}} f^2 (d e-c f) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} (c+d x)\right )}{2 b^{3/2} d^4}+\frac {f^3 \sin \left (b (c+d x)^2\right )}{2 b^2 d^4}-\frac {3 f^2 (c+d x) (d e-c f) \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac {\sqrt {\frac {\pi }{2}} (d e-c f)^3 S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}-\frac {3 f (d e-c f)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^3*Sin[b*(c + d*x)^2],x]

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 3379

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 3385

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 3433

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)^3 \sin \left (b (c+d x)^2\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \left (d^3 e^3 \left (1-\frac {c f \left (3 d^2 e^2-3 c d e f+c^2 f^2\right )}{d^3 e^3}\right ) \sin \left (b x^2\right )+3 d^2 e^2 f \left (1+\frac {c f (-2 d e+c f)}{d^2 e^2}\right ) x \sin \left (b x^2\right )+3 d e f^2 \left (1-\frac {c f}{d e}\right ) x^2 \sin \left (b x^2\right )+f^3 x^3 \sin \left (b x^2\right )\right ) \, dx,x,c+d x\right )}{d^4}\\ &=\frac {f^3 \operatorname {Subst}\left (\int x^3 \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^4}+\frac {\left (3 f^2 (d e-c f)\right ) \operatorname {Subst}\left (\int x^2 \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^4}+\frac {\left (3 f (d e-c f)^2\right ) \operatorname {Subst}\left (\int x \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^4}+\frac {(d e-c f)^3 \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^4}\\ &=-\frac {3 f^2 (d e-c f) (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac {(d e-c f)^3 \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}+\frac {f^3 \operatorname {Subst}\left (\int x \sin (b x) \, dx,x,(c+d x)^2\right )}{2 d^4}+\frac {\left (3 f^2 (d e-c f)\right ) \operatorname {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,c+d x\right )}{2 b d^4}+\frac {\left (3 f (d e-c f)^2\right ) \operatorname {Subst}\left (\int \sin (b x) \, dx,x,(c+d x)^2\right )}{2 d^4}\\ &=-\frac {3 f (d e-c f)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac {3 f^2 (d e-c f) (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac {3 f^2 (d e-c f) \sqrt {\frac {\pi }{2}} C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^4}+\frac {(d e-c f)^3 \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}+\frac {f^3 \operatorname {Subst}\left (\int \cos (b x) \, dx,x,(c+d x)^2\right )}{2 b d^4}\\ &=-\frac {3 f (d e-c f)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac {3 f^2 (d e-c f) (c+d x) \cos \left (b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (b (c+d x)^2\right )}{2 b d^4}+\frac {3 f^2 (d e-c f) \sqrt {\frac {\pi }{2}} C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^4}+\frac {(d e-c f)^3 \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}+\frac {f^3 \sin \left (b (c+d x)^2\right )}{2 b^2 d^4}\\ \end {align*}

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Mathematica [A]  time = 1.08, size = 173, normalized size = 0.78 \[ \frac {4 \sqrt {2 \pi } b^{3/2} (d e-c f)^3 S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )-4 b f \cos \left (b (c+d x)^2\right ) \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right )-6 \sqrt {2 \pi } \sqrt {b} f^2 (c f-d e) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )+4 f^3 \sin \left (b (c+d x)^2\right )}{8 b^2 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^3*Sin[b*(c + d*x)^2],x]

[Out]

(-4*b*f*(c^2*f^2 - c*d*f*(3*e + f*x) + d^2*(3*e^2 + 3*e*f*x + f^2*x^2))*Cos[b*(c + d*x)^2] - 6*Sqrt[b]*f^2*(-(
d*e) + c*f)*Sqrt[2*Pi]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)] + 4*b^(3/2)*(d*e - c*f)^3*Sqrt[2*Pi]*FresnelS[Sq
rt[b]*Sqrt[2/Pi]*(c + d*x)] + 4*f^3*Sin[b*(c + d*x)^2])/(8*b^2*d^4)

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fricas [A]  time = 0.71, size = 255, normalized size = 1.14 \[ \frac {2 \, d f^{3} \sin \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right ) + 3 \, \sqrt {2} \pi {\left (d e f^{2} - c f^{3}\right )} \sqrt {\frac {b d^{2}}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) + 2 \, \sqrt {2} \pi {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \sqrt {\frac {b d^{2}}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (b d^{3} f^{3} x^{2} + 3 \, b d^{3} e^{2} f - 3 \, b c d^{2} e f^{2} + b c^{2} d f^{3} + {\left (3 \, b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x\right )} \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )}{4 \, b^{2} d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(b*(d*x+c)^2),x, algorithm="fricas")

[Out]

1/4*(2*d*f^3*sin(b*d^2*x^2 + 2*b*c*d*x + b*c^2) + 3*sqrt(2)*pi*(d*e*f^2 - c*f^3)*sqrt(b*d^2/pi)*fresnel_cos(sq
rt(2)*sqrt(b*d^2/pi)*(d*x + c)/d) + 2*sqrt(2)*pi*(b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*s
qrt(b*d^2/pi)*fresnel_sin(sqrt(2)*sqrt(b*d^2/pi)*(d*x + c)/d) - 2*(b*d^3*f^3*x^2 + 3*b*d^3*e^2*f - 3*b*c*d^2*e
*f^2 + b*c^2*d*f^3 + (3*b*d^3*e*f^2 - b*c*d^2*f^3)*x)*cos(b*d^2*x^2 + 2*b*c*d*x + b*c^2))/(b^2*d^5)

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giac [C]  time = 0.73, size = 1023, normalized size = 4.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*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^3/(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^3/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)) - 1/4*(-3*I*sqrt(2)*sqrt(pi)*c*f*erf(-1/2*sqrt(2)*s
qrt(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)) + 3*f*e^(-I*b*
d^2*x^2 - 2*I*b*c*d*x - I*b*c^2 + 2)/(b*d))/d - 1/4*(3*I*sqrt(2)*sqrt(pi)*c*f*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)) + 3*f*e^(I*b*d^2*x^2 + 2*I
*b*c*d*x + I*b*c^2 + 2)/(b*d))/d - 1/8*(I*sqrt(2)*sqrt(pi)*(6*b*c^2*f^2 - 3*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))*e/(sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*b) + 2*I*(d*f^2*(-3*I*x -
3*I*c/d) + 6*I*c*f^2)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2 + 1)/(b*d))/d^2 - 1/8*(-I*sqrt(2)*sqrt(pi)*(6*b*
c^2*f^2 + 3*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))*e/(sqrt(b*d^2)*(-I*b*d
^2/sqrt(b^2*d^4) + 1)*b) + 2*I*(d*f^2*(-3*I*x - 3*I*c/d) + 6*I*c*f^2)*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2 +
1)/(b*d))/d^2 + 1/8*(sqrt(2)*sqrt(pi)*(2*I*b*c^3*f^3 + 3*c*f^3)*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*(b*d^2*f^3*(x + c/d)^2 - 3*b*c*d*f^3*(x
+ c/d) + 3*b*c^2*f^3 - I*f^3)*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)/(b^2*d))/d^3 + 1/8*(sqrt(2)*sqrt(pi)*(
-2*I*b*c^3*f^3 + 3*c*f^3)*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*(b*d^2*f^3*(x + c/d)^2 - 3*b*c*d*f^3*(x + c/d) + 3*b*c^2*f^3 + I*f^3)*e^(I*b
*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)/(b^2*d))/d^3

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maple [B]  time = 0.03, size = 586, normalized size = 2.63 \[ -\frac {f^{3} x^{2} \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {f^{3} c \left (-\frac {x \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {c \left (-\frac {\cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {c \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 d \sqrt {d^{2} b}}\right )}{d}+\frac {\sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{4 d^{2} b \sqrt {d^{2} b}}\right )}{d}+\frac {f^{3} \left (\frac {\sin \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {c \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 d \sqrt {d^{2} b}}\right )}{d^{2} b}-\frac {3 e \,f^{2} x \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {3 e \,f^{2} c \left (-\frac {\cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {c \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 d \sqrt {d^{2} b}}\right )}{d}+\frac {3 e \,f^{2} \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{4 d^{2} b \sqrt {d^{2} b}}-\frac {3 e^{2} f \cos \left (d^{2} x^{2} b +2 c d x b +b \,c^{2}\right )}{2 d^{2} b}-\frac {3 e^{2} f c \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 d \sqrt {d^{2} b}}+\frac {\sqrt {2}\, \sqrt {\pi }\, e^{3} \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {d^{2} b}}\right )}{2 \sqrt {d^{2} b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sin((d*x+c)^2*b),x)

[Out]

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

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maxima [C]  time = 2.28, size = 972, normalized size = 4.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(b*(d*x+c)^2),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*sqrt(pi)*e^3*((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) - 3/8*(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)))*e^2*f/(b*d^3*x + b*c*d^2) + 3/8*(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)))*e*f^2/(b^2*d^4*x +
b^2*c*d^3) - 1/16*(12*b*c^3*(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
)) + (12*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)) - 4*I*gamm
a(2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + 4*I*gamma(2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*d*x - c*(4*I*g
amma(2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 4*I*gamma(2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 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^3 + (-(3*I - 3)*sqrt(2)*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x
+ I*b*c^2) + (3*I + 3)*sqrt(2)*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*c)*sqrt(b*d^2*x^2 + 2*b*c*d*
x + b*c^2))*f^3/(b^2*d^5*x + b^2*c*d^4)

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mupad [B]  time = 4.71, size = 231, normalized size = 1.04 \[ \frac {f^3\,\sin \left (b\,{\left (c+d\,x\right )}^2\right )}{2\,b^2\,d^4}-\frac {\cos \left (b\,{\left (c+d\,x\right )}^2\right )\,\left (c^2\,f^3-3\,c\,d\,e\,f^2+3\,d^2\,e^2\,f\right )}{2\,b\,d^4}-\frac {f^3\,x^2\,\cos \left (b\,{\left (c+d\,x\right )}^2\right )}{2\,b\,d^2}+\frac {x\,\cos \left (b\,{\left (c+d\,x\right )}^2\right )\,\left (c\,f^3-3\,d\,e\,f^2\right )}{2\,b\,d^3}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{2\,\sqrt {b}\,d^4}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\sqrt {b}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )\,\left (3\,c\,f^3-3\,d\,e\,f^2\right )}{4\,b^{3/2}\,d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*(c + d*x)^2)*(e + f*x)^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e + f x\right )^{3} \sin {\left (b c^{2} + 2 b c d x + b d^{2} x^{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*sin(b*(d*x+c)**2),x)

[Out]

Integral((e + f*x)**3*sin(b*c**2 + 2*b*c*d*x + b*d**2*x**2), x)

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