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

Optimal. Leaf size=341 \[ \frac {3 \sqrt {\frac {\pi }{2}} f^2 \cos (a) (d e-c f) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^4}-\frac {3 \sqrt {\frac {\pi }{2}} f^2 \sin (a) (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^4}+\frac {f^3 \sin \left (a+b (c+d x)^2\right )}{2 b^2 d^4}-\frac {3 f^2 (c+d x) (d e-c f) \cos \left (a+b (c+d x)^2\right )}{2 b d^4}+\frac {\sqrt {\frac {\pi }{2}} \sin (a) (d e-c f)^3 C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) (d e-c f)^3 S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}-\frac {3 f (d e-c f)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4} \]

[Out]

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

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Rubi [A]  time = 0.57, antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3433, 3353, 3352, 3351, 3379, 2638, 3385, 3354, 3296, 2637} \[ \frac {3 \sqrt {\frac {\pi }{2}} f^2 \cos (a) (d e-c f) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} (c+d x)\right )}{2 b^{3/2} d^4}-\frac {3 \sqrt {\frac {\pi }{2}} f^2 \sin (a) (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{2 b^{3/2} d^4}+\frac {f^3 \sin \left (a+b (c+d x)^2\right )}{2 b^2 d^4}-\frac {3 f^2 (c+d x) (d e-c f) \cos \left (a+b (c+d x)^2\right )}{2 b d^4}+\frac {\sqrt {\frac {\pi }{2}} \sin (a) (d e-c f)^3 \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} (c+d x)\right )}{\sqrt {b} d^4}+\frac {\sqrt {\frac {\pi }{2}} \cos (a) (d e-c f)^3 S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}-\frac {3 f (d e-c f)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*f*(d*e - c*f)^2*Cos[a + b*(c + d*x)^2])/(2*b*d^4) - (3*f^2*(d*e - c*f)*(c + d*x)*Cos[a + b*(c + d*x)^2])/(
2*b*d^4) - (f^3*(c + d*x)^2*Cos[a + b*(c + d*x)^2])/(2*b*d^4) + (3*f^2*(d*e - c*f)*Sqrt[Pi/2]*Cos[a]*FresnelC[
Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(2*b^(3/2)*d^4) + ((d*e - c*f)^3*Sqrt[Pi/2]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*
(c + d*x)])/(Sqrt[b]*d^4) + ((d*e - c*f)^3*Sqrt[Pi/2]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)]*Sin[a])/(Sqrt[b]*
d^4) - (3*f^2*(d*e - c*f)*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)]*Sin[a])/(2*b^(3/2)*d^4) + (f^3*Sin
[a + 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 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, 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 (a+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 (a+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 (a+b x^2\right )+3 d e f^2 \left (1-\frac {c f}{d e}\right ) x^2 \sin \left (a+b x^2\right )+f^3 x^3 \sin \left (a+b x^2\right )\right ) \, dx,x,c+d x\right )}{d^4}\\ &=\frac {f^3 \operatorname {Subst}\left (\int x^3 \sin \left (a+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 (a+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 (a+b x^2\right ) \, dx,x,c+d x\right )}{d^4}+\frac {(d e-c f)^3 \operatorname {Subst}\left (\int \sin \left (a+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 (a+b (c+d x)^2\right )}{2 b d^4}+\frac {f^3 \operatorname {Subst}\left (\int x \sin (a+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 (a+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 (a+b x) \, dx,x,(c+d x)^2\right )}{2 d^4}+\frac {\left ((d e-c f)^3 \cos (a)\right ) \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{d^4}+\frac {\left ((d e-c f)^3 \sin (a)\right ) \operatorname {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,c+d x\right )}{d^4}\\ &=-\frac {3 f (d e-c f)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4}-\frac {3 f^2 (d e-c f) (c+d x) \cos \left (a+b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4}+\frac {(d e-c f)^3 \sqrt {\frac {\pi }{2}} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}+\frac {(d e-c f)^3 \sqrt {\frac {\pi }{2}} C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right ) \sin (a)}{\sqrt {b} d^4}+\frac {f^3 \operatorname {Subst}\left (\int \cos (a+b x) \, dx,x,(c+d x)^2\right )}{2 b d^4}+\frac {\left (3 f^2 (d e-c f) \cos (a)\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^2 (d e-c f) \sin (a)\right ) \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,c+d x\right )}{2 b d^4}\\ &=-\frac {3 f (d e-c f)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4}-\frac {3 f^2 (d e-c f) (c+d x) \cos \left (a+b (c+d x)^2\right )}{2 b d^4}-\frac {f^3 (c+d x)^2 \cos \left (a+b (c+d x)^2\right )}{2 b d^4}+\frac {3 f^2 (d e-c f) \sqrt {\frac {\pi }{2}} \cos (a) 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}} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d^4}+\frac {(d e-c f)^3 \sqrt {\frac {\pi }{2}} C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right ) \sin (a)}{\sqrt {b} d^4}-\frac {3 f^2 (d e-c f) \sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right ) \sin (a)}{2 b^{3/2} d^4}+\frac {f^3 \sin \left (a+b (c+d x)^2\right )}{2 b^2 d^4}\\ \end {align*}

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Mathematica [A]  time = 3.03, size = 218, normalized size = 0.64 \[ \frac {-4 b f \left (c^2 f^2-c d f (3 e+f x)+d^2 \left (3 e^2+3 e f x+f^2 x^2\right )\right ) \cos \left (a+b (c+d x)^2\right )+2 \sqrt {2 \pi } \sqrt {b} (d e-c f) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right ) \left (2 b \sin (a) (d e-c f)^2+3 f^2 \cos (a)\right )+2 \sqrt {2 \pi } \sqrt {b} (d e-c f) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right ) \left (2 b \cos (a) (d e-c f)^2-3 f^2 \sin (a)\right )+4 f^3 \sin \left (a+b (c+d x)^2\right )}{8 b^2 d^4} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.68, size = 328, normalized size = 0.96 \[ \frac {2 \, d f^{3} \sin \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right ) + \sqrt {2} {\left (3 \, \pi {\left (d e f^{2} - c f^{3}\right )} \cos \relax (a) + 2 \, \pi {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \sin \relax (a)\right )} \sqrt {\frac {b d^{2}}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) + \sqrt {2} {\left (2 \, \pi {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \cos \relax (a) - 3 \, \pi {\left (d e f^{2} - c f^{3}\right )} \sin \relax (a)\right )} \sqrt {\frac {b d^{2}}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right ) - 2 \, {\left (b d^{3} f^{3} x^{2} + 3 \, b d^{3} e^{2} f - 3 \, b c d^{2} e f^{2} + b c^{2} d f^{3} + {\left (3 \, b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x\right )} \cos \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a\right )}{4 \, b^{2} d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(a+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 + a) + sqrt(2)*(3*pi*(d*e*f^2 - c*f^3)*cos(a) + 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)*sin(a))*sqrt(b*d^2/pi)*fresnel_cos(sqrt(2)*sqrt(b*d^2/pi)*(d
*x + c)/d) + sqrt(2)*(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)*cos(a) - 3*pi*(d*e*f^2
- c*f^3)*sin(a))*sqrt(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 + a)
)/(b^2*d^5)

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giac [C]  time = 0.99, size = 1073, normalized size = 3.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(a+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^(I*a + 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^(-I*a + 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)*sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))*e^(I*a + 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 + I*a + 2)/(b*d))/d - 1/4*(-3*I*sqrt(2)*sqrt(pi)*c*f*e
rf(-1/2*sqrt(2)*sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))*e^(-I*a + 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 - I*a + 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^(I*a + 1)/(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 + I*a + 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^(-I*a + 1)/(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 - I*a + 1)/(b*d))/d^2 + 1/8*(sqrt(2)*sq
rt(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))*e^(I*a)
/(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 + I*a)/(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))*e^(-I*a)/(sqrt(b*d^2)*(I*b*d^2/sq
rt(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 - I*a)/(b^2*d))/d^3

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maple [B]  time = 0.03, size = 1248, normalized size = 3.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sin(a+(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+a)-f^3*c/d*(-1/2/d^2/b*x*cos(b*d^2*x^2+2*b*c*d*x+b*c^2+a)-c/d
*(-1/2/d^2/b*cos(b*d^2*x^2+2*b*c*d*x+b*c^2+a)-1/2*c/d*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*(cos((b^2*c^2*d^2-d^2*b*(
b*c^2+a))/d^2/b)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-sin((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^
2/b)*FresnelC(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)*(cos(
(b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/b)*FresnelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))+sin((b^2*c^2*d^2
-d^2*b*(b*c^2+a))/d^2/b)*FresnelS(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+a)-1/2*c/d*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*(cos((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/b)*F
resnelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))+sin((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/b)*FresnelS(2^(1
/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))))-3/2*f^2*e/d^2/b*x*cos(b*d^2*x^2+2*b*c*d*x+b*c^2+a)-3*f^2*e*c/d*(
-1/2/d^2/b*cos(b*d^2*x^2+2*b*c*d*x+b*c^2+a)-1/2*c/d*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*(cos((b^2*c^2*d^2-d^2*b*(b*
c^2+a))/d^2/b)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-sin((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/
b)*FresnelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))))+3/4*f^2*e/d^2/b*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*(
cos((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/b)*FresnelC(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))+sin((b^2*c^2
*d^2-d^2*b*(b*c^2+a))/d^2/b)*FresnelS(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+a)-3/2*e^2*f*c/d*2^(1/2)*Pi^(1/2)/(d^2*b)^(1/2)*(cos((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/
b)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-sin((b^2*c^2*d^2-d^2*b*(b*c^2+a))/d^2/b)*FresnelC(
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*(cos((b^2*c^2*d^2-d^2*
b*(b*c^2+a))/d^2/b)*FresnelS(2^(1/2)/Pi^(1/2)/(d^2*b)^(1/2)*(b*d^2*x+b*c*d))-sin((b^2*c^2*d^2-d^2*b*(b*c^2+a))
/d^2/b)*FresnelC(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 = 3.97, size = 1815, normalized size = 5.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(pi)*((-(I + 1)*cos(a) + (I - 1)*sin(a))*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (-(I - 1)*cos(a)
+ (I + 1)*sin(a))*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))*e^3/(sqrt(b)*d) - 3/4096*((1024*(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))*cos(a) + (1024*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*
b*c^2) - 1024*I*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*d*x + 2*sqrt(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*
(((256*I + 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) - (256*I - 256)*sqrt(2)*
sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*cos(a) + (-(256*I - 256)*sqrt(2)*sqrt(pi)*(erf
(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + (256*I + 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I
*b*c*d*x - I*b*c^2)) - 1))*sin(a))*c + (1024*(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))*cos(a) + (1024*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 1024*I*e^(-I*b*d^2*x^2 - 2*I*b*
c*d*x - I*b*c^2))*sin(a))*c)*e^2*f/(b*d^3*x + b*c*d^2) + 3/4096*((2048*(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))*cos(a) + (2048*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 2048*
I*e^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*b*c*d*x + (2048*(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))*cos(a) + (2048*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 2048*I*e
^(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*b*c^2 + 2*sqrt(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*((((256*I + 256
)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) - (256*I - 256)*sqrt(2)*sqrt(pi)*(erf(
sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*cos(a) + (-(256*I - 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*
x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) + (256*I + 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b
*c^2)) - 1))*sin(a))*b*c^2 + ((256*I - 256)*sqrt(2)*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - (256*I +
256)*sqrt(2)*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*cos(a) + ((256*I + 256)*sqrt(2)*gamma(3/2, I*b
*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - (256*I - 256)*sqrt(2)*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*si
n(a)))*e*f^2/(b^2*d^4*x + b^2*c*d^3) - 1/4096*((3072*(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))*cos(a) + (3072*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 3072*I*e^(-I*b*d^2*x^2
- 2*I*b*c*d*x - I*b*c^2))*sin(a))*b*c^3 + ((3072*(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))*cos(a) + (3072*I*e^(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - 3072*I*e^(-I*b*d^2*x^2 - 2*
I*b*c*d*x - I*b*c^2))*sin(a))*b*c^2 + (-1024*I*gamma(2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + 1024*I*gamma(2,
-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*cos(a) - 1024*(gamma(2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) + gamma(2
, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*d*x + ((-1024*I*gamma(2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)
+ 1024*I*gamma(2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*cos(a) - 1024*(gamma(2, I*b*d^2*x^2 + 2*I*b*c*d*x +
I*b*c^2) + gamma(2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*c + 2*((((256*I + 256)*sqrt(2)*sqrt(pi)*(er
f(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2)) - 1) - (256*I - 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*
I*b*c*d*x - I*b*c^2)) - 1))*cos(a) + (-(256*I - 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b*d^2*x^2 + 2*I*b*c*d*x + I*
b*c^2)) - 1) + (256*I + 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2)) - 1))*sin(a))*b
*c^3 + (((768*I - 768)*sqrt(2)*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*x + I*b*c^2) - (768*I + 768)*sqrt(2)*gamma(3
/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*cos(a) + ((768*I + 768)*sqrt(2)*gamma(3/2, I*b*d^2*x^2 + 2*I*b*c*d*
x + I*b*c^2) - (768*I - 768)*sqrt(2)*gamma(3/2, -I*b*d^2*x^2 - 2*I*b*c*d*x - I*b*c^2))*sin(a))*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 [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sin \left (a+b\,{\left (c+d\,x\right )}^2\right )\,{\left (e+f\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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