3.223 \(\int (e+f x) \sin (a+\frac {b}{(c+d x)^{2/3}}) \, dx\)

Optimal. Leaf size=318 \[ \frac {2 \sqrt {2 \pi } b^{3/2} \sin (a) (d e-c f) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d^2}+\frac {2 \sqrt {2 \pi } b^{3/2} \cos (a) (d e-c f) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d^2}+\frac {b^3 f \cos (a) \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}-\frac {b^3 f \sin (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}-\frac {b^2 f (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {(c+d x) (d e-c f) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {2 b \sqrt [3]{c+d x} (d e-c f) \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}+\frac {b f (c+d x)^{4/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3433, 3409, 3387, 3388, 3353, 3352, 3351, 3379, 3297, 3303, 3299, 3302} \[ \frac {b^3 f \cos (a) \text {CosIntegral}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {2 \sqrt {2 \pi } b^{3/2} \sin (a) (d e-c f) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{\sqrt [3]{c+d x}}\right )}{d^2}+\frac {2 \sqrt {2 \pi } b^{3/2} \cos (a) (d e-c f) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d^2}-\frac {b^3 f \sin (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}-\frac {b^2 f (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {(c+d x) (d e-c f) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {2 b \sqrt [3]{c+d x} (d e-c f) \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}+\frac {b f (c+d x)^{4/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(d*e - c*f)*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)])/d^2 + (b*f*(c + d*x)^(4/3)*Cos[a + b/(c + d*x)^(2
/3)])/(4*d^2) + (b^3*f*Cos[a]*CosIntegral[b/(c + d*x)^(2/3)])/(4*d^2) + (2*b^(3/2)*(d*e - c*f)*Sqrt[2*Pi]*Cos[
a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/d^2 + (2*b^(3/2)*(d*e - c*f)*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*S
qrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/d^2 - (b^2*f*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)])/(4*d^2) + ((d*e -
c*f)*(c + d*x)*Sin[a + b/(c + d*x)^(2/3)])/d^2 + (f*(c + d*x)^2*Sin[a + b/(c + d*x)^(2/3)])/(2*d^2) - (b^3*f*
Sin[a]*SinIntegral[b/(c + d*x)^(2/3)])/(4*d^2)

Rule 3297

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 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 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 3387

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

Rule 3388

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

Rule 3409

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sin[c + d/x^
n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2
]

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) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right ) \, dx &=\frac {3 \operatorname {Subst}\left (\int \left ((d e-c f) x^2 \sin \left (a+\frac {b}{x^2}\right )+f x^5 \sin \left (a+\frac {b}{x^2}\right )\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}\\ &=\frac {(3 f) \operatorname {Subst}\left (\int x^5 \sin \left (a+\frac {b}{x^2}\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}+\frac {(3 (d e-c f)) \operatorname {Subst}\left (\int x^2 \sin \left (a+\frac {b}{x^2}\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}\\ &=-\frac {(3 f) \operatorname {Subst}\left (\int \frac {\sin (a+b x)}{x^4} \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{2 d^2}-\frac {(3 (d e-c f)) \operatorname {Subst}\left (\int \frac {\sin \left (a+b x^2\right )}{x^4} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d^2}\\ &=\frac {(d e-c f) (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}-\frac {(b f) \operatorname {Subst}\left (\int \frac {\cos (a+b x)}{x^3} \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{2 d^2}-\frac {(2 b (d e-c f)) \operatorname {Subst}\left (\int \frac {\cos \left (a+b x^2\right )}{x^2} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d^2}\\ &=\frac {2 b (d e-c f) \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {b f (c+d x)^{4/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {(d e-c f) (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}+\frac {\left (b^2 f\right ) \operatorname {Subst}\left (\int \frac {\sin (a+b x)}{x^2} \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {\left (4 b^2 (d e-c f)\right ) \operatorname {Subst}\left (\int \sin \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d^2}\\ &=\frac {2 b (d e-c f) \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {b f (c+d x)^{4/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}-\frac {b^2 f (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {(d e-c f) (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}+\frac {\left (b^3 f\right ) \operatorname {Subst}\left (\int \frac {\cos (a+b x)}{x} \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {\left (4 b^2 (d e-c f) \cos (a)\right ) \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d^2}+\frac {\left (4 b^2 (d e-c f) \sin (a)\right ) \operatorname {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d^2}\\ &=\frac {2 b (d e-c f) \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {b f (c+d x)^{4/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {2 b^{3/2} (d e-c f) \sqrt {2 \pi } \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d^2}+\frac {2 b^{3/2} (d e-c f) \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d^2}-\frac {b^2 f (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {(d e-c f) (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}+\frac {\left (b^3 f \cos (a)\right ) \operatorname {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{4 d^2}-\frac {\left (b^3 f \sin (a)\right ) \operatorname {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,\frac {1}{(c+d x)^{2/3}}\right )}{4 d^2}\\ &=\frac {2 b (d e-c f) \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {b f (c+d x)^{4/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {b^3 f \cos (a) \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {2 b^{3/2} (d e-c f) \sqrt {2 \pi } \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d^2}+\frac {2 b^{3/2} (d e-c f) \sqrt {2 \pi } C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{d^2}-\frac {b^2 f (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}+\frac {(d e-c f) (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d^2}+\frac {f (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d^2}-\frac {b^3 f \sin (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2}\\ \end {align*}

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Mathematica [A]  time = 1.18, size = 378, normalized size = 1.19 \[ \frac {8 \sqrt {2 \pi } b^{3/2} \cos (a) (d e-c f) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+8 \sqrt {2 \pi } b^{3/2} d e \sin (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )-8 \sqrt {2 \pi } b^{3/2} c f \sin (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+b^3 f \cos (a) \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right )-b^3 f \sin (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )-b^2 f (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )-2 c^2 f \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+4 d^2 e x \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+2 d^2 f x^2 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+4 c d e \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+8 b d e \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )-7 b c f \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+b d f x \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d*e*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)] - 7*b*c*f*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)] + b*d
*f*x*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)] + b^3*f*Cos[a]*CosIntegral[b/(c + d*x)^(2/3)] + 8*b^(3/2)*(d*e
- c*f)*Sqrt[2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] + 8*b^(3/2)*d*e*Sqrt[2*Pi]*FresnelC[(
Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] - 8*b^(3/2)*c*f*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)
^(1/3)]*Sin[a] + 4*c*d*e*Sin[a + b/(c + d*x)^(2/3)] - 2*c^2*f*Sin[a + b/(c + d*x)^(2/3)] + 4*d^2*e*x*Sin[a + b
/(c + d*x)^(2/3)] + 2*d^2*f*x^2*Sin[a + b/(c + d*x)^(2/3)] - b^2*f*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)]
- b^3*f*Sin[a]*SinIntegral[b/(c + d*x)^(2/3)])/(4*d^2)

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fricas [A]  time = 0.77, size = 267, normalized size = 0.84 \[ \frac {b^{3} f \cos \relax (a) \operatorname {Ci}\left (\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + b^{3} f \cos \relax (a) \operatorname {Ci}\left (-\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) - 2 \, b^{3} f \sin \relax (a) \operatorname {Si}\left (\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + 16 \, \sqrt {2} \pi {\left (b d e - b c f\right )} \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + 16 \, \sqrt {2} \pi {\left (b d e - b c f\right )} \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) \sin \relax (a) + 2 \, {\left (b d f x + 8 \, b d e - 7 \, b c f\right )} {\left (d x + c\right )}^{\frac {1}{3}} \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right ) + 2 \, {\left (2 \, d^{2} f x^{2} + 4 \, d^{2} e x - {\left (d x + c\right )}^{\frac {2}{3}} b^{2} f + 4 \, c d e - 2 \, c^{2} f\right )} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )}{8 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b^3*f*cos(a)*cos_integral(b/(d*x + c)^(2/3)) + b^3*f*cos(a)*cos_integral(-b/(d*x + c)^(2/3)) - 2*b^3*f*si
n(a)*sin_integral(b/(d*x + c)^(2/3)) + 16*sqrt(2)*pi*(b*d*e - b*c*f)*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*sqr
t(b/pi)/(d*x + c)^(1/3)) + 16*sqrt(2)*pi*(b*d*e - b*c*f)*sqrt(b/pi)*fresnel_cos(sqrt(2)*sqrt(b/pi)/(d*x + c)^(
1/3))*sin(a) + 2*(b*d*f*x + 8*b*d*e - 7*b*c*f)*(d*x + c)^(1/3)*cos((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c)
) + 2*(2*d^2*f*x^2 + 4*d^2*e*x - (d*x + c)^(2/3)*b^2*f + 4*c*d*e - 2*c^2*f)*sin((a*d*x + a*c + (d*x + c)^(1/3)
*b)/(d*x + c)))/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 225, normalized size = 0.71 \[ \frac {-\left (c f -d e \right ) \left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )+2 \left (c f -d e \right ) b \left (-\left (d x +c \right )^{\frac {1}{3}} \cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )+\sin \relax (a ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )\right )\right )+\frac {f \left (d x +c \right )^{2} \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{2}-f b \left (-\frac {\left (d x +c \right )^{\frac {4}{3}} \cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{4}-\frac {b \left (-\frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right ) \left (d x +c \right )^{\frac {2}{3}}}{2}+b \left (\frac {\cos \relax (a ) \Ci \left (\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{2}-\frac {\sin \relax (a ) \Si \left (\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{2}\right )\right )}{2}\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/d^2*(-1/3*(c*f-d*e)*(d*x+c)*sin(a+b/(d*x+c)^(2/3))+2/3*(c*f-d*e)*b*(-(d*x+c)^(1/3)*cos(a+b/(d*x+c)^(2/3))-b^
(1/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))+sin(a)*FresnelC(b^(1/2)*2^(1/2
)/Pi^(1/2)/(d*x+c)^(1/3))))+1/6*f*(d*x+c)^2*sin(a+b/(d*x+c)^(2/3))-1/3*f*b*(-1/4*(d*x+c)^(4/3)*cos(a+b/(d*x+c)
^(2/3))-1/2*b*(-1/2*sin(a+b/(d*x+c)^(2/3))*(d*x+c)^(2/3)+b*(1/2*cos(a)*Ci(b/(d*x+c)^(2/3))-1/2*sin(a)*Si(b/(d*
x+c)^(2/3))))))

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maxima [C]  time = 0.77, size = 584, normalized size = 1.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*sqrt(2)*(2*sqrt(2)*(d*x + c)^(2/3)*sqrt((d*x + c)^(-4/3))*b^2*cos(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/
3)) + sqrt(2)*(d*x + c)^(4/3)*sqrt((d*x + c)^(-4/3))*b*sin(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + (((I + 1
)*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) - (I - 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(
a) + (-(I - 1)*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) + (I + 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3)
)) - 1))*sin(a))*b^2*(b^2/(d*x + c)^(4/3))^(1/4))*sqrt((d*x + c)^(4/3))*e/((d*x + c)^(1/3)*b) - 4*sqrt(2)*(2*s
qrt(2)*(d*x + c)^(2/3)*sqrt((d*x + c)^(-4/3))*b^2*cos(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + sqrt(2)*(d*x
+ c)^(4/3)*sqrt((d*x + c)^(-4/3))*b*sin(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + (((I + 1)*sqrt(pi)*(erf(sqr
t(I*b/(d*x + c)^(2/3))) - 1) - (I - 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(a) + (-(I - 1)*sqrt
(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) + (I + 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*sin(a))*b^
2*(b^2/(d*x + c)^(4/3))^(1/4))*sqrt((d*x + c)^(4/3))*c*f/((d*x + c)^(1/3)*b*d) + (((Ei(I*b/(d*x + c)^(2/3)) +
Ei(-I*b/(d*x + c)^(2/3)))*cos(a) + (I*Ei(I*b/(d*x + c)^(2/3)) - I*Ei(-I*b/(d*x + c)^(2/3)))*sin(a))*b^3 + 2*(d
*x + c)^(4/3)*b*cos(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) - 2*((d*x + c)^(2/3)*b^2 - 2*(d*x + c)^2)*sin(((d
*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)))*f/d)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )\,\left (e+f\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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