3.99 \(\int \frac{\cos (a+b \sqrt [3]{c+d x})}{x^2} \, dx\)

Optimal. Leaf size=332 \[ -\frac{b d \sin \left (a+b \sqrt [3]{c}\right ) \text{CosIntegral}\left (b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{\sqrt [3]{-1} b d \sin \left (a-\sqrt [3]{-1} b \sqrt [3]{c}\right ) \text{CosIntegral}\left (b \sqrt [3]{c+d x}+\sqrt [3]{-1} b \sqrt [3]{c}\right )}{3 c^{2/3}}-\frac{(-1)^{2/3} b d \sin \left (a+(-1)^{2/3} b \sqrt [3]{c}\right ) \text{CosIntegral}\left ((-1)^{2/3} b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{b d \cos \left (a+b \sqrt [3]{c}\right ) \text{Si}\left (b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{(-1)^{2/3} b d \cos \left (a+(-1)^{2/3} b \sqrt [3]{c}\right ) \text{Si}\left ((-1)^{2/3} b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{\sqrt [3]{-1} b d \cos \left (a-\sqrt [3]{-1} b \sqrt [3]{c}\right ) \text{Si}\left (\sqrt [3]{-1} \sqrt [3]{c} b+\sqrt [3]{c+d x} b\right )}{3 c^{2/3}}-\frac{\cos \left (a+b \sqrt [3]{c+d x}\right )}{x} \]

[Out]

-(Cos[a + b*(c + d*x)^(1/3)]/x) - (b*d*CosIntegral[b*c^(1/3) - b*(c + d*x)^(1/3)]*Sin[a + b*c^(1/3)])/(3*c^(2/
3)) + ((-1)^(1/3)*b*d*CosIntegral[(-1)^(1/3)*b*c^(1/3) + b*(c + d*x)^(1/3)]*Sin[a - (-1)^(1/3)*b*c^(1/3)])/(3*
c^(2/3)) - ((-1)^(2/3)*b*d*CosIntegral[(-1)^(2/3)*b*c^(1/3) - b*(c + d*x)^(1/3)]*Sin[a + (-1)^(2/3)*b*c^(1/3)]
)/(3*c^(2/3)) + (b*d*Cos[a + b*c^(1/3)]*SinIntegral[b*c^(1/3) - b*(c + d*x)^(1/3)])/(3*c^(2/3)) + ((-1)^(2/3)*
b*d*Cos[a + (-1)^(2/3)*b*c^(1/3)]*SinIntegral[(-1)^(2/3)*b*c^(1/3) - b*(c + d*x)^(1/3)])/(3*c^(2/3)) + ((-1)^(
1/3)*b*d*Cos[a - (-1)^(1/3)*b*c^(1/3)]*SinIntegral[(-1)^(1/3)*b*c^(1/3) + b*(c + d*x)^(1/3)])/(3*c^(2/3))

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Rubi [A]  time = 0.741003, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3432, 3342, 3333, 3303, 3299, 3302} \[ -\frac{b d \sin \left (a+b \sqrt [3]{c}\right ) \text{CosIntegral}\left (b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{\sqrt [3]{-1} b d \sin \left (a-\sqrt [3]{-1} b \sqrt [3]{c}\right ) \text{CosIntegral}\left (b \sqrt [3]{c+d x}+\sqrt [3]{-1} b \sqrt [3]{c}\right )}{3 c^{2/3}}-\frac{(-1)^{2/3} b d \sin \left (a+(-1)^{2/3} b \sqrt [3]{c}\right ) \text{CosIntegral}\left ((-1)^{2/3} b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{b d \cos \left (a+b \sqrt [3]{c}\right ) \text{Si}\left (b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{(-1)^{2/3} b d \cos \left (a+(-1)^{2/3} b \sqrt [3]{c}\right ) \text{Si}\left ((-1)^{2/3} b \sqrt [3]{c}-b \sqrt [3]{c+d x}\right )}{3 c^{2/3}}+\frac{\sqrt [3]{-1} b d \cos \left (a-\sqrt [3]{-1} b \sqrt [3]{c}\right ) \text{Si}\left (\sqrt [3]{-1} \sqrt [3]{c} b+\sqrt [3]{c+d x} b\right )}{3 c^{2/3}}-\frac{\cos \left (a+b \sqrt [3]{c+d x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*(c + d*x)^(1/3)]/x^2,x]

[Out]

-(Cos[a + b*(c + d*x)^(1/3)]/x) - (b*d*CosIntegral[b*c^(1/3) - b*(c + d*x)^(1/3)]*Sin[a + b*c^(1/3)])/(3*c^(2/
3)) + ((-1)^(1/3)*b*d*CosIntegral[(-1)^(1/3)*b*c^(1/3) + b*(c + d*x)^(1/3)]*Sin[a - (-1)^(1/3)*b*c^(1/3)])/(3*
c^(2/3)) - ((-1)^(2/3)*b*d*CosIntegral[(-1)^(2/3)*b*c^(1/3) - b*(c + d*x)^(1/3)]*Sin[a + (-1)^(2/3)*b*c^(1/3)]
)/(3*c^(2/3)) + (b*d*Cos[a + b*c^(1/3)]*SinIntegral[b*c^(1/3) - b*(c + d*x)^(1/3)])/(3*c^(2/3)) + ((-1)^(2/3)*
b*d*Cos[a + (-1)^(2/3)*b*c^(1/3)]*SinIntegral[(-1)^(2/3)*b*c^(1/3) - b*(c + d*x)^(1/3)])/(3*c^(2/3)) + ((-1)^(
1/3)*b*d*Cos[a - (-1)^(1/3)*b*c^(1/3)]*SinIntegral[(-1)^(1/3)*b*c^(1/3) + b*(c + d*x)^(1/3)])/(3*c^(2/3))

Rule 3432

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Cos[c + d*x])^p, x^(1/n - 1)*(g - (e*h)/f + (h*x^(1/n))/f)^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rule 3342

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

Rule 3333

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

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 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]

Rubi steps

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

Mathematica [C]  time = 0.620772, size = 138, normalized size = 0.42 \[ -\frac{1}{6} i b d \text{RootSum}\left [c-\text{$\#$1}^3\& ,\frac{e^{-i \text{$\#$1} b-i a} \text{Ei}\left (-i b \left (\sqrt [3]{c+d x}-\text{$\#$1}\right )\right )}{\text{$\#$1}^2}\& \right ]+\frac{1}{6} i b d \text{RootSum}\left [c-\text{$\#$1}^3\& ,\frac{e^{i \text{$\#$1} b+i a} \text{Ei}\left (i b \left (\sqrt [3]{c+d x}-\text{$\#$1}\right )\right )}{\text{$\#$1}^2}\& \right ]-\frac{\cos \left (a+b \sqrt [3]{c+d x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*(c + d*x)^(1/3)]/x^2,x]

[Out]

-(Cos[a + b*(c + d*x)^(1/3)]/x) - (I/6)*b*d*RootSum[c - #1^3 & , (E^((-I)*a - I*b*#1)*ExpIntegralEi[(-I)*b*((c
+ d*x)^(1/3) - #1)])/#1^2 & ] + (I/6)*b*d*RootSum[c - #1^3 & , (E^(I*a + I*b*#1)*ExpIntegralEi[I*b*((c + d*x)
^(1/3) - #1)])/#1^2 & ]

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Maple [C]  time = 0.32, size = 933, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a+b*(d*x+c)^(1/3))/x^2,x)

[Out]

3*d/b^3*(cos(a+b*(d*x+c)^(1/3))*(-2/3*a*b^3/c*(a+b*(d*x+c)^(1/3))^2+a^2*b^3/c*(a+b*(d*x+c)^(1/3))-1/3*b^3*(b^3
*c+a^3)/c)/(-b^3*c+(a+b*(d*x+c)^(1/3))^3-3*(a+b*(d*x+c)^(1/3))^2*a+3*a^2*(a+b*(d*x+c)^(1/3))-a^3)-2/9*a*b^3/c*
sum(_R1/(_R1^2-2*_R1*a+a^2)*(Si(-b*(d*x+c)^(1/3)+_R1-a)*sin(_R1)+Ci(b*(d*x+c)^(1/3)-_R1+a)*cos(_R1)),_R1=RootO
f(-b^3*c+_Z^3-3*_Z^2*a+3*_Z*a^2-a^3))-1/9*b^3/c*sum((b^3*c+2*_RR1^2*a-3*_RR1*a^2+a^3)/(_RR1^2-2*_RR1*a+a^2)*(-
Si(-b*(d*x+c)^(1/3)+_RR1-a)*cos(_RR1)+Ci(b*(d*x+c)^(1/3)-_RR1+a)*sin(_RR1)),_RR1=RootOf(-b^3*c+_Z^3-3*_Z^2*a+3
*_Z*a^2-a^3))+cos(a+b*(d*x+c)^(1/3))*(2/3*a*b^3/c*(a+b*(d*x+c)^(1/3))^2-2/3*a^2*b^3/c*(a+b*(d*x+c)^(1/3)))/(-b
^3*c+(a+b*(d*x+c)^(1/3))^3-3*(a+b*(d*x+c)^(1/3))^2*a+3*a^2*(a+b*(d*x+c)^(1/3))-a^3)+2/9*a*b^3/c*sum((_R1+a)/(_
R1^2-2*_R1*a+a^2)*(Si(-b*(d*x+c)^(1/3)+_R1-a)*sin(_R1)+Ci(b*(d*x+c)^(1/3)-_R1+a)*cos(_R1)),_R1=RootOf(-b^3*c+_
Z^3-3*_Z^2*a+3*_Z*a^2-a^3))+2/9*a*b^3/c*sum(_RR1/(_RR1-a)*(-Si(-b*(d*x+c)^(1/3)+_RR1-a)*cos(_RR1)+Ci(b*(d*x+c)
^(1/3)-_RR1+a)*sin(_RR1)),_RR1=RootOf(-b^3*c+_Z^3-3*_Z^2*a+3*_Z*a^2-a^3))+a^2*b^6*(cos(a+b*(d*x+c)^(1/3))*(-1/
3/b^3/c*(a+b*(d*x+c)^(1/3))+1/3*a/b^3/c)/(-b^3*c+(a+b*(d*x+c)^(1/3))^3-3*(a+b*(d*x+c)^(1/3))^2*a+3*a^2*(a+b*(d
*x+c)^(1/3))-a^3)-2/9/b^3/c*sum(1/(_R1^2-2*_R1*a+a^2)*(Si(-b*(d*x+c)^(1/3)+_R1-a)*sin(_R1)+Ci(b*(d*x+c)^(1/3)-
_R1+a)*cos(_R1)),_R1=RootOf(-b^3*c+_Z^3-3*_Z^2*a+3*_Z*a^2-a^3))-1/9/b^3/c*sum(1/(_RR1-a)*(-Si(-b*(d*x+c)^(1/3)
+_RR1-a)*cos(_RR1)+Ci(b*(d*x+c)^(1/3)-_RR1+a)*sin(_RR1)),_RR1=RootOf(-b^3*c+_Z^3-3*_Z^2*a+3*_Z*a^2-a^3))))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)^(1/3))/x^2,x, algorithm="maxima")

[Out]

integrate(cos((d*x + c)^(1/3)*b + a)/x^2, x)

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Fricas [C]  time = 2.05983, size = 1137, normalized size = 3.42 \begin{align*} -\frac{2 \, \left (i \, b^{3} c\right )^{\frac{1}{3}} d x{\rm Ei}\left (i \,{\left (d x + c\right )}^{\frac{1}{3}} b + \left (i \, b^{3} c\right )^{\frac{1}{3}}\right ) e^{\left (i \, a - \left (i \, b^{3} c\right )^{\frac{1}{3}}\right )} + 2 \, \left (-i \, b^{3} c\right )^{\frac{1}{3}} d x{\rm Ei}\left (-i \,{\left (d x + c\right )}^{\frac{1}{3}} b + \left (-i \, b^{3} c\right )^{\frac{1}{3}}\right ) e^{\left (-i \, a - \left (-i \, b^{3} c\right )^{\frac{1}{3}}\right )} - \left (i \, b^{3} c\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} d x + d x\right )}{\rm Ei}\left (i \,{\left (d x + c\right )}^{\frac{1}{3}} b + \frac{1}{2} \, \left (i \, b^{3} c\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (i \, b^{3} c\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} + 1\right )} + i \, a\right )} - \left (-i \, b^{3} c\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} d x + d x\right )}{\rm Ei}\left (-i \,{\left (d x + c\right )}^{\frac{1}{3}} b + \frac{1}{2} \, \left (-i \, b^{3} c\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (-i \, b^{3} c\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} + 1\right )} - i \, a\right )} - \left (i \, b^{3} c\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} d x + d x\right )}{\rm Ei}\left (i \,{\left (d x + c\right )}^{\frac{1}{3}} b + \frac{1}{2} \, \left (i \, b^{3} c\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (i \, b^{3} c\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} + 1\right )} + i \, a\right )} - \left (-i \, b^{3} c\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} d x + d x\right )}{\rm Ei}\left (-i \,{\left (d x + c\right )}^{\frac{1}{3}} b + \frac{1}{2} \, \left (-i \, b^{3} c\right )^{\frac{1}{3}}{\left (i \, \sqrt{3} - 1\right )}\right ) e^{\left (\frac{1}{2} \, \left (-i \, b^{3} c\right )^{\frac{1}{3}}{\left (-i \, \sqrt{3} + 1\right )} - i \, a\right )} + 12 \, c \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}{12 \, c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)^(1/3))/x^2,x, algorithm="fricas")

[Out]

-1/12*(2*(I*b^3*c)^(1/3)*d*x*Ei(I*(d*x + c)^(1/3)*b + (I*b^3*c)^(1/3))*e^(I*a - (I*b^3*c)^(1/3)) + 2*(-I*b^3*c
)^(1/3)*d*x*Ei(-I*(d*x + c)^(1/3)*b + (-I*b^3*c)^(1/3))*e^(-I*a - (-I*b^3*c)^(1/3)) - (I*b^3*c)^(1/3)*(I*sqrt(
3)*d*x + d*x)*Ei(I*(d*x + c)^(1/3)*b + 1/2*(I*b^3*c)^(1/3)*(-I*sqrt(3) - 1))*e^(1/2*(I*b^3*c)^(1/3)*(I*sqrt(3)
+ 1) + I*a) - (-I*b^3*c)^(1/3)*(I*sqrt(3)*d*x + d*x)*Ei(-I*(d*x + c)^(1/3)*b + 1/2*(-I*b^3*c)^(1/3)*(-I*sqrt(
3) - 1))*e^(1/2*(-I*b^3*c)^(1/3)*(I*sqrt(3) + 1) - I*a) - (I*b^3*c)^(1/3)*(-I*sqrt(3)*d*x + d*x)*Ei(I*(d*x + c
)^(1/3)*b + 1/2*(I*b^3*c)^(1/3)*(I*sqrt(3) - 1))*e^(1/2*(I*b^3*c)^(1/3)*(-I*sqrt(3) + 1) + I*a) - (-I*b^3*c)^(
1/3)*(-I*sqrt(3)*d*x + d*x)*Ei(-I*(d*x + c)^(1/3)*b + 1/2*(-I*b^3*c)^(1/3)*(I*sqrt(3) - 1))*e^(1/2*(-I*b^3*c)^
(1/3)*(-I*sqrt(3) + 1) - I*a) + 12*c*cos((d*x + c)^(1/3)*b + a))/(c*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (a + b \sqrt [3]{c + d x} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)**(1/3))/x**2,x)

[Out]

Integral(cos(a + b*(c + d*x)**(1/3))/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)^(1/3))/x^2,x, algorithm="giac")

[Out]

integrate(cos((d*x + c)^(1/3)*b + a)/x^2, x)