\(\int \cos (a+b \sqrt [3]{c+d x}) \, dx\) [97]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 85 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {6 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[Out]

6*(d*x+c)^(1/3)*cos(a+b*(d*x+c)^(1/3))/b^2/d-6*sin(a+b*(d*x+c)^(1/3))/b^3/d+3*(d*x+c)^(2/3)*sin(a+b*(d*x+c)^(1
/3))/b/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3443, 3377, 2717} \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=-\frac {6 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac {3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[In]

Int[Cos[a + b*(c + d*x)^(1/3)],x]

[Out]

(6*(c + d*x)^(1/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^2*d) - (6*Sin[a + b*(c + d*x)^(1/3)])/(b^3*d) + (3*(c + d*x)
^(2/3)*Sin[a + b*(c + d*x)^(1/3)])/(b*d)

Rule 2717

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

Rule 3377

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 3443

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

Rubi steps \begin{align*} \text {integral}& = \frac {3 \text {Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d} \\ & = \frac {3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {6 \text {Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d} \\ & = \frac {6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac {3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac {6 \text {Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d} \\ & = \frac {6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac {6 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac {3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6 b \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )+3 \left (-2+b^2 (c+d x)^{2/3}\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.54

method result size
derivativedivides \(\frac {3 a^{2} \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 a \left (\cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+\left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )\right )+3 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )^{2} \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+6 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )}{b^{3} d}\) \(131\)
default \(\frac {3 a^{2} \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 a \left (\cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+\left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )\right )+3 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )^{2} \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+6 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )}{b^{3} d}\) \(131\)

[In]

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

[Out]

3/d/b^3*(a^2*sin(a+b*(d*x+c)^(1/3))-2*a*(cos(a+b*(d*x+c)^(1/3))+(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))+(a
+b*(d*x+c)^(1/3))^2*sin(a+b*(d*x+c)^(1/3))-2*sin(a+b*(d*x+c)^(1/3))+2*(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3
)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left (2 \, {\left (d x + c\right )}^{\frac {1}{3}} b \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) + {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} - 2\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{3} d} \]

[In]

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

[Out]

3*(2*(d*x + c)^(1/3)*b*cos((d*x + c)^(1/3)*b + a) + ((d*x + c)^(2/3)*b^2 - 2)*sin((d*x + c)^(1/3)*b + a))/(b^3
*d)

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\begin {cases} x \cos {\left (a \right )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \cos {\left (a + b \sqrt [3]{c} \right )} & \text {for}\: d = 0 \\\frac {3 \left (c + d x\right )^{\frac {2}{3}} \sin {\left (a + b \sqrt [3]{c + d x} \right )}}{b d} + \frac {6 \sqrt [3]{c + d x} \cos {\left (a + b \sqrt [3]{c + d x} \right )}}{b^{2} d} - \frac {6 \sin {\left (a + b \sqrt [3]{c + d x} \right )}}{b^{3} d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*cos(a), Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x*cos(a + b*c**(1/3)), Eq(d, 0)), (3*(c + d*x)**(2/3)
*sin(a + b*(c + d*x)**(1/3))/(b*d) + 6*(c + d*x)**(1/3)*cos(a + b*(c + d*x)**(1/3))/(b**2*d) - 6*sin(a + b*(c
+ d*x)**(1/3))/(b**3*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.39 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left (a^{2} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - 2 \, {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) + \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )} a + 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) + {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{3} d} \]

[In]

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

[Out]

3*(a^2*sin((d*x + c)^(1/3)*b + a) - 2*(((d*x + c)^(1/3)*b + a)*sin((d*x + c)^(1/3)*b + a) + cos((d*x + c)^(1/3
)*b + a))*a + 2*((d*x + c)^(1/3)*b + a)*cos((d*x + c)^(1/3)*b + a) + (((d*x + c)^(1/3)*b + a)^2 - 2)*sin((d*x
+ c)^(1/3)*b + a))/(b^3*d)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.95 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left (\frac {2 \, {\left (d x + c\right )}^{\frac {1}{3}} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b} + \frac {{\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a + a^{2} - 2\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b^{2}}\right )}}{b d} \]

[In]

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

[Out]

3*(2*(d*x + c)^(1/3)*cos((d*x + c)^(1/3)*b + a)/b + (((d*x + c)^(1/3)*b + a)^2 - 2*((d*x + c)^(1/3)*b + a)*a +
 a^2 - 2)*sin((d*x + c)^(1/3)*b + a)/b^2)/(b*d)

Mupad [B] (verification not implemented)

Time = 14.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {6\,b\,\cos \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c+d\,x\right )}^{1/3}-6\,\sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )+3\,b^2\,\sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c+d\,x\right )}^{2/3}}{b^3\,d} \]

[In]

int(cos(a + b*(c + d*x)^(1/3)),x)

[Out]

(6*b*cos(a + b*(c + d*x)^(1/3))*(c + d*x)^(1/3) - 6*sin(a + b*(c + d*x)^(1/3)) + 3*b^2*sin(a + b*(c + d*x)^(1/
3))*(c + d*x)^(2/3))/(b^3*d)