\(\int x^2 \cos (a+b \sqrt [3]{c+d x}) \, dx\) [95]

Optimal result

Integrand size = 18, antiderivative size = 537 \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=-\frac {720 c \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^3}-\frac {120960 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^8 d^3}+\frac {6 c^2 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^3}+\frac {360 c (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^3}+\frac {20160 (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^3}-\frac {30 c (c+d x)^{4/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^3}-\frac {1008 (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^3}+\frac {24 (c+d x)^{7/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^3}+\frac {120960 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^9 d^3}-\frac {6 c^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^3}-\frac {720 c \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^3}-\frac {60480 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^7 d^3}+\frac {3 c^2 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d^3}+\frac {120 c (c+d x) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^3}+\frac {5040 (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^3}-\frac {6 c (c+d x)^{5/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d^3}-\frac {168 (c+d x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^3}+\frac {3 (c+d x)^{8/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d^3} \] Output:

-720*c*cos(a+b*(d*x+c)^(1/3))/b^6/d^3-120960*(d*x+c)^(1/3)*cos(a+b*(d*x+c) 
^(1/3))/b^8/d^3+6*c^2*(d*x+c)^(1/3)*cos(a+b*(d*x+c)^(1/3))/b^2/d^3+360*c*( 
d*x+c)^(2/3)*cos(a+b*(d*x+c)^(1/3))/b^4/d^3+20160*(d*x+c)*cos(a+b*(d*x+c)^ 
(1/3))/b^6/d^3-30*c*(d*x+c)^(4/3)*cos(a+b*(d*x+c)^(1/3))/b^2/d^3-1008*(d*x 
+c)^(5/3)*cos(a+b*(d*x+c)^(1/3))/b^4/d^3+24*(d*x+c)^(7/3)*cos(a+b*(d*x+c)^ 
(1/3))/b^2/d^3+120960*sin(a+b*(d*x+c)^(1/3))/b^9/d^3-6*c^2*sin(a+b*(d*x+c) 
^(1/3))/b^3/d^3-720*c*(d*x+c)^(1/3)*sin(a+b*(d*x+c)^(1/3))/b^5/d^3-60480*( 
d*x+c)^(2/3)*sin(a+b*(d*x+c)^(1/3))/b^7/d^3+3*c^2*(d*x+c)^(2/3)*sin(a+b*(d 
*x+c)^(1/3))/b/d^3+120*c*(d*x+c)*sin(a+b*(d*x+c)^(1/3))/b^3/d^3+5040*(d*x+ 
c)^(4/3)*sin(a+b*(d*x+c)^(1/3))/b^5/d^3-6*c*(d*x+c)^(5/3)*sin(a+b*(d*x+c)^ 
(1/3))/b/d^3-168*(d*x+c)^2*sin(a+b*(d*x+c)^(1/3))/b^3/d^3+3*(d*x+c)^(8/3)* 
sin(a+b*(d*x+c)^(1/3))/b/d^3
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.26 (sec) , antiderivative size = 382, normalized size of antiderivative = 0.71 \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 e^{-i \left (a+b \sqrt [3]{c+d x}\right )} \left (-40320 i \left (-1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right )-40320 b \left (1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) \sqrt [3]{c+d x}+20160 i b^2 \left (-1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) (c+d x)^{2/3}-i b^8 d^2 \left (-1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) x^2 (c+d x)^{2/3}+2 b^7 d \left (1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) x \sqrt [3]{c+d x} (3 c+4 d x)-240 i b^4 \left (-1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) \sqrt [3]{c+d x} (6 c+7 d x)-24 b^5 \left (1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) (c+d x)^{2/3} (9 c+14 d x)+240 b^3 \left (1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) (27 c+28 d x)+2 i b^6 \left (-1+e^{2 i \left (a+b \sqrt [3]{c+d x}\right )}\right ) \left (9 c^2+36 c d x+28 d^2 x^2\right )\right )}{2 b^9 d^3} \] Input:

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

Output:

(3*((-40320*I)*(-1 + E^((2*I)*(a + b*(c + d*x)^(1/3)))) - 40320*b*(1 + E^( 
(2*I)*(a + b*(c + d*x)^(1/3))))*(c + d*x)^(1/3) + (20160*I)*b^2*(-1 + E^(( 
2*I)*(a + b*(c + d*x)^(1/3))))*(c + d*x)^(2/3) - I*b^8*d^2*(-1 + E^((2*I)* 
(a + b*(c + d*x)^(1/3))))*x^2*(c + d*x)^(2/3) + 2*b^7*d*(1 + E^((2*I)*(a + 
 b*(c + d*x)^(1/3))))*x*(c + d*x)^(1/3)*(3*c + 4*d*x) - (240*I)*b^4*(-1 + 
E^((2*I)*(a + b*(c + d*x)^(1/3))))*(c + d*x)^(1/3)*(6*c + 7*d*x) - 24*b^5* 
(1 + E^((2*I)*(a + b*(c + d*x)^(1/3))))*(c + d*x)^(2/3)*(9*c + 14*d*x) + 2 
40*b^3*(1 + E^((2*I)*(a + b*(c + d*x)^(1/3))))*(27*c + 28*d*x) + (2*I)*b^6 
*(-1 + E^((2*I)*(a + b*(c + d*x)^(1/3))))*(9*c^2 + 36*c*d*x + 28*d^2*x^2)) 
)/(2*b^9*d^3*E^(I*(a + b*(c + d*x)^(1/3))))
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 540, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3913, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3913

\(\displaystyle \frac {3 \int \left (\frac {\cos \left (a+b \sqrt [3]{c+d x}\right ) (c+d x)^{8/3}}{d^2}-\frac {2 c \cos \left (a+b \sqrt [3]{c+d x}\right ) (c+d x)^{5/3}}{d^2}+\frac {c^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) (c+d x)^{2/3}}{d^2}\right )d\sqrt [3]{c+d x}}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \left (\frac {40320 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^9 d^2}-\frac {40320 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^8 d^2}-\frac {20160 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^7 d^2}+\frac {6720 (c+d x) \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}-\frac {240 c \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^6 d^2}+\frac {1680 (c+d x)^{4/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac {240 c \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^5 d^2}-\frac {336 (c+d x)^{5/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}+\frac {120 c (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^4 d^2}-\frac {2 c^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}-\frac {56 (c+d x)^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac {40 c (c+d x) \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d^2}+\frac {2 c^2 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}+\frac {8 (c+d x)^{7/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}-\frac {10 c (c+d x)^{4/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d^2}+\frac {c^2 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}+\frac {(c+d x)^{8/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}-\frac {2 c (c+d x)^{5/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d^2}\right )}{d}\)

Input:

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

Output:

(3*((-240*c*Cos[a + b*(c + d*x)^(1/3)])/(b^6*d^2) - (40320*(c + d*x)^(1/3) 
*Cos[a + b*(c + d*x)^(1/3)])/(b^8*d^2) + (2*c^2*(c + d*x)^(1/3)*Cos[a + b* 
(c + d*x)^(1/3)])/(b^2*d^2) + (120*c*(c + d*x)^(2/3)*Cos[a + b*(c + d*x)^( 
1/3)])/(b^4*d^2) + (6720*(c + d*x)*Cos[a + b*(c + d*x)^(1/3)])/(b^6*d^2) - 
 (10*c*(c + d*x)^(4/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^2*d^2) - (336*(c + d 
*x)^(5/3)*Cos[a + b*(c + d*x)^(1/3)])/(b^4*d^2) + (8*(c + d*x)^(7/3)*Cos[a 
 + b*(c + d*x)^(1/3)])/(b^2*d^2) + (40320*Sin[a + b*(c + d*x)^(1/3)])/(b^9 
*d^2) - (2*c^2*Sin[a + b*(c + d*x)^(1/3)])/(b^3*d^2) - (240*c*(c + d*x)^(1 
/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^5*d^2) - (20160*(c + d*x)^(2/3)*Sin[a + 
 b*(c + d*x)^(1/3)])/(b^7*d^2) + (c^2*(c + d*x)^(2/3)*Sin[a + b*(c + d*x)^ 
(1/3)])/(b*d^2) + (40*c*(c + d*x)*Sin[a + b*(c + d*x)^(1/3)])/(b^3*d^2) + 
(1680*(c + d*x)^(4/3)*Sin[a + b*(c + d*x)^(1/3)])/(b^5*d^2) - (2*c*(c + d* 
x)^(5/3)*Sin[a + b*(c + d*x)^(1/3)])/(b*d^2) - (56*(c + d*x)^2*Sin[a + b*( 
c + d*x)^(1/3)])/(b^3*d^2) + ((c + d*x)^(8/3)*Sin[a + b*(c + d*x)^(1/3)])/ 
(b*d^2)))/d
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3913

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_ 
.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[1/(n*f)   Subst[Int[ExpandIntegra 
nd[(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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1808\) vs. \(2(477)=954\).

Time = 1.55 (sec) , antiderivative size = 1809, normalized size of antiderivative = 3.37

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1809\)
default	\(\text {Expression too large to display}\)	\(1809\)
parts	\(\text {Expression too large to display}\)	\(2944\)

Input:

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

Output:

3/d^3/b^3*(a^2*c^2*sin(a+b*(d*x+c)^(1/3))-2*a*c^2*(cos(a+b*(d*x+c)^(1/3))+ 
(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))+c^2*((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)))+2/b^3*a^5*c*sin(a+b*(d*x+c)^(1/3))-10/b^3*a^4*c*(cos(a+b* 
(d*x+c)^(1/3))+(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))+20/b^3*a^3*c*(( 
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)))-20/b^3*a^2*c*((a+b*(d*x+c)^(1/3)) 
^3*sin(a+b*(d*x+c)^(1/3))+3*(a+b*(d*x+c)^(1/3))^2*cos(a+b*(d*x+c)^(1/3))-6 
*cos(a+b*(d*x+c)^(1/3))-6*(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))+10/b 
^3*a*c*((a+b*(d*x+c)^(1/3))^4*sin(a+b*(d*x+c)^(1/3))+4*(a+b*(d*x+c)^(1/3)) 
^3*cos(a+b*(d*x+c)^(1/3))-12*(a+b*(d*x+c)^(1/3))^2*sin(a+b*(d*x+c)^(1/3))+ 
24*sin(a+b*(d*x+c)^(1/3))-24*(a+b*(d*x+c)^(1/3))*cos(a+b*(d*x+c)^(1/3)))-2 
/b^3*c*((a+b*(d*x+c)^(1/3))^5*sin(a+b*(d*x+c)^(1/3))+5*(a+b*(d*x+c)^(1/3)) 
^4*cos(a+b*(d*x+c)^(1/3))-20*(a+b*(d*x+c)^(1/3))^3*sin(a+b*(d*x+c)^(1/3))- 
60*(a+b*(d*x+c)^(1/3))^2*cos(a+b*(d*x+c)^(1/3))+120*cos(a+b*(d*x+c)^(1/3)) 
+120*(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))+1/b^6*a^8*sin(a+b*(d*x+c) 
^(1/3))-8/b^6*a^7*(cos(a+b*(d*x+c)^(1/3))+(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x 
+c)^(1/3)))+28/b^6*a^6*((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)))-56/b^6*a 
^5*((a+b*(d*x+c)^(1/3))^3*sin(a+b*(d*x+c)^(1/3))+3*(a+b*(d*x+c)^(1/3))^...
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.34 \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {3 \, {\left (2 \, {\left (3360 \, b^{3} d x + 3240 \, b^{3} c - 12 \, {\left (14 \, b^{5} d x + 9 \, b^{5} c\right )} {\left (d x + c\right )}^{\frac {2}{3}} + {\left (4 \, b^{7} d^{2} x^{2} + 3 \, b^{7} c d x - 20160 \, b\right )} {\left (d x + c\right )}^{\frac {1}{3}}\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - {\left (56 \, b^{6} d^{2} x^{2} + 72 \, b^{6} c d x + 18 \, b^{6} c^{2} - {\left (b^{8} d^{2} x^{2} - 20160 \, b^{2}\right )} {\left (d x + c\right )}^{\frac {2}{3}} - 240 \, {\left (7 \, b^{4} d x + 6 \, b^{4} c\right )} {\left (d x + c\right )}^{\frac {1}{3}} - 40320\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{9} d^{3}} \] Input:

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

Output:

3*(2*(3360*b^3*d*x + 3240*b^3*c - 12*(14*b^5*d*x + 9*b^5*c)*(d*x + c)^(2/3 
) + (4*b^7*d^2*x^2 + 3*b^7*c*d*x - 20160*b)*(d*x + c)^(1/3))*cos((d*x + c) 
^(1/3)*b + a) - (56*b^6*d^2*x^2 + 72*b^6*c*d*x + 18*b^6*c^2 - (b^8*d^2*x^2 
 - 20160*b^2)*(d*x + c)^(2/3) - 240*(7*b^4*d*x + 6*b^4*c)*(d*x + c)^(1/3) 
- 40320)*sin((d*x + c)^(1/3)*b + a))/(b^9*d^3)
 

Sympy [F]

\[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\int x^{2} \cos {\left (a + b \sqrt [3]{c + d x} \right )}\, dx \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1349 vs. \(2 (477) = 954\).

Time = 0.09 (sec) , antiderivative size = 1349, normalized size of antiderivative = 2.51 \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

3*(a^2*c^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*c^2 + 2*a^5*c*sin((d*x 
 + c)^(1/3)*b + a)/b^3 - 10*(((d*x + c)^(1/3)*b + a)*sin((d*x + c)^(1/3)*b 
 + a) + cos((d*x + c)^(1/3)*b + a))*a^4*c/b^3 + (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))*c^2 + a^8*sin((d*x + c)^(1/3)*b + a)/b^6 - 8*(((d*x + c)^( 
1/3)*b + a)*sin((d*x + c)^(1/3)*b + a) + cos((d*x + c)^(1/3)*b + a))*a^7/b 
^6 + 20*(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))*a^3*c/b^3 + 28*(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))*a^6/b^6 - 20*(3*(((d*x + c)^(1/3)*b + a)^ 
2 - 2)*cos((d*x + c)^(1/3)*b + a) + (((d*x + c)^(1/3)*b + a)^3 - 6*(d*x + 
c)^(1/3)*b - 6*a)*sin((d*x + c)^(1/3)*b + a))*a^2*c/b^3 - 56*(3*(((d*x + c 
)^(1/3)*b + a)^2 - 2)*cos((d*x + c)^(1/3)*b + a) + (((d*x + c)^(1/3)*b + a 
)^3 - 6*(d*x + c)^(1/3)*b - 6*a)*sin((d*x + c)^(1/3)*b + a))*a^5/b^6 + 10* 
(4*(((d*x + c)^(1/3)*b + a)^3 - 6*(d*x + c)^(1/3)*b - 6*a)*cos((d*x + c)^( 
1/3)*b + a) + (((d*x + c)^(1/3)*b + a)^4 - 12*((d*x + c)^(1/3)*b + a)^2 + 
24)*sin((d*x + c)^(1/3)*b + a))*a*c/b^3 + 70*(4*(((d*x + c)^(1/3)*b + a)^3 
 - 6*(d*x + c)^(1/3)*b - 6*a)*cos((d*x + c)^(1/3)*b + a) + (((d*x + c)^(1/ 
3)*b + a)^4 - 12*((d*x + c)^(1/3)*b + a)^2 + 24)*sin((d*x + c)^(1/3)*b ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1098 vs. \(2 (477) = 954\).

Time = 0.36 (sec) , antiderivative size = 1098, normalized size of antiderivative = 2.04 \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

3*(2*(((d*x + c)^(1/3)*b + a)*b^6*c^2 - a*b^6*c^2 - 5*((d*x + c)^(1/3)*b + 
 a)^4*b^3*c + 20*((d*x + c)^(1/3)*b + a)^3*a*b^3*c - 30*((d*x + c)^(1/3)*b 
 + a)^2*a^2*b^3*c + 20*((d*x + c)^(1/3)*b + a)*a^3*b^3*c - 5*a^4*b^3*c + 4 
*((d*x + c)^(1/3)*b + a)^7 - 28*((d*x + c)^(1/3)*b + a)^6*a + 84*((d*x + c 
)^(1/3)*b + a)^5*a^2 - 140*((d*x + c)^(1/3)*b + a)^4*a^3 + 140*((d*x + c)^ 
(1/3)*b + a)^3*a^4 - 84*((d*x + c)^(1/3)*b + a)^2*a^5 + 28*((d*x + c)^(1/3 
)*b + a)*a^6 - 4*a^7 + 60*((d*x + c)^(1/3)*b + a)^2*b^3*c - 120*((d*x + c) 
^(1/3)*b + a)*a*b^3*c + 60*a^2*b^3*c - 168*((d*x + c)^(1/3)*b + a)^5 + 840 
*((d*x + c)^(1/3)*b + a)^4*a - 1680*((d*x + c)^(1/3)*b + a)^3*a^2 + 1680*( 
(d*x + c)^(1/3)*b + a)^2*a^3 - 840*((d*x + c)^(1/3)*b + a)*a^4 + 168*a^5 - 
 120*b^3*c + 3360*((d*x + c)^(1/3)*b + a)^3 - 10080*((d*x + c)^(1/3)*b + a 
)^2*a + 10080*((d*x + c)^(1/3)*b + a)*a^2 - 3360*a^3 - 20160*(d*x + c)^(1/ 
3)*b)*cos((d*x + c)^(1/3)*b + a)/b^8 + (((d*x + c)^(1/3)*b + a)^2*b^6*c^2 
- 2*((d*x + c)^(1/3)*b + a)*a*b^6*c^2 + a^2*b^6*c^2 - 2*((d*x + c)^(1/3)*b 
 + a)^5*b^3*c + 10*((d*x + c)^(1/3)*b + a)^4*a*b^3*c - 20*((d*x + c)^(1/3) 
*b + a)^3*a^2*b^3*c + 20*((d*x + c)^(1/3)*b + a)^2*a^3*b^3*c - 10*((d*x + 
c)^(1/3)*b + a)*a^4*b^3*c + 2*a^5*b^3*c + ((d*x + c)^(1/3)*b + a)^8 - 8*(( 
d*x + c)^(1/3)*b + a)^7*a + 28*((d*x + c)^(1/3)*b + a)^6*a^2 - 56*((d*x + 
c)^(1/3)*b + a)^5*a^3 + 70*((d*x + c)^(1/3)*b + a)^4*a^4 - 56*((d*x + c)^( 
1/3)*b + a)^3*a^5 + 28*((d*x + c)^(1/3)*b + a)^2*a^6 - 8*((d*x + c)^(1/...
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\int x^2\,\cos \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right ) \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 357, normalized size of antiderivative = 0.66 \[ \int x^2 \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx=\frac {-648 \left (d x +c \right )^{\frac {2}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{5} c -1008 \left (d x +c \right )^{\frac {2}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{5} d x +18 \left (d x +c \right )^{\frac {1}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{7} c d x +24 \left (d x +c \right )^{\frac {1}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{7} d^{2} x^{2}-120960 \left (d x +c \right )^{\frac {1}{3}} \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b +19440 \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{3} c +20160 \cos \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{3} d x +3 \left (d x +c \right )^{\frac {2}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{8} d^{2} x^{2}-60480 \left (d x +c \right )^{\frac {2}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{2}+4320 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{4} c +5040 \left (d x +c \right )^{\frac {1}{3}} \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{4} d x -54 \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{6} c^{2}-216 \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{6} c d x -168 \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right ) b^{6} d^{2} x^{2}+120960 \sin \left (\left (d x +c \right )^{\frac {1}{3}} b +a \right )}{b^{9} d^{3}} \] Input:

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

Output:

(3*( - 216*(c + d*x)**(2/3)*cos((c + d*x)**(1/3)*b + a)*b**5*c - 336*(c + 
d*x)**(2/3)*cos((c + d*x)**(1/3)*b + a)*b**5*d*x + 6*(c + d*x)**(1/3)*cos( 
(c + d*x)**(1/3)*b + a)*b**7*c*d*x + 8*(c + d*x)**(1/3)*cos((c + d*x)**(1/ 
3)*b + a)*b**7*d**2*x**2 - 40320*(c + d*x)**(1/3)*cos((c + d*x)**(1/3)*b + 
 a)*b + 6480*cos((c + d*x)**(1/3)*b + a)*b**3*c + 6720*cos((c + d*x)**(1/3 
)*b + a)*b**3*d*x + (c + d*x)**(2/3)*sin((c + d*x)**(1/3)*b + a)*b**8*d**2 
*x**2 - 20160*(c + d*x)**(2/3)*sin((c + d*x)**(1/3)*b + a)*b**2 + 1440*(c 
+ d*x)**(1/3)*sin((c + d*x)**(1/3)*b + a)*b**4*c + 1680*(c + d*x)**(1/3)*s 
in((c + d*x)**(1/3)*b + a)*b**4*d*x - 18*sin((c + d*x)**(1/3)*b + a)*b**6* 
c**2 - 72*sin((c + d*x)**(1/3)*b + a)*b**6*c*d*x - 56*sin((c + d*x)**(1/3) 
*b + a)*b**6*d**2*x**2 + 40320*sin((c + d*x)**(1/3)*b + a)))/(b**9*d**3)