\(\int x^2 (a+b \tan (c+d \sqrt {x})) \, dx\) [26]

Optimal result

Integrand size = 18, antiderivative size = 195 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {15 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6} \] Output:

1/3*a*x^3+1/3*I*b*x^3-2*b*x^(5/2)*ln(1+exp(2*I*(c+d*x^(1/2))))/d+5*I*b*x^2 
*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^2-10*b*x^(3/2)*polylog(3,-exp(2*I*(c 
+d*x^(1/2))))/d^3-15*I*b*x*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^4+15*b*x^( 
1/2)*polylog(5,-exp(2*I*(c+d*x^(1/2))))/d^5+15/2*I*b*polylog(6,-exp(2*I*(c 
+d*x^(1/2))))/d^6
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {5 i b x^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {10 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {15 i b x \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {15 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {15 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6} \] Input:

Integrate[x^2*(a + b*Tan[c + d*Sqrt[x]]),x]
 

Output:

(a*x^3)/3 + (I/3)*b*x^3 - (2*b*x^(5/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))]) 
/d + ((5*I)*b*x^2*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 - (10*b*x^(3 
/2)*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - ((15*I)*b*x*PolyLog[4, - 
E^((2*I)*(c + d*Sqrt[x]))])/d^4 + (15*b*Sqrt[x]*PolyLog[5, -E^((2*I)*(c + 
d*Sqrt[x]))])/d^5 + (((15*I)/2)*b*PolyLog[6, -E^((2*I)*(c + d*Sqrt[x]))])/ 
d^6
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2010, 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.

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

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (a x^2+b x^2 \tan \left (c+d \sqrt {x}\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x^3}{3}+\frac {15 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {15 b \sqrt {x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 i b x \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {10 b x^{3/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {5 i b x^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{3} i b x^3\)

Input:

Int[x^2*(a + b*Tan[c + d*Sqrt[x]]),x]
 

Output:

(a*x^3)/3 + (I/3)*b*x^3 - (2*b*x^(5/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))]) 
/d + ((5*I)*b*x^2*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 - (10*b*x^(3 
/2)*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - ((15*I)*b*x*PolyLog[4, - 
E^((2*I)*(c + d*Sqrt[x]))])/d^4 + (15*b*Sqrt[x]*PolyLog[5, -E^((2*I)*(c + 
d*Sqrt[x]))])/d^5 + (((15*I)/2)*b*PolyLog[6, -E^((2*I)*(c + d*Sqrt[x]))])/ 
d^6
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

Maple [F]

\[\int x^{2} \left (a +b \tan \left (c +d \sqrt {x}\right )\right )d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(b*x^2*tan(d*sqrt(x) + c) + a*x^2, x)
 

Sympy [F]

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

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

Output:

Integral(x**2*(a + b*tan(c + d*sqrt(x))), x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 618 vs. \(2 (150) = 300\).

Time = 0.19 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.17 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/15*(5*(d*sqrt(x) + c)^6*a + 5*I*(d*sqrt(x) + c)^6*b - 30*(d*sqrt(x) + c) 
^5*a*c - 30*I*(d*sqrt(x) + c)^5*b*c + 75*(d*sqrt(x) + c)^4*a*c^2 + 75*I*(d 
*sqrt(x) + c)^4*b*c^2 - 100*(d*sqrt(x) + c)^3*a*c^3 - 100*I*(d*sqrt(x) + c 
)^3*b*c^3 + 75*(d*sqrt(x) + c)^2*a*c^4 + 75*I*(d*sqrt(x) + c)^2*b*c^4 - 30 
*(d*sqrt(x) + c)*a*c^5 - 30*b*c^5*log(sec(d*sqrt(x) + c)) + 2*(-48*I*(d*sq 
rt(x) + c)^5*b + 150*I*(d*sqrt(x) + c)^4*b*c - 200*I*(d*sqrt(x) + c)^3*b*c 
^2 + 150*I*(d*sqrt(x) + c)^2*b*c^3 - 75*I*(d*sqrt(x) + c)*b*c^4)*arctan2(s 
in(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) + 15*(16*I*(d*sqrt(x) + 
 c)^4*b - 40*I*(d*sqrt(x) + c)^3*b*c + 40*I*(d*sqrt(x) + c)^2*b*c^2 - 20*I 
*(d*sqrt(x) + c)*b*c^3 + 5*I*b*c^4)*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) - (4 
8*(d*sqrt(x) + c)^5*b - 150*(d*sqrt(x) + c)^4*b*c + 200*(d*sqrt(x) + c)^3* 
b*c^2 - 150*(d*sqrt(x) + c)^2*b*c^3 + 75*(d*sqrt(x) + c)*b*c^4)*log(cos(2* 
d*sqrt(x) + 2*c)^2 + sin(2*d*sqrt(x) + 2*c)^2 + 2*cos(2*d*sqrt(x) + 2*c) + 
 1) + 360*I*b*polylog(6, -e^(2*I*d*sqrt(x) + 2*I*c)) + 90*(8*(d*sqrt(x) + 
c)*b - 5*b*c)*polylog(5, -e^(2*I*d*sqrt(x) + 2*I*c)) + 60*(-12*I*(d*sqrt(x 
) + c)^2*b + 15*I*(d*sqrt(x) + c)*b*c - 5*I*b*c^2)*polylog(4, -e^(2*I*d*sq 
rt(x) + 2*I*c)) - 30*(16*(d*sqrt(x) + c)^3*b - 30*(d*sqrt(x) + c)^2*b*c + 
20*(d*sqrt(x) + c)*b*c^2 - 5*b*c^3)*polylog(3, -e^(2*I*d*sqrt(x) + 2*I*c)) 
)/d^6
 

Giac [F]

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

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

Output:

integrate((b*tan(d*sqrt(x) + c) + a)*x^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(3*int(tan(sqrt(x)*d + c)*x**2,x)*b + a*x**3)/3