\(\int x (a+b \tan (c+d \sqrt {x}))^2 \, dx\) [32]

Optimal result

Integrand size = 18, antiderivative size = 274 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}+i a b x^2-\frac {b^2 x^2}{2}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 i a b x \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 a b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {3 i a b \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 1.53 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.33 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {b \left (4 i b d^3 x^{3/2}-2 i a d^4 x^2+6 b d^2 x \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )+6 b d^2 e^{2 i c} x \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-4 a d^3 x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-4 a d^3 e^{2 i c} x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-6 i d \left (1+e^{2 i c}\right ) \left (-b+a d \sqrt {x}\right ) \sqrt {x} \operatorname {PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 \left (1+e^{2 i c}\right ) \left (b-2 a d \sqrt {x}\right ) \operatorname {PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 i a \operatorname {PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )+3 i a e^{2 i c} \operatorname {PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt {x}\right )}\right )\right )}{d^4 \left (1+e^{2 i c}\right )}+\frac {2 b^2 x^{3/2} \sec (c) \sec \left (c+d \sqrt {x}\right ) \sin \left (d \sqrt {x}\right )}{d}+\frac {1}{2} x^2 \left (a^2-b^2+2 a b \tan (c)\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4234, 3042, 4205, 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.

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

\(\Big \downarrow \) 4234

\(\displaystyle 2 \int x^{3/2} \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int x^{3/2} \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\)

\(\Big \downarrow \) 4205

\(\displaystyle 2 \int \left (x^{3/2} a^2+2 b x^{3/2} \tan \left (c+d \sqrt {x}\right ) a+b^2 x^{3/2} \tan ^2\left (c+d \sqrt {x}\right )\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4205

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

rule 4234

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x*(a + b*tan(c + d*sqrt(x)))**2, 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. 1290 vs. \(2 (218) = 436\).

Time = 0.22 (sec) , antiderivative size = 1290, normalized size of antiderivative = 4.71 \[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((d*sqrt(x) + c)^4*a^2 - 4*(d*sqrt(x) + c)^3*a^2*c + 6*(d*sqrt(x) + c) 
^2*a^2*c^2 - 4*(d*sqrt(x) + c)*a^2*c^3 - 8*a*b*c^3*log(sec(d*sqrt(x) + c)) 
 - 4*(12*I*(d*sqrt(x) + c)*b^2*c^3 - 3*(2*a*b + I*b^2)*(d*sqrt(x) + c)^4 + 
 12*(2*a*b + I*b^2)*(d*sqrt(x) + c)^3*c - 18*(2*a*b + I*b^2)*(d*sqrt(x) + 
c)^2*c^2 + 24*b^2*c^3 + 4*(8*(d*sqrt(x) + c)^3*a*b - 9*b^2*c^2 - 9*(2*a*b* 
c + b^2)*(d*sqrt(x) + c)^2 + 18*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c) + (8*(d* 
sqrt(x) + c)^3*a*b - 9*b^2*c^2 - 9*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 18* 
(a*b*c^2 + b^2*c)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (-8*I*(d*sqrt( 
x) + c)^3*a*b + 9*I*b^2*c^2 + 9*(2*I*a*b*c + I*b^2)*(d*sqrt(x) + c)^2 + 18 
*(-I*a*b*c^2 - I*b^2*c)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(s 
in(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) - 3*((2*a*b + I*b^2)*(d 
*sqrt(x) + c)^4 - 4*(2*b^2 + (2*a*b + I*b^2)*c)*(d*sqrt(x) + c)^3 + 6*(4*b 
^2*c + (2*a*b + I*b^2)*c^2)*(d*sqrt(x) + c)^2 + 4*(-I*b^2*c^3 - 6*b^2*c^2) 
*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - 12*(4*(d*sqrt(x) + c)^2*a*b + 3 
*a*b*c^2 + 3*b^2*c - 3*(2*a*b*c + b^2)*(d*sqrt(x) + c) + (4*(d*sqrt(x) + c 
)^2*a*b + 3*a*b*c^2 + 3*b^2*c - 3*(2*a*b*c + b^2)*(d*sqrt(x) + c))*cos(2*d 
*sqrt(x) + 2*c) + (4*I*(d*sqrt(x) + c)^2*a*b + 3*I*a*b*c^2 + 3*I*b^2*c + 3 
*(-2*I*a*b*c - I*b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(2 
*I*d*sqrt(x) + 2*I*c)) - 2*(8*I*(d*sqrt(x) + c)^3*a*b - 9*I*b^2*c^2 + 9*(- 
2*I*a*b*c - I*b^2)*(d*sqrt(x) + c)^2 + 18*(I*a*b*c^2 + I*b^2*c)*(d*sqrt...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {4 \sqrt {x}\, \tan \left (\sqrt {x}\, d +c \right ) b^{2} x -6 \left (\int \sqrt {x}\, \tan \left (\sqrt {x}\, d +c \right )d x \right ) b^{2}+4 \left (\int \tan \left (\sqrt {x}\, d +c \right ) x d x \right ) a b d +a^{2} d \,x^{2}-b^{2} d \,x^{2}}{2 d} \] Input:

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

Output:

(4*sqrt(x)*tan(sqrt(x)*d + c)*b**2*x - 6*int(sqrt(x)*tan(sqrt(x)*d + c),x) 
*b**2 + 4*int(tan(sqrt(x)*d + c)*x,x)*a*b*d + a**2*d*x**2 - b**2*d*x**2)/( 
2*d)