\(\int \frac {x}{(a+b \tan (c+d \sqrt {x}))^2} \, dx\) [43]

Optimal result

Integrand size = 18, antiderivative size = 787 \[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=-\frac {4 i b^2 x^{3/2}}{\left (a^2+b^2\right )^2 d}+\frac {4 b^2 x^{3/2}}{(a+i b) (i a+b)^2 d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt {x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {6 b^2 x \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {4 b x^{3/2} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {4 i b^2 x^{3/2} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {6 b x \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {6 b^2 x \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {3 b^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}+\frac {6 b \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {6 i b^2 \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}-\frac {3 b \operatorname {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {3 b^2 \operatorname {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt {x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.46 (sec) , antiderivative size = 662, normalized size of antiderivative = 0.84 \[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {-\frac {2 i b \left (4 (a+i b) b (i a+b) d^3 x^{3/2}+2 a (a+i b) (i a+b) d^4 x^2+6 (a-i b) b d^2 \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) x \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+4 a (a-i b) d^3 \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) x^{3/2} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+3 (a-i b) b \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \left (2 i d \sqrt {x} \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+\operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )\right )+3 a (a-i b) \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \left (2 i d^2 x \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+2 d \sqrt {x} \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-i \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )\right )\right )}{d^4 \left (b-b e^{2 i c}-i a \left (1+e^{2 i c}\right )\right )}+\frac {(a-i b)^2 (a+i b) x^2 (a \cos (c)-b \sin (c))}{a \cos (c)+b \sin (c)}+\frac {4 (a-i b)^2 (a+i b) b^2 x^{3/2} \sin \left (d \sqrt {x}\right )}{d (a \cos (c)+b \sin (c)) \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )}}{2 (a-i b)^3 (a+i b)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.93 (sec) , antiderivative size = 830, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4234, 3042, 4217, 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.

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

\(\Big \downarrow \) 4234

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4217

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 4217

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(a - I*b) - 2*I*(b/(a^2 + 
 b^2 + (a - I*b)^2*E^(2*I*(e + f*x)))))^(-n), x], x] /; FreeQ[{a, b, c, d, 
e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[n, 0] && IGtQ[m, 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]

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

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]

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

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

Output:

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

Sympy [F]

\[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {x}{\left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )^{2}}\, dx \] 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. 2477 vs. \(2 (638) = 1276\).

Time = 0.57 (sec) , antiderivative size = 2477, normalized size of antiderivative = 3.15 \[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/6*(12*(2*a*b*log(b*tan(d*sqrt(x) + c) + a)/(a^4 + 2*a^2*b^2 + b^4) - a* 
b*log(tan(d*sqrt(x) + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + (a^2 - b^2)*(d*s 
qrt(x) + c)/(a^4 + 2*a^2*b^2 + b^4) - b/(a^3 + a*b^2 + (a^2*b + b^3)*tan(d 
*sqrt(x) + c)))*c^3 - (3*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*sqrt(x) + c)^4 
 - 12*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*sqrt(x) + c)^3*c + 18*(a^3 - I*a^ 
2*b + a*b^2 - I*b^3)*(d*sqrt(x) + c)^2*c^2 - 36*((-I*a*b^2 - b^3)*c^2*cos( 
2*d*sqrt(x) + 2*c) + (a*b^2 - I*b^3)*c^2*sin(2*d*sqrt(x) + 2*c) + (-I*a*b^ 
2 + b^3)*c^2)*arctan2(-b*cos(2*d*sqrt(x) + 2*c) + a*sin(2*d*sqrt(x) + 2*c) 
 + b, a*cos(2*d*sqrt(x) + 2*c) + b*sin(2*d*sqrt(x) + 2*c) + a) - 4*(8*(I*a 
^2*b - a*b^2)*(d*sqrt(x) + c)^3 + 9*(I*a*b^2 - b^3 + 2*(-I*a^2*b + a*b^2)* 
c)*(d*sqrt(x) + c)^2 + 18*((I*a^2*b - a*b^2)*c^2 + (-I*a*b^2 + b^3)*c)*(d* 
sqrt(x) + c) + (8*(I*a^2*b + a*b^2)*(d*sqrt(x) + c)^3 + 9*(I*a*b^2 + b^3 + 
 2*(-I*a^2*b - a*b^2)*c)*(d*sqrt(x) + c)^2 + 18*((I*a^2*b + a*b^2)*c^2 + ( 
-I*a*b^2 - b^3)*c)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (8*(a^2*b - I 
*a*b^2)*(d*sqrt(x) + c)^3 + 9*(a*b^2 - I*b^3 - 2*(a^2*b - I*a*b^2)*c)*(d*s 
qrt(x) + c)^2 + 18*((a^2*b - I*a*b^2)*c^2 - (a*b^2 - I*b^3)*c)*(d*sqrt(x) 
+ c))*sin(2*d*sqrt(x) + 2*c))*arctan2((2*a*b*cos(2*d*sqrt(x) + 2*c) - (a^2 
 - b^2)*sin(2*d*sqrt(x) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*sqrt(x) + 2*c) 
 + a^2 + b^2 + (a^2 - b^2)*cos(2*d*sqrt(x) + 2*c))/(a^2 + b^2)) + 3*((a^3 
- 3*I*a^2*b - 3*a*b^2 + I*b^3)*(d*sqrt(x) + c)^4 - 4*(2*I*a*b^2 + 2*b^3...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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