\(\int \frac {1}{(a+b \tan (c+d \sqrt {x}))^2} \, dx\) [44]

Optimal result

Integrand size = 16, antiderivative size = 204 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {\left (b+2 a d \sqrt {x}\right )^2}{2 a (a+i b) \left (a^2+b^2\right ) d^2}-\frac {x}{a^2+b^2}+\frac {2 b \left (b+2 a d \sqrt {x}\right ) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {2 i a b \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {2 b \sqrt {x}}{\left (a^2+b^2\right ) d \left (a+b \tan \left (c+d \sqrt {x}\right )\right )} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(517\) vs. \(2(204)=408\).

Time = 5.31 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.53 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {\sec ^2\left (c+d \sqrt {x}\right ) \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right ) \left (2 b^2 \left (a^2+b^2\right ) d \sqrt {x} \sin \left (c+d \sqrt {x}\right )-a \left (a^2+b^2\right ) \left (c-d \sqrt {x}\right ) \left (c+d \sqrt {x}\right ) \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )-2 b^2 \left (b \left (c+d \sqrt {x}\right )-a \log \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )\right ) \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )+4 a b c \left (b \left (c+d \sqrt {x}\right )-a \log \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )\right ) \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )-2 a b \left (\sqrt {1+\frac {a^2}{b^2}} b e^{i \arctan \left (\frac {a}{b}\right )} \left (c+d \sqrt {x}\right )^2+a \left (-i \left (c+d \sqrt {x}\right ) \left (\pi -2 \arctan \left (\frac {a}{b}\right )\right )-\pi \log \left (1+e^{-2 i \left (c+d \sqrt {x}\right )}\right )-2 \left (c+d \sqrt {x}+\arctan \left (\frac {a}{b}\right )\right ) \log \left (1-e^{2 i \left (c+d \sqrt {x}+\arctan \left (\frac {a}{b}\right )\right )}\right )+\pi \log \left (\cos \left (c+d \sqrt {x}\right )\right )+2 \arctan \left (\frac {a}{b}\right ) \log \left (\sin \left (c+d \sqrt {x}+\arctan \left (\frac {a}{b}\right )\right )\right )+i \operatorname {PolyLog}\left (2,e^{2 i \left (c+d \sqrt {x}+\arctan \left (\frac {a}{b}\right )\right )}\right )\right )\right ) \left (a \cos \left (c+d \sqrt {x}\right )+b \sin \left (c+d \sqrt {x}\right )\right )\right )}{a \left (a^2+b^2\right )^2 d^2 \left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \] Input:

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

Output:

(Sec[c + d*Sqrt[x]]^2*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]])*(2*b^2 
*(a^2 + b^2)*d*Sqrt[x]*Sin[c + d*Sqrt[x]] - a*(a^2 + b^2)*(c - d*Sqrt[x])* 
(c + d*Sqrt[x])*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]) - 2*b^2*(b*( 
c + d*Sqrt[x]) - a*Log[a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]])*(a*Co 
s[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]) + 4*a*b*c*(b*(c + d*Sqrt[x]) - a* 
Log[a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]])*(a*Cos[c + d*Sqrt[x]] + 
b*Sin[c + d*Sqrt[x]]) - 2*a*b*(Sqrt[1 + a^2/b^2]*b*E^(I*ArcTan[a/b])*(c + 
d*Sqrt[x])^2 + a*((-I)*(c + d*Sqrt[x])*(Pi - 2*ArcTan[a/b]) - Pi*Log[1 + E 
^((-2*I)*(c + d*Sqrt[x]))] - 2*(c + d*Sqrt[x] + ArcTan[a/b])*Log[1 - E^((2 
*I)*(c + d*Sqrt[x] + ArcTan[a/b]))] + Pi*Log[Cos[c + d*Sqrt[x]]] + 2*ArcTa 
n[a/b]*Log[Sin[c + d*Sqrt[x] + ArcTan[a/b]]] + I*PolyLog[2, E^((2*I)*(c + 
d*Sqrt[x] + ArcTan[a/b]))]))*(a*Cos[c + d*Sqrt[x]] + b*Sin[c + d*Sqrt[x]]) 
))/(a*(a^2 + b^2)^2*d^2*(a + b*Tan[c + d*Sqrt[x]])^2)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4226, 3042, 4216, 3042, 4215, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx\)

\(\Big \downarrow \) 4226

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4215

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 4215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp[2*I*b   In 
t[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Simp[2 
*I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2 
, 0] && IGtQ[m, 0]
 

rule 4216

Int[((c_.) + (d_.)*(x_))/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol 
] :> Simp[-(c + d*x)^2/(2*d*(a^2 + b^2)), x] + (Simp[1/(f*(a^2 + b^2))   In 
t[(b*d + 2*a*c*f + 2*a*d*f*x)/(a + b*Tan[e + f*x]), x], x] - Simp[b*((c + d 
*x)/(f*(a^2 + b^2)*(a + b*Tan[e + f*x]))), x]) /; FreeQ[{a, b, c, d, e, f}, 
 x] && NeQ[a^2 + b^2, 0]
 

rule 4226

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

Maple [F]

\[\int \frac {1}{\left (a +b \tan \left (c +d \sqrt {x}\right )\right )^{2}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 828 vs. \(2 (177) = 354\).

Time = 0.11 (sec) , antiderivative size = 828, normalized size of antiderivative = 4.06 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-(2*b^3*d*sqrt(x) - (a^3 - a*b^2)*d^2*x + (a^3 - a*b^2)*d^2 - (I*a*b^2*tan 
(d*sqrt(x) + c) + I*a^2*b)*dilog(2*((I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - a 
^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan( 
d*sqrt(x) + c)^2 + a^2 + b^2) + 1) - (-I*a*b^2*tan(d*sqrt(x) + c) - I*a^2* 
b)*dilog(2*((-I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - a^2 + I*a*b + (-I*a^2 - 
2*a*b + I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sqrt(x) + c)^2 + a^2 
 + b^2) + 1) - 2*(a^2*b*d*sqrt(x) + a^2*b*c + (a*b^2*d*sqrt(x) + a*b^2*c)* 
tan(d*sqrt(x) + c))*log(-2*((I*a*b - b^2)*tan(d*sqrt(x) + c)^2 - a^2 - I*a 
*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)*tan(d*sqrt(x 
) + c)^2 + a^2 + b^2)) - 2*(a^2*b*d*sqrt(x) + a^2*b*c + (a*b^2*d*sqrt(x) + 
 a*b^2*c)*tan(d*sqrt(x) + c))*log(-2*((-I*a*b - b^2)*tan(d*sqrt(x) + c)^2 
- a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*sqrt(x) + c))/((a^2 + b^2)* 
tan(d*sqrt(x) + c)^2 + a^2 + b^2)) + (2*a^2*b*c - a*b^2 + (2*a*b^2*c - b^3 
)*tan(d*sqrt(x) + c))*log(((I*a*b + b^2)*tan(d*sqrt(x) + c)^2 - a^2 + I*a* 
b + (I*a^2 + I*b^2)*tan(d*sqrt(x) + c))/(tan(d*sqrt(x) + c)^2 + 1)) + (2*a 
^2*b*c - a*b^2 + (2*a*b^2*c - b^3)*tan(d*sqrt(x) + c))*log(((I*a*b - b^2)* 
tan(d*sqrt(x) + c)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*sqrt(x) + c))/( 
tan(d*sqrt(x) + c)^2 + 1)) - (2*a*b^2*d*sqrt(x) + (a^2*b - b^3)*d^2*x - (a 
^2*b - b^3)*d^2)*tan(d*sqrt(x) + c))/((a^4*b + 2*a^2*b^3 + b^5)*d^2*tan(d* 
sqrt(x) + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d^2)
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d \sqrt {x} \right )}\right )^{2}}\, dx \] Input:

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

Output:

Integral((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. 994 vs. \(2 (177) = 354\).

Time = 0.34 (sec) , antiderivative size = 994, normalized size of antiderivative = 4.87 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((a^3 - I*a^2*b + a*b^2 - I*b^3)*d^2*x - 2*(-I*a*b^2 + b^3 + (-I*a*b^2 - b 
^3)*cos(2*d*sqrt(x) + 2*c) + (a*b^2 - I*b^3)*sin(2*d*sqrt(x) + 2*c))*arcta 
n2(-b*cos(2*d*sqrt(x) + 2*c) + a*sin(2*d*sqrt(x) + 2*c) + b, a*cos(2*d*sqr 
t(x) + 2*c) + b*sin(2*d*sqrt(x) + 2*c) + a) - 4*((I*a^2*b + a*b^2)*d*sqrt( 
x)*cos(2*d*sqrt(x) + 2*c) - (a^2*b - I*a*b^2)*d*sqrt(x)*sin(2*d*sqrt(x) + 
2*c) + (I*a^2*b - a*b^2)*d*sqrt(x))*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)) + (( 
a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*d^2*x - 4*(I*a*b^2 + b^3)*d*sqrt(x))*co 
s(2*d*sqrt(x) + 2*c) - 2*(I*a^2*b - a*b^2 + (I*a^2*b + a*b^2)*cos(2*d*sqrt 
(x) + 2*c) - (a^2*b - I*a*b^2)*sin(2*d*sqrt(x) + 2*c))*dilog((I*a + b)*e^( 
2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + (a*b^2 + I*b^3 + (a*b^2 - I*b^3)*cos( 
2*d*sqrt(x) + 2*c) + (I*a*b^2 + b^3)*sin(2*d*sqrt(x) + 2*c))*log((a^2 + b^ 
2)*cos(2*d*sqrt(x) + 2*c)^2 + 4*a*b*sin(2*d*sqrt(x) + 2*c) + (a^2 + b^2)*s 
in(2*d*sqrt(x) + 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*sqrt(x) + 2*c) 
) + 2*((a^2*b - I*a*b^2)*d*sqrt(x)*cos(2*d*sqrt(x) + 2*c) - (-I*a^2*b - a* 
b^2)*d*sqrt(x)*sin(2*d*sqrt(x) + 2*c) + (a^2*b + I*a*b^2)*d*sqrt(x))*log(( 
(a^2 + b^2)*cos(2*d*sqrt(x) + 2*c)^2 + 4*a*b*sin(2*d*sqrt(x) + 2*c) + (a^2 
 + b^2)*sin(2*d*sqrt(x) + 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*sqrt( 
x) + 2*c))/(a^2 + b^2)) + ((I*a^3 + 3*a^2*b - 3*I*a*b^2 - b^3)*d^2*x + ...
 

Giac [F]

\[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int { \frac {1}{{\left (b \tan \left (d \sqrt {x} + c\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {1}{\tan \left (\sqrt {x}\, d +c \right )^{2} b^{2}+2 \tan \left (\sqrt {x}\, d +c \right ) a b +a^{2}}d x \] Input:

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

Output:

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