\(\int \frac {x}{a+b \tan (c+d \sqrt {x})} \, dx\) [38]

Optimal result

Integrand size = 18, antiderivative size = 234 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {x^2}{2 (a+i b)}+\frac {2 b x^{3/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b x \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 b \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {3 i b \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.91 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {a d^4 x^2+i b d^4 x^2+4 b d^3 x^{3/2} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+6 i b d^2 x \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+6 b d \sqrt {x} \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-3 i b \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )}{2 \left (a^2+b^2\right ) d^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4234, 3042, 4215, 2620, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 4234

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4215

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

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 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]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x/(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. 555 vs. \(2 (195) = 390\).

Time = 0.26 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.37 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/6*(6*(2*(d*sqrt(x) + c)*a/(a^2 + b^2) + 2*b*log(b*tan(d*sqrt(x) + c) + 
a)/(a^2 + b^2) - b*log(tan(d*sqrt(x) + c)^2 + 1)/(a^2 + b^2))*c^3 - (3*(d* 
sqrt(x) + c)^4*(a - I*b) - 12*(d*sqrt(x) + c)^3*(a - I*b)*c + 18*(d*sqrt(x 
) + c)^2*(a - I*b)*c^2 - 4*(4*I*(d*sqrt(x) + c)^3*b - 9*I*(d*sqrt(x) + c)^ 
2*b*c + 9*I*(d*sqrt(x) + c)*b*c^2)*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)) - 6*( 
4*I*(d*sqrt(x) + c)^2*b - 6*I*(d*sqrt(x) + c)*b*c + 3*I*b*c^2)*dilog((I*a 
+ b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 2*(4*(d*sqrt(x) + c)^3*b - 9* 
(d*sqrt(x) + c)^2*b*c + 9*(d*sqrt(x) + c)*b*c^2)*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)) + 12*I*b*polylog(4, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 
6*(4*(d*sqrt(x) + c)*b - 3*b*c)*polylog(3, (I*a + b)*e^(2*I*d*sqrt(x) + 2* 
I*c)/(-I*a + b)))/(a^2 + b^2))/d^4
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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