Integrand size = 20, antiderivative size = 344 \[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {x^3}{3 (a+i b)}+\frac {2 b x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {5 i b x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {10 b x^{3/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {15 i b x \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^4}-\frac {15 b \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^5}-\frac {15 i b \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^6} \]
1/3*x^3/(a+I*b)+2*b*x^(5/2)*ln(1+(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^ 2)/(a^2+b^2)/d-5*I*b*x^2*polylog(2,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I* b)^2)/(a^2+b^2)/d^2+10*b*x^(3/2)*polylog(3,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2) ))/(a+I*b)^2)/(a^2+b^2)/d^3+15*I*b*x*polylog(4,-(a^2+b^2)*exp(2*I*(c+d*x^( 1/2)))/(a+I*b)^2)/(a^2+b^2)/d^4-15/2*I*b*polylog(6,-(a^2+b^2)*exp(2*I*(c+d *x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^6-15*b*polylog(5,-(a^2+b^2)*exp(2*I*(c+d *x^(1/2)))/(a+I*b)^2)*x^(1/2)/(a^2+b^2)/d^5
Time = 1.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.90 \[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {2 a d^6 x^3+2 i b d^6 x^3+12 b d^5 x^{5/2} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+30 i b d^4 x^2 \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+60 b d^3 x^{3/2} \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-90 i b d^2 x \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-90 b d \sqrt {x} \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+45 i b \operatorname {PolyLog}\left (6,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )}{6 \left (a^2+b^2\right ) d^6} \]
(2*a*d^6*x^3 + (2*I)*b*d^6*x^3 + 12*b*d^5*x^(5/2)*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + (30*I)*b*d^4*x^2*PolyLog[2, (-a - I*b)/ ((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + 60*b*d^3*x^(3/2)*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] - (90*I)*b*d^2*x*PolyLog[4, ( -a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] - 90*b*d*Sqrt[x]*PolyLog[ 5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + (45*I)*b*PolyLog[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))])/(6*(a^2 + b^2)*d^6)
Time = 1.40 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4234, 3042, 4215, 2620, 3011, 7163, 7163, 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.
\(\displaystyle \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx\) |
\(\Big \downarrow \) 4234 |
\(\displaystyle 2 \int \frac {x^{5/2}}{a+b \tan \left (c+d \sqrt {x}\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \frac {x^{5/2}}{a+b \tan \left (c+d \sqrt {x}\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 4215 |
\(\displaystyle 2 \left (2 i b \int \frac {e^{2 i \left (c+d \sqrt {x}\right )} x^{5/2}}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}d\sqrt {x}+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \int x^2 \log \left (\frac {e^{2 i \left (c+d \sqrt {x}\right )} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )d\sqrt {x}}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \int x^{3/2} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \int x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \int \sqrt {x} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{d}-\frac {i x \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{2 d}-\frac {i \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {\int \frac {\operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\sqrt {x}}de^{2 i \left (c+d \sqrt {x}\right )}}{4 d^2}-\frac {i \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle 2 \left (2 i b \left (\frac {5 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {\operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{4 d^2}-\frac {i \sqrt {x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^3}{6 (a+i b)}\right )\) |
2*(x^3/(6*(a + I*b)) + (2*I)*b*(((-1/2*I)*x^(5/2)*Log[1 + ((a^2 + b^2)*E^( (2*I)*(c + d*Sqrt[x])))/(a + I*b)^2])/((a^2 + b^2)*d) + (((5*I)/2)*(((I/2) *x^2*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d - ((2*I)*(((-1/2*I)*x^(3/2)*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqr t[x])))/(a + I*b)^2)])/d + (((3*I)/2)*(((-1/2*I)*x*PolyLog[4, -(((a^2 + b^ 2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d + (I*(((-1/2*I)*Sqrt[x]*Pol yLog[5, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d + PolyL og[6, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)]/(4*d^2)))/d) )/d))/d))/((a^2 + b^2)*d)))
3.1.37.3.1 Defintions of rubi rules used
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]
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]]]
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]
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]
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]
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]
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]
\[\int \frac {x^{2}}{a +b \tan \left (c +d \sqrt {x}\right )}d x\]
\[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{2}}{b \tan \left (d \sqrt {x} + c\right ) + a} \,d x } \]
\[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^{2}}{a + b \tan {\left (c + d \sqrt {x} \right )}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 813 vs. \(2 (289) = 578\).
Time = 0.53 (sec) , antiderivative size = 813, normalized size of antiderivative = 2.36 \[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\text {Too large to display} \]
-1/15*(15*(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^5 - (5*( d*sqrt(x) + c)^6*(a - I*b) - 30*(d*sqrt(x) + c)^5*(a - I*b)*c + 75*(d*sqrt (x) + c)^4*(a - I*b)*c^2 - 100*(d*sqrt(x) + c)^3*(a - I*b)*c^3 + 75*(d*sqr t(x) + c)^2*(a - I*b)*c^4 - 2*(48*I*(d*sqrt(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((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)) - 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((I*a + b)*e^(2*I* d*sqrt(x) + 2*I*c)/(-I*a + b)) + (48*(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 + 7 5*(d*sqrt(x) + c)*b*c^4)*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)) - 360*I*b*polylog(6, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) - 90*(8*(d*sqrt(x) + c)*b - 5*b*c)*polylog(5, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) - 60* (-12*I*(d*sqrt(x) + c)^2*b + 15*I*(d*sqrt(x) + c)*b*c - 5*I*b*c^2)*polylog (4, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 30*(16*(d*sqrt(x)...
\[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{2}}{b \tan \left (d \sqrt {x} + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^2}{a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )} \,d x \]