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

Optimal. Leaf size=344 \[ -\frac{5 i b x^2 \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac{10 b x^{3/2} \text{PolyLog}\left (3,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}+\frac{15 i b x \text{PolyLog}\left (4,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^4 \left (a^2+b^2\right )}-\frac{15 b \sqrt{x} \text{PolyLog}\left (5,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^5 \left (a^2+b^2\right )}-\frac{15 i b \text{PolyLog}\left (6,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{2 d^6 \left (a^2+b^2\right )}+\frac{2 b x^{5/2} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x^3}{3 (a+i b)} \]

[Out]

x^3/(3*(a + I*b)) + (2*b*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)*b*x^2*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^2) + (10*b*
x^(3/2)*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^3) + ((15*I)*b*x*Po
lyLog[4, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^4) - (15*b*Sqrt[x]*PolyLog[5,
-(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^5) - (((15*I)/2)*b*PolyLog[6, -(((a^2
+ b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^6)

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Rubi [A]  time = 0.453081, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3747, 3732, 2190, 2531, 6609, 2282, 6589} \[ -\frac{5 i b x^2 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac{10 b x^{3/2} \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}+\frac{15 i b x \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^4 \left (a^2+b^2\right )}-\frac{15 b \sqrt{x} \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d^5 \left (a^2+b^2\right )}-\frac{15 i b \text{Li}_6\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{2 d^6 \left (a^2+b^2\right )}+\frac{2 b x^{5/2} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x^3}{3 (a+i b)} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

x^3/(3*(a + I*b)) + (2*b*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)*b*x^2*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^2) + (10*b*
x^(3/2)*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^3) + ((15*I)*b*x*Po
lyLog[4, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^4) - (15*b*Sqrt[x]*PolyLog[5,
-(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^5) - (((15*I)/2)*b*PolyLog[6, -(((a^2
+ b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^6)

Rule 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2531

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] + Dist[(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 6609

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] - Dist[(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]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rubi steps

\begin{align*} \int \frac{x^2}{a+b \tan \left (c+d \sqrt{x}\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^5}{a+b \tan (c+d x)} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^3}{3 (a+i b)}+(4 i b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^5}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )\\ &=\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{(10 b) \operatorname{Subst}\left (\int x^4 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d}\\ &=\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 \text{Li}_2\left (-\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{(20 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d^2}\\ &=\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 \text{Li}_2\left (-\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} \text{Li}_3\left (-\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{(30 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d^3}\\ &=\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 \text{Li}_2\left (-\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} \text{Li}_3\left (-\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 \text{Li}_4\left (-\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{(30 i b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d^4}\\ &=\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 \text{Li}_2\left (-\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} \text{Li}_3\left (-\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 \text{Li}_4\left (-\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} \text{Li}_5\left (-\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 b) \operatorname{Subst}\left (\int \text{Li}_5\left (-\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt{x}\right )}{\left (a^2+b^2\right ) d^5}\\ &=\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 \text{Li}_2\left (-\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} \text{Li}_3\left (-\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 \text{Li}_4\left (-\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} \text{Li}_5\left (-\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{Subst}\left (\int \frac{\text{Li}_5\left (-\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{2 \left (a^2+b^2\right ) d^6}\\ &=\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 \text{Li}_2\left (-\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} \text{Li}_3\left (-\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 \text{Li}_4\left (-\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} \text{Li}_5\left (-\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 \text{Li}_6\left (-\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}\\ \end{align*}

Mathematica [A]  time = 1.12991, size = 308, normalized size = 0.9 \[ \frac{30 i b d^4 x^2 \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )+60 b d^3 x^{3/2} \text{PolyLog}\left (3,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )-90 i b d^2 x \text{PolyLog}\left (4,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )-90 b d \sqrt{x} \text{PolyLog}\left (5,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )+45 i b \text{PolyLog}\left (6,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )+12 b d^5 x^{5/2} \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d \sqrt{x}\right )}}{a-i b}\right )+2 a d^6 x^3+2 i b d^6 x^3}{6 d^6 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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*PolyLo
g[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))])/(6*(a^2 + b^2)*d^6)

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Maple [F]  time = 0.166, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ( a+b\tan \left ( c+d\sqrt{x} \right ) \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 3.4792, size = 1094, normalized size = 3.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*sq
rt(x) + c)^4*(a - I*b)*c^2 - 100*(d*sqrt(x) + c)^3*(a - I*b)*c^3 + 75*(d*sqrt(x) + c)^2*(a - I*b)*c^4 + (-96*I
*(d*sqrt(x) + c)^5*b + 300*I*(d*sqrt(x) + c)^4*b*c - 400*I*(d*sqrt(x) + c)^3*b*c^2 + 300*I*(d*sqrt(x) + c)^2*b
*c^3 - 150*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)) +
(-240*I*(d*sqrt(x) + c)^4*b + 600*I*(d*sqrt(x) + c)^3*b*c - 600*I*(d*sqrt(x) + c)^2*b*c^2 + 300*I*(d*sqrt(x) +
c)*b*c^3 - 75*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 + 75*(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*sqr
t(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)) + (720*I*(d*sqrt(x) + c)^2*b - 900*I*(d*sqrt(x) + c)*b*c + 300*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) + c)^3*b - 30*(d*sqrt(x) + c)^2*b*c + 20*(d*sqrt(x) + c)*b*
c^2 - 5*b*c^3)*polylog(3, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)))/(a^2 + b^2))/d^6

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{b \tan \left (d \sqrt{x} + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{b \tan \left (d \sqrt{x} + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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