3.1.37 \(\int \frac {x^2}{a+b \tan (c+d \sqrt {x})} \, dx\) [37]

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

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3832, 3813, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\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 {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 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 {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 {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 {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)} \end {gather*}

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 2221

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]
- 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 2320

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 2611

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 3813

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^Si
mp[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 3832

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 6724

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]

Rule 6744

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]

Rubi steps

\begin {align*} \int \frac {x^2}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx &=2 \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 1.89, size = 303, normalized size = 0.88 \begin {gather*} \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 \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 )}{6 \left (a^2+b^2\right ) d^6} \end {gather*}

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)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)] - (
30*I)*b*d^4*x^2*PolyLog[2, -(((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b))] + 60*b*d^3*x^(3/2)*PolyLog[3, -
(((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b))] + (90*I)*b*d^2*x*PolyLog[4, -(((a - I*b)*E^((2*I)*(c + d*Sq
rt[x])))/(a + I*b))] - 90*b*d*Sqrt[x]*PolyLog[5, -(((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b))] - (45*I)*
b*PolyLog[6, -(((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b))])/(6*(a^2 + b^2)*d^6)

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Maple [F]
time = 0.82, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{a +b \tan \left (c +d \sqrt {x}\right )}\, dx\]

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] 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 = 0.68, size = 813, normalized size = 2.36 \begin {gather*} -\frac {15 \, {\left (\frac {2 \, {\left (d \sqrt {x} + c\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (d \sqrt {x} + c\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d \sqrt {x} + c\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{5} - \frac {5 \, {\left (d \sqrt {x} + c\right )}^{6} {\left (a - i \, b\right )} - 30 \, {\left (d \sqrt {x} + c\right )}^{5} {\left (a - i \, b\right )} c + 75 \, {\left (d \sqrt {x} + c\right )}^{4} {\left (a - i \, b\right )} c^{2} - 100 \, {\left (d \sqrt {x} + c\right )}^{3} {\left (a - i \, b\right )} c^{3} + 75 \, {\left (d \sqrt {x} + c\right )}^{2} {\left (a - i \, b\right )} c^{4} - 2 \, {\left (48 i \, {\left (d \sqrt {x} + c\right )}^{5} b - 150 i \, {\left (d \sqrt {x} + c\right )}^{4} b c + 200 i \, {\left (d \sqrt {x} + c\right )}^{3} b c^{2} - 150 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{3} + 75 i \, {\left (d \sqrt {x} + c\right )} b c^{4}\right )} \arctan \left (\frac {2 \, a b \cos \left (2 \, d \sqrt {x} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d \sqrt {x} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - 15 \, {\left (16 i \, {\left (d \sqrt {x} + c\right )}^{4} b - 40 i \, {\left (d \sqrt {x} + c\right )}^{3} b c + 40 i \, {\left (d \sqrt {x} + c\right )}^{2} b c^{2} - 20 i \, {\left (d \sqrt {x} + c\right )} b c^{3} + 5 i \, b c^{4}\right )} {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}\right ) + {\left (48 \, {\left (d \sqrt {x} + c\right )}^{5} b - 150 \, {\left (d \sqrt {x} + c\right )}^{4} b c + 200 \, {\left (d \sqrt {x} + c\right )}^{3} b c^{2} - 150 \, {\left (d \sqrt {x} + c\right )}^{2} b c^{3} + 75 \, {\left (d \sqrt {x} + c\right )} b c^{4}\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d \sqrt {x} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d \sqrt {x} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d \sqrt {x} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - 360 i \, b {\rm Li}_{6}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}) - 90 \, {\left (8 \, {\left (d \sqrt {x} + c\right )} b - 5 \, b c\right )} {\rm Li}_{5}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}) - 60 \, {\left (-12 i \, {\left (d \sqrt {x} + c\right )}^{2} b + 15 i \, {\left (d \sqrt {x} + c\right )} b c - 5 i \, b c^{2}\right )} {\rm Li}_{4}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b}) + 30 \, {\left (16 \, {\left (d \sqrt {x} + c\right )}^{3} b - 30 \, {\left (d \sqrt {x} + c\right )}^{2} b c + 20 \, {\left (d \sqrt {x} + c\right )} b c^{2} - 5 \, b c^{3}\right )} {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d \sqrt {x} + 2 i \, c\right )}}{-i \, a + b})}{a^{2} + b^{2}}}{15 \, d^{6}} \end {gather*}

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 - 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 + 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*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*s
qrt(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{a + b \tan {\left (c + d \sqrt {x} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )} \,d x \end {gather*}

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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