\(\int \frac {x^3}{a+b \tan (c+d \sqrt {x})} \, dx\) [36]

Optimal result

Integrand size = 20, antiderivative size = 460 \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {x^4}{4 (a+i b)}+\frac {2 b x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^5}-\frac {315 i b x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^6}+\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^7}+\frac {315 i b \operatorname {PolyLog}\left (8,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^8} \]

[Out]

1/4*x^4/(a+I*b)+2*b*x^(7/2)*ln(1+(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d-7*I*b*x^3*polylog(2,-
(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^2+21*b*x^(5/2)*polylog(3,-(a^2+b^2)*exp(2*I*(c+d*x^(1/
2)))/(a+I*b)^2)/(a^2+b^2)/d^3+105/2*I*b*x^2*polylog(4,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d
^4-105*b*x^(3/2)*polylog(5,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^5-315/2*I*b*x*polylog(6,-(
a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^6+315/4*I*b*polylog(8,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))
/(a+I*b)^2)/(a^2+b^2)/d^8+315/2*b*polylog(7,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)*x^(1/2)/(a^2+b^2)/d^7

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3832, 3813, 2221, 2611, 6744, 2320, 6724} \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {315 i b \operatorname {PolyLog}\left (8,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{4 d^8 \left (a^2+b^2\right )}+\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d^7 \left (a^2+b^2\right )}-\frac {315 i b x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d^6 \left (a^2+b^2\right )}-\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^5 \left (a^2+b^2\right )}+\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x^4}{4 (a+i b)} \]

[In]

Int[x^3/(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

x^4/(4*(a + I*b)) + (2*b*x^(7/2)*Log[1 + ((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2])/((a^2 + b^2)*d)
- ((7*I)*b*x^3*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^2) + (21*b*
x^(5/2)*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^3) + (((105*I)/2)*b
*x^2*PolyLog[4, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^4) - (105*b*x^(3/2)*Po
lyLog[5, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^5) - (((315*I)/2)*b*x*PolyLog
[6, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^6) + (315*b*Sqrt[x]*PolyLog[7, -((
(a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/(2*(a^2 + b^2)*d^7) + (((315*I)/4)*b*PolyLog[8, -(((a^2
+ b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/((a^2 + b^2)*d^8)

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*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^7}{a+b \tan (c+d x)} \, dx,x,\sqrt {x}\right ) \\ & = \frac {x^4}{4 (a+i b)}+(4 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^7}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {x^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {(14 b) \text {Subst}\left (\int x^6 \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^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {(42 i b) \text {Subst}\left (\int x^5 \operatorname {PolyLog}\left (2,-\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^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {(105 b) \text {Subst}\left (\int x^4 \operatorname {PolyLog}\left (3,-\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^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {105 i b x^2 \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}-\frac {(210 i b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (4,-\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^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {105 i b x^2 \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}-\frac {105 b x^{3/2} \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 {(315 b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (5,-\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^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {105 i b x^2 \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}-\frac {105 b x^{3/2} \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 {315 i b x \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}+\frac {(315 i b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (6,-\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^6} \\ & = \frac {x^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {105 i b x^2 \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}-\frac {105 b x^{3/2} \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 {315 i b x \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}+\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-\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^7}-\frac {(315 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt {x}\right )}{2 \left (a^2+b^2\right ) d^7} \\ & = \frac {x^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {105 i b x^2 \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}-\frac {105 b x^{3/2} \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 {315 i b x \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}+\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-\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^7}+\frac {(315 i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (7,-\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 )}{4 \left (a^2+b^2\right ) d^8} \\ & = \frac {x^4}{4 (a+i b)}+\frac {2 b x^{7/2} \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 {7 i b x^3 \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 {21 b x^{5/2} \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 {105 i b x^2 \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}-\frac {105 b x^{3/2} \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 {315 i b x \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}+\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-\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^7}+\frac {315 i b \operatorname {PolyLog}\left (8,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^8} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 401, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {a d^8 x^4+i b d^8 x^4+8 b d^7 x^{7/2} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+28 i b d^6 x^3 \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+84 b d^5 x^{5/2} \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-210 i b d^4 x^2 \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-420 b d^3 x^{3/2} \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+630 i b d^2 x \operatorname {PolyLog}\left (6,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+630 b d \sqrt {x} \operatorname {PolyLog}\left (7,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-315 i b \operatorname {PolyLog}\left (8,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )}{4 \left (a^2+b^2\right ) d^8} \]

[In]

Integrate[x^3/(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

(a*d^8*x^4 + I*b*d^8*x^4 + 8*b*d^7*x^(7/2)*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + (28*I)*b
*d^6*x^3*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + 84*b*d^5*x^(5/2)*PolyLog[3, (-a - I*b)
/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] - (210*I)*b*d^4*x^2*PolyLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*S
qrt[x])))] - 420*b*d^3*x^(3/2)*PolyLog[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + (630*I)*b*d^2*x*
PolyLog[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + 630*b*d*Sqrt[x]*PolyLog[7, (-a - I*b)/((a - I*b
)*E^((2*I)*(c + d*Sqrt[x])))] - (315*I)*b*PolyLog[8, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))])/(4*(a^
2 + b^2)*d^8)

Maple [F]

\[\int \frac {x^{3}}{a +b \tan \left (c +d \sqrt {x}\right )}d x\]

[In]

int(x^3/(a+b*tan(c+d*x^(1/2))),x)

[Out]

int(x^3/(a+b*tan(c+d*x^(1/2))),x)

Fricas [F]

\[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{3}}{b \tan \left (d \sqrt {x} + c\right ) + a} \,d x } \]

[In]

integrate(x^3/(a+b*tan(c+d*x^(1/2))),x, algorithm="fricas")

[Out]

integral(x^3/(b*tan(d*sqrt(x) + c) + a), x)

Sympy [F]

\[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^{3}}{a + b \tan {\left (c + d \sqrt {x} \right )}}\, dx \]

[In]

integrate(x**3/(a+b*tan(c+d*x**(1/2))),x)

[Out]

Integral(x**3/(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. 1133 vs. \(2 (383) = 766\).

Time = 0.59 (sec) , antiderivative size = 1133, normalized size of antiderivative = 2.46 \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\text {Too large to display} \]

[In]

integrate(x^3/(a+b*tan(c+d*x^(1/2))),x, algorithm="maxima")

[Out]

-1/420*(420*(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*sqr
t(x) + c)^2 + 1)/(a^2 + b^2))*c^7 - (105*(d*sqrt(x) + c)^8*(a - I*b) - 840*(d*sqrt(x) + c)^7*(a - I*b)*c + 294
0*(d*sqrt(x) + c)^6*(a - I*b)*c^2 - 5880*(d*sqrt(x) + c)^5*(a - I*b)*c^3 + 7350*(d*sqrt(x) + c)^4*(a - I*b)*c^
4 - 5880*(d*sqrt(x) + c)^3*(a - I*b)*c^5 + 2940*(d*sqrt(x) + c)^2*(a - I*b)*c^6 - 8*(960*I*(d*sqrt(x) + c)^7*b
- 3920*I*(d*sqrt(x) + c)^6*b*c + 7056*I*(d*sqrt(x) + c)^5*b*c^2 - 7350*I*(d*sqrt(x) + c)^4*b*c^3 + 4900*I*(d*
sqrt(x) + c)^3*b*c^4 - 2205*I*(d*sqrt(x) + c)^2*b*c^5 + 735*I*(d*sqrt(x) + c)*b*c^6)*arctan2((2*a*b*cos(2*d*sq
rt(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)) - 420*(64*I*(d*sqrt(x) + c)^6*b - 224*I*(d*sqrt(x) + c)^5*b*c +
336*I*(d*sqrt(x) + c)^4*b*c^2 - 280*I*(d*sqrt(x) + c)^3*b*c^3 + 140*I*(d*sqrt(x) + c)^2*b*c^4 - 42*I*(d*sqrt(
x) + c)*b*c^5 + 7*I*b*c^6)*dilog((I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 4*(960*(d*sqrt(x) + c)^7*b
- 3920*(d*sqrt(x) + c)^6*b*c + 7056*(d*sqrt(x) + c)^5*b*c^2 - 7350*(d*sqrt(x) + c)^4*b*c^3 + 4900*(d*sqrt(x) +
c)^3*b*c^4 - 2205*(d*sqrt(x) + c)^2*b*c^5 + 735*(d*sqrt(x) + c)*b*c^6)*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)) + 302400*I*b*polylog(8, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 50400*(
12*(d*sqrt(x) + c)*b - 7*b*c)*polylog(7, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) - 10080*(60*I*(d*sqrt
(x) + c)^2*b - 70*I*(d*sqrt(x) + c)*b*c + 21*I*b*c^2)*polylog(6, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b
)) - 2520*(160*(d*sqrt(x) + c)^3*b - 280*(d*sqrt(x) + c)^2*b*c + 168*(d*sqrt(x) + c)*b*c^2 - 35*b*c^3)*polylog
(5, (I*a + b)*e^(2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) - 840*(-240*I*(d*sqrt(x) + c)^4*b + 560*I*(d*sqrt(x) + c)^
3*b*c - 504*I*(d*sqrt(x) + c)^2*b*c^2 + 210*I*(d*sqrt(x) + c)*b*c^3 - 35*I*b*c^4)*polylog(4, (I*a + b)*e^(2*I*
d*sqrt(x) + 2*I*c)/(-I*a + b)) + 420*(192*(d*sqrt(x) + c)^5*b - 560*(d*sqrt(x) + c)^4*b*c + 672*(d*sqrt(x) + c
)^3*b*c^2 - 420*(d*sqrt(x) + c)^2*b*c^3 + 140*(d*sqrt(x) + c)*b*c^4 - 21*b*c^5)*polylog(3, (I*a + b)*e^(2*I*d*
sqrt(x) + 2*I*c)/(-I*a + b)))/(a^2 + b^2))/d^8

Giac [F]

\[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int { \frac {x^{3}}{b \tan \left (d \sqrt {x} + c\right ) + a} \,d x } \]

[In]

integrate(x^3/(a+b*tan(c+d*x^(1/2))),x, algorithm="giac")

[Out]

integrate(x^3/(b*tan(d*sqrt(x) + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\int \frac {x^3}{a+b\,\mathrm {tan}\left (c+d\,\sqrt {x}\right )} \,d x \]

[In]

int(x^3/(a + b*tan(c + d*x^(1/2))),x)

[Out]

int(x^3/(a + b*tan(c + d*x^(1/2))), x)