\(\int x^3 (a+b \tan (c+d \sqrt {x})) \, dx\) [25]

Optimal result

Integrand size = 18, antiderivative size = 261 \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \]

[Out]

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {14, 3832, 3800, 2221, 2611, 6744, 2320, 6724} \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {1}{4} i b x^4 \]

[In]

Int[x^3*(a + b*Tan[c + d*Sqrt[x]]),x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[2*I, Int[(c + d*x)^m*(E^(2*I*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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}& = \int \left (a x^3+b x^3 \tan \left (c+d \sqrt {x}\right )\right ) \, dx \\ & = \frac {a x^4}{4}+b \int x^3 \tan \left (c+d \sqrt {x}\right ) \, dx \\ & = \frac {a x^4}{4}+(2 b) \text {Subst}\left (\int x^7 \tan (c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-(4 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^7}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {(14 b) \text {Subst}\left (\int x^6 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(42 i b) \text {Subst}\left (\int x^5 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(105 b) \text {Subst}\left (\int x^4 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {(210 i b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (4,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(315 b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (5,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {(315 i b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (6,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^6} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}+\frac {(315 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (7,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{2 d^7} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {(315 i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(7,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \\ & = \frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} i b x^4-\frac {2 b x^{7/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}+\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {315 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}-\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 i b \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8} \]

[In]

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

[Out]

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

Maple [F]

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

[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]

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

[In]

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

[Out]

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

Sympy [F]

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

[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. 937 vs. \(2 (198) = 396\).

Time = 0.46 (sec) , antiderivative size = 937, normalized size of antiderivative = 3.59 \[ \int x^3 \left (a+b \tan \left (c+d \sqrt {x}\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/420*(105*(d*sqrt(x) + c)^8*a + 105*I*(d*sqrt(x) + c)^8*b - 840*(d*sqrt(x) + c)^7*a*c - 840*I*(d*sqrt(x) + c)
^7*b*c + 2940*(d*sqrt(x) + c)^6*a*c^2 + 2940*I*(d*sqrt(x) + c)^6*b*c^2 - 5880*(d*sqrt(x) + c)^5*a*c^3 - 5880*I
*(d*sqrt(x) + c)^5*b*c^3 + 7350*(d*sqrt(x) + c)^4*a*c^4 + 7350*I*(d*sqrt(x) + c)^4*b*c^4 - 5880*(d*sqrt(x) + c
)^3*a*c^5 - 5880*I*(d*sqrt(x) + c)^3*b*c^5 + 2940*(d*sqrt(x) + c)^2*a*c^6 + 2940*I*(d*sqrt(x) + c)^2*b*c^6 - 8
40*(d*sqrt(x) + c)*a*c^7 - 840*b*c^7*log(sec(d*sqrt(x) + c)) + 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(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqr
t(x) + 2*c) + 1) + 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)*dil
og(-e^(2*I*d*sqrt(x) + 2*I*c)) - 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(cos(2*d*sqrt(x) + 2*c)^2 + sin(2*d*sqrt(x) + 2*c)^2 + 2*cos(2*d*sqrt(x) + 2*c) + 1)
- 302400*I*b*polylog(8, -e^(2*I*d*sqrt(x) + 2*I*c)) - 50400*(12*(d*sqrt(x) + c)*b - 7*b*c)*polylog(7, -e^(2*I*
d*sqrt(x) + 2*I*c)) + 10080*(60*I*(d*sqrt(x) + c)^2*b - 70*I*(d*sqrt(x) + c)*b*c + 21*I*b*c^2)*polylog(6, -e^(
2*I*d*sqrt(x) + 2*I*c)) + 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, -e^(2*I*d*sqrt(x) + 2*I*c)) + 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, -e^(2*I*d*sqrt(x
) + 2*I*c)) - 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, -e^(2*I*d*sqrt(x) + 2*I*c)))/d^8

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)