3.31 \(\int x^3 (a+b \sec (c+d \sqrt {x})) \, dx\)

Optimal. Leaf size=476 \[ \frac {a x^4}{4}-\frac {10080 i b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {10080 b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {5040 i b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]

[Out]

1/4*a*x^4-420*I*b*x^2*polylog(4,-I*exp(I*(c+d*x^(1/2))))/d^4+420*I*b*x^2*polylog(4,I*exp(I*(c+d*x^(1/2))))/d^4
-4*I*b*x^(7/2)*arctan(exp(I*(c+d*x^(1/2))))/d-84*b*x^(5/2)*polylog(3,-I*exp(I*(c+d*x^(1/2))))/d^3+84*b*x^(5/2)
*polylog(3,I*exp(I*(c+d*x^(1/2))))/d^3+5040*I*b*x*polylog(6,-I*exp(I*(c+d*x^(1/2))))/d^6-5040*I*b*x*polylog(6,
I*exp(I*(c+d*x^(1/2))))/d^6+1680*b*x^(3/2)*polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5-1680*b*x^(3/2)*polylog(5,I*e
xp(I*(c+d*x^(1/2))))/d^5+14*I*b*x^3*polylog(2,-I*exp(I*(c+d*x^(1/2))))/d^2-14*I*b*x^3*polylog(2,I*exp(I*(c+d*x
^(1/2))))/d^2-10080*I*b*polylog(8,-I*exp(I*(c+d*x^(1/2))))/d^8+10080*I*b*polylog(8,I*exp(I*(c+d*x^(1/2))))/d^8
-10080*b*polylog(7,-I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^7+10080*b*polylog(7,I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^7

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 476, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {14, 4204, 4181, 2531, 6609, 2282, 6589} \[ \frac {14 i b x^3 \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \text {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \text {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 b \sqrt {x} \text {PolyLog}\left (7,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \text {PolyLog}\left (7,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {10080 i b \text {PolyLog}\left (8,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \text {PolyLog}\left (8,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sec[c + d*Sqrt[x]]),x]

[Out]

(a*x^4)/4 - ((4*I)*b*x^(7/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((14*I)*b*x^3*PolyLog[2, (-I)*E^(I*(c + d*Sqrt
[x]))])/d^2 - ((14*I)*b*x^3*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (84*b*x^(5/2)*PolyLog[3, (-I)*E^(I*(c +
d*Sqrt[x]))])/d^3 + (84*b*x^(5/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((420*I)*b*x^2*PolyLog[4, (-I)*E
^(I*(c + d*Sqrt[x]))])/d^4 + ((420*I)*b*x^2*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (1680*b*x^(3/2)*PolyLog
[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (1680*b*x^(3/2)*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((5040*I)*b*
x*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((5040*I)*b*x*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 - (1008
0*b*Sqrt[x]*PolyLog[7, (-I)*E^(I*(c + d*Sqrt[x]))])/d^7 + (10080*b*Sqrt[x]*PolyLog[7, I*E^(I*(c + d*Sqrt[x]))]
)/d^7 - ((10080*I)*b*PolyLog[8, (-I)*E^(I*(c + d*Sqrt[x]))])/d^8 + ((10080*I)*b*PolyLog[8, I*E^(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 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 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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4204

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

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]

Rubi steps

\begin {align*} \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^3+b x^3 \sec \left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \sec \left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^4}{4}+(2 b) \operatorname {Subst}\left (\int x^7 \sec (c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {(14 b) \operatorname {Subst}\left (\int x^6 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(14 b) \operatorname {Subst}\left (\int x^6 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(84 i b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(84 i b) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(420 b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(420 b) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(1680 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(1680 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(5040 b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(5040 b) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {(10080 i b) \operatorname {Subst}\left (\int x \text {Li}_6\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^6}+\frac {(10080 i b) \operatorname {Subst}\left (\int x \text {Li}_6\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^6}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {(10080 b) \operatorname {Subst}\left (\int \text {Li}_7\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^7}-\frac {(10080 b) \operatorname {Subst}\left (\int \text {Li}_7\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^7}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {(10080 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {(10080 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_7(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}\\ &=\frac {a x^4}{4}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {5040 i b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {10080 b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}-\frac {10080 i b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 479, normalized size = 1.01 \[ \frac {a x^4}{4}-\frac {10080 i b \text {Li}_8\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i b \text {Li}_8\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {10080 b \sqrt {x} \text {Li}_7\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 b \sqrt {x} \text {Li}_7\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {5040 i b x \text {Li}_6\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i b x \text {Li}_6\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {1680 b x^{3/2} \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {420 i b x^2 \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i b x^2 \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {84 b x^{5/2} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 b x^{5/2} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {14 i b x^3 \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i b x^3 \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {4 i b x^{7/2} \tan ^{-1}\left (e^{i c+i d \sqrt {x}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^4)/4 - ((4*I)*b*x^(7/2)*ArcTan[E^(I*c + I*d*Sqrt[x])])/d + ((14*I)*b*x^3*PolyLog[2, (-I)*E^(I*(c + d*Sqrt
[x]))])/d^2 - ((14*I)*b*x^3*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (84*b*x^(5/2)*PolyLog[3, (-I)*E^(I*(c +
d*Sqrt[x]))])/d^3 + (84*b*x^(5/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((420*I)*b*x^2*PolyLog[4, (-I)*E
^(I*(c + d*Sqrt[x]))])/d^4 + ((420*I)*b*x^2*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (1680*b*x^(3/2)*PolyLog
[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (1680*b*x^(3/2)*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((5040*I)*b*
x*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((5040*I)*b*x*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 - (1008
0*b*Sqrt[x]*PolyLog[7, (-I)*E^(I*(c + d*Sqrt[x]))])/d^7 + (10080*b*Sqrt[x]*PolyLog[7, I*E^(I*(c + d*Sqrt[x]))]
)/d^7 - ((10080*I)*b*PolyLog[8, (-I)*E^(I*(c + d*Sqrt[x]))])/d^8 + ((10080*I)*b*PolyLog[8, I*E^(I*(c + d*Sqrt[
x]))])/d^8

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fricas [F]  time = 2.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{3} \sec \left (d \sqrt {x} + c\right ) + a x^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.29, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.07, size = 1512, normalized size = 3.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((d*sqrt(x) + c)^8*a - 8*(d*sqrt(x) + c)^7*a*c + 28*(d*sqrt(x) + c)^6*a*c^2 - 56*(d*sqrt(x) + c)^5*a*c^3 +
70*(d*sqrt(x) + c)^4*a*c^4 - 56*(d*sqrt(x) + c)^3*a*c^5 + 28*(d*sqrt(x) + c)^2*a*c^6 - 8*(d*sqrt(x) + c)*a*c^
7 - 8*b*c^7*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) + 4*(-2*I*(d*sqrt(x) + c)^7*b + 14*I*(d*sqrt(x) + c)^
6*b*c - 42*I*(d*sqrt(x) + c)^5*b*c^2 + 70*I*(d*sqrt(x) + c)^4*b*c^3 - 70*I*(d*sqrt(x) + c)^3*b*c^4 + 42*I*(d*s
qrt(x) + c)^2*b*c^5 - 14*I*(d*sqrt(x) + c)*b*c^6)*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1) + 4*(-2*
I*(d*sqrt(x) + c)^7*b + 14*I*(d*sqrt(x) + c)^6*b*c - 42*I*(d*sqrt(x) + c)^5*b*c^2 + 70*I*(d*sqrt(x) + c)^4*b*c
^3 - 70*I*(d*sqrt(x) + c)^3*b*c^4 + 42*I*(d*sqrt(x) + c)^2*b*c^5 - 14*I*(d*sqrt(x) + c)*b*c^6)*arctan2(cos(d*s
qrt(x) + c), -sin(d*sqrt(x) + c) + 1) + 4*(-14*I*(d*sqrt(x) + c)^6*b + 84*I*(d*sqrt(x) + c)^5*b*c - 210*I*(d*s
qrt(x) + c)^4*b*c^2 + 280*I*(d*sqrt(x) + c)^3*b*c^3 - 210*I*(d*sqrt(x) + c)^2*b*c^4 + 84*I*(d*sqrt(x) + c)*b*c
^5 - 14*I*b*c^6)*dilog(I*e^(I*d*sqrt(x) + I*c)) + 4*(14*I*(d*sqrt(x) + c)^6*b - 84*I*(d*sqrt(x) + c)^5*b*c + 2
10*I*(d*sqrt(x) + c)^4*b*c^2 - 280*I*(d*sqrt(x) + c)^3*b*c^3 + 210*I*(d*sqrt(x) + c)^2*b*c^4 - 84*I*(d*sqrt(x)
+ c)*b*c^5 + 14*I*b*c^6)*dilog(-I*e^(I*d*sqrt(x) + I*c)) + 4*((d*sqrt(x) + c)^7*b - 7*(d*sqrt(x) + c)^6*b*c +
21*(d*sqrt(x) + c)^5*b*c^2 - 35*(d*sqrt(x) + c)^4*b*c^3 + 35*(d*sqrt(x) + c)^3*b*c^4 - 21*(d*sqrt(x) + c)^2*b
*c^5 + 7*(d*sqrt(x) + c)*b*c^6)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*sin(d*sqrt(x) + c) + 1) -
4*((d*sqrt(x) + c)^7*b - 7*(d*sqrt(x) + c)^6*b*c + 21*(d*sqrt(x) + c)^5*b*c^2 - 35*(d*sqrt(x) + c)^4*b*c^3 + 3
5*(d*sqrt(x) + c)^3*b*c^4 - 21*(d*sqrt(x) + c)^2*b*c^5 + 7*(d*sqrt(x) + c)*b*c^6)*log(cos(d*sqrt(x) + c)^2 + s
in(d*sqrt(x) + c)^2 - 2*sin(d*sqrt(x) + c) + 1) + 40320*I*b*polylog(8, I*e^(I*d*sqrt(x) + I*c)) - 40320*I*b*po
lylog(8, -I*e^(I*d*sqrt(x) + I*c)) + 40320*((d*sqrt(x) + c)*b - b*c)*polylog(7, I*e^(I*d*sqrt(x) + I*c)) - 403
20*((d*sqrt(x) + c)*b - b*c)*polylog(7, -I*e^(I*d*sqrt(x) + I*c)) + 4*(-5040*I*(d*sqrt(x) + c)^2*b + 10080*I*(
d*sqrt(x) + c)*b*c - 5040*I*b*c^2)*polylog(6, I*e^(I*d*sqrt(x) + I*c)) + 4*(5040*I*(d*sqrt(x) + c)^2*b - 10080
*I*(d*sqrt(x) + c)*b*c + 5040*I*b*c^2)*polylog(6, -I*e^(I*d*sqrt(x) + I*c)) - 6720*((d*sqrt(x) + c)^3*b - 3*(d
*sqrt(x) + c)^2*b*c + 3*(d*sqrt(x) + c)*b*c^2 - b*c^3)*polylog(5, I*e^(I*d*sqrt(x) + I*c)) + 6720*((d*sqrt(x)
+ c)^3*b - 3*(d*sqrt(x) + c)^2*b*c + 3*(d*sqrt(x) + c)*b*c^2 - b*c^3)*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) + 4
*(420*I*(d*sqrt(x) + c)^4*b - 1680*I*(d*sqrt(x) + c)^3*b*c + 2520*I*(d*sqrt(x) + c)^2*b*c^2 - 1680*I*(d*sqrt(x
) + c)*b*c^3 + 420*I*b*c^4)*polylog(4, I*e^(I*d*sqrt(x) + I*c)) + 4*(-420*I*(d*sqrt(x) + c)^4*b + 1680*I*(d*sq
rt(x) + c)^3*b*c - 2520*I*(d*sqrt(x) + c)^2*b*c^2 + 1680*I*(d*sqrt(x) + c)*b*c^3 - 420*I*b*c^4)*polylog(4, -I*
e^(I*d*sqrt(x) + I*c)) + 336*((d*sqrt(x) + c)^5*b - 5*(d*sqrt(x) + c)^4*b*c + 10*(d*sqrt(x) + c)^3*b*c^2 - 10*
(d*sqrt(x) + c)^2*b*c^3 + 5*(d*sqrt(x) + c)*b*c^4 - b*c^5)*polylog(3, I*e^(I*d*sqrt(x) + I*c)) - 336*((d*sqrt(
x) + c)^5*b - 5*(d*sqrt(x) + c)^4*b*c + 10*(d*sqrt(x) + c)^3*b*c^2 - 10*(d*sqrt(x) + c)^2*b*c^3 + 5*(d*sqrt(x)
+ c)*b*c^4 - b*c^5)*polylog(3, -I*e^(I*d*sqrt(x) + I*c)))/d^8

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/cos(c + d*x^(1/2))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*sec(c + d*sqrt(x))), x)

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