3.32 \(\int x^2 (a+b \sec (c+d \sqrt{x})) \, dx\)

Optimal. Leaf size=348 \[ \frac{10 i b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 b \sqrt{x} \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{240 b \sqrt{x} \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{240 i b \text{PolyLog}\left (6,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{240 i b \text{PolyLog}\left (6,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

[Out]

(a*x^3)/3 - ((4*I)*b*x^(5/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((10*I)*b*x^2*PolyLog[2, (-I)*E^(I*(c + d*Sqrt
[x]))])/d^2 - ((10*I)*b*x^2*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (40*b*x^(3/2)*PolyLog[3, (-I)*E^(I*(c +
d*Sqrt[x]))])/d^3 + (40*b*x^(3/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((120*I)*b*x*PolyLog[4, (-I)*E^(
I*(c + d*Sqrt[x]))])/d^4 + ((120*I)*b*x*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (240*b*Sqrt[x]*PolyLog[5, (
-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (240*b*Sqrt[x]*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((240*I)*b*PolyLog
[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((240*I)*b*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6

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Rubi [A]  time = 0.310915, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 16, 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} \[ \frac{10 i b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 b \sqrt{x} \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{240 b \sqrt{x} \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{240 i b \text{PolyLog}\left (6,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{240 i b \text{PolyLog}\left (6,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*Sec[c + d*Sqrt[x]]),x]

[Out]

(a*x^3)/3 - ((4*I)*b*x^(5/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + ((10*I)*b*x^2*PolyLog[2, (-I)*E^(I*(c + d*Sqrt
[x]))])/d^2 - ((10*I)*b*x^2*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (40*b*x^(3/2)*PolyLog[3, (-I)*E^(I*(c +
d*Sqrt[x]))])/d^3 + (40*b*x^(3/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((120*I)*b*x*PolyLog[4, (-I)*E^(
I*(c + d*Sqrt[x]))])/d^4 + ((120*I)*b*x*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (240*b*Sqrt[x]*PolyLog[5, (
-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (240*b*Sqrt[x]*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((240*I)*b*PolyLog
[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((240*I)*b*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6

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 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 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 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 x^2 \left (a+b \sec \left (c+d \sqrt{x}\right )\right ) \, dx &=\int \left (a x^2+b x^2 \sec \left (c+d \sqrt{x}\right )\right ) \, dx\\ &=\frac{a x^3}{3}+b \int x^2 \sec \left (c+d \sqrt{x}\right ) \, dx\\ &=\frac{a x^3}{3}+(2 b) \operatorname{Subst}\left (\int x^5 \sec (c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{(10 b) \operatorname{Subst}\left (\int x^4 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(10 b) \operatorname{Subst}\left (\int x^4 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(40 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(40 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{(120 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{(120 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{(240 i b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(240 i b) \operatorname{Subst}\left (\int x \text{Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{240 b \sqrt{x} \text{Li}_5\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{(240 b) \operatorname{Subst}\left (\int \text{Li}_5\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{(240 b) \operatorname{Subst}\left (\int \text{Li}_5\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{240 b \sqrt{x} \text{Li}_5\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{(240 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{(240 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}\\ &=\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{240 b \sqrt{x} \text{Li}_5\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{240 i b \text{Li}_6\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{240 i b \text{Li}_6\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}\\ \end{align*}

Mathematica [A]  time = 0.0810967, size = 351, normalized size = 1.01 \[ \frac{10 i b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{10 i b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{40 b x^{3/2} \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{40 b x^{3/2} \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{120 i b x \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{120 i b x \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 b \sqrt{x} \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{240 b \sqrt{x} \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{240 i b \text{PolyLog}\left (6,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{240 i b \text{PolyLog}\left (6,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a x^3}{3}-\frac{4 i b x^{5/2} \tan ^{-1}\left (e^{i c+i d \sqrt{x}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^3)/3 - ((4*I)*b*x^(5/2)*ArcTan[E^(I*c + I*d*Sqrt[x])])/d + ((10*I)*b*x^2*PolyLog[2, (-I)*E^(I*(c + d*Sqrt
[x]))])/d^2 - ((10*I)*b*x^2*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - (40*b*x^(3/2)*PolyLog[3, (-I)*E^(I*(c +
d*Sqrt[x]))])/d^3 + (40*b*x^(3/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/d^3 - ((120*I)*b*x*PolyLog[4, (-I)*E^(
I*(c + d*Sqrt[x]))])/d^4 + ((120*I)*b*x*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (240*b*Sqrt[x]*PolyLog[5, (
-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (240*b*Sqrt[x]*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 + ((240*I)*b*PolyLog
[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((240*I)*b*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6

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Maple [F]  time = 0.112, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b\sec \left ( c+d\sqrt{x} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.18431, size = 1304, normalized size = 3.75 \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*sec(c+d*x^(1/2))),x, algorithm="maxima")

[Out]

1/3*((d*sqrt(x) + c)^6*a - 6*(d*sqrt(x) + c)^5*a*c + 15*(d*sqrt(x) + c)^4*a*c^2 - 20*(d*sqrt(x) + c)^3*a*c^3 +
15*(d*sqrt(x) + c)^2*a*c^4 - 6*(d*sqrt(x) + c)*a*c^5 - 6*b*c^5*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) +
3*(-2*I*(d*sqrt(x) + c)^5*b + 10*I*(d*sqrt(x) + c)^4*b*c - 20*I*(d*sqrt(x) + c)^3*b*c^2 + 20*I*(d*sqrt(x) + c
)^2*b*c^3 - 10*I*(d*sqrt(x) + c)*b*c^4)*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1) + 3*(-2*I*(d*sqrt(
x) + c)^5*b + 10*I*(d*sqrt(x) + c)^4*b*c - 20*I*(d*sqrt(x) + c)^3*b*c^2 + 20*I*(d*sqrt(x) + c)^2*b*c^3 - 10*I*
(d*sqrt(x) + c)*b*c^4)*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1) + 3*(-10*I*(d*sqrt(x) + c)^4*b + 4
0*I*(d*sqrt(x) + c)^3*b*c - 60*I*(d*sqrt(x) + c)^2*b*c^2 + 40*I*(d*sqrt(x) + c)*b*c^3 - 10*I*b*c^4)*dilog(I*e^
(I*d*sqrt(x) + I*c)) + 3*(10*I*(d*sqrt(x) + c)^4*b - 40*I*(d*sqrt(x) + c)^3*b*c + 60*I*(d*sqrt(x) + c)^2*b*c^2
- 40*I*(d*sqrt(x) + c)*b*c^3 + 10*I*b*c^4)*dilog(-I*e^(I*d*sqrt(x) + I*c)) + 3*((d*sqrt(x) + c)^5*b - 5*(d*sq
rt(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)*log(cos(
d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*sin(d*sqrt(x) + c) + 1) - 3*((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)*log(cos(d*sqrt(x
) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*sin(d*sqrt(x) + c) + 1) - 720*I*b*polylog(6, I*e^(I*d*sqrt(x) + I*c)) + 72
0*I*b*polylog(6, -I*e^(I*d*sqrt(x) + I*c)) - 720*((d*sqrt(x) + c)*b - b*c)*polylog(5, I*e^(I*d*sqrt(x) + I*c))
+ 720*((d*sqrt(x) + c)*b - b*c)*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) + 3*(120*I*(d*sqrt(x) + c)^2*b - 240*I*(
d*sqrt(x) + c)*b*c + 120*I*b*c^2)*polylog(4, I*e^(I*d*sqrt(x) + I*c)) + 3*(-120*I*(d*sqrt(x) + c)^2*b + 240*I*
(d*sqrt(x) + c)*b*c - 120*I*b*c^2)*polylog(4, -I*e^(I*d*sqrt(x) + I*c)) + 120*((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(3, I*e^(I*d*sqrt(x) + I*c)) - 120*((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(3, -I*e^(I*d*sqrt(x) + I*c)))/d^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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