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

Optimal. Leaf size=355 \[ \frac {a^2 x^2}{2}-\frac {24 i a b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {24 i a b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 a b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 i a b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {3 b^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 i b^2 \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{3/2}}{d} \]

[Out]

-2*I*b^2*x^(3/2)/d+1/2*a^2*x^2-8*I*a*b*x^(3/2)*arctan(exp(I*(c+d*x^(1/2))))/d+6*b^2*x*ln(1+exp(2*I*(c+d*x^(1/2
))))/d^2+12*I*a*b*x*polylog(2,-I*exp(I*(c+d*x^(1/2))))/d^2-12*I*a*b*x*polylog(2,I*exp(I*(c+d*x^(1/2))))/d^2+3*
b^2*polylog(3,-exp(2*I*(c+d*x^(1/2))))/d^4-24*I*a*b*polylog(4,-I*exp(I*(c+d*x^(1/2))))/d^4+24*I*a*b*polylog(4,
I*exp(I*(c+d*x^(1/2))))/d^4-6*I*b^2*polylog(2,-exp(2*I*(c+d*x^(1/2))))*x^(1/2)/d^3-24*a*b*polylog(3,-I*exp(I*(
c+d*x^(1/2))))*x^(1/2)/d^3+24*a*b*polylog(3,I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^3+2*b^2*x^(3/2)*tan(c+d*x^(1/2))
/d

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Rubi [A]  time = 0.45, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4204, 4190, 4181, 2531, 6609, 2282, 6589, 4184, 3719, 2190} \[ \frac {12 i a b x \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 a b \sqrt {x} \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {24 i a b \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {24 i a b \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {6 i b^2 \sqrt {x} \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {3 b^2 \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Sec[c + d*Sqrt[x]])^2,x]

[Out]

((-2*I)*b^2*x^(3/2))/d + (a^2*x^2)/2 - ((8*I)*a*b*x^(3/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + (6*b^2*x*Log[1 +
E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((12*I)*a*b*x*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((12*I)*a*b*x*Po
lyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((6*I)*b^2*Sqrt[x]*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (24*
a*b*Sqrt[x]*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/d^3 + (24*a*b*Sqrt[x]*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])
/d^3 + (3*b^2*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((24*I)*a*b*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))]
)/d^4 + ((24*I)*a*b*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + (2*b^2*x^(3/2)*Tan[c + d*Sqrt[x]])/d

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 3719

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 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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^3 (a+b \sec (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^3+2 a b x^3 \sec (c+d x)+b^2 x^3 \sec ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^2}{2}+(4 a b) \operatorname {Subst}\left (\int x^3 \sec (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^3 \sec ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(12 a b) \operatorname {Subst}\left (\int x^2 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(12 a b) \operatorname {Subst}\left (\int x^2 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {12 i a b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(24 i a b) \operatorname {Subst}\left (\int x \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(24 i a b) \operatorname {Subst}\left (\int x \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (12 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {12 i a b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {24 a b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(24 a b) \operatorname {Subst}\left (\int \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(24 a b) \operatorname {Subst}\left (\int \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {12 i a b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b^2 \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(24 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(24 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {12 i a b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b^2 \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {24 i a b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {24 i a b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}\\ &=-\frac {2 i b^2 x^{3/2}}{d}+\frac {a^2 x^2}{2}-\frac {8 i a b x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {12 i a b x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i a b x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {6 i b^2 \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {24 a b \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {24 a b \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {3 b^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {24 i a b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {24 i a b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^{3/2} \tan \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 347, normalized size = 0.98 \[ \frac {a^2 d^4 x^2-16 i a b d^3 x^{3/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+24 i a b d^2 x \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )-24 i a b d^2 x \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-48 a b d \sqrt {x} \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+48 a b d \sqrt {x} \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )-48 i a b \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )+48 i a b \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )+4 b^2 d^3 x^{3/2} \tan \left (c+d \sqrt {x}\right )+12 b^2 d^2 x \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )-12 i b^2 d \sqrt {x} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )+6 b^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )-4 i b^2 d^3 x^{3/2}}{2 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)*b^2*d^3*x^(3/2) + a^2*d^4*x^2 - (16*I)*a*b*d^3*x^(3/2)*ArcTan[E^(I*(c + d*Sqrt[x]))] + 12*b^2*d^2*x*Lo
g[1 + E^((2*I)*(c + d*Sqrt[x]))] + (24*I)*a*b*d^2*x*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (24*I)*a*b*d^2*x*
PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (12*I)*b^2*d*Sqrt[x]*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))] - 48*a*b*d*S
qrt[x]*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))] + 48*a*b*d*Sqrt[x]*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] + 6*b^2*P
olyLog[3, -E^((2*I)*(c + d*Sqrt[x]))] - (48*I)*a*b*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))] + (48*I)*a*b*PolyLog
[4, I*E^(I*(c + d*Sqrt[x]))] + 4*b^2*d^3*x^(3/2)*Tan[c + d*Sqrt[x]])/(2*d^4)

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x \sec \left (d \sqrt {x} + c\right )^{2} + 2 \, a b x \sec \left (d \sqrt {x} + c\right ) + a^{2} x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.38, size = 0, normalized size = 0.00 \[ \int x \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.92, size = 2006, normalized size = 5.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((d*sqrt(x) + c)^4*a^2 - 4*(d*sqrt(x) + c)^3*a^2*c + 6*(d*sqrt(x) + c)^2*a^2*c^2 - 4*(d*sqrt(x) + c)*a^2*c
^3 - 8*a*b*c^3*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) - 4*(4*b^2*c^3 + (4*(d*sqrt(x) + c)^3*a*b - 12*(d*
sqrt(x) + c)^2*a*b*c + 12*(d*sqrt(x) + c)*a*b*c^2 + 4*((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c + 3*(
d*sqrt(x) + c)*a*b*c^2)*cos(2*d*sqrt(x) + 2*c) - (-4*I*(d*sqrt(x) + c)^3*a*b + 12*I*(d*sqrt(x) + c)^2*a*b*c -
12*I*(d*sqrt(x) + c)*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1) + (4
*(d*sqrt(x) + c)^3*a*b - 12*(d*sqrt(x) + c)^2*a*b*c + 12*(d*sqrt(x) + c)*a*b*c^2 + 4*((d*sqrt(x) + c)^3*a*b -
3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2)*cos(2*d*sqrt(x) + 2*c) - (-4*I*(d*sqrt(x) + c)^3*a*b +
12*I*(d*sqrt(x) + c)^2*a*b*c - 12*I*(d*sqrt(x) + c)*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c
), -sin(d*sqrt(x) + c) + 1) - (6*(d*sqrt(x) + c)^2*b^2 - 12*(d*sqrt(x) + c)*b^2*c + 6*b^2*c^2 + 6*((d*sqrt(x)
+ c)^2*b^2 - 2*(d*sqrt(x) + c)*b^2*c + b^2*c^2)*cos(2*d*sqrt(x) + 2*c) + (6*I*(d*sqrt(x) + c)^2*b^2 - 12*I*(d*
sqrt(x) + c)*b^2*c + 6*I*b^2*c^2)*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*
c) + 1) + 4*((d*sqrt(x) + c)^3*b^2 - 3*(d*sqrt(x) + c)^2*b^2*c + 3*(d*sqrt(x) + c)*b^2*c^2)*cos(2*d*sqrt(x) +
2*c) + (6*(d*sqrt(x) + c)*b^2 - 6*b^2*c + 6*((d*sqrt(x) + c)*b^2 - b^2*c)*cos(2*d*sqrt(x) + 2*c) - (-6*I*(d*sq
rt(x) + c)*b^2 + 6*I*b^2*c)*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) + (12*(d*sqrt(x) + c)^2*
a*b - 24*(d*sqrt(x) + c)*a*b*c + 12*a*b*c^2 + 12*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + a*b*c^2)*c
os(2*d*sqrt(x) + 2*c) - (-12*I*(d*sqrt(x) + c)^2*a*b + 24*I*(d*sqrt(x) + c)*a*b*c - 12*I*a*b*c^2)*sin(2*d*sqrt
(x) + 2*c))*dilog(I*e^(I*d*sqrt(x) + I*c)) - (12*(d*sqrt(x) + c)^2*a*b - 24*(d*sqrt(x) + c)*a*b*c + 12*a*b*c^2
+ 12*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + a*b*c^2)*cos(2*d*sqrt(x) + 2*c) + (12*I*(d*sqrt(x) +
c)^2*a*b - 24*I*(d*sqrt(x) + c)*a*b*c + 12*I*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*dilog(-I*e^(I*d*sqrt(x) + I*c))
- (-3*I*(d*sqrt(x) + c)^2*b^2 + 6*I*(d*sqrt(x) + c)*b^2*c - 3*I*b^2*c^2 + (-3*I*(d*sqrt(x) + c)^2*b^2 + 6*I*(d
*sqrt(x) + c)*b^2*c - 3*I*b^2*c^2)*cos(2*d*sqrt(x) + 2*c) + 3*((d*sqrt(x) + c)^2*b^2 - 2*(d*sqrt(x) + c)*b^2*c
+ b^2*c^2)*sin(2*d*sqrt(x) + 2*c))*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) - (-2*I*(d*sqrt(x) + c)^3*a*b + 6*I*(d*sqrt(x) + c)^2*a*b*c - 6*I*(d*sqrt(x) + c)*a*b*c^2 + (-2*
I*(d*sqrt(x) + c)^3*a*b + 6*I*(d*sqrt(x) + c)^2*a*b*c - 6*I*(d*sqrt(x) + c)*a*b*c^2)*cos(2*d*sqrt(x) + 2*c) +
2*((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*log(
cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*sin(d*sqrt(x) + c) + 1) - (2*I*(d*sqrt(x) + c)^3*a*b - 6*I*(d*
sqrt(x) + c)^2*a*b*c + 6*I*(d*sqrt(x) + c)*a*b*c^2 + (2*I*(d*sqrt(x) + c)^3*a*b - 6*I*(d*sqrt(x) + c)^2*a*b*c
+ 6*I*(d*sqrt(x) + c)*a*b*c^2)*cos(2*d*sqrt(x) + 2*c) - 2*((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c +
3*(d*sqrt(x) + c)*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*sin(d*
sqrt(x) + c) + 1) - 24*(a*b*cos(2*d*sqrt(x) + 2*c) + I*a*b*sin(2*d*sqrt(x) + 2*c) + a*b)*polylog(4, I*e^(I*d*s
qrt(x) + I*c)) + 24*(a*b*cos(2*d*sqrt(x) + 2*c) + I*a*b*sin(2*d*sqrt(x) + 2*c) + a*b)*polylog(4, -I*e^(I*d*sqr
t(x) + I*c)) - (-3*I*b^2*cos(2*d*sqrt(x) + 2*c) + 3*b^2*sin(2*d*sqrt(x) + 2*c) - 3*I*b^2)*polylog(3, -e^(2*I*d
*sqrt(x) + 2*I*c)) - (-24*I*(d*sqrt(x) + c)*a*b + 24*I*a*b*c + (-24*I*(d*sqrt(x) + c)*a*b + 24*I*a*b*c)*cos(2*
d*sqrt(x) + 2*c) + 24*((d*sqrt(x) + c)*a*b - a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(3, I*e^(I*d*sqrt(x) + I*c)
) - (24*I*(d*sqrt(x) + c)*a*b - 24*I*a*b*c + (24*I*(d*sqrt(x) + c)*a*b - 24*I*a*b*c)*cos(2*d*sqrt(x) + 2*c) -
24*((d*sqrt(x) + c)*a*b - a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(3, -I*e^(I*d*sqrt(x) + I*c)) - (-4*I*(d*sqrt(
x) + c)^3*b^2 + 12*I*(d*sqrt(x) + c)^2*b^2*c - 12*I*(d*sqrt(x) + c)*b^2*c^2)*sin(2*d*sqrt(x) + 2*c))/(-2*I*cos
(2*d*sqrt(x) + 2*c) + 2*sin(2*d*sqrt(x) + 2*c) - 2*I))/d^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/cos(c + d*x^(1/2)))^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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