\(\int \sqrt {x} (a+b \sec (c+d \sqrt {x}))^2 \, dx\) [57]

Optimal result

Integrand size = 22, antiderivative size = 255 \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=-\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {8 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4289, 4275, 4266, 2611, 2320, 6724, 4269, 3800, 2221, 2317, 2438} \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {8 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x}{d} \]

[In]

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

[Out]

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

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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 4266

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 4269

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 4275

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 4289

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 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]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (a^2 x^2+2 a b x^2 \sec (c+d x)+b^2 x^2 \sec ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {2}{3} a^2 x^{3/2}+(4 a b) \text {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^2 \sec ^2(c+d x) \, dx,x,\sqrt {x}\right ) \\ & = \frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(8 a b) \text {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(8 a b) \text {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (4 b^2\right ) \text {Subst}\left (\int x \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d} \\ & = -\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(8 i a b) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(8 i a b) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (8 i b^2\right ) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d} \\ & = -\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(8 a b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(8 a b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {\left (4 b^2\right ) \text {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2} \\ & = -\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d}+\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3} \\ & = -\frac {2 i b^2 x}{d}+\frac {2}{3} a^2 x^{3/2}-\frac {8 i a b x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {4 b^2 \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 i a b \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {8 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x \tan \left (c+d \sqrt {x}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.97 \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\frac {2 \left (-3 i b^2 d^2 x+a^2 d^3 x^{3/2}-12 i a b d^2 x \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )+6 b^2 d \sqrt {x} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )+12 i a b d \sqrt {x} \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )-12 i a b d \sqrt {x} \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )-3 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-12 a b \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )+12 a b \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )+3 b^2 d^2 x \tan \left (c+d \sqrt {x}\right )\right )}{3 d^3} \]

[In]

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

[Out]

(2*((-3*I)*b^2*d^2*x + a^2*d^3*x^(3/2) - (12*I)*a*b*d^2*x*ArcTan[E^(I*(c + d*Sqrt[x]))] + 6*b^2*d*Sqrt[x]*Log[
1 + E^((2*I)*(c + d*Sqrt[x]))] + (12*I)*a*b*d*Sqrt[x]*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (12*I)*a*b*d*Sq
rt[x]*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (3*I)*b^2*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))] - 12*a*b*PolyLog[
3, (-I)*E^(I*(c + d*Sqrt[x]))] + 12*a*b*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] + 3*b^2*d^2*x*Tan[c + d*Sqrt[x]]))
/(3*d^3)

Maple [F]

\[\int \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2} \sqrt {x}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} \sqrt {x} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int \sqrt {x} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]

[In]

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

[Out]

Integral(sqrt(x)*(a + b*sec(c + d*sqrt(x)))**2, 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. 1272 vs. \(2 (194) = 388\).

Time = 0.45 (sec) , antiderivative size = 1272, normalized size of antiderivative = 4.99 \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} \sqrt {x} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {x} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int \sqrt {x}\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]

[In]

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

[Out]

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