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

Optimal. Leaf size=255 \[ \frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{2 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x}{d} \]

[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

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Rubi [A]  time = 0.324528, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{2 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x}{d} \]

Antiderivative was successfully verified.

[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 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 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 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 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]

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 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 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 2279

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 2391

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]

Rubi steps

\begin{align*} \int \sqrt{x} \left (a+b \sec \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{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) \operatorname{Subst}\left (\int x^2 \sec (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{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 \tan ^{-1}\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) \operatorname{Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(8 a b) \operatorname{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 ) \operatorname{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 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (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) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(8 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{\left (8 i b^2\right ) \operatorname{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 \tan ^{-1}\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} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (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) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{(8 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{\left (4 b^2\right ) \operatorname{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 \tan ^{-1}\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} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (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 ) \operatorname{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 \tan ^{-1}\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} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 i b^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{8 a b \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (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]  time = 0.694471, size = 247, normalized size = 0.97 \[ \frac{2 \left (12 i a b d \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )-12 i a b d \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )-12 a b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )+12 a b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )-3 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+a^2 d^3 x^{3/2}-12 i a b d^2 x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )+3 b^2 d^2 x \tan \left (c+d \sqrt{x}\right )+6 b^2 d \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )-3 i b^2 d^2 x\right )}{3 d^3} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.081, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( c+d\sqrt{x} \right ) \right ) ^{2}\sqrt{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [B]  time = 2.10303, size = 1727, normalized size = 6.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 - 4*(d*sqrt(x) + c)*a*b*c + 2*((d*sqrt
(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c)*cos(2*d*sqrt(x) + 2*c) - (-2*I*(d*sqrt(x) + c)^2*a*b + 4*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 - 4*(d*sqrt(x) + c)*a*b*c + 2*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c)*cos(2*d*sqrt(x) + 2*c)
- (-2*I*(d*sqrt(x) + c)^2*a*b + 4*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)*b^2 - 2*b^2*c + 2*((d*sqrt(x) + c)*b^2 - b^2*c)*cos(2*d*sqrt(x)
+ 2*c) + (2*I*(d*sqrt(x) + c)*b^2 - 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 - 4*a*b*c + 4*((d*sqrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) - (-4*I*(d*sqrt(x) + c)*a*b + 4*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 - 4*a*b*c + 4*((d*sqrt(x)
+ c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) + (4*I*(d*sqrt(x) + c)*a*b - 4*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*sq
rt(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*sq
rt(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*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) + (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(c
os(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) + 4*
a*b*sin(2*d*sqrt(x) + 2*c) - 4*I*a*b)*polylog(3, I*e^(I*d*sqrt(x) + I*c)) + (4*I*a*b*cos(2*d*sqrt(x) + 2*c) -
4*a*b*sin(2*d*sqrt(x) + 2*c) + 4*I*a*b)*polylog(3, -I*e^(I*d*sqrt(x) + I*c)) + (-2*I*(d*sqrt(x) + c)^2*b^2 + 4
*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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 )}^{2} \sqrt{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)