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

Optimal. Leaf size=241 \[ \frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{2 i b^2 \text{PolyLog}\left (2,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x}{d} \]

[Out]

((-2*I)*b^2*x)/d + (2*a^2*x^(3/2))/3 - (8*a*b*x*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d - (2*b^2*x*Cot[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, -E^(I*(c + d*Sqr
t[x]))])/d^2 - ((8*I)*a*b*Sqrt[x]*PolyLog[2, 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, -E^(I*(c + d*Sqrt[x]))])/d^3 + (8*a*b*PolyLog[3, E^(I*(c + d*Sqrt[x]))]
)/d^3

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Rubi [A]  time = 0.325918, antiderivative size = 241, 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 = {4205, 4190, 4183, 2531, 2282, 6589, 4184, 3717, 2190, 2279, 2391} \[ \frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{2 i b^2 \text{PolyLog}\left (2,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*b^2*x)/d + (2*a^2*x^(3/2))/3 - (8*a*b*x*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d - (2*b^2*x*Cot[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, -E^(I*(c + d*Sqr
t[x]))])/d^2 - ((8*I)*a*b*Sqrt[x]*PolyLog[2, 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, -E^(I*(c + d*Sqrt[x]))])/d^3 + (8*a*b*PolyLog[3, E^(I*(c + d*Sqrt[x]))]
)/d^3

Rule 4205

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Csc[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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && 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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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 \csc \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^2 (a+b \csc (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \csc (c+d x)+b^2 x^2 \csc ^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 \csc (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^2 \csc ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a^2 x^{3/2}-\frac{8 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\right )}{d}-\frac{(8 a b) \operatorname{Subst}\left (\int x \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(8 a b) \operatorname{Subst}\left (\int x \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int x \cot (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 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\right )}{d}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(8 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-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 (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 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(8 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-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(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 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\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 a b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}\\ \end{align*}

Mathematica [B]  time = 3.80401, size = 681, normalized size = 2.83 \[ \frac{12 i b \left (-1+e^{2 i c}\right ) \left (b-2 a d \sqrt{x}\right ) \text{PolyLog}\left (2,-e^{-i \left (c+d \sqrt{x}\right )}\right )+12 i b \left (-1+e^{2 i c}\right ) \left (2 a d \sqrt{x}+b\right ) \text{PolyLog}\left (2,e^{-i \left (c+d \sqrt{x}\right )}\right )-24 a b e^{2 i c} \text{PolyLog}\left (3,-e^{-i \left (c+d \sqrt{x}\right )}\right )+24 a b \text{PolyLog}\left (3,-e^{-i \left (c+d \sqrt{x}\right )}\right )+24 a b e^{2 i c} \text{PolyLog}\left (3,e^{-i \left (c+d \sqrt{x}\right )}\right )-24 a b \text{PolyLog}\left (3,e^{-i \left (c+d \sqrt{x}\right )}\right )+2 a^2 e^{2 i c} d^3 x^{3/2}-2 a^2 d^3 x^{3/2}+12 a b e^{2 i c} d^2 x \log \left (1-e^{-i \left (c+d \sqrt{x}\right )}\right )-12 a b d^2 x \log \left (1-e^{-i \left (c+d \sqrt{x}\right )}\right )-12 a b e^{2 i c} d^2 x \log \left (1+e^{-i \left (c+d \sqrt{x}\right )}\right )+12 a b d^2 x \log \left (1+e^{-i \left (c+d \sqrt{x}\right )}\right )+3 b^2 e^{2 i c} d^2 x \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d \sqrt{x}}{2}\right ) \csc \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )-3 b^2 d^2 x \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d \sqrt{x}}{2}\right ) \csc \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )+3 b^2 e^{2 i c} d^2 x \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d \sqrt{x}}{2}\right ) \sec \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )-3 b^2 d^2 x \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d \sqrt{x}}{2}\right ) \sec \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )+12 b^2 e^{2 i c} d \sqrt{x} \log \left (1-e^{-i \left (c+d \sqrt{x}\right )}\right )-12 b^2 d \sqrt{x} \log \left (1-e^{-i \left (c+d \sqrt{x}\right )}\right )+12 b^2 e^{2 i c} d \sqrt{x} \log \left (1+e^{-i \left (c+d \sqrt{x}\right )}\right )-12 b^2 d \sqrt{x} \log \left (1+e^{-i \left (c+d \sqrt{x}\right )}\right )-12 i b^2 d^2 x}{3 \left (-1+e^{2 i c}\right ) d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*I)*b^2*d^2*x - 2*a^2*d^3*x^(3/2) + 2*a^2*d^3*E^((2*I)*c)*x^(3/2) - 12*b^2*d*Sqrt[x]*Log[1 - E^((-I)*(c +
d*Sqrt[x]))] + 12*b^2*d*E^((2*I)*c)*Sqrt[x]*Log[1 - E^((-I)*(c + d*Sqrt[x]))] - 12*a*b*d^2*x*Log[1 - E^((-I)*
(c + d*Sqrt[x]))] + 12*a*b*d^2*E^((2*I)*c)*x*Log[1 - E^((-I)*(c + d*Sqrt[x]))] - 12*b^2*d*Sqrt[x]*Log[1 + E^((
-I)*(c + d*Sqrt[x]))] + 12*b^2*d*E^((2*I)*c)*Sqrt[x]*Log[1 + E^((-I)*(c + d*Sqrt[x]))] + 12*a*b*d^2*x*Log[1 +
E^((-I)*(c + d*Sqrt[x]))] - 12*a*b*d^2*E^((2*I)*c)*x*Log[1 + E^((-I)*(c + d*Sqrt[x]))] + (12*I)*b*(-1 + E^((2*
I)*c))*(b - 2*a*d*Sqrt[x])*PolyLog[2, -E^((-I)*(c + d*Sqrt[x]))] + (12*I)*b*(-1 + E^((2*I)*c))*(b + 2*a*d*Sqrt
[x])*PolyLog[2, E^((-I)*(c + d*Sqrt[x]))] + 24*a*b*PolyLog[3, -E^((-I)*(c + d*Sqrt[x]))] - 24*a*b*E^((2*I)*c)*
PolyLog[3, -E^((-I)*(c + d*Sqrt[x]))] - 24*a*b*PolyLog[3, E^((-I)*(c + d*Sqrt[x]))] + 24*a*b*E^((2*I)*c)*PolyL
og[3, E^((-I)*(c + d*Sqrt[x]))] - 3*b^2*d^2*x*Csc[c/2]*Csc[(c + d*Sqrt[x])/2]*Sin[(d*Sqrt[x])/2] + 3*b^2*d^2*E
^((2*I)*c)*x*Csc[c/2]*Csc[(c + d*Sqrt[x])/2]*Sin[(d*Sqrt[x])/2] - 3*b^2*d^2*x*Sec[c/2]*Sec[(c + d*Sqrt[x])/2]*
Sin[(d*Sqrt[x])/2] + 3*b^2*d^2*E^((2*I)*c)*x*Sec[c/2]*Sec[(c + d*Sqrt[x])/2]*Sin[(d*Sqrt[x])/2])/(3*d^3*(-1 +
E^((2*I)*c)))

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Maple [F]  time = 0.197, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\csc \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*csc(c+d*x^(1/2)))^2*x^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.44613, size = 1648, normalized size = 6.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(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(cot(d*sqrt(
x) + c) + csc(d*sqrt(x) + c)) - 3*(2*b^2*c^2 - (2*(d*sqrt(x) + c)^2*a*b + 2*b^2*c - 2*(2*a*b*c + b^2)*(d*sqrt(
x) + c) - 2*((d*sqrt(x) + c)^2*a*b + b^2*c - (2*a*b*c + b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (2*I*(d
*sqrt(x) + c)^2*a*b + 2*I*b^2*c + (-4*I*a*b*c - 2*I*b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(
d*sqrt(x) + c), cos(d*sqrt(x) + c) + 1) + 2*(b^2*c*cos(2*d*sqrt(x) + 2*c) + I*b^2*c*sin(2*d*sqrt(x) + 2*c) - b
^2*c)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) - 1) - (2*(d*sqrt(x) + c)^2*a*b - 2*(2*a*b*c - b^2)*(d*sq
rt(x) + c) - 2*((d*sqrt(x) + c)^2*a*b - (2*a*b*c - b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (2*I*(d*sqrt
(x) + c)^2*a*b + (-4*I*a*b*c + 2*I*b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(d*sqrt(x) + c), -
cos(d*sqrt(x) + 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) + (4*(d*s
qrt(x) + c)*a*b - 4*a*b*c - 2*b^2 - 2*(2*(d*sqrt(x) + c)*a*b - 2*a*b*c - b^2)*cos(2*d*sqrt(x) + 2*c) + (-4*I*(
d*sqrt(x) + c)*a*b + 4*I*a*b*c + 2*I*b^2)*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(I*d*sqrt(x) + I*c)) - (4*(d*sqrt(x
) + c)*a*b - 4*a*b*c + 2*b^2 - 2*(2*(d*sqrt(x) + c)*a*b - 2*a*b*c + b^2)*cos(2*d*sqrt(x) + 2*c) - (4*I*(d*sqrt
(x) + c)*a*b - 4*I*a*b*c + 2*I*b^2)*sin(2*d*sqrt(x) + 2*c))*dilog(e^(I*d*sqrt(x) + I*c)) + (I*(d*sqrt(x) + c)^
2*a*b + I*b^2*c + (-2*I*a*b*c - I*b^2)*(d*sqrt(x) + c) + (-I*(d*sqrt(x) + c)^2*a*b - I*b^2*c + (2*I*a*b*c + I*
b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + ((d*sqrt(x) + c)^2*a*b + b^2*c - (2*a*b*c + b^2)*(d*sqrt(x) + c
))*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) + (-I*(
d*sqrt(x) + c)^2*a*b + I*b^2*c + (2*I*a*b*c - I*b^2)*(d*sqrt(x) + c) + (I*(d*sqrt(x) + c)^2*a*b - I*b^2*c + (-
2*I*a*b*c + I*b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - ((d*sqrt(x) + c)^2*a*b - b^2*c - (2*a*b*c - b^2)*
(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*cos(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, -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, 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} \csc \left (d \sqrt{x} + c\right )^{2} + 2 \, a b \sqrt{x} \csc \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*csc(c+d*x^(1/2)))^2*x^(1/2),x, algorithm="fricas")

[Out]

integral(b^2*sqrt(x)*csc(d*sqrt(x) + c)^2 + 2*a*b*sqrt(x)*csc(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 \csc{\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*csc(c+d*x**(1/2)))**2*x**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \csc \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*csc(c+d*x^(1/2)))^2*x^(1/2),x, algorithm="giac")

[Out]

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