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

Optimal. Leaf size=144 \[ \frac{4 i b \sqrt{x} \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 b \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{4 b \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a x^{3/2}-\frac{4 b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

[Out]

(2*a*x^(3/2))/3 - (4*b*x*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((4*I)*b*Sqrt[x]*PolyLog[2, -E^(I*(c + d*Sqrt[x])
)])/d^2 - ((4*I)*b*Sqrt[x]*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - (4*b*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d
^3 + (4*b*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3

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Rubi [A]  time = 0.125561, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {14, 4205, 4183, 2531, 2282, 6589} \[ \frac{4 i b \sqrt{x} \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 b \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{4 b \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a x^{3/2}-\frac{4 b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*x^(3/2))/3 - (4*b*x*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((4*I)*b*Sqrt[x]*PolyLog[2, -E^(I*(c + d*Sqrt[x])
)])/d^2 - ((4*I)*b*Sqrt[x]*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - (4*b*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d
^3 + (4*b*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3

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 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 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]

Rubi steps

\begin{align*} \int \sqrt{x} \left (a+b \csc \left (c+d \sqrt{x}\right )\right ) \, dx &=\int \left (a \sqrt{x}+b \sqrt{x} \csc \left (c+d \sqrt{x}\right )\right ) \, dx\\ &=\frac{2}{3} a x^{3/2}+b \int \sqrt{x} \csc \left (c+d \sqrt{x}\right ) \, dx\\ &=\frac{2}{3} a x^{3/2}+(2 b) \operatorname{Subst}\left (\int x^2 \csc (c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a x^{3/2}-\frac{4 b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{(4 b) \operatorname{Subst}\left (\int x \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(4 b) \operatorname{Subst}\left (\int x \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2}{3} a x^{3/2}-\frac{4 b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 i b \sqrt{x} \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(4 i b) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(4 i b) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{2}{3} a x^{3/2}-\frac{4 b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 i b \sqrt{x} \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(4 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{(4 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{2}{3} a x^{3/2}-\frac{4 b x \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 i b \sqrt{x} \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 i b \sqrt{x} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 b \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{4 b \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 3.54977, size = 191, normalized size = 1.33 \[ \frac{2 \left (6 i b d \sqrt{x} \text{PolyLog}\left (2,-\cos \left (c+d \sqrt{x}\right )-i \sin \left (c+d \sqrt{x}\right )\right )-6 i b d \sqrt{x} \text{PolyLog}\left (2,\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )-6 b \text{PolyLog}\left (3,-\cos \left (c+d \sqrt{x}\right )-i \sin \left (c+d \sqrt{x}\right )\right )+6 b \text{PolyLog}\left (3,\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )+a d^3 x^{3/2}-6 b d^2 x \tanh ^{-1}\left (\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )\right )}{3 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(a*d^3*x^(3/2) - 6*b*d^2*x*ArcTanh[Cos[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]] + (6*I)*b*d*Sqrt[x]*PolyLog[2
, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] - (6*I)*b*d*Sqrt[x]*PolyLog[2, Cos[c + d*Sqrt[x]] + I*Sin[c + d*
Sqrt[x]]] - 6*b*PolyLog[3, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] + 6*b*PolyLog[3, Cos[c + d*Sqrt[x]] + I
*Sin[c + d*Sqrt[x]]]))/(3*d^3)

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Maple [F]  time = 0.116, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\csc \left ( c+d\sqrt{x} \right ) \right ) \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)))*x^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.38325, size = 500, normalized size = 3.47 \begin{align*} \frac{2 \,{\left (d \sqrt{x} + c\right )}^{3} a - 6 \,{\left (d \sqrt{x} + c\right )}^{2} a c + 6 \,{\left (d \sqrt{x} + c\right )} a c^{2} - 6 \, b c^{2} \log \left (\cot \left (d \sqrt{x} + c\right ) + \csc \left (d \sqrt{x} + c\right )\right ) - 3 \,{\left (2 i \,{\left (d \sqrt{x} + c\right )}^{2} b - 4 i \,{\left (d \sqrt{x} + c\right )} b c\right )} \arctan \left (\sin \left (d \sqrt{x} + c\right ), \cos \left (d \sqrt{x} + c\right ) + 1\right ) - 3 \,{\left (2 i \,{\left (d \sqrt{x} + c\right )}^{2} b - 4 i \,{\left (d \sqrt{x} + c\right )} b c\right )} \arctan \left (\sin \left (d \sqrt{x} + c\right ), -\cos \left (d \sqrt{x} + c\right ) + 1\right ) - 3 \,{\left (-4 i \,{\left (d \sqrt{x} + c\right )} b + 4 i \, b c\right )}{\rm Li}_2\left (-e^{\left (i \, d \sqrt{x} + i \, c\right )}\right ) - 3 \,{\left (4 i \,{\left (d \sqrt{x} + c\right )} b - 4 i \, b c\right )}{\rm Li}_2\left (e^{\left (i \, d \sqrt{x} + i \, c\right )}\right ) - 3 \,{\left ({\left (d \sqrt{x} + c\right )}^{2} b - 2 \,{\left (d \sqrt{x} + c\right )} b c\right )} \log \left (\cos \left (d \sqrt{x} + c\right )^{2} + \sin \left (d \sqrt{x} + c\right )^{2} + 2 \, \cos \left (d \sqrt{x} + c\right ) + 1\right ) + 3 \,{\left ({\left (d \sqrt{x} + c\right )}^{2} b - 2 \,{\left (d \sqrt{x} + c\right )} b c\right )} \log \left (\cos \left (d \sqrt{x} + c\right )^{2} + \sin \left (d \sqrt{x} + c\right )^{2} - 2 \, \cos \left (d \sqrt{x} + c\right ) + 1\right ) - 12 \, b{\rm Li}_{3}(-e^{\left (i \, d \sqrt{x} + i \, c\right )}) + 12 \, b{\rm Li}_{3}(e^{\left (i \, d \sqrt{x} + i \, c\right )})}{3 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*(d*sqrt(x) + c)^3*a - 6*(d*sqrt(x) + c)^2*a*c + 6*(d*sqrt(x) + c)*a*c^2 - 6*b*c^2*log(cot(d*sqrt(x) + c
) + csc(d*sqrt(x) + c)) - 3*(2*I*(d*sqrt(x) + c)^2*b - 4*I*(d*sqrt(x) + c)*b*c)*arctan2(sin(d*sqrt(x) + c), co
s(d*sqrt(x) + c) + 1) - 3*(2*I*(d*sqrt(x) + c)^2*b - 4*I*(d*sqrt(x) + c)*b*c)*arctan2(sin(d*sqrt(x) + c), -cos
(d*sqrt(x) + c) + 1) - 3*(-4*I*(d*sqrt(x) + c)*b + 4*I*b*c)*dilog(-e^(I*d*sqrt(x) + I*c)) - 3*(4*I*(d*sqrt(x)
+ c)*b - 4*I*b*c)*dilog(e^(I*d*sqrt(x) + I*c)) - 3*((d*sqrt(x) + c)^2*b - 2*(d*sqrt(x) + c)*b*c)*log(cos(d*sqr
t(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) + 3*((d*sqrt(x) + c)^2*b - 2*(d*sqrt(x) + c)*b*
c)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*cos(d*sqrt(x) + c) + 1) - 12*b*polylog(3, -e^(I*d*sqrt(
x) + I*c)) + 12*b*polylog(3, e^(I*d*sqrt(x) + I*c)))/d^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b \sqrt{x} \csc \left (d \sqrt{x} + c\right ) + a \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)))*x^(1/2),x, algorithm="fricas")

[Out]

integral(b*sqrt(x)*csc(d*sqrt(x) + c) + a*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 )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x)*(a + b*csc(c + d*sqrt(x))), 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 )} \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)))*x^(1/2),x, algorithm="giac")

[Out]

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