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

Optimal. Leaf size=200 \[ \frac{6 i b x \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 i b x \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 b \sqrt{x} \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b \sqrt{x} \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{12 i b \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{12 i b \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

[Out]

(a*x^2)/2 - (4*b*x^(3/2)*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((6*I)*b*x*PolyLog[2, -E^(I*(c + d*Sqrt[x]))])/d^
2 - ((6*I)*b*x*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - (12*b*Sqrt[x]*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d^3
+ (12*b*Sqrt[x]*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 - ((12*I)*b*PolyLog[4, -E^(I*(c + d*Sqrt[x]))])/d^4 + (
(12*I)*b*PolyLog[4, E^(I*(c + d*Sqrt[x]))])/d^4

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Rubi [A]  time = 0.177666, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {14, 4205, 4183, 2531, 6609, 2282, 6589} \[ \frac{6 i b x \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 i b x \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 b \sqrt{x} \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b \sqrt{x} \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{12 i b \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{12 i b \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Csc[c + d*Sqrt[x]]),x]

[Out]

(a*x^2)/2 - (4*b*x^(3/2)*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((6*I)*b*x*PolyLog[2, -E^(I*(c + d*Sqrt[x]))])/d^
2 - ((6*I)*b*x*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - (12*b*Sqrt[x]*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d^3
+ (12*b*Sqrt[x]*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 - ((12*I)*b*PolyLog[4, -E^(I*(c + d*Sqrt[x]))])/d^4 + (
(12*I)*b*PolyLog[4, E^(I*(c + d*Sqrt[x]))])/d^4

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 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]

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 x \left (a+b \csc \left (c+d \sqrt{x}\right )\right ) \, dx &=\int \left (a x+b x \csc \left (c+d \sqrt{x}\right )\right ) \, dx\\ &=\frac{a x^2}{2}+b \int x \csc \left (c+d \sqrt{x}\right ) \, dx\\ &=\frac{a x^2}{2}+(2 b) \operatorname{Subst}\left (\int x^3 \csc (c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{(6 b) \operatorname{Subst}\left (\int x^2 \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(6 b) \operatorname{Subst}\left (\int x^2 \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{6 i b x \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 i b x \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(12 i b) \operatorname{Subst}\left (\int x \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(12 i b) \operatorname{Subst}\left (\int x \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{6 i b x \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 i b x \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 b \sqrt{x} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b \sqrt{x} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{(12 b) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{(12 b) \operatorname{Subst}\left (\int \text{Li}_3\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{6 i b x \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 i b x \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 b \sqrt{x} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b \sqrt{x} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{(12 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{(12 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}\\ &=\frac{a x^2}{2}-\frac{4 b x^{3/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{6 i b x \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 i b x \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 b \sqrt{x} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b \sqrt{x} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{12 i b \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{12 i b \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}\\ \end{align*}

Mathematica [A]  time = 0.376526, size = 260, normalized size = 1.3 \[ \frac{a x^2}{2}-\frac{2 b \left (-3 i d^2 x \text{PolyLog}\left (2,-\cos \left (c+d \sqrt{x}\right )-i \sin \left (c+d \sqrt{x}\right )\right )+3 i d^2 x \text{PolyLog}\left (2,\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )+6 d \sqrt{x} \text{PolyLog}\left (3,-\cos \left (c+d \sqrt{x}\right )-i \sin \left (c+d \sqrt{x}\right )\right )-6 d \sqrt{x} \text{PolyLog}\left (3,\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )+6 i \text{PolyLog}\left (4,-\cos \left (c+d \sqrt{x}\right )-i \sin \left (c+d \sqrt{x}\right )\right )-6 i \text{PolyLog}\left (4,\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )+2 d^3 x^{3/2} \tanh ^{-1}\left (\cos \left (c+d \sqrt{x}\right )+i \sin \left (c+d \sqrt{x}\right )\right )\right )}{d^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x^2)/2 - (2*b*(2*d^3*x^(3/2)*ArcTanh[Cos[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]] - (3*I)*d^2*x*PolyLog[2, -C
os[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] + (3*I)*d^2*x*PolyLog[2, Cos[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]]
+ 6*d*Sqrt[x]*PolyLog[3, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] - 6*d*Sqrt[x]*PolyLog[3, Cos[c + d*Sqrt[x
]] + I*Sin[c + d*Sqrt[x]]] + (6*I)*PolyLog[4, -Cos[c + d*Sqrt[x]] - I*Sin[c + d*Sqrt[x]]] - (6*I)*PolyLog[4, C
os[c + d*Sqrt[x]] + I*Sin[c + d*Sqrt[x]]]))/d^4

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Maple [F]  time = 0.11, size = 0, normalized size = 0. \begin{align*} \int 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]

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

[Out]

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

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Maxima [B]  time = 1.38328, size = 721, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(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 )} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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