3.1 \(\int x^5 (a+b \csc (c+d x^2)) \, dx\)

Optimal. Leaf size=129 \[ \frac{i b x^2 \text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i b x^2 \text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{b \text{PolyLog}\left (3,-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{b \text{PolyLog}\left (3,e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{a x^6}{6}-\frac{b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.149944, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {14, 4205, 4183, 2531, 2282, 6589} \[ \frac{i b x^2 \text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i b x^2 \text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{b \text{PolyLog}\left (3,-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{b \text{PolyLog}\left (3,e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{a x^6}{6}-\frac{b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Csc[c + d*x^2]),x]

[Out]

(a*x^6)/6 - (b*x^4*ArcTanh[E^(I*(c + d*x^2))])/d + (I*b*x^2*PolyLog[2, -E^(I*(c + d*x^2))])/d^2 - (I*b*x^2*Pol
yLog[2, E^(I*(c + d*x^2))])/d^2 - (b*PolyLog[3, -E^(I*(c + d*x^2))])/d^3 + (b*PolyLog[3, E^(I*(c + d*x^2))])/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 x^5 \left (a+b \csc \left (c+d x^2\right )\right ) \, dx &=\int \left (a x^5+b x^5 \csc \left (c+d x^2\right )\right ) \, dx\\ &=\frac{a x^6}{6}+b \int x^5 \csc \left (c+d x^2\right ) \, dx\\ &=\frac{a x^6}{6}+\frac{1}{2} b \operatorname{Subst}\left (\int x^2 \csc (c+d x) \, dx,x,x^2\right )\\ &=\frac{a x^6}{6}-\frac{b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac{b \operatorname{Subst}\left (\int x \log \left (1-e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac{b \operatorname{Subst}\left (\int x \log \left (1+e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}\\ &=\frac{a x^6}{6}-\frac{b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{i b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{(i b) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}+\frac{(i b) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}\\ &=\frac{a x^6}{6}-\frac{b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{i b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{b \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{b \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}\\ &=\frac{a x^6}{6}-\frac{b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{i b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{b \text{Li}_3\left (-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{b \text{Li}_3\left (e^{i \left (c+d x^2\right )}\right )}{d^3}\\ \end{align*}

Mathematica [A]  time = 0.25874, size = 159, normalized size = 1.23 \[ \frac{a x^6}{6}-\frac{b \left (-i d x^2 \text{PolyLog}\left (2,-\cos \left (c+d x^2\right )-i \sin \left (c+d x^2\right )\right )+i d x^2 \text{PolyLog}\left (2,\cos \left (c+d x^2\right )+i \sin \left (c+d x^2\right )\right )+\text{PolyLog}\left (3,-\cos \left (c+d x^2\right )-i \sin \left (c+d x^2\right )\right )-\text{PolyLog}\left (3,\cos \left (c+d x^2\right )+i \sin \left (c+d x^2\right )\right )+d^2 x^4 \tanh ^{-1}\left (\cos \left (c+d x^2\right )+i \sin \left (c+d x^2\right )\right )\right )}{d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^5*(a + b*Csc[c + d*x^2]),x]

[Out]

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

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Maple [F]  time = 0.15, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \left ( a+b\csc \left ( d{x}^{2}+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*csc(d*x^2+c)),x)

[Out]

int(x^5*(a+b*csc(d*x^2+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, a x^{6} + b{\left (\int \frac{x^{5} \sin \left (d x^{2} + c\right )}{\cos \left (d x^{2} + c\right )^{2} + \sin \left (d x^{2} + c\right )^{2} + 2 \, \cos \left (d x^{2} + c\right ) + 1}\,{d x} + \int \frac{x^{5} \sin \left (d x^{2} + c\right )}{\cos \left (d x^{2} + c\right )^{2} + \sin \left (d x^{2} + c\right )^{2} - 2 \, \cos \left (d x^{2} + c\right ) + 1}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*csc(d*x^2+c)),x, algorithm="maxima")

[Out]

1/6*a*x^6 + b*(integrate(x^5*sin(d*x^2 + c)/(cos(d*x^2 + c)^2 + sin(d*x^2 + c)^2 + 2*cos(d*x^2 + c) + 1), x) +
integrate(x^5*sin(d*x^2 + c)/(cos(d*x^2 + c)^2 + sin(d*x^2 + c)^2 - 2*cos(d*x^2 + c) + 1), x))

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Fricas [C]  time = 0.567168, size = 1126, normalized size = 8.73 \begin{align*} \frac{2 \, a d^{3} x^{6} - 3 \, b d^{2} x^{4} \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, b d^{2} x^{4} \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + 1\right ) - 6 i \, b d x^{2}{\rm Li}_2\left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right )\right ) + 6 i \, b d x^{2}{\rm Li}_2\left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right )\right ) - 6 i \, b d x^{2}{\rm Li}_2\left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right )\right ) + 6 i \, b d x^{2}{\rm Li}_2\left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right )\right ) + 3 \, b c^{2} \log \left (-\frac{1}{2} \, \cos \left (d x^{2} + c\right ) + \frac{1}{2} i \, \sin \left (d x^{2} + c\right ) + \frac{1}{2}\right ) + 3 \, b c^{2} \log \left (-\frac{1}{2} \, \cos \left (d x^{2} + c\right ) - \frac{1}{2} i \, \sin \left (d x^{2} + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \,{\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + 1\right ) + 6 \, b{\rm polylog}\left (3, \cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right )\right ) + 6 \, b{\rm polylog}\left (3, \cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right )\right ) - 6 \, b{\rm polylog}\left (3, -\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right )\right ) - 6 \, b{\rm polylog}\left (3, -\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right )\right )}{12 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*csc(d*x^2+c)),x, algorithm="fricas")

[Out]

1/12*(2*a*d^3*x^6 - 3*b*d^2*x^4*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + 1) - 3*b*d^2*x^4*log(cos(d*x^2 + c) -
I*sin(d*x^2 + c) + 1) - 6*I*b*d*x^2*dilog(cos(d*x^2 + c) + I*sin(d*x^2 + c)) + 6*I*b*d*x^2*dilog(cos(d*x^2 + c
) - I*sin(d*x^2 + c)) - 6*I*b*d*x^2*dilog(-cos(d*x^2 + c) + I*sin(d*x^2 + c)) + 6*I*b*d*x^2*dilog(-cos(d*x^2 +
c) - I*sin(d*x^2 + c)) + 3*b*c^2*log(-1/2*cos(d*x^2 + c) + 1/2*I*sin(d*x^2 + c) + 1/2) + 3*b*c^2*log(-1/2*cos
(d*x^2 + c) - 1/2*I*sin(d*x^2 + c) + 1/2) + 3*(b*d^2*x^4 - b*c^2)*log(-cos(d*x^2 + c) + I*sin(d*x^2 + c) + 1)
+ 3*(b*d^2*x^4 - b*c^2)*log(-cos(d*x^2 + c) - I*sin(d*x^2 + c) + 1) + 6*b*polylog(3, cos(d*x^2 + c) + I*sin(d*
x^2 + c)) + 6*b*polylog(3, cos(d*x^2 + c) - I*sin(d*x^2 + c)) - 6*b*polylog(3, -cos(d*x^2 + c) + I*sin(d*x^2 +
c)) - 6*b*polylog(3, -cos(d*x^2 + c) - I*sin(d*x^2 + c)))/d^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \left (a + b \csc{\left (c + d x^{2} \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*csc(d*x**2+c)),x)

[Out]

Integral(x**5*(a + b*csc(c + d*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \csc \left (d x^{2} + c\right ) + a\right )} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*csc(d*x^2+c)),x, algorithm="giac")

[Out]

integrate((b*csc(d*x^2 + c) + a)*x^5, x)