3.3 \(\int x^3 (a+b \csc (c+d x^2)) \, dx\)

Optimal. Leaf size=84 \[ \frac{i b \text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac{i b \text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )}{2 d^2}+\frac{a x^4}{4}-\frac{b x^2 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0825487, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {14, 4205, 4183, 2279, 2391} \[ \frac{i b \text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac{i b \text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )}{2 d^2}+\frac{a x^4}{4}-\frac{b x^2 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Csc[c + d*x^2]),x]

[Out]

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

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

Mathematica [A]  time = 0.093025, size = 118, normalized size = 1.4 \[ \frac{a x^4}{4}+\frac{b \left (i \left (\text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )-\text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )\right )+\left (c+d x^2\right ) \left (\log \left (1-e^{i \left (c+d x^2\right )}\right )-\log \left (1+e^{i \left (c+d x^2\right )}\right )\right )-c \log \left (\tan \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Csc[c + d*x^2]),x]

[Out]

(a*x^4)/4 + (b*((c + d*x^2)*(Log[1 - E^(I*(c + d*x^2))] - Log[1 + E^(I*(c + d*x^2))]) - c*Log[Tan[(c + d*x^2)/
2]] + I*(PolyLog[2, -E^(I*(c + d*x^2))] - PolyLog[2, E^(I*(c + d*x^2))])))/(2*d^2)

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Maple [F]  time = 0.118, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \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^3*(a+b*csc(d*x^2+c)),x)

[Out]

int(x^3*(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}{4} \, a x^{4} + b{\left (\int \frac{x^{3} \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^{3} \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^3*(a+b*csc(d*x^2+c)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + b*(integrate(x^3*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^3*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 [B]  time = 0.547048, size = 757, normalized size = 9.01 \begin{align*} \frac{a d^{2} x^{4} - b d x^{2} \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + 1\right ) - b d x^{2} \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + 1\right ) - b c \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 ) - b c \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 ) - i \, b{\rm Li}_2\left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right )\right ) + i \, b{\rm Li}_2\left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right )\right ) - i \, b{\rm Li}_2\left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right )\right ) + i \, b{\rm Li}_2\left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right )\right ) +{\left (b d x^{2} + b c\right )} \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + 1\right ) +{\left (b d x^{2} + b c\right )} \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + 1\right )}{4 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*csc(d*x^2+c)),x, algorithm="fricas")

[Out]

1/4*(a*d^2*x^4 - b*d*x^2*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + 1) - b*d*x^2*log(cos(d*x^2 + c) - I*sin(d*x^2
+ c) + 1) - b*c*log(-1/2*cos(d*x^2 + c) + 1/2*I*sin(d*x^2 + c) + 1/2) - b*c*log(-1/2*cos(d*x^2 + c) - 1/2*I*s
in(d*x^2 + c) + 1/2) - I*b*dilog(cos(d*x^2 + c) + I*sin(d*x^2 + c)) + I*b*dilog(cos(d*x^2 + c) - I*sin(d*x^2 +
c)) - I*b*dilog(-cos(d*x^2 + c) + I*sin(d*x^2 + c)) + I*b*dilog(-cos(d*x^2 + c) - I*sin(d*x^2 + c)) + (b*d*x^
2 + b*c)*log(-cos(d*x^2 + c) + I*sin(d*x^2 + c) + 1) + (b*d*x^2 + b*c)*log(-cos(d*x^2 + c) - I*sin(d*x^2 + c)
+ 1))/d^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**3*(a+b*csc(d*x**2+c)),x)

[Out]

Integral(x**3*(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^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*csc(d*x^2+c)),x, algorithm="giac")

[Out]

integrate((b*csc(d*x^2 + c) + a)*x^3, x)