3.48 \(\int \frac{a+b \csc ^{-1}(c x)}{d+e x} \, dx\)

Optimal. Leaf size=257 \[ -\frac{i b \text{PolyLog}\left (2,\frac{i \left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{PolyLog}\left (2,\frac{i \left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )}{2 e}+\frac{\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i \left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i \left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}-\frac{\log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \left (a+b \csc ^{-1}(c x)\right )}{e} \]

[Out]

((a + b*ArcCsc[c*x])*Log[1 - (I*(e - Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e + ((a + b*ArcCsc[c*x
])*Log[1 - (I*(e + Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e - ((a + b*ArcCsc[c*x])*Log[1 - E^((2*I
)*ArcCsc[c*x])])/e - (I*b*PolyLog[2, (I*(e - Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e - (I*b*PolyL
og[2, (I*(e + Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e + ((I/2)*b*PolyLog[2, E^((2*I)*ArcCsc[c*x])
])/e

________________________________________________________________________________________

Rubi [A]  time = 0.405979, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5225, 2518} \[ -\frac{i b \text{PolyLog}\left (2,\frac{i \left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{PolyLog}\left (2,\frac{i \left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )}{2 e}+\frac{\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i \left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \csc ^{-1}(c x)\right ) \log \left (1-\frac{i \left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \csc ^{-1}(c x)}}{c d}\right )}{e}-\frac{\log \left (1-e^{2 i \csc ^{-1}(c x)}\right ) \left (a+b \csc ^{-1}(c x)\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsc[c*x])/(d + e*x),x]

[Out]

((a + b*ArcCsc[c*x])*Log[1 - (I*(e - Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e + ((a + b*ArcCsc[c*x
])*Log[1 - (I*(e + Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e - ((a + b*ArcCsc[c*x])*Log[1 - E^((2*I
)*ArcCsc[c*x])])/e - (I*b*PolyLog[2, (I*(e - Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e - (I*b*PolyL
og[2, (I*(e + Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e + ((I/2)*b*PolyLog[2, E^((2*I)*ArcCsc[c*x])
])/e

Rule 5225

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*ArcCsc[c*x])*Log[1 - (I
*(e - Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e, x] + (Dist[b/(c*e), Int[Log[1 - (I*(e - Sqrt[-(c^2
*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)]/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x] + Dist[b/(c*e), Int[Log[1 - (I*(e
+ Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)]/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x] - Dist[b/(c*e), Int[Lo
g[1 - E^(2*I*ArcCsc[c*x])]/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x] + Simp[((a + b*ArcCsc[c*x])*Log[1 - (I*(e + Sqr
t[-(c^2*d^2) + e^2])*E^(I*ArcCsc[c*x]))/(c*d)])/e, x] - Simp[((a + b*ArcCsc[c*x])*Log[1 - E^(2*I*ArcCsc[c*x])]
)/e, x]) /; FreeQ[{a, b, c, d, e}, x]

Rule 2518

Int[Log[v_]*(u_), x_Symbol] :> With[{w = DerivativeDivides[v, u*(1 - v), x]}, Simp[w*PolyLog[2, 1 - v], x] /;
 !FalseQ[w]]

Rubi steps

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

Mathematica [A]  time = 0.658497, size = 411, normalized size = 1.6 \[ \frac{a \log (d+e x)}{e}+\frac{b \left (8 i \left (\text{PolyLog}\left (2,\frac{i \left (\sqrt{e^2-c^2 d^2}-e\right ) e^{-i \csc ^{-1}(c x)}}{c d}\right )+\text{PolyLog}\left (2,-\frac{i \left (\sqrt{e^2-c^2 d^2}+e\right ) e^{-i \csc ^{-1}(c x)}}{c d}\right )\right )+4 i \left (\csc ^{-1}(c x)^2+\text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )\right )-4 \log \left (1+\frac{i \left (e-\sqrt{e^2-c^2 d^2}\right ) e^{-i \csc ^{-1}(c x)}}{c d}\right ) \left (4 \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right )-2 \csc ^{-1}(c x)+\pi \right )-4 \log \left (1+\frac{i \left (\sqrt{e^2-c^2 d^2}+e\right ) e^{-i \csc ^{-1}(c x)}}{c d}\right ) \left (-4 \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right )-2 \csc ^{-1}(c x)+\pi \right )+32 i \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(c d-e) \cot \left (\frac{1}{4} \left (2 \csc ^{-1}(c x)+\pi \right )\right )}{\sqrt{e^2-c^2 d^2}}\right )+4 \left (\pi -2 \csc ^{-1}(c x)\right ) \log \left (\frac{d}{x}+e\right )+8 \csc ^{-1}(c x) \log \left (\frac{d}{x}+e\right )+i \left (\pi -2 \csc ^{-1}(c x)\right )^2-8 \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )\right )}{8 e} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCsc[c*x])/(d + e*x),x]

[Out]

(a*Log[d + e*x])/e + (b*(I*(Pi - 2*ArcCsc[c*x])^2 + (32*I)*ArcSin[Sqrt[1 + e/(c*d)]/Sqrt[2]]*ArcTan[((c*d - e)
*Cot[(Pi + 2*ArcCsc[c*x])/4])/Sqrt[-(c^2*d^2) + e^2]] - 4*(Pi - 2*ArcCsc[c*x] + 4*ArcSin[Sqrt[1 + e/(c*d)]/Sqr
t[2]])*Log[1 + (I*(e - Sqrt[-(c^2*d^2) + e^2]))/(c*d*E^(I*ArcCsc[c*x]))] - 4*(Pi - 2*ArcCsc[c*x] - 4*ArcSin[Sq
rt[1 + e/(c*d)]/Sqrt[2]])*Log[1 + (I*(e + Sqrt[-(c^2*d^2) + e^2]))/(c*d*E^(I*ArcCsc[c*x]))] - 8*ArcCsc[c*x]*Lo
g[1 - E^((2*I)*ArcCsc[c*x])] + 4*(Pi - 2*ArcCsc[c*x])*Log[e + d/x] + 8*ArcCsc[c*x]*Log[e + d/x] + (8*I)*(PolyL
og[2, (I*(-e + Sqrt[-(c^2*d^2) + e^2]))/(c*d*E^(I*ArcCsc[c*x]))] + PolyLog[2, ((-I)*(e + Sqrt[-(c^2*d^2) + e^2
]))/(c*d*E^(I*ArcCsc[c*x]))]) + (4*I)*(ArcCsc[c*x]^2 + PolyLog[2, E^((2*I)*ArcCsc[c*x])])))/(8*e)

________________________________________________________________________________________

Maple [B]  time = 0.417, size = 881, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsc(c*x))/(e*x+d),x)

[Out]

a*ln(c*e*x+c*d)/e-b*e*arccsc(c*x)/(c^2*d^2-e^2)*ln((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e-(c^2*d^2-e^2)^(1/2))/(
I*e-(c^2*d^2-e^2)^(1/2)))-b*e*arccsc(c*x)/(c^2*d^2-e^2)*ln((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e+(c^2*d^2-e^2)^
(1/2))/(I*e+(c^2*d^2-e^2)^(1/2)))+I*b/e*dilog(1+I/c/x+(1-1/c^2/x^2)^(1/2))-b/e*arccsc(c*x)*ln(1+I/c/x+(1-1/c^2
/x^2)^(1/2))-I*c^2*b/e/(c^2*d^2-e^2)*dilog((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e-(c^2*d^2-e^2)^(1/2))/(I*e-(c^2
*d^2-e^2)^(1/2)))*d^2-I*c^2*b/e/(c^2*d^2-e^2)*dilog((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e+(c^2*d^2-e^2)^(1/2))/
(I*e+(c^2*d^2-e^2)^(1/2)))*d^2+c^2*b*d^2/e*arccsc(c*x)/(c^2*d^2-e^2)*ln((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e-(
c^2*d^2-e^2)^(1/2))/(I*e-(c^2*d^2-e^2)^(1/2)))+c^2*b*d^2/e*arccsc(c*x)/(c^2*d^2-e^2)*ln((d*c*(I/c/x+(1-1/c^2/x
^2)^(1/2))+I*e+(c^2*d^2-e^2)^(1/2))/(I*e+(c^2*d^2-e^2)^(1/2)))-I*b/e*dilog(I/c/x+(1-1/c^2/x^2)^(1/2))+I*b*e/(c
^2*d^2-e^2)*dilog((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e+(c^2*d^2-e^2)^(1/2))/(I*e+(c^2*d^2-e^2)^(1/2)))+I*b*e/(
c^2*d^2-e^2)*dilog((d*c*(I/c/x+(1-1/c^2/x^2)^(1/2))+I*e-(c^2*d^2-e^2)^(1/2))/(I*e-(c^2*d^2-e^2)^(1/2)))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\arctan \left (1, \sqrt{c x + 1} \sqrt{c x - 1}\right )}{e x + d}\,{d x} + \frac{a \log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))/(e*x+d),x, algorithm="maxima")

[Out]

b*integrate(arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1))/(e*x + d), x) + a*log(e*x + d)/e

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*arccsc(c*x) + a)/(e*x + d), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{acsc}{\left (c x \right )}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsc(c*x))/(e*x+d),x)

[Out]

Integral((a + b*acsc(c*x))/(d + e*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arccsc}\left (c x\right ) + a}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)/(e*x + d), x)