3.45 \(\int \frac {\cos ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\)

Optimal. Leaf size=68 \[ -\frac {i \text {Li}_2\left (-e^{2 i \cos ^{-1}(a+b x)}\right )}{2 d}-\frac {i \cos ^{-1}(a+b x)^2}{2 d}+\frac {\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d} \]

[Out]

-1/2*I*arccos(b*x+a)^2/d+arccos(b*x+a)*ln(1+(b*x+a+I*(1-(b*x+a)^2)^(1/2))^2)/d-1/2*I*polylog(2,-(b*x+a+I*(1-(b
*x+a)^2)^(1/2))^2)/d

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Rubi [A]  time = 0.08, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {4806, 12, 4626, 3719, 2190, 2279, 2391} \[ -\frac {i \text {PolyLog}\left (2,-e^{2 i \cos ^{-1}(a+b x)}\right )}{2 d}-\frac {i \cos ^{-1}(a+b x)^2}{2 d}+\frac {\cos ^{-1}(a+b x) \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4806

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 59, normalized size = 0.87 \[ -\frac {i \left (\text {Li}_2\left (-e^{2 i \cos ^{-1}(a+b x)}\right )+\cos ^{-1}(a+b x) \left (\cos ^{-1}(a+b x)+2 i \log \left (1+e^{2 i \cos ^{-1}(a+b x)}\right )\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arccos \left (b x + a\right )}{b d x + a d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(b*x+a)/(a*d/b+d*x),x, algorithm="fricas")

[Out]

integral(b*arccos(b*x + a)/(b*d*x + a*d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arccos \left (b x + a\right )}{d x + \frac {a d}{b}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(b*x+a)/(a*d/b+d*x),x, algorithm="giac")

[Out]

integrate(arccos(b*x + a)/(d*x + a*d/b), x)

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maple [A]  time = 0.16, size = 85, normalized size = 1.25 \[ -\frac {i \arccos \left (b x +a \right )^{2}}{2 d}+\frac {\arccos \left (b x +a \right ) \ln \left (1+\left (b x +a +i \sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{d}-\frac {i \polylog \left (2, -\left (b x +a +i \sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(b*x+a)/(a*d/b+d*x),x)

[Out]

-1/2*I*arccos(b*x+a)^2/d+arccos(b*x+a)*ln(1+(b*x+a+I*(1-(b*x+a)^2)^(1/2))^2)/d-1/2*I*polylog(2,-(b*x+a+I*(1-(b
*x+a)^2)^(1/2))^2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arccos \left (b x + a\right )}{d x + \frac {a d}{b}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(b*x+a)/(a*d/b+d*x),x, algorithm="maxima")

[Out]

integrate(arccos(b*x + a)/(d*x + a*d/b), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (a+b\,x\right )}{d\,x+\frac {a\,d}{b}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a + b*x)/(d*x + (a*d)/b),x)

[Out]

int(acos(a + b*x)/(d*x + (a*d)/b), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b \int \frac {\operatorname {acos}{\left (a + b x \right )}}{a + b x}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(b*x+a)/(a*d/b+d*x),x)

[Out]

b*Integral(acos(a + b*x)/(a + b*x), x)/d

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