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

Optimal. Leaf size=89 \[ -\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e} \]

[Out]

-1/2*I*(a+b*arcsin(d*x+c))^2/b/d/e+(a+b*arcsin(d*x+c))*ln(1-(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))^2)/d/e-1/2*I*b*pol
ylog(2,(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))^2)/d/e

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Rubi [A]  time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4805, 12, 4625, 3717, 2190, 2279, 2391} \[ -\frac {i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])/(c*e + d*e*x),x]

[Out]

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

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4625

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

Rule 4805

Int[((a_.) + ArcSin[(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*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 71, normalized size = 0.80 \[ \frac {a \log (c+d x)-\frac {1}{2} i b \left (\sin ^{-1}(c+d x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )\right )+b \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (d x + c\right ) + a}{d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 182, normalized size = 2.04 \[ \frac {a \ln \left (d x +c \right )}{d e}-\frac {i b \arcsin \left (d x +c \right )^{2}}{2 d e}+\frac {b \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {b \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {i b \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {i b \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a/e*ln(d*x+c)-1/2*I/d*b/e*arcsin(d*x+c)^2+1/d*b/e*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+1/d*b/
e*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-I/d*b/e*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-I/d*b/e*p
olylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d e x + c e}\,{d x} + \frac {a \log \left (d e x + c e\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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