3.2.94 \(\int \frac {(a+b \text {ArcSin}(c+d x))^2}{(c e+d e x)^2} \, dx\) [194]

Optimal. Leaf size=116 \[ -\frac {(a+b \text {ArcSin}(c+d x))^2}{d e^2 (c+d x)}-\frac {4 b (a+b \text {ArcSin}(c+d x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(c+d x)}\right )}{d e^2}+\frac {2 i b^2 \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c+d x)}\right )}{d e^2}-\frac {2 i b^2 \text {PolyLog}\left (2,e^{i \text {ArcSin}(c+d x)}\right )}{d e^2} \]

[Out]

-(a+b*arcsin(d*x+c))^2/d/e^2/(d*x+c)-4*b*(a+b*arcsin(d*x+c))*arctanh(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^2+2*I*
b^2*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))/d/e^2-2*I*b^2*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^2

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Rubi [A]
time = 0.12, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4889, 12, 4723, 4803, 4268, 2317, 2438} \begin {gather*} -\frac {(a+b \text {ArcSin}(c+d x))^2}{d e^2 (c+d x)}-\frac {4 b \tanh ^{-1}\left (e^{i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))}{d e^2}+\frac {2 i b^2 \text {Li}_2\left (-e^{i \text {ArcSin}(c+d x)}\right )}{d e^2}-\frac {2 i b^2 \text {Li}_2\left (e^{i \text {ArcSin}(c+d x)}\right )}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^2,x]

[Out]

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

Rule 12

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

Rule 2317

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 2438

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 4268

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 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 -
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4803

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; Free
Q[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4889

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

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Mathematica [A]
time = 0.41, size = 176, normalized size = 1.52 \begin {gather*} \frac {-\frac {a^2}{c+d x}-2 a b \left (\frac {\text {ArcSin}(c+d x)}{c+d x}+\log \left (\frac {1}{2} (c+d x) \csc \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} \text {ArcSin}(c+d x)\right )\right )\right )+b^2 \left (\text {ArcSin}(c+d x) \left (-\frac {\text {ArcSin}(c+d x)}{c+d x}+2 \log \left (1-e^{i \text {ArcSin}(c+d x)}\right )-2 \log \left (1+e^{i \text {ArcSin}(c+d x)}\right )\right )+2 i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c+d x)}\right )-2 i \text {PolyLog}\left (2,e^{i \text {ArcSin}(c+d x)}\right )\right )}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^2,x]

[Out]

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

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Maple [A]
time = 0.15, size = 229, normalized size = 1.97

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arcsin \left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}+\frac {2 b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}-\frac {2 b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}+\frac {2 i b^{2} \dilog \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}-\frac {2 i b^{2} \dilog \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}+\frac {2 a b \left (-\frac {\arcsin \left (d x +c \right )}{d x +c}-\arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )\right )}{e^{2}}}{d}\)	\(229\)
default	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arcsin \left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}+\frac {2 b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}-\frac {2 b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}+\frac {2 i b^{2} \dilog \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}-\frac {2 i b^{2} \dilog \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{2}}+\frac {2 a b \left (-\frac {\arcsin \left (d x +c \right )}{d x +c}-\arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )\right )}{e^{2}}}{d}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^2,x, algorithm="maxima")

[Out]

-2*a*b*(e^(-2)*log(2*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)/abs(d^2*x*e^2 + c*d*e^2) + 2/abs(d^2*x*e^2 + c*d*e^2))
/d + arcsin(d*x + c)/(d^2*x*e^2 + c*d*e^2)) - (arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 + 2*(d
^2*x*e^2 + c*d*e^2)*integrate(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*
x - c + 1))/(d^3*x^3*e^2 + 3*c*d^2*x^2*e^2 + c^3*e^2 + (3*c^2*e^2 - e^2)*d*x - c*e^2), x))*b^2/(d^2*x*e^2 + c*
d*e^2) - a^2/(d^2*x*e^2 + c*d*e^2)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^2,x, algorithm="fricas")

[Out]

integral((b^2*arcsin(d*x + c)^2 + 2*a*b*arcsin(d*x + c) + a^2)*e^(-2)/(d^2*x^2 + 2*c*d*x + c^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))**2/(d*e*x+c*e)**2,x)

[Out]

(Integral(a**2/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b**2*asin(c + d*x)**2/(c**2 + 2*c*d*x + d**2*x**2),
x) + Integral(2*a*b*asin(c + d*x)/(c**2 + 2*c*d*x + d**2*x**2), x))/e**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^2,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)^2/(d*e*x + c*e)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^2,x)

[Out]

int((a + b*asin(c + d*x))^2/(c*e + d*e*x)^2, x)

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