3.194 \(\int \frac{(a+b \sin ^{-1}(c+d x))^2}{(c e+d e x)^2} \, dx\)

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

[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)

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Rubi [A]  time = 0.163736, antiderivative size = 116, normalized size of antiderivative = 1., 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 = {4805, 12, 4627, 4709, 4183, 2279, 2391} \[ \frac{2 i b^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{2 i b^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )}{d e^2}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{4 b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2} \]

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 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]

Rule 12

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

Rule 4627

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 4709

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

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 \frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{(c e+d e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{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) \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.574653, size = 176, normalized size = 1.52 \[ \frac{b^2 \left (2 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right )-2 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right )+\sin ^{-1}(c+d x) \left (-\frac{\sin ^{-1}(c+d x)}{c+d x}+2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )\right )\right )-\frac{a^2}{c+d x}-2 a b \left (\frac{\sin ^{-1}(c+d x)}{c+d x}-\log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c+d x)\right )\right )+\log \left (\frac{1}{2} (c+d x) \csc \left (\frac{1}{2} \sin ^{-1}(c+d x)\right )\right )\right )}{d e^2} \]

Warning: Unable to verify antiderivative.

[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 = 0.069, size = 251, normalized size = 2.2 \begin{align*} -{\frac{{a}^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{d{e}^{2} \left ( dx+c \right ) }}+2\,{\frac{{b}^{2}\arcsin \left ( dx+c \right ) \ln \left ( 1-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}-2\,{\frac{{b}^{2}\arcsin \left ( dx+c \right ) \ln \left ( 1+i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }{d{e}^{2}}}+{\frac{2\,i{b}^{2}}{d{e}^{2}}{\it polylog} \left ( 2,-i \left ( dx+c \right ) -\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-{\frac{2\,i{b}^{2}}{d{e}^{2}}{\it polylog} \left ( 2,i \left ( dx+c \right ) +\sqrt{1- \left ( dx+c \right ) ^{2}} \right ) }-2\,{\frac{ab\arcsin \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-2\,{\frac{ab{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( dx+c \right ) ^{2}}}} \right ) }{d{e}^{2}}} \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \end{align*}

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)/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}

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)