3.2.93 \(\int \frac {(a+b \text {ArcSin}(c+d x))^2}{c e+d e x} \, dx\) [193]

Optimal. Leaf size=126 \[ -\frac {i (a+b \text {ArcSin}(c+d x))^3}{3 b d e}+\frac {(a+b \text {ArcSin}(c+d x))^2 \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right )}{d e}-\frac {i b (a+b \text {ArcSin}(c+d x)) \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c+d x)}\right )}{d e}+\frac {b^2 \text {PolyLog}\left (3,e^{2 i \text {ArcSin}(c+d x)}\right )}{2 d e} \]

[Out]

-1/3*I*(a+b*arcsin(d*x+c))^3/b/d/e+(a+b*arcsin(d*x+c))^2*ln(1-(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))^2)/d/e-I*b*(a+b*
arcsin(d*x+c))*polylog(2,(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))^2)/d/e+1/2*b^2*polylog(3,(I*(d*x+c)+(1-(d*x+c)^2)^(1/
2))^2)/d/e

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Rubi [A]
time = 0.13, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4889, 12, 4721, 3798, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {i b \text {Li}_2\left (e^{2 i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))}{d e}-\frac {i (a+b \text {ArcSin}(c+d x))^3}{3 b d e}+\frac {\log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))^2}{d e}+\frac {b^2 \text {Li}_3\left (e^{2 i \text {ArcSin}(c+d x)}\right )}{2 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*(a + b*ArcSin[c + d*x])^3)/(b*d*e) + ((a + b*ArcSin[c + d*x])^2*Log[1 - E^((2*I)*ArcSin[c + d*x])])/
(d*e) - (I*b*(a + b*ArcSin[c + d*x])*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(d*e) + (b^2*PolyLog[3, E^((2*I)*A
rcSin[c + d*x])])/(2*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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3798

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 4721

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

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]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {\text {Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^3}{3 b d e}-\frac {(2 i) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{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 )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {(2 b) \text {Subst}\left (\int (a+b x) \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 )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {i b \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (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 )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {i b \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {b^2 \text {Subst}\left (\int \frac {\text {Li}_2(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 )^3}{3 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {i b \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac {b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 170, normalized size = 1.35 \begin {gather*} \frac {2 a b \text {ArcSin}(c+d x) \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right )+a^2 \log (c+d x)-i a b \left (\text {ArcSin}(c+d x)^2+\text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c+d x)}\right )\right )+b^2 \left (-\frac {i \pi ^3}{24}+\frac {1}{3} i \text {ArcSin}(c+d x)^3+\text {ArcSin}(c+d x)^2 \log \left (1-e^{-2 i \text {ArcSin}(c+d x)}\right )+i \text {ArcSin}(c+d x) \text {PolyLog}\left (2,e^{-2 i \text {ArcSin}(c+d x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,e^{-2 i \text {ArcSin}(c+d x)}\right )\right )}{d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*ArcSin[c + d*x]*Log[1 - E^((2*I)*ArcSin[c + d*x])] + a^2*Log[c + d*x] - I*a*b*(ArcSin[c + d*x]^2 + Poly
Log[2, E^((2*I)*ArcSin[c + d*x])]) + b^2*((-1/24*I)*Pi^3 + (I/3)*ArcSin[c + d*x]^3 + ArcSin[c + d*x]^2*Log[1 -
E^((-2*I)*ArcSin[c + d*x])] + I*ArcSin[c + d*x]*PolyLog[2, E^((-2*I)*ArcSin[c + d*x])] + PolyLog[3, E^((-2*I)
*ArcSin[c + d*x])]/2))/(d*e)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (156 ) = 312\).
time = 0.20, size = 420, normalized size = 3.33

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \ln \left (d x +c \right )}{e}-\frac {i b^{2} \arcsin \left (d x +c \right )^{3}}{3 e}+\frac {b^{2} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i b^{2} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {b^{2} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i b^{2} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {i a b \arcsin \left (d x +c \right )^{2}}{e}+\frac {2 a b \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i a b \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i a b \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}}{d}\)	\(420\)
default	\(\frac {\frac {a^{2} \ln \left (d x +c \right )}{e}-\frac {i b^{2} \arcsin \left (d x +c \right )^{3}}{3 e}+\frac {b^{2} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i b^{2} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {b^{2} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i b^{2} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {2 b^{2} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {i a b \arcsin \left (d x +c \right )^{2}}{e}+\frac {2 a b \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}+\frac {2 a b \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i a b \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}-\frac {2 i a b \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e}}{d}\)	\(420\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2/e*ln(d*x+c)-1/3*I*b^2/e*arcsin(d*x+c)^3+b^2/e*arcsin(d*x+c)^2*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-2*I
*b^2/e*arcsin(d*x+c)*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+2*b^2/e*polylog(3,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2
))+b^2/e*arcsin(d*x+c)^2*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-2*I*b^2/e*arcsin(d*x+c)*polylog(2,I*(d*x+c)+(1-(d
*x+c)^2)^(1/2))+2*b^2/e*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-I*a*b/e*arcsin(d*x+c)^2+2*a*b/e*arcsin(d*x+c)
*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+2*a*b/e*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-2*I*a*b/e*polyl
og(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-2*I*a*b/e*polylog(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 \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),x, algorithm="maxima")

[Out]

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

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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),x, algorithm="fricas")

[Out]

integral((b^2*arcsin(d*x + c)^2 + 2*a*b*arcsin(d*x + c) + a^2)*e^(-1)/(d*x + c), x)

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

[Out]

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

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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),x, algorithm="giac")

[Out]

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

[Out]

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

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