3.1.13 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{d+e x} \, dx\) [13]

Optimal. Leaf size=347 \[ -\frac {i (a+b \text {ArcSin}(c x))^3}{3 b e}+\frac {(a+b \text {ArcSin}(c x))^2 \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {(a+b \text {ArcSin}(c x))^2 \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {PolyLog}\left (3,\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {PolyLog}\left (3,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e} \]

[Out]

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

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Rubi [A]
time = 0.35, antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4825, 4615, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {2 i b (a+b \text {ArcSin}(c x)) \text {Li}_2\left (\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}-\frac {2 i b (a+b \text {ArcSin}(c x)) \text {Li}_2\left (\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {(a+b \text {ArcSin}(c x))^2 \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {(a+b \text {ArcSin}(c x))^2 \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e}-\frac {i (a+b \text {ArcSin}(c x))^3}{3 b e}+\frac {2 b^2 \text {Li}_3\left (\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e}+\frac {2 b^2 \text {Li}_3\left (\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4615

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(-I)*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] - I*
b*E^(I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x)))), x])
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4825

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

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

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Mathematica [A]
time = 0.26, size = 330, normalized size = 0.95 \begin {gather*} \frac {-\frac {i (a+b \text {ArcSin}(c x))^3}{b}+3 (a+b \text {ArcSin}(c x))^2 \log \left (1+\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )+3 (a+b \text {ArcSin}(c x))^2 \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+6 b \left (-i (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )+b \text {PolyLog}\left (3,-\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )\right )+6 b \left (-i (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+b \text {PolyLog}\left (3,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )}{3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*(a + b*ArcSin[c*x])^3)/b + 3*(a + b*ArcSin[c*x])^2*Log[1 + (I*e*E^(I*ArcSin[c*x]))/(-(c*d) + Sqrt[c^2*d
^2 - e^2])] + 3*(a + b*ArcSin[c*x])^2*Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])] + 6*b*((-I)
*(a + b*ArcSin[c*x])*PolyLog[2, ((-I)*e*E^(I*ArcSin[c*x]))/(-(c*d) + Sqrt[c^2*d^2 - e^2])] + b*PolyLog[3, ((-I
)*e*E^(I*ArcSin[c*x]))/(-(c*d) + Sqrt[c^2*d^2 - e^2])]) + 6*b*((-I)*(a + b*ArcSin[c*x])*PolyLog[2, (I*e*E^(I*A
rcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])] + b*PolyLog[3, (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])
)/(3*e)

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{e x +d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

a^2*e^(-1)*log(x*e + d) + integrate((b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*arctan2(c*x, sqr
t(c*x + 1)*sqrt(-c*x + 1)))/(x*e + d), 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(c*x))^2/(e*x+d),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(c*x))^2/(e*x+d),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))^2/(d + e*x),x)

[Out]

int((a + b*asin(c*x))^2/(d + e*x), x)

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