3.1.15 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{(d+e x)^3} \, dx\) [15]

Optimal. Leaf size=401 \[ \frac {b c \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {(a+b \text {ArcSin}(c x))^2}{2 e (d+e x)^2}-\frac {i b c^3 d (a+b \text {ArcSin}(c x)) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d (a+b \text {ArcSin}(c x)) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )}-\frac {b^2 c^3 d \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b^2 c^3 d \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.47, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4827, 4857, 3405, 3404, 2296, 2221, 2317, 2438, 2747, 31} \begin {gather*} \frac {b c \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b c^3 d (a+b \text {ArcSin}(c x)) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {i b c^3 d (a+b \text {ArcSin}(c x)) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {(a+b \text {ArcSin}(c x))^2}{2 e (d+e x)^2}-\frac {b^2 c^3 d \text {Li}_2\left (\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b^2 c^3 d \text {Li}_2\left (\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^2/(d + e*x)^3,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

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 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3404

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

Rule 3405

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

Rule 4827

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

Rule 4857

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

Rubi steps

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

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Mathematica [A]
time = 0.75, size = 314, normalized size = 0.78 \begin {gather*} \frac {\frac {2 b c e \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {(a+b \text {ArcSin}(c x))^2}{(d+e x)^2}-\frac {2 b^2 c^2 \log (d+e x)}{c^2 d^2-e^2}+\frac {2 b c^3 d \left (-i (a+b \text {ArcSin}(c x)) \left (\log \left (1+\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )-\log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )-b \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 (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )}{\left (c^2 d^2-e^2\right )^{3/2}}}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])^2/(d + e*x)^3,x]

[Out]

((2*b*c*e*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/((c^2*d^2 - e^2)*(d + e*x)) - (a + b*ArcSin[c*x])^2/(d + e*x)
^2 - (2*b^2*c^2*Log[d + e*x])/(c^2*d^2 - e^2) + (2*b*c^3*d*((-I)*(a + b*ArcSin[c*x])*(Log[1 + (I*e*E^(I*ArcSin
[c*x]))/(-(c*d) + Sqrt[c^2*d^2 - e^2])] - Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])]) - b*Po
lyLog[2, ((-I)*e*E^(I*ArcSin[c*x]))/(-(c*d) + Sqrt[c^2*d^2 - e^2])] + b*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*
d + Sqrt[c^2*d^2 - e^2])]))/(c^2*d^2 - e^2)^(3/2))/(2*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. 1178 vs. \(2 (407 ) = 814\).
time = 1.48, size = 1179, normalized size = 2.94 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^2/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

1/c*(-1/2*a^2*c^3/(c*e*x+c*d)^2/e-1/2*b^2*c^5*arcsin(c*x)^2/(c*e*x+c*d)^2/(c^2*d^2-e^2)/e*d^2+I*b^2*c^4/e*(-c^
2*d^2+e^2)^(1/2)/(c^2*d^2-e^2)^2*d*dilog((I*d*c+e*(I*c*x+(-c^2*x^2+1)^(1/2))-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(-c^
2*d^2+e^2)^(1/2)))-I*b^2*c^5*arcsin(c*x)/(c*e*x+c*d)^2/(c^2*d^2-e^2)/e*d^2-2*I*b^2*c^5*arcsin(c*x)/(c*e*x+c*d)
^2/(c^2*d^2-e^2)*d*x+b^2*c^4*arcsin(c*x)/(c*e*x+c*d)^2/(c^2*d^2-e^2)*(-c^2*x^2+1)^(1/2)*d+b^2*c^4*arcsin(c*x)/
(c*e*x+c*d)^2/(c^2*d^2-e^2)*e*(-c^2*x^2+1)^(1/2)*x+1/2*b^2*c^3*arcsin(c*x)^2/(c*e*x+c*d)^2/(c^2*d^2-e^2)*e+2*b
^2*c^3/e/(c^2*d^2-e^2)*ln(I*c*x+(-c^2*x^2+1)^(1/2))-b^2*c^3/e/(c^2*d^2-e^2)*ln(I*e*(I*c*x+(-c^2*x^2+1)^(1/2))^
2-2*d*c*(I*c*x+(-c^2*x^2+1)^(1/2))-I*e)-I*b^2*c^5*arcsin(c*x)/(c*e*x+c*d)^2/(c^2*d^2-e^2)*e*x^2-b^2*c^4/e*(-c^
2*d^2+e^2)^(1/2)/(c^2*d^2-e^2)^2*d*arcsin(c*x)*ln((I*d*c+e*(I*c*x+(-c^2*x^2+1)^(1/2))-(-c^2*d^2+e^2)^(1/2))/(I
*d*c-(-c^2*d^2+e^2)^(1/2)))-I*b^2*c^4/e*(-c^2*d^2+e^2)^(1/2)/(c^2*d^2-e^2)^2*d*dilog((I*d*c+e*(I*c*x+(-c^2*x^2
+1)^(1/2))+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))+b^2*c^4/e*(-c^2*d^2+e^2)^(1/2)/(c^2*d^2-e^2)^2*
d*arcsin(c*x)*ln((I*d*c+e*(I*c*x+(-c^2*x^2+1)^(1/2))+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))-a*b*c
^3/(c*e*x+c*d)^2/e*arcsin(c*x)+a*b*c^3/e/(c^2*d^2-e^2)/(c*x+d*c/e)*(-(c*x+d*c/e)^2+2*d*c/e*(c*x+d*c/e)-(c^2*d^
2-e^2)/e^2)^(1/2)-a*b*c^4/e^2*d/(c^2*d^2-e^2)/(-(c^2*d^2-e^2)/e^2)^(1/2)*ln((-2*(c^2*d^2-e^2)/e^2+2*d*c/e*(c*x
+d*c/e)+2*(-(c^2*d^2-e^2)/e^2)^(1/2)*(-(c*x+d*c/e)^2+2*d*c/e*(c*x+d*c/e)-(c^2*d^2-e^2)/e^2)^(1/2))/(c*x+d*c/e)
))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*d-%e>0)', see `assume?` for
more details

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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)^3,x, algorithm="fricas")

[Out]

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3), 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}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))**2/(e*x+d)**3,x)

[Out]

Integral((a + b*asin(c*x))**2/(d + e*x)**3, 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)^3,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^2/(e*x + d)^3, 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}{{\left (d+e\,x\right )}^3} \,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)^3,x)

[Out]

int((a + b*asin(c*x))^2/(d + e*x)^3, x)

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