3.249 \(\int \frac {(a+b \sin ^{-1}(c x))^2}{(d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=195 \[ \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d \sqrt {d-c^2 d x^2}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}} \]

[Out]

x*(a+b*arcsin(c*x))^2/d/(-c^2*d*x^2+d)^(1/2)-I*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/c/d/(-c^2*d*x^2+d)^(1/2)
+2*b*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c/d/(-c^2*d*x^2+d)^(1/2)-I*b^2*po
lylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c/d/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4653, 4675, 3719, 2190, 2279, 2391} \[ -\frac {i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2190

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4653

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 - c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSin[c*x
])^(n - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4675

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b c \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt {d-c^2 d x^2}}+\frac {\left (4 i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}}+\frac {\left (i b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 165, normalized size = 0.85 \[ \frac {a \left (a c x+b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )\right )+2 b \sin ^{-1}(c x) \left (a c x+b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )-i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+b^2 \left (c x-i \sqrt {1-c^2 x^2}\right ) \sin ^{-1}(c x)^2}{c d \sqrt {d-c^2 d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2
), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^4-1)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [B]  time = 0.16, size = 425, normalized size = 2.18 \[ \frac {a^{2} x}{d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} \sqrt {-c^{2} x^{2}+1}}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{c \,d^{2} \left (c^{2} x^{2}-1\right )}+\frac {i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{c \,d^{2} \left (c^{2} x^{2}-1\right )}+\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{c \,d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{c \,d^{2} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, a b x \arcsin \left (c x\right )}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {\frac {b^{2} \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2}}{{\left (c x + 1\right )}^{\frac {3}{2}} {\left (c x - 1\right )} \sqrt {-c x + 1}}\,{d x}}{d}}{\sqrt {d}} + \frac {a^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {a b \log \left (x^{2} - \frac {1}{c^{2}}\right )}{c d^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))**2/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)

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