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

Optimal. Leaf size=142 \[ \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}+\frac {a+b \sin ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {d-c^2 d x^2} \tanh ^{-1}(c x)}{c^4 d^2 \sqrt {1-c^2 x^2}}-\frac {b x \sqrt {d-c^2 d x^2}}{c^3 d^2 \sqrt {1-c^2 x^2}} \]

[Out]

(a+b*arcsin(c*x))/c^4/d/(-c^2*d*x^2+d)^(1/2)+(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4/d^2-b*x*(-c^2*d*x^2+d)
^(1/2)/c^3/d^2/(-c^2*x^2+1)^(1/2)-b*arctanh(c*x)*(-c^2*d*x^2+d)^(1/2)/c^4/d^2/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 146, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4703, 4677, 8, 321, 206} \[ \frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*x*Sqrt[1 - c^2*x^2])/(c^3*d*Sqrt[d - c^2*d*x^2])) + (x^2*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2])
+ (2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(c^4*d^2) - (b*Sqrt[1 - c^2*x^2]*ArcTanh[c*x])/(c^4*d*Sqrt[d -
c^2*d*x^2])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4677

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

Rule 4703

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p +
1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*f*n*d^IntPart[p]*(d + e*x^2
)^FracPart[p])/(2*c*(p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {2 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx}{c^2 d}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {x^2}{1-c^2 x^2} \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{c^3 d \sqrt {d-c^2 d x^2}}\\ &=-\frac {b x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^4 d^2}-\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{c^4 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 136, normalized size = 0.96 \[ \frac {\sqrt {d-c^2 d x^2} \left (\sqrt {-c^2} \left (a c^2 x^2-2 a+b c x \sqrt {1-c^2 x^2}+b \left (c^2 x^2-2\right ) \sin ^{-1}(c x)\right )-i b c \sqrt {1-c^2 x^2} F\left (\left .i \sinh ^{-1}\left (\sqrt {-c^2} x\right )\right |1\right )\right )}{c^4 \sqrt {-c^2} d^2 \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(-2*a + a*c^2*x^2 + b*c*x*Sqrt[1 - c^2*x^2] + b*(-2 + c^2*x^2)*ArcSin[c*x]) -
I*b*c*Sqrt[1 - c^2*x^2]*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1]))/(c^4*Sqrt[-c^2]*d^2*(-1 + c^2*x^2))

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fricas [A]  time = 0.72, size = 382, normalized size = 2.69 \[ \left [\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x + {\left (b c^{2} x^{2} - b\right )} \sqrt {d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} + 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} \sqrt {d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) + 4 \, {\left (a c^{2} x^{2} + {\left (b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{4 \, {\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}, \frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - {\left (b c^{2} x^{2} - b\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} c \sqrt {-d} x}{c^{4} d x^{4} - d}\right ) + 2 \, {\left (a c^{2} x^{2} + {\left (b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{2 \, {\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x + (b*c^2*x^2 - b)*sqrt(d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4
- 5*c^2*d*x^2 + 4*(c^3*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*sqrt(d) - d)/(c^6*x^6 - 3*c^4*x^4 +
3*c^2*x^2 - 1)) + 4*(a*c^2*x^2 + (b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4
*d^2), 1/2*(2*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - (b*c^2*x^2 - b)*sqrt(-d)*arctan(2*sqrt(-c^2*d*x^
2 + d)*sqrt(-c^2*x^2 + 1)*c*sqrt(-d)*x/(c^4*d*x^4 - d)) + 2*(a*c^2*x^2 + (b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*
sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4*d^2)]

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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(x^3*(a+b*arcsin(c*x))/(-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:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.27, size = 306, normalized size = 2.15 \[ -\frac {a \,x^{2}}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 a}{d \,c^{4} \sqrt {-c^{2} d \,x^{2}+d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{c^{3} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{2}}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{c^{4} d^{2} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.67, size = 142, normalized size = 1.00 \[ -\frac {1}{2} \, b c {\left (\frac {2 \, x}{c^{4} d^{\frac {3}{2}}} + \frac {\log \left (c x + 1\right )}{c^{5} d^{\frac {3}{2}}} - \frac {\log \left (c x - 1\right )}{c^{5} d^{\frac {3}{2}}}\right )} - b {\left (\frac {x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} - \frac {2}{\sqrt {-c^{2} d x^{2} + d} c^{4} d}\right )} \arcsin \left (c x\right ) - a {\left (\frac {x^{2}}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} - \frac {2}{\sqrt {-c^{2} d x^{2} + d} c^{4} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*c*(2*x/(c^4*d^(3/2)) + log(c*x + 1)/(c^5*d^(3/2)) - log(c*x - 1)/(c^5*d^(3/2))) - b*(x^2/(sqrt(-c^2*d*x
^2 + d)*c^2*d) - 2/(sqrt(-c^2*d*x^2 + d)*c^4*d))*arcsin(c*x) - a*(x^2/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 2/(sqrt(-
c^2*d*x^2 + d)*c^4*d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError

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