3.132 \(\int \frac {x^3 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=150 \[ -\frac {a+b \sin ^{-1}(c x)}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {a+b \sin ^{-1}(c x)}{3 c^4 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {5 b \sqrt {d-c^2 d x^2} \tanh ^{-1}(c x)}{6 c^4 d^3 \sqrt {1-c^2 x^2}}-\frac {b x \sqrt {d-c^2 d x^2}}{6 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

1/3*(a+b*arcsin(c*x))/c^4/d/(-c^2*d*x^2+d)^(3/2)+(-a-b*arcsin(c*x))/c^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b*x*(-c^2
*d*x^2+d)^(1/2)/c^3/d^3/(-c^2*x^2+1)^(3/2)+5/6*b*arctanh(c*x)*(-c^2*d*x^2+d)^(1/2)/c^4/d^3/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 155, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {4703, 4677, 206, 288} \[ -\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b x}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {5 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^4 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-(b*x)/(6*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^2*(a + b*ArcSin[c*x]))/(3*c^2*d*(d - c^2*d*x^2)^
(3/2)) - (2*(a + b*ArcSin[c*x]))/(3*c^4*d^2*Sqrt[d - c^2*d*x^2]) + (5*b*Sqrt[1 - c^2*x^2]*ArcTanh[c*x])/(6*c^4
*d^2*Sqrt[d - c^2*d*x^2])

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 )^{5/2}} \, dx &=\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 c^2 d}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {x^2}{\left (1-c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b x}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{6 c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {1}{1-c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b x}{6 c^3 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {5 b \sqrt {1-c^2 x^2} \tanh ^{-1}(c x)}{6 c^4 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 143, normalized size = 0.95 \[ \frac {\sqrt {d-c^2 d x^2} \left (\sqrt {-c^2} \left (6 a c^2 x^2-4 a-b c x \sqrt {1-c^2 x^2}+2 b \left (3 c^2 x^2-2\right ) \sin ^{-1}(c x)\right )-5 i b c \left (1-c^2 x^2\right )^{3/2} F\left (\left .i \sinh ^{-1}\left (\sqrt {-c^2} x\right )\right |1\right )\right )}{6 c^4 \sqrt {-c^2} d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(Sqrt[-c^2]*(-4*a + 6*a*c^2*x^2 - b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(-2 + 3*c^2*x^2)*ArcSin[c
*x]) - (5*I)*b*c*(1 - c^2*x^2)^(3/2)*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], 1]))/(6*c^4*Sqrt[-c^2]*d^3*(-1 + c^2*x
^2)^2)

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fricas [A]  time = 2.89, size = 421, normalized size = 2.81 \[ \left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - 5 \, {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} - 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} \sqrt {d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) - 8 \, {\left (3 \, a c^{2} x^{2} + {\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{24 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}, -\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} b c x - 5 \, {\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt {-d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} c \sqrt {-d} x}{c^{4} d x^{4} - d}\right ) - 4 \, {\left (3 \, a c^{2} x^{2} + {\left (3 \, b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{12 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}}\right ] \]

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)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(4*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - 5*(b*c^4*x^4 - 2*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)) - 8*(3*a*c^2*x^2 + (3*b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 +
d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3), -1/12*(2*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*b*c*x - 5*(b*c^4
*x^4 - 2*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)
) - 4*(3*a*c^2*x^2 + (3*b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^8*d^3*x^4 - 2*c^6*d^3*x^2
+ c^4*d^3)]

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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)^(5/2),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 = 0.28, size = 307, normalized size = 2.05 \[ \frac {a \,x^{2}}{c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2 a}{3 d \,c^{4} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{2}}{d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{2}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x}{6 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{3}}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{3 d^{3} \left (c^{2} x^{2}-1\right )^{2} c^{4}}-\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}+i\right )}{6 c^{4} d^{3} \left (c^{2} x^{2}-1\right )}+\frac {5 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-i\right )}{6 c^{4} d^{3} \left (c^{2} x^{2}-1\right )} \]

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)^(5/2),x)

[Out]

a*x^2/c^2/d/(-c^2*d*x^2+d)^(3/2)-2/3*a/d/c^4/(-c^2*d*x^2+d)^(3/2)+b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c
^2*arcsin(c*x)*x^2-1/6*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(c^2*x^2-1)^2/c^3*(-c^2*x^2+1)^(1/2)*x-2/3*b*(-d*(c^2*x^2-
1))^(1/2)/d^3/(c^2*x^2-1)^2/c^4*arcsin(c*x)-5/6*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^4/d^3/(c^2*x^2-1
)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)+5/6*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^4/d^3/(c^2*x^2-1)*ln(I*c*x+
(-c^2*x^2+1)^(1/2)-I)

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maxima [A]  time = 0.57, size = 160, normalized size = 1.07 \[ \frac {1}{12} \, b c {\left (\frac {2 \, x}{c^{6} d^{\frac {5}{2}} x^{2} - c^{4} d^{\frac {5}{2}}} + \frac {5 \, \log \left (c x + 1\right )}{c^{5} d^{\frac {5}{2}}} - \frac {5 \, \log \left (c x - 1\right )}{c^{5} d^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/12*b*c*(2*x/(c^6*d^(5/2)*x^2 - c^4*d^(5/2)) + 5*log(c*x + 1)/(c^5*d^(5/2)) - 5*log(c*x - 1)/(c^5*d^(5/2))) +
1/3*b*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))*arcsin(c*x) + 1/3*a*(3*x^2/((
-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

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)^(5/2),x)

[Out]

int((x^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(5/2), 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)**(5/2),x)

[Out]

Exception raised: TypeError

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