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

Optimal. Leaf size=185 \[ -\frac {3}{2} c^2 d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 c d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {b c d \log (x) \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}}+\frac {b c^3 d x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4695, 4647, 4641, 30, 14} \[ -\frac {3}{2} c^2 d x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 c d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {b c^3 d x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}+\frac {b c d \log (x) \sqrt {d-c^2 d x^2}}{\sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4641

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

Rule 4647

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

Rule 4695

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

Rubi steps

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

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Mathematica [A]  time = 0.62, size = 222, normalized size = 1.20 \[ \frac {3}{2} a c d^{3/2} \tan ^{-1}\left (\frac {c x \sqrt {-d \left (c^2 x^2-1\right )}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+\sqrt {-d \left (c^2 x^2-1\right )} \left (-\frac {1}{2} a c^2 d x-\frac {a d}{x}\right )-\frac {b c d \sqrt {d \left (1-c^2 x^2\right )} \left (\frac {2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{c x}-2 \log (c x)+\sin ^{-1}(c x)^2\right )}{2 \sqrt {1-c^2 x^2}}-\frac {b c d \sqrt {d \left (1-c^2 x^2\right )} \left (2 \sin ^{-1}(c x) \left (\sin ^{-1}(c x)+\sin \left (2 \sin ^{-1}(c x)\right )\right )+\cos \left (2 \sin ^{-1}(c x)\right )\right )}{8 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*d)/x) - (a*c^2*d*x)/2)*Sqrt[-(d*(-1 + c^2*x^2))] + (3*a*c*d^(3/2)*ArcTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])
/(Sqrt[d]*(-1 + c^2*x^2))])/2 - (b*c*d*Sqrt[d*(1 - c^2*x^2)]*((2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*x) + ArcSin
[c*x]^2 - 2*Log[c*x]))/(2*Sqrt[1 - c^2*x^2]) - (b*c*d*Sqrt[d*(1 - c^2*x^2)]*(Cos[2*ArcSin[c*x]] + 2*ArcSin[c*x
]*(ArcSin[c*x] + Sin[2*ArcSin[c*x]])))/(8*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*c^2*d*x^2 - a*d + (b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)/x^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((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^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 = 0.31, size = 464, normalized size = 2.51 \[ -\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{d x}-a \,c^{2} x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}-\frac {3 a \,c^{2} d x \sqrt {-c^{2} d \,x^{2}+d}}{2}-\frac {3 a \,c^{2} d^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 \sqrt {c^{2} d}}+\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} d c}{4 \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d \,c^{3} \sqrt {-c^{2} x^{2}+1}\, x^{2}}{4 \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d \,c^{4} \arcsin \left (c x \right ) x^{3}}{2 \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d c \sqrt {-c^{2} x^{2}+1}}{8 c^{2} x^{2}-8}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) d c}{c^{2} x^{2}-1}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, d \,c^{2} \arcsin \left (c x \right ) x}{2 \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) d}{\left (c^{2} x^{2}-1\right ) x}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}-1\right ) d c}{c^{2} x^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d/x*(-c^2*d*x^2+d)^(5/2)-a*c^2*x*(-c^2*d*x^2+d)^(3/2)-3/2*a*c^2*d*x*(-c^2*d*x^2+d)^(1/2)-3/2*a*c^2*d^2/(c^2
*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+3/4*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^
2-1)*arcsin(c*x)^2*d*c-1/4*b*(-d*(c^2*x^2-1))^(1/2)*d*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-1/2*b*(-d*(c^2*x^
2-1))^(1/2)*d*c^4/(c^2*x^2-1)*arcsin(c*x)*x^3+1/8*b*(-d*(c^2*x^2-1))^(1/2)*d*c/(c^2*x^2-1)*(-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^2*x^2-1)*arcsin(c*x)*d*c-1/2*b*(-d*(c^2*x^2-1))^(1/2)*d*c^2/(
c^2*x^2-1)*arcsin(c*x)*x+b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*d/(c^2*x^2-1)/x-b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x
^2+1)^(1/2)/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*d*c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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