3.72 \(\int \frac {(d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=191 \[ \frac {c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac {c^3 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b \sqrt {1-c^2 x^2}}-\frac {b c d \sqrt {d-c^2 d x^2}}{6 x^2 \sqrt {1-c^2 x^2}}-\frac {4 b c^3 d \log (x) \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 191, 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, 4693, 29, 4641, 14} \[ \frac {c^3 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b \sqrt {1-c^2 x^2}}+\frac {c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {b c d \sqrt {d-c^2 d x^2}}{6 x^2 \sqrt {1-c^2 x^2}}-\frac {4 b c^3 d \log (x) \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

-(b*c*d*Sqrt[d - c^2*d*x^2])/(6*x^2*Sqrt[1 - c^2*x^2]) + (c^2*d*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/x - (
(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(3*x^3) + (c^3*d*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*Sq
rt[1 - c^2*x^2]) - (4*b*c^3*d*Sqrt[d - c^2*d*x^2]*Log[x])/(3*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 4693

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

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^4} \, dx &=-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\left (c^2 d\right ) \int \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \frac {1-c^2 x^2}{x^3} \, dx}{3 \sqrt {1-c^2 x^2}}\\ &=\frac {c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \left (\frac {1}{x^3}-\frac {c^2}{x}\right ) \, dx}{3 \sqrt {1-c^2 x^2}}-\frac {\left (b c^3 d \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{x} \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (c^4 d \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {1-c^2 x^2}}\\ &=-\frac {b c d \sqrt {d-c^2 d x^2}}{6 x^2 \sqrt {1-c^2 x^2}}+\frac {c^2 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac {c^3 d \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b \sqrt {1-c^2 x^2}}-\frac {4 b c^3 d \sqrt {d-c^2 d x^2} \log (x)}{3 \sqrt {1-c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.85, size = 211, normalized size = 1.10 \[ -\frac {d \sqrt {d-c^2 d x^2} \left (2 a \left (1-4 c^2 x^2\right ) \sqrt {1-c^2 x^2}+8 b c^3 x^3 \log (c x)+b c x\right )}{6 x^3 \sqrt {1-c^2 x^2}}-a c^3 d^{3/2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+\frac {b d \left (4 c^2 x^2-1\right ) \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{3 x^3}+\frac {b c^3 d \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^2}{2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

(b*d*(-1 + 4*c^2*x^2)*Sqrt[d - c^2*d*x^2]*ArcSin[c*x])/(3*x^3) + (b*c^3*d*Sqrt[d - c^2*d*x^2]*ArcSin[c*x]^2)/(
2*Sqrt[1 - c^2*x^2]) - a*c^3*d^(3/2)*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] - (d*Sqrt[d -
c^2*d*x^2]*(b*c*x + 2*a*(1 - 4*c^2*x^2)*Sqrt[1 - c^2*x^2] + 8*b*c^3*x^3*Log[c*x]))/(6*x^3*Sqrt[1 - c^2*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{4}}, x\right ) \]

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^4,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^4, x)

________________________________________________________________________________________

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^4,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

________________________________________________________________________________________

maple [C]  time = 0.41, size = 1289, normalized size = 6.75 \[ \text {result too large to display} \]

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^4,x)

[Out]

-1/3*a/d/x^3*(-c^2*d*x^2+d)^(5/2)+2/3*a*c^2/d/x*(-c^2*d*x^2+d)^(5/2)+2/3*a*c^4*x*(-c^2*d*x^2+d)^(3/2)+a*c^4*d*
x*(-c^2*d*x^2+d)^(1/2)+a*c^4*d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/2*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^3-8/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c
^2*x^2+1)*x^5/(c^2*x^2-1)*c^8+32*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x^4/(c^2*x^2-1)*(-c^2*x
^2+1)^(1/2)*arcsin(c*x)*c^7-8/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x^3/(c^2*x^2-1)*(-c^2*x^
2+1)*c^6+32*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^8+2/3*I*b*(-d*(c
^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x/(c^2*x^2-1)*(-c^2*x^2+1)*c^4-12*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(24
*c^4*x^4-9*c^2*x^2+1)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5-8*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^
2+1)^(1/2)*arcsin(c*x)*d*c^3/(3*c^2*x^2-3)-52*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x^3/(c^2*x^2
-1)*arcsin(c*x)*c^6+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*a
rcsin(c*x)*c^3+4*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^5+10
/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x^3/(c^2*x^2-1)*c^6-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)*d/
(24*c^4*x^4-9*c^2*x^2+1)*x/(c^2*x^2-1)*c^4+73/3*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)*x/(c^2*x^2
-1)*arcsin(c*x)*c^4-3/2*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)/(c^2*x^2-1)*c^3*(-c^2*x^2+1)^(1/2)
-14/3*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)/x/(c^2*x^2-1)*arcsin(c*x)*c^2+1/6*b*(-d*(c^2*x^2-1))
^(1/2)*d/(24*c^4*x^4-9*c^2*x^2+1)/x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+1/3*b*(-d*(c^2*x^2-1))^(1/2)*d/(24*c^4*
x^4-9*c^2*x^2+1)/x^3/(c^2*x^2-1)*arcsin(c*x)+4/3*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^3

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -b \sqrt {d} \int \frac {{\left (c^{2} d x^{2} - d\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x^{4}}\,{d x} + \frac {1}{3} \, {\left (3 \, \sqrt {-c^{2} d x^{2} + d} c^{4} d x + 3 \, c^{3} d^{\frac {3}{2}} \arcsin \left (c x\right ) + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{x} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{d x^{3}}\right )} a \]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2}}{x^4} \,d x \]

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^4,x)

[Out]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(3/2))/x^4, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{4}}\, dx \]

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**4,x)

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(3/2)*(a + b*asin(c*x))/x**4, x)

________________________________________________________________________________________