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

Optimal. Leaf size=147 \[ -\frac {2 c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2 \sqrt {d-c^2 d x^2}}+\frac {2 b c^3 \sqrt {1-c^2 x^2} \log (x)}{3 \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {4701, 4681, 29, 30} \[ -\frac {2 c^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^3}-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2 \sqrt {d-c^2 d x^2}}+\frac {2 b c^3 \sqrt {1-c^2 x^2} \log (x)}{3 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

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

Rule 30

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

Rule 4681

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 + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), 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*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 4701

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 + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 152, normalized size = 1.03 \[ \frac {\sqrt {d-c^2 d x^2} \left (a \left (-4 c^4 x^4+2 c^2 x^2+2\right )+b c x \sqrt {1-c^2 x^2} \left (6 c^2 x^2+1\right )+2 b \left (-2 c^4 x^4+c^2 x^2+1\right ) \sin ^{-1}(c x)\right )}{6 d x^3 \left (c^2 x^2-1\right )}+\frac {2 b c^3 \log (x) \sqrt {d-c^2 d x^2}}{3 d \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.52, size = 433, normalized size = 2.95 \[ \left [\frac {2 \, {\left (b c^{5} x^{5} - b c^{3} x^{3}\right )} \sqrt {d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} - \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {d} - d}{c^{2} x^{4} - x^{2}}\right ) - \sqrt {-c^{2} d x^{2} + d} {\left (b c x^{3} - b c x\right )} \sqrt {-c^{2} x^{2} + 1} - 2 \, {\left (2 \, a c^{4} x^{4} - a c^{2} x^{2} + {\left (2 \, b c^{4} x^{4} - b c^{2} x^{2} - b\right )} \arcsin \left (c x\right ) - a\right )} \sqrt {-c^{2} d x^{2} + d}}{6 \, {\left (c^{2} d x^{5} - d x^{3}\right )}}, \frac {4 \, {\left (b c^{5} x^{5} - b c^{3} x^{3}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{2} + 1\right )} \sqrt {-d}}{c^{2} d x^{4} - {\left (c^{2} + 1\right )} d x^{2} + d}\right ) - \sqrt {-c^{2} d x^{2} + d} {\left (b c x^{3} - b c x\right )} \sqrt {-c^{2} x^{2} + 1} - 2 \, {\left (2 \, a c^{4} x^{4} - a c^{2} x^{2} + {\left (2 \, b c^{4} x^{4} - b c^{2} x^{2} - b\right )} \arcsin \left (c x\right ) - a\right )} \sqrt {-c^{2} d x^{2} + d}}{6 \, {\left (c^{2} d x^{5} - d x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.49, size = 849, normalized size = 5.78 \[ -\frac {a \sqrt {-c^{2} d \,x^{2}+d}}{3 d \,x^{3}}-\frac {2 a \,c^{2} \sqrt {-c^{2} d \,x^{2}+d}}{3 d x}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \left (-c^{2} x^{2}+1\right ) c^{4}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}-\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{5} c^{8}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}-\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{5}}{\left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}+\frac {4 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{3}}{3 d \left (c^{2} x^{2}-1\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} \arcsin \left (c x \right ) c^{6}}{\left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} c^{6}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}-\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{3}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \,c^{4}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \arcsin \left (c x \right ) c^{4}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}-\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x^{3} \left (-c^{2} x^{2}+1\right ) c^{6}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{3} \sqrt {-c^{2} x^{2}+1}}{2 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d}+\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c^{2}}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d x}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, c}{6 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d \,x^{2}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{3 \left (3 c^{4} x^{4}-2 c^{2} x^{2}-1\right ) d \,x^{3}}-\frac {2 b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}-1\right ) c^{3}}{3 d \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/d/x^3*(-c^2*d*x^2+d)^(1/2)-2/3*a*c^2/d/x*(-c^2*d*x^2+d)^(1/2)-1/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4
-2*c^2*x^2-1)/d*x*(-c^2*x^2+1)*c^4-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^5*c^8-2*I*b*(-d*
(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5+4/3*I*b*(-d*(c^2*x^2-1))^(
1/2)*(-c^2*x^2+1)^(1/2)/d/(c^2*x^2-1)*arcsin(c*x)*c^3-2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^3
*arcsin(c*x)*c^6+1/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^3*c^6-2/3*I*b*(-d*(c^2*x^2-1))^(1/
2)/(3*c^4*x^4-2*c^2*x^2-1)/d*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^3+1/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^
2*x^2-1)/d*x*c^4+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x*arcsin(c*x)*c^4-2/3*I*b*(-d*(c^2*x^2
-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d*x^3*(-c^2*x^2+1)*c^6+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)
/d*c^3*(-c^2*x^2+1)^(1/2)+4/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d/x*arcsin(c*x)*c^2+1/6*b*(-d*(
c^2*x^2-1))^(1/2)/(3*c^4*x^4-2*c^2*x^2-1)/d/x^2*(-c^2*x^2+1)^(1/2)*c+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^4*x^4-2
*c^2*x^2-1)/d/x^3*arcsin(c*x)-2/3*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^
2+1)^(1/2))^2-1)*c^3

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maxima [A]  time = 0.75, size = 124, normalized size = 0.84 \[ \frac {1}{6} \, {\left (\frac {4 \, c^{2} \log \relax (x)}{\sqrt {d}} - \frac {1}{\sqrt {d} x^{2}}\right )} b c - \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} c^{2}}{d x} + \frac {\sqrt {-c^{2} d x^{2} + d}}{d x^{3}}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} c^{2}}{d x} + \frac {\sqrt {-c^{2} d x^{2} + d}}{d x^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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