3.73 \(\int \frac {(d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x))}{x^6} \, dx\)
Optimal. Leaf size=154 \[ -\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}-\frac {b c d \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^5 d \log (x) \sqrt {d-c^2 d x^2}}{5 \sqrt {1-c^2 x^2}}+\frac {b c^3 d \sqrt {d-c^2 d x^2}}{5 x^2 \sqrt {1-c^2 x^2}} \]
[Out]
-1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/d/x^5-1/20*b*c*d*(-c^2*d*x^2+d)^(1/2)/x^4/(-c^2*x^2+1)^(1/2)+1/5*b
*c^3*d*(-c^2*d*x^2+d)^(1/2)/x^2/(-c^2*x^2+1)^(1/2)+1/5*b*c^5*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 = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used =
{4681, 266, 43} \[ -\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}+\frac {b c^3 d \sqrt {d-c^2 d x^2}}{5 x^2 \sqrt {1-c^2 x^2}}-\frac {b c d \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^5 d \log (x) \sqrt {d-c^2 d x^2}}{5 \sqrt {1-c^2 x^2}} \]
Antiderivative was successfully verified.
[In]
Int[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^6,x]
[Out]
-(b*c*d*Sqrt[d - c^2*d*x^2])/(20*x^4*Sqrt[1 - c^2*x^2]) + (b*c^3*d*Sqrt[d - c^2*d*x^2])/(5*x^2*Sqrt[1 - c^2*x^
2]) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*d*x^5) + (b*c^5*d*Sqrt[d - c^2*d*x^2]*Log[x])/(5*Sqrt[1 -
c^2*x^2])
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rule 266
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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]
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^6} \, dx &=-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (1-c^2 x^2\right )^2}{x^5} \, dx}{5 \sqrt {1-c^2 x^2}}\\ &=-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-c^2 x\right )^2}{x^3} \, dx,x,x^2\right )}{10 \sqrt {1-c^2 x^2}}\\ &=-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}+\frac {\left (b c d \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^3}-\frac {2 c^2}{x^2}+\frac {c^4}{x}\right ) \, dx,x,x^2\right )}{10 \sqrt {1-c^2 x^2}}\\ &=-\frac {b c d \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {b c^3 d \sqrt {d-c^2 d x^2}}{5 x^2 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 d x^5}+\frac {b c^5 d \sqrt {d-c^2 d x^2} \log (x)}{5 \sqrt {1-c^2 x^2}}\\ \end {align*}
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Mathematica [A] time = 0.26, size = 144, normalized size = 0.94 \[ \frac {b c^5 d \log (x) \sqrt {d-c^2 d x^2}}{5 \sqrt {1-c^2 x^2}}-\frac {d \sqrt {d-c^2 d x^2} \left (12 a \left (c^2 x^2-1\right )^3+12 b \left (c^2 x^2-1\right )^3 \sin ^{-1}(c x)+b c x \sqrt {1-c^2 x^2} \left (-25 c^4 x^4+12 c^2 x^2-3\right )\right )}{60 x^5 \left (c^2 x^2-1\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/x^6,x]
[Out]
-1/60*(d*Sqrt[d - c^2*d*x^2]*(12*a*(-1 + c^2*x^2)^3 + b*c*x*Sqrt[1 - c^2*x^2]*(-3 + 12*c^2*x^2 - 25*c^4*x^4) +
12*b*(-1 + c^2*x^2)^3*ArcSin[c*x]))/(x^5*(-1 + c^2*x^2)) + (b*c^5*d*Sqrt[d - c^2*d*x^2]*Log[x])/(5*Sqrt[1 - c
^2*x^2])
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fricas [A] time = 0.77, size = 525, normalized size = 3.41 \[ \left [\frac {2 \, {\left (b c^{7} d x^{7} - b c^{5} d x^{5}\right )} \sqrt {d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} - \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {d} - d}{c^{2} x^{4} - x^{2}}\right ) - {\left (4 \, b c^{3} d x^{3} - {\left (4 \, b c^{3} - b c\right )} d x^{5} - b c d x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (a c^{6} d x^{6} - 3 \, a c^{4} d x^{4} + 3 \, a c^{2} d x^{2} - a d + {\left (b c^{6} d x^{6} - 3 \, b c^{4} d x^{4} + 3 \, b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{20 \, {\left (c^{2} x^{7} - x^{5}\right )}}, \frac {4 \, {\left (b c^{7} d x^{7} - b c^{5} d x^{5}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{2} + 1\right )} \sqrt {-d}}{c^{2} d x^{4} - {\left (c^{2} + 1\right )} d x^{2} + d}\right ) - {\left (4 \, b c^{3} d x^{3} - {\left (4 \, b c^{3} - b c\right )} d x^{5} - b c d x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (a c^{6} d x^{6} - 3 \, a c^{4} d x^{4} + 3 \, a c^{2} d x^{2} - a d + {\left (b c^{6} d x^{6} - 3 \, b c^{4} d x^{4} + 3 \, b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{20 \, {\left (c^{2} x^{7} - x^{5}\right )}}\right ] \]
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^6,x, algorithm="fricas")
[Out]
[1/20*(2*(b*c^7*d*x^7 - b*c^5*d*x^5)*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)) - (4*b*c^3*d*x^3 - (4*b*c^3 - b*c)*d*x^5 - b*c*d*x)*sqrt(-
c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(a*c^6*d*x^6 - 3*a*c^4*d*x^4 + 3*a*c^2*d*x^2 - a*d + (b*c^6*d*x^6 - 3*b*
c^4*d*x^4 + 3*b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^7 - x^5), 1/20*(4*(b*c^7*d*x^7 - b*
c^5*d*x^5)*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)) - (4*b*c^3*d*x^3 - (4*b*c^3 - b*c)*d*x^5 - b*c*d*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(a*
c^6*d*x^6 - 3*a*c^4*d*x^4 + 3*a*c^2*d*x^2 - a*d + (b*c^6*d*x^6 - 3*b*c^4*d*x^4 + 3*b*c^2*d*x^2 - b*d)*arcsin(c
*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^7 - x^5)]
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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^6,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 = 0.47, size = 2350, normalized size = 15.26 \[ \text {result too large to display} \]
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^6,x)
[Out]
-1/5*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)
*arcsin(c*x)*c^5+1/5*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^7/(c^2*x^2-1
)*(-c^2*x^2+1)*c^12-9/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^5/(c^2*x
^2-1)*(-c^2*x^2+1)*c^10+3/10*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^3/(c
^2*x^2-1)*(-c^2*x^2+1)*c^8-1/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x/(
c^2*x^2-1)*(-c^2*x^2+1)*c^6-1/5*a/d/x^5*(-c^2*d*x^2+d)^(5/2)+I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^
6+10*c^4*x^4-5*c^2*x^2+1)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^7-I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c
^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^8/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^13+2*I*b*(-d*(c^2*x
^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^6/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^
11-2*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(
1/2)*arcsin(c*x)*c^9+1/5*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x^5/(c^2*x^2
-1)*arcsin(c*x)-1/5*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^5+3/2*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/(c^2*x^2-1)*c^5*(-c^2*x^2+
1)^(1/2)-7/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^3/(c^2*x^2-1)*c^8+1
/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x/(c^2*x^2-1)*c^6+9/4*b*(-d*(c^
2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^9+14*b*(-
d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^3/(c^2*x^2-1)*arcsin(c*x)*c^8-5/2*b*(-d
*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^2/(c^2*x^2-1)*c^7*(-c^2*x^2+1)^(1/2)-56/
5*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x/(c^2*x^2-1)*arcsin(c*x)*c^6+28/5*
b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x/(c^2*x^2-1)*arcsin(c*x)*c^4-b*(-d*(
c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^6/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^11-b*(-
d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^9/(c^2*x^2-1)*arcsin(c*x)*c^14+5*b*(-d*
(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^7/(c^2*x^2-1)*arcsin(c*x)*c^12-11*b*(-d*(
c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^10-9/20*b*(-d*
(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x^2/(c^2*x^2-1)*c^3*(-c^2*x^2+1)^(1/2)-8/5*
b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x^3/(c^2*x^2-1)*arcsin(c*x)*c^2+1/20*
b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+
2*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*d*c^5/(5*c^2*x^2-5)+1/5*I*b*(-d*(c^2*x^2-1))^(1/2)
*d/(5*c^8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^9/(c^2*x^2-1)*c^14-13/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^
8*x^8-10*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^7/(c^2*x^2-1)*c^12+3/4*I*b*(-d*(c^2*x^2-1))^(1/2)*d/(5*c^8*x^8-10*c
^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^5/(c^2*x^2-1)*c^10
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maxima [A] time = 0.72, size = 172, normalized size = 1.12 \[ -\frac {{\left (2 \, \left (-1\right )^{-2 \, c^{2} d x^{2} + 2 \, d} c^{4} d^{\frac {5}{2}} \log \left (-2 \, c^{2} d + \frac {2 \, d}{x^{2}}\right ) + 2 \, c^{4} d^{\frac {5}{2}} \log \left (x^{2} - \frac {1}{c^{2}}\right ) - \frac {3 \, \sqrt {c^{4} d x^{4} - 2 \, c^{2} d x^{2} + d} c^{2} d^{2}}{x^{2}} + \frac {\sqrt {c^{4} d x^{4} - 2 \, c^{2} d x^{2} + d} d^{2}}{x^{4}}\right )} b c}{20 \, d} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b \arcsin \left (c x\right )}{5 \, d x^{5}} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a}{5 \, d x^{5}} \]
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^6,x, algorithm="maxima")
[Out]
-1/20*(2*(-1)^(-2*c^2*d*x^2 + 2*d)*c^4*d^(5/2)*log(-2*c^2*d + 2*d/x^2) + 2*c^4*d^(5/2)*log(x^2 - 1/c^2) - 3*sq
rt(c^4*d*x^4 - 2*c^2*d*x^2 + d)*c^2*d^2/x^2 + sqrt(c^4*d*x^4 - 2*c^2*d*x^2 + d)*d^2/x^4)*b*c/d - 1/5*(-c^2*d*x
^2 + d)^(5/2)*b*arcsin(c*x)/(d*x^5) - 1/5*(-c^2*d*x^2 + d)^(5/2)*a/(d*x^5)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2}}{x^6} \,d x \]
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^6,x)
[Out]
int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(3/2))/x^6, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{6}}\, dx \]
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**6,x)
[Out]
Integral((-d*(c*x - 1)*(c*x + 1))**(3/2)*(a + b*asin(c*x))/x**6, x)
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