3.89 \(\int \frac {(d-c^2 d x^2)^{5/2} (a+b \sin ^{-1}(c x))}{x^6} \, dx\)
Optimal. Leaf size=277 \[ -\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 x^5}+\frac {c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {c^5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b \sqrt {1-c^2 x^2}}-\frac {c^4 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {23 b c^5 d^2 \log (x) \sqrt {d-c^2 d x^2}}{15 \sqrt {1-c^2 x^2}}+\frac {11 b c^3 d^2 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}} \]
[Out]
1/3*c^2*d*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/x^3-1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^5-c^4*d^2*(a
+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/x-1/20*b*c*d^2*(-c^2*d*x^2+d)^(1/2)/x^4/(-c^2*x^2+1)^(1/2)+11/30*b*c^3*d^
2*(-c^2*d*x^2+d)^(1/2)/x^2/(-c^2*x^2+1)^(1/2)-1/2*c^5*d^2*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/(-c^2*x^2
+1)^(1/2)+23/15*b*c^5*d^2*ln(x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)
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Rubi [A] time = 0.35, antiderivative size = 277, normalized size of antiderivative = 1.00,
number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used
= {4695, 4693, 29, 4641, 14, 266, 43} \[ -\frac {c^5 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b \sqrt {1-c^2 x^2}}-\frac {c^4 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 x^5}+\frac {11 b c^3 d^2 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}}-\frac {b c d^2 \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {23 b c^5 d^2 \log (x) \sqrt {d-c^2 d x^2}}{15 \sqrt {1-c^2 x^2}} \]
Antiderivative was successfully verified.
[In]
Int[((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x^6,x]
[Out]
-(b*c*d^2*Sqrt[d - c^2*d*x^2])/(20*x^4*Sqrt[1 - c^2*x^2]) + (11*b*c^3*d^2*Sqrt[d - c^2*d*x^2])/(30*x^2*Sqrt[1
- c^2*x^2]) - (c^4*d^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/x + (c^2*d*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin
[c*x]))/(3*x^3) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*x^5) - (c^5*d^2*Sqrt[d - c^2*d*x^2]*(a + b*Ar
cSin[c*x])^2)/(2*b*Sqrt[1 - c^2*x^2]) + (23*b*c^5*d^2*Sqrt[d - c^2*d*x^2]*Log[x])/(15*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 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 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 )^{5/2} \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 x^5}-\left (c^2 d\right ) \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 (b c d^2 \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 {c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 x^5}+\left (c^4 d^2\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^2 \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 (b c^3 d^2 \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^4 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 x^5}+\frac {\left (b c d^2 \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 {\left (b c^3 d^2 \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^5 d^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{x} \, dx}{\sqrt {1-c^2 x^2}}-\frac {\left (c^6 d^2 \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^2 \sqrt {d-c^2 d x^2}}{20 x^4 \sqrt {1-c^2 x^2}}+\frac {11 b c^3 d^2 \sqrt {d-c^2 d x^2}}{30 x^2 \sqrt {1-c^2 x^2}}-\frac {c^4 d^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {c^2 d \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 x^5}-\frac {c^5 d^2 \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 {23 b c^5 d^2 \sqrt {d-c^2 d x^2} \log (x)}{15 \sqrt {1-c^2 x^2}}\\ \end {align*}
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Mathematica [A] time = 1.60, size = 234, normalized size = 0.84 \[ \frac {1}{60} d^2 \left (\frac {\sqrt {d-c^2 d x^2} \left (-4 a \sqrt {1-c^2 x^2} \left (23 c^4 x^4-11 c^2 x^2+3\right )+92 b c^5 x^5 \log (c x)+b c x \left (22 c^2 x^2-3\right )\right )}{x^5 \sqrt {1-c^2 x^2}}+60 a c^5 \sqrt {d} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )-\frac {30 b c^5 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)^2}{\sqrt {1-c^2 x^2}}-\frac {4 b \left (23 c^4 x^4-11 c^2 x^2+3\right ) \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{x^5}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/x^6,x]
[Out]
(d^2*((-4*b*Sqrt[d - c^2*d*x^2]*(3 - 11*c^2*x^2 + 23*c^4*x^4)*ArcSin[c*x])/x^5 - (30*b*c^5*Sqrt[d - c^2*d*x^2]
*ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] + 60*a*c^5*Sqrt[d]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))
] + (Sqrt[d - c^2*d*x^2]*(b*c*x*(-3 + 22*c^2*x^2) - 4*a*Sqrt[1 - c^2*x^2]*(3 - 11*c^2*x^2 + 23*c^4*x^4) + 92*b
*c^5*x^5*Log[c*x]))/(x^5*Sqrt[1 - c^2*x^2])))/60
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fricas [F] time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{x^{6}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="fricas")
[Out]
integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*sqr
t(-c^2*d*x^2 + d)/x^6, 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)^(5/2)*(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 = 0.51, size = 2615, normalized size = 9.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^6,x)
[Out]
-69/5*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)/(c^2*x^2-1)*arcsin(c*
x)*(-c^2*x^2+1)^(1/2)*c^5-7153/60*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*
x^2+9)*x^5/(c^2*x^2-1)*(-c^2*x^2+1)*c^10+759/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c
^4*x^4-75*c^2*x^2+9)*x^3/(c^2*x^2-1)*(-c^2*x^2+1)*c^8-69/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c
^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x/(c^2*x^2-1)*(-c^2*x^2+1)*c^6+5819/30*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c
^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^7/(c^2*x^2-1)*(-c^2*x^2+1)*c^12+9/5*b*(-d*(c^2*x^2-1))^(1/2)*d^
2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)/x^5/(c^2*x^2-1)*arcsin(c*x)-23/15*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^2*c^5+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^2*c^5+175/4*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+
325*c^4*x^4-75*c^2*x^2+9)/(c^2*x^2-1)*c^5*(-c^2*x^2+1)^(1/2)-1/5*a/d/x^5*(-c^2*d*x^2+d)^(7/2)-8/15*a*c^6*x*(-c
^2*d*x^2+d)^(5/2)-8/15*a*c^4/d/x*(-c^2*d*x^2+d)^(7/2)-2/3*a*c^6*d*x*(-c^2*d*x^2+d)^(3/2)-a*c^6*d^2*x*(-c^2*d*x
^2+d)^(1/2)-1587*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^9/(c^2*x^2
-1)*arcsin(c*x)*c^14+46*I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)*d^2*c^5/(15*c^2*x^2-15)+5819
/30*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^9/(c^2*x^2-1)*c^14-18
791/60*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^7/(c^2*x^2-1)*c^12
+943/6*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^5/(c^2*x^2-1)*c^10
-207/5*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^3/(c^2*x^2-1)*c^8-
759/2*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^6/(c^2*x^2-1)*(-c^2*x
^2+1)^(1/2)*c^11+777/5*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)/x/(c^2
*x^2-1)*arcsin(c*x)*c^4-141/20*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9
)/x^2/(c^2*x^2-1)*c^3*(-c^2*x^2+1)^(1/2)-117/5*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*
x^4-75*c^2*x^2+9)/x^3/(c^2*x^2-1)*arcsin(c*x)*c^2+9/20*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+
325*c^4*x^4-75*c^2*x^2+9)/x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c+69/20*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*
x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x/(c^2*x^2-1)*c^6+3519*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-76
5*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^7/(c^2*x^2-1)*arcsin(c*x)*c^12-9595/3*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035
*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^5/(c^2*x^2-1)*arcsin(c*x)*c^10+1329/4*b*(-d*(c^2*x^2-1))^(1/2
)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^9+5318/3*b*(-d*
(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^3/(c^2*x^2-1)*arcsin(c*x)*c^8-188
9/12*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^2/(c^2*x^2-1)*c^7*(-c^
2*x^2+1)^(1/2)-9602/15*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x/(c^2
*x^2-1)*arcsin(c*x)*c^6+115*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)
*x^2/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^7+1173*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*
x^6+325*c^4*x^4-75*c^2*x^2+9)*x^6/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^11-1495/3*I*b*(-d*(c^2*x^2-1))^
(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^4/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c
^9-1587*I*b*(-d*(c^2*x^2-1))^(1/2)*d^2/(1035*c^8*x^8-765*c^6*x^6+325*c^4*x^4-75*c^2*x^2+9)*x^8/(c^2*x^2-1)*arc
sin(c*x)*(-c^2*x^2+1)^(1/2)*c^13+2/15*a*c^2/d/x^3*(-c^2*d*x^2+d)^(7/2)-a*c^6*d^3/(c^2*d)^(1/2)*arctan((c^2*d)^
(1/2)*x/(-c^2*d*x^2+d)^(1/2))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \int \frac {{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x^{6}}\,{d x} - \frac {1}{15} \, {\left (10 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{6} d x + 15 \, \sqrt {-c^{2} d x^{2} + d} c^{6} d^{2} x + 15 \, c^{5} d^{\frac {5}{2}} \arcsin \left (c x\right ) + \frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} c^{4}}{x} - \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}} c^{2}}{d x^{3}} + \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}}}{d x^{5}}\right )} a \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/x^6,x, algorithm="maxima")
[Out]
b*sqrt(d)*integrate((c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1
)*sqrt(-c*x + 1))/x^6, x) - 1/15*(10*(-c^2*d*x^2 + d)^(3/2)*c^6*d*x + 15*sqrt(-c^2*d*x^2 + d)*c^6*d^2*x + 15*c
^5*d^(5/2)*arcsin(c*x) + 8*(-c^2*d*x^2 + d)^(5/2)*c^4/x - 2*(-c^2*d*x^2 + d)^(7/2)*c^2/(d*x^3) + 3*(-c^2*d*x^2
+ d)^(7/2)/(d*x^5))*a
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^6} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(5/2))/x^6,x)
[Out]
int(((a + b*asin(c*x))*(d - c^2*d*x^2)^(5/2))/x^6, x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-c**2*d*x**2+d)**(5/2)*(a+b*asin(c*x))/x**6,x)
[Out]
Timed out
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