3.135 \(\int \frac {a+b \sin ^{-1}(c x)}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac {2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4655, 4653, 260, 261} \[ \frac {2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 4653

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 - c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSin[c*x
])^(n - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(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])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(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] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {2 \int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^{3/2}} \, dx}{3 d}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \int \frac {x}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}-\frac {\left (2 b c \sqrt {1-c^2 x^2}\right ) \int \frac {x}{1-c^2 x^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {2 x \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 113, normalized size = 0.73 \[ -\frac {\sqrt {d-c^2 d x^2} \left (4 a c^3 x^3-6 a c x+b \sqrt {1-c^2 x^2}-2 b \left (1-c^2 x^2\right )^{3/2} \log \left (c^2 x^2-1\right )+2 b c x \left (2 c^2 x^2-3\right ) \sin ^{-1}(c x)\right )}{6 c d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/(-c^2*d*x^2 + d)^(5/2), x)

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maple [C]  time = 0.16, size = 1071, normalized size = 6.95 \[ \frac {a x}{3 d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 a x}{3 d^{2} \sqrt {-c^{2} d \,x^{2}+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 )}{3 c \,d^{3} \left (c^{2} x^{2}-1\right )}+\frac {8 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} x^{3}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}+\frac {14 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x^{2}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x}{d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{3} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x^{4}}{d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{4} \arcsin \left (c x \right ) x^{5}}{d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-c^{2} x^{2}+1\right ) x}{d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \sqrt {-c^{2} x^{2}+1}\, x^{2}}{2 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {5 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \left (-c^{2} x^{2}+1\right ) x^{3}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}+\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{4} \left (-c^{2} x^{2}+1\right ) x^{5}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}+\frac {17 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \arcsin \left (c x \right ) x^{3}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {7 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{4} x^{5}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}+\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right ) c}+\frac {2 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{6} x^{7}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {8 i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right ) c}-\frac {4 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{d^{3} \left (3 c^{6} x^{6}-10 c^{4} x^{4}+11 c^{2} x^{2}-4\right )}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{3 c \,d^{3} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

1/3*a*x/d/(-c^2*d*x^2+d)^(3/2)+2/3*a/d^2*x/(-c^2*d*x^2+d)^(1/2)+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1
/2)/c/d^3/(c^2*x^2-1)*arcsin(c*x)+8/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*x
^3+14/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^
2-I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*x-2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6
*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^4-2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6
-10*c^4*x^4+11*c^2*x^2-4)*c^4*arcsin(c*x)*x^5+I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-
4)*(-c^2*x^2+1)*x-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c*(-c^2*x^2+1)^(1/2)*x^
2-5/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*(-c^2*x^2+1)*x^3+2/3*I*b*(-d*(c^2
*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*(-c^2*x^2+1)*x^5+17/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/
(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*arcsin(c*x)*x^3-7/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x
^4+11*c^2*x^2-4)*c^4*x^5+2/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*(-c^2*x^2+1)^(
1/2)+2/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^6*x^7-8/3*I*b*(-d*(c^2*x^2-1))^(
1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-4*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(
3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*x-2/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^3/(c^2*x^
2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)

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maxima [A]  time = 0.71, size = 141, normalized size = 0.92 \[ \frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} x^{2} - c^{2} d^{\frac {5}{2}}} + \frac {2 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}}} + \frac {2 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {2 \, x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((a + b*asin(c*x))/(d - c^2*d*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral((a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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