3.4.91 \(\int \frac {a+b \text {ArcSin}(c+d x^2)}{x^5} \, dx\) [391]

Optimal. Leaf size=137 \[ -\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right ) x^2}-\frac {a+b \text {ArcSin}\left (c+d x^2\right )}{4 x^4}-\frac {b c d^2 \tanh ^{-1}\left (\frac {1-c^2-c d x^2}{\sqrt {1-c^2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\right )}{4 \left (1-c^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4926, 12, 1128, 744, 738, 212} \begin {gather*} -\frac {a+b \text {ArcSin}\left (c+d x^2\right )}{4 x^4}-\frac {b d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{4 \left (1-c^2\right ) x^2}-\frac {b c d^2 \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}\right )}{4 \left (1-c^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x^5,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*
((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0]

Rule 1128

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 4926

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcSin[
u])/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 - u^2]), x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}\left (c+d x^2\right )}{x^5} \, dx &=-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{4 x^4}+\frac {1}{4} b \int \frac {2 d}{x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{4 x^4}+\frac {1}{2} (b d) \int \frac {1}{x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{4 x^4}+\frac {1}{4} (b d) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )\\ &=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right ) x^2}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{4 x^4}+\frac {\left (b c d^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{4 \left (1-c^2\right )}\\ &=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right ) x^2}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{4 x^4}-\frac {\left (b c d^2\right ) \text {Subst}\left (\int \frac {1}{4 \left (1-c^2\right )-x^2} \, dx,x,\frac {2 \left (1-c^2-c d x^2\right )}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}}\right )}{2 \left (1-c^2\right )}\\ &=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right ) x^2}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{4 x^4}-\frac {b c d^2 \tanh ^{-1}\left (\frac {1-c^2-c d x^2}{\sqrt {1-c^2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\right )}{4 \left (1-c^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 148, normalized size = 1.08 \begin {gather*} -\frac {a}{4 x^4}+\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (-1+c^2\right ) x^2}-\frac {b \text {ArcSin}\left (c+d x^2\right )}{4 x^4}+\frac {b c d^2 \text {ArcTan}\left (\frac {\sqrt {-d^2} x^2-\sqrt {1-c^2-2 c d x^2-d^2 x^4}}{\sqrt {-1+c^2}}\right )}{2 (-1+c) (1+c) \sqrt {-1+c^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 132, normalized size = 0.96

method	result	size
default	\(-\frac {a}{4 x^{4}}-\frac {b \arcsin \left (d \,x^{2}+c \right )}{4 x^{4}}-\frac {b d \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{4 \left (-c^{2}+1\right ) x^{2}}-\frac {b \,d^{2} c \ln \left (\frac {-2 c^{2}+2-2 c d \,x^{2}+2 \sqrt {-c^{2}+1}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{x^{2}}\right )}{4 \left (-c^{2}+1\right )^{\frac {3}{2}}}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for mor
e details)Is

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Fricas [A]
time = 3.18, size = 392, normalized size = 2.86 \begin {gather*} \left [-\frac {\sqrt {-c^{2} + 1} b c d^{2} x^{4} \log \left (\frac {{\left (2 \, c^{2} - 1\right )} d^{2} x^{4} + 2 \, c^{4} + 4 \, {\left (c^{3} - c\right )} d x^{2} - 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {-c^{2} + 1} - 4 \, c^{2} + 2}{x^{4}}\right ) + 2 \, a c^{4} - 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (b c^{2} - b\right )} d x^{2} - 4 \, a c^{2} + 2 \, {\left (b c^{4} - 2 \, b c^{2} + b\right )} \arcsin \left (d x^{2} + c\right ) + 2 \, a}{8 \, {\left (c^{4} - 2 \, c^{2} + 1\right )} x^{4}}, -\frac {\sqrt {c^{2} - 1} b c d^{2} x^{4} \arctan \left (\frac {\sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {c^{2} - 1}}{{\left (c^{2} - 1\right )} d^{2} x^{4} + c^{4} + 2 \, {\left (c^{3} - c\right )} d x^{2} - 2 \, c^{2} + 1}\right ) + a c^{4} - \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (b c^{2} - b\right )} d x^{2} - 2 \, a c^{2} + {\left (b c^{4} - 2 \, b c^{2} + b\right )} \arcsin \left (d x^{2} + c\right ) + a}{4 \, {\left (c^{4} - 2 \, c^{2} + 1\right )} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(-c^2 + 1)*b*c*d^2*x^4*log(((2*c^2 - 1)*d^2*x^4 + 2*c^4 + 4*(c^3 - c)*d*x^2 - 2*sqrt(-d^2*x^4 - 2*c
*d*x^2 - c^2 + 1)*(c*d*x^2 + c^2 - 1)*sqrt(-c^2 + 1) - 4*c^2 + 2)/x^4) + 2*a*c^4 - 2*sqrt(-d^2*x^4 - 2*c*d*x^2
- c^2 + 1)*(b*c^2 - b)*d*x^2 - 4*a*c^2 + 2*(b*c^4 - 2*b*c^2 + b)*arcsin(d*x^2 + c) + 2*a)/((c^4 - 2*c^2 + 1)*
x^4), -1/4*(sqrt(c^2 - 1)*b*c*d^2*x^4*arctan(sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1)*(c*d*x^2 + c^2 - 1)*sqrt(c^2
- 1)/((c^2 - 1)*d^2*x^4 + c^4 + 2*(c^3 - c)*d*x^2 - 2*c^2 + 1)) + a*c^4 - sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1
)*(b*c^2 - b)*d*x^2 - 2*a*c^2 + (b*c^4 - 2*b*c^2 + b)*arcsin(d*x^2 + c) + a)/((c^4 - 2*c^2 + 1)*x^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x^2))/x^5,x)

[Out]

int((a + b*asin(c + d*x^2))/x^5, x)

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