3.392 \(\int \frac {a+b \sin ^{-1}(c+d x^2)}{x^7} \, dx\)

Optimal. Leaf size=190 \[ -\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}-\frac {b c d^2 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{4 \left (1-c^2\right )^2 x^2}-\frac {b d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{12 \left (1-c^2\right ) x^4}-\frac {b \left (2 c^2+1\right ) d^3 \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}\right )}{12 \left (1-c^2\right )^{5/2}} \]

[Out]

1/6*(-a-b*arcsin(d*x^2+c))/x^6-1/12*b*(2*c^2+1)*d^3*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)^(5/2)-1/12*b*d*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)/x^4-1/4*b*c*d^2*(-d^2*x^4-2*
c*d*x^2-c^2+1)^(1/2)/(-c^2+1)^2/x^2

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Rubi [A]  time = 0.22, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4842, 12, 1114, 744, 806, 724, 206} \[ -\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}-\frac {b c d^2 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{4 \left (1-c^2\right )^2 x^2}-\frac {b d \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{12 \left (1-c^2\right ) x^4}-\frac {b \left (2 c^2+1\right ) d^3 \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}\right )}{12 \left (1-c^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x^2])/x^7,x]

[Out]

-(b*d*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(12*(1 - c^2)*x^4) - (b*c*d^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/
(4*(1 - c^2)^2*x^2) - (a + b*ArcSin[c + d*x^2])/(6*x^6) - (b*(1 + 2*c^2)*d^3*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])])/(12*(1 - c^2)^(5/2))

Rule 12

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

Rule 206

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

Rule 724

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[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 1114

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 4842

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^7} \, dx &=-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}+\frac {1}{6} b \int \frac {2 d}{x^5 \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 )}{6 x^6}+\frac {1}{3} (b d) \int \frac {1}{x^5 \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 )}{6 x^6}+\frac {1}{6} (b d) \operatorname {Subst}\left (\int \frac {1}{x^3 \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}}{12 \left (1-c^2\right ) x^4}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}-\frac {(b d) \operatorname {Subst}\left (\int \frac {-3 c d-d^2 x}{x^2 \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{12 \left (1-c^2\right )}\\ &=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{12 \left (1-c^2\right ) x^4}-\frac {b c d^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right )^2 x^2}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}+\frac {\left (b \left (1+2 c^2\right ) d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{12 \left (1-c^2\right )^2}\\ &=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{12 \left (1-c^2\right ) x^4}-\frac {b c d^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right )^2 x^2}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}-\frac {\left (b \left (1+2 c^2\right ) d^3\right ) \operatorname {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 )}{6 \left (1-c^2\right )^2}\\ &=-\frac {b d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{12 \left (1-c^2\right ) x^4}-\frac {b c d^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{4 \left (1-c^2\right )^2 x^2}-\frac {a+b \sin ^{-1}\left (c+d x^2\right )}{6 x^6}-\frac {b \left (1+2 c^2\right ) d^3 \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 )}{12 \left (1-c^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 176, normalized size = 0.93 \[ -\frac {a}{6 x^6}+b \left (\frac {d}{12 \left (c^2-1\right ) x^4}-\frac {c d^2}{4 \left (c^2-1\right )^2 x^2}\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}-\frac {b \left (2 c^2+1\right ) d^3 \tanh ^{-1}\left (\frac {-c^2-c d x^2+1}{\sqrt {1-c^2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}\right )}{12 (c-1)^2 (c+1)^2 \sqrt {1-c^2}}-\frac {b \sin ^{-1}\left (c+d x^2\right )}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x^2])/x^7,x]

[Out]

-1/6*a/x^6 + b*(d/(12*(-1 + c^2)*x^4) - (c*d^2)/(4*(-1 + c^2)^2*x^2))*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] - (b
*ArcSin[c + d*x^2])/(6*x^6) - (b*(1 + 2*c^2)*d^3*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])])/(12*(-1 + c)^2*(1 + c)^2*Sqrt[1 - c^2])

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fricas [A]  time = 0.65, size = 496, normalized size = 2.61 \[ \left [-\frac {{\left (2 \, b c^{2} + b\right )} \sqrt {-c^{2} + 1} d^{3} x^{6} \log \left (\frac {{\left (2 \, c^{2} - 1\right )} d^{2} x^{4} + 2 \, c^{4} + 4 \, {\left (c^{3} - c\right )} d x^{2} + 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {-c^{2} + 1} - 4 \, c^{2} + 2}{x^{4}}\right ) + 4 \, a c^{6} - 12 \, a c^{4} + 12 \, a c^{2} + 4 \, {\left (b c^{6} - 3 \, b c^{4} + 3 \, b c^{2} - b\right )} \arcsin \left (d x^{2} + c\right ) + 2 \, {\left (3 \, {\left (b c^{3} - b c\right )} d^{2} x^{4} - {\left (b c^{4} - 2 \, b c^{2} + b\right )} d x^{2}\right )} \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} - 4 \, a}{24 \, {\left (c^{6} - 3 \, c^{4} + 3 \, c^{2} - 1\right )} x^{6}}, \frac {{\left (2 \, b c^{2} + b\right )} \sqrt {c^{2} - 1} d^{3} x^{6} \arctan \left (\frac {\sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (c d x^{2} + c^{2} - 1\right )} \sqrt {c^{2} - 1}}{{\left (c^{2} - 1\right )} d^{2} x^{4} + c^{4} + 2 \, {\left (c^{3} - c\right )} d x^{2} - 2 \, c^{2} + 1}\right ) - 2 \, a c^{6} + 6 \, a c^{4} - 6 \, a c^{2} - 2 \, {\left (b c^{6} - 3 \, b c^{4} + 3 \, b c^{2} - b\right )} \arcsin \left (d x^{2} + c\right ) - {\left (3 \, {\left (b c^{3} - b c\right )} d^{2} x^{4} - {\left (b c^{4} - 2 \, b c^{2} + b\right )} d x^{2}\right )} \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} + 2 \, a}{12 \, {\left (c^{6} - 3 \, c^{4} + 3 \, c^{2} - 1\right )} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^7,x, algorithm="fricas")

[Out]

[-1/24*((2*b*c^2 + b)*sqrt(-c^2 + 1)*d^3*x^6*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) + 4*a*c^6 - 12*a*c^4 + 12*a*
c^2 + 4*(b*c^6 - 3*b*c^4 + 3*b*c^2 - b)*arcsin(d*x^2 + c) + 2*(3*(b*c^3 - b*c)*d^2*x^4 - (b*c^4 - 2*b*c^2 + b)
*d*x^2)*sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1) - 4*a)/((c^6 - 3*c^4 + 3*c^2 - 1)*x^6), 1/12*((2*b*c^2 + b)*sqrt(
c^2 - 1)*d^3*x^6*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)) - 2*a*c^6 + 6*a*c^4 - 6*a*c^2 - 2*(b*c^6 - 3*b*c^4 + 3*b*c^2 - b)*
arcsin(d*x^2 + c) - (3*(b*c^3 - b*c)*d^2*x^4 - (b*c^4 - 2*b*c^2 + b)*d*x^2)*sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 +
1) + 2*a)/((c^6 - 3*c^4 + 3*c^2 - 1)*x^6)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (d x^{2} + c\right ) + a}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^7,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + c) + a)/x^7, x)

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maple [A]  time = 0.02, size = 246, normalized size = 1.29 \[ -\frac {a}{6 x^{6}}-\frac {b \arcsin \left (d \,x^{2}+c \right )}{6 x^{6}}-\frac {b d \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{12 \left (-c^{2}+1\right ) x^{4}}-\frac {b c \,d^{2} \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{4 \left (-c^{2}+1\right )^{2} x^{2}}-\frac {b \,d^{3} c^{2} \ln \left (\frac {-2 c^{2}+2-2 c d \,x^{2}+2 \sqrt {-c^{2}+1}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{x^{2}}\right )}{4 \left (-c^{2}+1\right )^{\frac {5}{2}}}-\frac {b \,d^{3} \ln \left (\frac {-2 c^{2}+2-2 c d \,x^{2}+2 \sqrt {-c^{2}+1}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{x^{2}}\right )}{12 \left (-c^{2}+1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2+c))/x^7,x)

[Out]

-1/6*a/x^6-1/6*b/x^6*arcsin(d*x^2+c)-1/12*b*d*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)/x^4-1/4*b*c*d^2*(-d^2*
x^4-2*c*d*x^2-c^2+1)^(1/2)/(-c^2+1)^2/x^2-1/4*b*d^3*c^2/(-c^2+1)^(5/2)*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)-1/12*b*d^3/(-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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2+c))/x^7,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 c-1 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x^2))/x^7,x)

[Out]

int((a + b*asin(c + d*x^2))/x^7, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2+c))/x**7,x)

[Out]

Integral((a + b*asin(c + d*x**2))/x**7, x)

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