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

Optimal. Leaf size=64 \[ -\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}-\frac {b c \sqrt {1-c^2 x^4}}{12 x^4}-\frac {1}{12} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^4}\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4842, 12, 266, 51, 63, 208} \[ -\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}-\frac {b c \sqrt {1-c^2 x^4}}{12 x^4}-\frac {1}{12} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^7,x]

[Out]

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

Rule 12

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

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 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 x^2\right )}{x^7} \, dx &=-\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{6} b \int \frac {2 c}{x^5 \sqrt {1-c^2 x^4}} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{3} (b c) \int \frac {1}{x^5 \sqrt {1-c^2 x^4}} \, dx\\ &=-\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{12} (b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1-c^2 x}} \, dx,x,x^4\right )\\ &=-\frac {b c \sqrt {1-c^2 x^4}}{12 x^4}-\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}+\frac {1}{24} \left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^4\right )\\ &=-\frac {b c \sqrt {1-c^2 x^4}}{12 x^4}-\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}-\frac {1}{12} (b c) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^4}\right )\\ &=-\frac {b c \sqrt {1-c^2 x^4}}{12 x^4}-\frac {a+b \sin ^{-1}\left (c x^2\right )}{6 x^6}-\frac {1}{12} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^4}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 69, normalized size = 1.08 \[ -\frac {a}{6 x^6}-\frac {b c \sqrt {1-c^2 x^4}}{12 x^4}-\frac {1}{12} b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^4}\right )-\frac {b \sin ^{-1}\left (c x^2\right )}{6 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 84, normalized size = 1.31 \[ -\frac {b c^{3} x^{6} \log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right ) - b c^{3} x^{6} \log \left (\sqrt {-c^{2} x^{4} + 1} - 1\right ) + 2 \, \sqrt {-c^{2} x^{4} + 1} b c x^{2} + 4 \, b \arcsin \left (c x^{2}\right ) + 4 \, a}{24 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(b*c^3*x^6*log(sqrt(-c^2*x^4 + 1) + 1) - b*c^3*x^6*log(sqrt(-c^2*x^4 + 1) - 1) + 2*sqrt(-c^2*x^4 + 1)*b*
c*x^2 + 4*b*arcsin(c*x^2) + 4*a)/x^6

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giac [B]  time = 0.60, size = 301, normalized size = 4.70 \[ -\frac {\frac {b c^{7} x^{6} \arcsin \left (c x^{2}\right )}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{3}} + \frac {a c^{7} x^{6}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{3}} - \frac {b c^{6} x^{4}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac {3 \, b c^{5} x^{2} \arcsin \left (c x^{2}\right )}{\sqrt {-c^{2} x^{4} + 1} + 1} + \frac {3 \, a c^{5} x^{2}}{\sqrt {-c^{2} x^{4} + 1} + 1} - 4 \, b c^{4} \log \left (x^{2} {\left | c \right |}\right ) + 4 \, b c^{4} \log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right ) + \frac {3 \, b c^{3} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )} \arcsin \left (c x^{2}\right )}{x^{2}} + \frac {3 \, a c^{3} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}}{x^{2}} + \frac {b c^{2} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}}{x^{4}} + \frac {b c {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{3} \arcsin \left (c x^{2}\right )}{x^{6}} + \frac {a c {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{3}}{x^{6}}}{48 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(b*c^7*x^6*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^3 + a*c^7*x^6/(sqrt(-c^2*x^4 + 1) + 1)^3 - b*c^6*x^4/(
sqrt(-c^2*x^4 + 1) + 1)^2 + 3*b*c^5*x^2*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1) + 3*a*c^5*x^2/(sqrt(-c^2*x^4 +
1) + 1) - 4*b*c^4*log(x^2*abs(c)) + 4*b*c^4*log(sqrt(-c^2*x^4 + 1) + 1) + 3*b*c^3*(sqrt(-c^2*x^4 + 1) + 1)*arc
sin(c*x^2)/x^2 + 3*a*c^3*(sqrt(-c^2*x^4 + 1) + 1)/x^2 + b*c^2*(sqrt(-c^2*x^4 + 1) + 1)^2/x^4 + b*c*(sqrt(-c^2*
x^4 + 1) + 1)^3*arcsin(c*x^2)/x^6 + a*c*(sqrt(-c^2*x^4 + 1) + 1)^3/x^6)/c

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maple [A]  time = 0.02, size = 61, normalized size = 0.95 \[ -\frac {a}{6 x^{6}}+b \left (-\frac {\arcsin \left (c \,x^{2}\right )}{6 x^{6}}+\frac {c \left (-\frac {\sqrt {-c^{2} x^{4}+1}}{4 x^{4}}-\frac {c^{2} \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{4}+1}}\right )}{4}\right )}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 81, normalized size = 1.27 \[ -\frac {1}{24} \, {\left ({\left (c^{2} \log \left (\sqrt {-c^{2} x^{4} + 1} + 1\right ) - c^{2} \log \left (\sqrt {-c^{2} x^{4} + 1} - 1\right ) + \frac {2 \, \sqrt {-c^{2} x^{4} + 1}}{x^{4}}\right )} c + \frac {4 \, \arcsin \left (c x^{2}\right )}{x^{6}}\right )} b - \frac {a}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((c^2*log(sqrt(-c^2*x^4 + 1) + 1) - c^2*log(sqrt(-c^2*x^4 + 1) - 1) + 2*sqrt(-c^2*x^4 + 1)/x^4)*c + 4*ar
csin(c*x^2)/x^6)*b - 1/6*a/x^6

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x^2\right )}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x^2))/x^7,x)

[Out]

int((a + b*asin(c*x^2))/x^7, x)

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sympy [A]  time = 3.35, size = 128, normalized size = 2.00 \[ - \frac {a}{6 x^{6}} + \frac {b c \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x^{2}} \right )}}{4} - \frac {c \sqrt {-1 + \frac {1}{c^{2} x^{4}}}}{4 x^{2}} & \text {for}\: \frac {1}{\left |{c^{2} x^{4}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x^{2}} \right )}}{4} - \frac {i c}{4 x^{2} \sqrt {1 - \frac {1}{c^{2} x^{4}}}} + \frac {i}{4 c x^{6} \sqrt {1 - \frac {1}{c^{2} x^{4}}}} & \text {otherwise} \end {cases}\right )}{3} - \frac {b \operatorname {asin}{\left (c x^{2} \right )}}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(6*x**6) + b*c*Piecewise((-c**2*acosh(1/(c*x**2))/4 - c*sqrt(-1 + 1/(c**2*x**4))/(4*x**2), 1/Abs(c**2*x**4)
> 1), (I*c**2*asin(1/(c*x**2))/4 - I*c/(4*x**2*sqrt(1 - 1/(c**2*x**4))) + I/(4*c*x**6*sqrt(1 - 1/(c**2*x**4))
), True))/3 - b*asin(c*x**2)/(6*x**6)

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