3.14 \(\int \frac {a+b \text {csch}^{-1}(c x)}{x^7} \, dx\)

Optimal. Leaf size=98 \[ -\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}-\frac {5}{96} b c^6 \text {csch}^{-1}(c x)+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1}}{36 x^5}+\frac {5 b c^5 \sqrt {\frac {1}{c^2 x^2}+1}}{96 x}-\frac {5 b c^3 \sqrt {\frac {1}{c^2 x^2}+1}}{144 x^3} \]

[Out]

-5/96*b*c^6*arccsch(c*x)+1/6*(-a-b*arccsch(c*x))/x^6+1/36*b*c*(1+1/c^2/x^2)^(1/2)/x^5-5/144*b*c^3*(1+1/c^2/x^2
)^(1/2)/x^3+5/96*b*c^5*(1+1/c^2/x^2)^(1/2)/x

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6284, 335, 321, 215} \[ -\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}+\frac {5 b c^5 \sqrt {\frac {1}{c^2 x^2}+1}}{96 x}-\frac {5 b c^3 \sqrt {\frac {1}{c^2 x^2}+1}}{144 x^3}+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1}}{36 x^5}-\frac {5}{96} b c^6 \text {csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/x^7,x]

[Out]

(b*c*Sqrt[1 + 1/(c^2*x^2)])/(36*x^5) - (5*b*c^3*Sqrt[1 + 1/(c^2*x^2)])/(144*x^3) + (5*b*c^5*Sqrt[1 + 1/(c^2*x^
2)])/(96*x) - (5*b*c^6*ArcCsch[c*x])/96 - (a + b*ArcCsch[c*x])/(6*x^6)

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 6284

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCsch[c*
x]))/(d*(m + 1)), x] + Dist[(b*d)/(c*(m + 1)), Int[(d*x)^(m - 1)/Sqrt[1 + 1/(c^2*x^2)], x], x] /; FreeQ[{a, b,
c, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \text {csch}^{-1}(c x)}{x^7} \, dx &=-\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}-\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^8} \, dx}{6 c}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}+\frac {b \operatorname {Subst}\left (\int \frac {x^6}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{6 c}\\ &=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{36 x^5}-\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}-\frac {1}{36} (5 b c) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{36 x^5}-\frac {5 b c^3 \sqrt {1+\frac {1}{c^2 x^2}}}{144 x^3}-\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}+\frac {1}{48} \left (5 b c^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{36 x^5}-\frac {5 b c^3 \sqrt {1+\frac {1}{c^2 x^2}}}{144 x^3}+\frac {5 b c^5 \sqrt {1+\frac {1}{c^2 x^2}}}{96 x}-\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}-\frac {1}{96} \left (5 b c^5\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{36 x^5}-\frac {5 b c^3 \sqrt {1+\frac {1}{c^2 x^2}}}{144 x^3}+\frac {5 b c^5 \sqrt {1+\frac {1}{c^2 x^2}}}{96 x}-\frac {5}{96} b c^6 \text {csch}^{-1}(c x)-\frac {a+b \text {csch}^{-1}(c x)}{6 x^6}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 88, normalized size = 0.90 \[ -\frac {a}{6 x^6}-\frac {5}{96} b c^6 \sinh ^{-1}\left (\frac {1}{c x}\right )+b \left (\frac {5 c^5}{96 x}-\frac {5 c^3}{144 x^3}+\frac {c}{36 x^5}\right ) \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}-\frac {b \text {csch}^{-1}(c x)}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/x^7,x]

[Out]

-1/6*a/x^6 + b*(c/(36*x^5) - (5*c^3)/(144*x^3) + (5*c^5)/(96*x))*Sqrt[(1 + c^2*x^2)/(c^2*x^2)] - (b*ArcCsch[c*
x])/(6*x^6) - (5*b*c^6*ArcSinh[1/(c*x)])/96

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fricas [A]  time = 1.12, size = 99, normalized size = 1.01 \[ -\frac {3 \, {\left (5 \, b c^{6} x^{6} + 16 \, b\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (15 \, b c^{5} x^{5} - 10 \, b c^{3} x^{3} + 8 \, b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 48 \, a}{288 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/288*(3*(5*b*c^6*x^6 + 16*b)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - (15*b*c^5*x^5 - 10*b*c^3*x
^3 + 8*b*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 48*a)/x^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccsch(c*x) + a)/x^7, x)

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maple [A]  time = 0.06, size = 139, normalized size = 1.42 \[ c^{6} \left (-\frac {a}{6 c^{6} x^{6}}+b \left (-\frac {\mathrm {arccsch}\left (c x \right )}{6 c^{6} x^{6}}-\frac {\sqrt {c^{2} x^{2}+1}\, \left (15 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) c^{6} x^{6}-15 c^{4} x^{4} \sqrt {c^{2} x^{2}+1}+10 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}-8 \sqrt {c^{2} x^{2}+1}\right )}{288 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{7} x^{7}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^6*(-1/6*a/c^6/x^6+b*(-1/6/c^6/x^6*arccsch(c*x)-1/288*(c^2*x^2+1)^(1/2)*(15*arctanh(1/(c^2*x^2+1)^(1/2))*c^6*
x^6-15*c^4*x^4*(c^2*x^2+1)^(1/2)+10*c^2*x^2*(c^2*x^2+1)^(1/2)-8*(c^2*x^2+1)^(1/2))/((c^2*x^2+1)/c^2/x^2)^(1/2)
/c^7/x^7))

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maxima [B]  time = 0.31, size = 185, normalized size = 1.89 \[ -\frac {1}{576} \, b {\left (\frac {15 \, c^{7} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right ) - 15 \, c^{7} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right ) - \frac {2 \, {\left (15 \, c^{12} x^{5} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 40 \, c^{10} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 33 \, c^{8} x \sqrt {\frac {1}{c^{2} x^{2}} + 1}\right )}}{c^{6} x^{6} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{3} - 3 \, c^{4} x^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{2} + 3 \, c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} - 1}}{c} + \frac {96 \, \operatorname {arcsch}\left (c x\right )}{x^{6}}\right )} - \frac {a}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/576*b*((15*c^7*log(c*x*sqrt(1/(c^2*x^2) + 1) + 1) - 15*c^7*log(c*x*sqrt(1/(c^2*x^2) + 1) - 1) - 2*(15*c^12*
x^5*(1/(c^2*x^2) + 1)^(5/2) - 40*c^10*x^3*(1/(c^2*x^2) + 1)^(3/2) + 33*c^8*x*sqrt(1/(c^2*x^2) + 1))/(c^6*x^6*(
1/(c^2*x^2) + 1)^3 - 3*c^4*x^4*(1/(c^2*x^2) + 1)^2 + 3*c^2*x^2*(1/(c^2*x^2) + 1) - 1))/c + 96*arccsch(c*x)/x^6
) - 1/6*a/x^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/x^7,x)

[Out]

int((a + b*asinh(1/(c*x)))/x^7, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))/x**7, x)

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