3.2.94 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{x^4} \, dx\) [194]

Optimal. Leaf size=73 \[ -\frac {b c}{15 x^{5/2}}-\frac {b c^3}{9 x^{3/2}}-\frac {b c^5}{3 \sqrt {x}}+\frac {1}{3} b c^6 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3} \]

[Out]

-1/15*b*c/x^(5/2)-1/9*b*c^3/x^(3/2)+1/3*b*c^6*arctanh(c*x^(1/2))+1/3*(-a-b*arctanh(c*x^(1/2)))/x^3-1/3*b*c^5/x
^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6037, 53, 65, 212} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{3} b c^6 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c^5}{3 \sqrt {x}}-\frac {b c^3}{9 x^{3/2}}-\frac {b c}{15 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/x^4,x]

[Out]

-1/15*(b*c)/x^(5/2) - (b*c^3)/(9*x^(3/2)) - (b*c^5)/(3*Sqrt[x]) + (b*c^6*ArcTanh[c*Sqrt[x]])/3 - (a + b*ArcTan
h[c*Sqrt[x]])/(3*x^3)

Rule 53

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 65

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 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 6037

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

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x^4} \, dx &=-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{6} (b c) \int \frac {1}{x^{7/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac {b c}{15 x^{5/2}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{6} \left (b c^3\right ) \int \frac {1}{x^{5/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac {b c}{15 x^{5/2}}-\frac {b c^3}{9 x^{3/2}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{6} \left (b c^5\right ) \int \frac {1}{x^{3/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac {b c}{15 x^{5/2}}-\frac {b c^3}{9 x^{3/2}}-\frac {b c^5}{3 \sqrt {x}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{6} \left (b c^7\right ) \int \frac {1}{\sqrt {x} \left (1-c^2 x\right )} \, dx\\ &=-\frac {b c}{15 x^{5/2}}-\frac {b c^3}{9 x^{3/2}}-\frac {b c^5}{3 \sqrt {x}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{3} \left (b c^7\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{15 x^{5/2}}-\frac {b c^3}{9 x^{3/2}}-\frac {b c^5}{3 \sqrt {x}}+\frac {1}{3} b c^6 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 99, normalized size = 1.36 \begin {gather*} -\frac {a}{3 x^3}-\frac {b c}{15 x^{5/2}}-\frac {b c^3}{9 x^{3/2}}-\frac {b c^5}{3 \sqrt {x}}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-\frac {1}{6} b c^6 \log \left (1-c \sqrt {x}\right )+\frac {1}{6} b c^6 \log \left (1+c \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/x^4,x]

[Out]

-1/3*a/x^3 - (b*c)/(15*x^(5/2)) - (b*c^3)/(9*x^(3/2)) - (b*c^5)/(3*Sqrt[x]) - (b*ArcTanh[c*Sqrt[x]])/(3*x^3) -
(b*c^6*Log[1 - c*Sqrt[x]])/6 + (b*c^6*Log[1 + c*Sqrt[x]])/6

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Maple [A]
time = 0.07, size = 80, normalized size = 1.10

method	result	size
derivativedivides	\(2 c^{6} \left (-\frac {a}{6 c^{6} x^{3}}-\frac {b \arctanh \left (c \sqrt {x}\right )}{6 c^{6} x^{3}}-\frac {b}{30 c^{5} x^{\frac {5}{2}}}-\frac {b}{18 c^{3} x^{\frac {3}{2}}}-\frac {b}{6 c \sqrt {x}}-\frac {b \ln \left (c \sqrt {x}-1\right )}{12}+\frac {b \ln \left (1+c \sqrt {x}\right )}{12}\right )\)	\(80\)
default	\(2 c^{6} \left (-\frac {a}{6 c^{6} x^{3}}-\frac {b \arctanh \left (c \sqrt {x}\right )}{6 c^{6} x^{3}}-\frac {b}{30 c^{5} x^{\frac {5}{2}}}-\frac {b}{18 c^{3} x^{\frac {3}{2}}}-\frac {b}{6 c \sqrt {x}}-\frac {b \ln \left (c \sqrt {x}-1\right )}{12}+\frac {b \ln \left (1+c \sqrt {x}\right )}{12}\right )\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x^4,x,method=_RETURNVERBOSE)

[Out]

2*c^6*(-1/6*a/c^6/x^3-1/6*b/c^6/x^3*arctanh(c*x^(1/2))-1/30*b/c^5/x^(5/2)-1/18*b/c^3/x^(3/2)-1/6*b/c/x^(1/2)-1
/12*b*ln(c*x^(1/2)-1)+1/12*b*ln(1+c*x^(1/2)))

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Maxima [A]
time = 0.26, size = 72, normalized size = 0.99 \begin {gather*} \frac {1}{90} \, {\left ({\left (15 \, c^{5} \log \left (c \sqrt {x} + 1\right ) - 15 \, c^{5} \log \left (c \sqrt {x} - 1\right ) - \frac {2 \, {\left (15 \, c^{4} x^{2} + 5 \, c^{2} x + 3\right )}}{x^{\frac {5}{2}}}\right )} c - \frac {30 \, \operatorname {artanh}\left (c \sqrt {x}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^4,x, algorithm="maxima")

[Out]

1/90*((15*c^5*log(c*sqrt(x) + 1) - 15*c^5*log(c*sqrt(x) - 1) - 2*(15*c^4*x^2 + 5*c^2*x + 3)/x^(5/2))*c - 30*ar
ctanh(c*sqrt(x))/x^3)*b - 1/3*a/x^3

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Fricas [A]
time = 0.35, size = 74, normalized size = 1.01 \begin {gather*} \frac {15 \, {\left (b c^{6} x^{3} - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) - 2 \, {\left (15 \, b c^{5} x^{2} + 5 \, b c^{3} x + 3 \, b c\right )} \sqrt {x} - 30 \, a}{90 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^4,x, algorithm="fricas")

[Out]

1/90*(15*(b*c^6*x^3 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) - 2*(15*b*c^5*x^2 + 5*b*c^3*x + 3*b*c)*sq
rt(x) - 30*a)/x^3

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (68) = 136\).
time = 25.83, size = 371, normalized size = 5.08 \begin {gather*} \begin {cases} - \frac {a}{3 x^{3}} + \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{3 x^{3}} & \text {for}\: c = - \sqrt {\frac {1}{x}} \\- \frac {a}{3 x^{3}} - \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{3 x^{3}} & \text {for}\: c = \sqrt {\frac {1}{x}} \\- \frac {15 a c^{2} x^{\frac {3}{2}}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} + \frac {15 a \sqrt {x}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} + \frac {15 b c^{8} x^{\frac {9}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} - \frac {15 b c^{7} x^{4}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} - \frac {15 b c^{6} x^{\frac {7}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} + \frac {10 b c^{5} x^{3}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} + \frac {2 b c^{3} x^{2}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} - \frac {15 b c^{2} x^{\frac {3}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} + \frac {3 b c x}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} + \frac {15 b \sqrt {x} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{45 c^{2} x^{\frac {9}{2}} - 45 x^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x**4,x)

[Out]

Piecewise((-a/(3*x**3) + b*atanh(sqrt(x)*sqrt(1/x))/(3*x**3), Eq(c, -sqrt(1/x))), (-a/(3*x**3) - b*atanh(sqrt(
x)*sqrt(1/x))/(3*x**3), Eq(c, sqrt(1/x))), (-15*a*c**2*x**(3/2)/(45*c**2*x**(9/2) - 45*x**(7/2)) + 15*a*sqrt(x
)/(45*c**2*x**(9/2) - 45*x**(7/2)) + 15*b*c**8*x**(9/2)*atanh(c*sqrt(x))/(45*c**2*x**(9/2) - 45*x**(7/2)) - 15
*b*c**7*x**4/(45*c**2*x**(9/2) - 45*x**(7/2)) - 15*b*c**6*x**(7/2)*atanh(c*sqrt(x))/(45*c**2*x**(9/2) - 45*x**
(7/2)) + 10*b*c**5*x**3/(45*c**2*x**(9/2) - 45*x**(7/2)) + 2*b*c**3*x**2/(45*c**2*x**(9/2) - 45*x**(7/2)) - 15
*b*c**2*x**(3/2)*atanh(c*sqrt(x))/(45*c**2*x**(9/2) - 45*x**(7/2)) + 3*b*c*x/(45*c**2*x**(9/2) - 45*x**(7/2))
+ 15*b*sqrt(x)*atanh(c*sqrt(x))/(45*c**2*x**(9/2) - 45*x**(7/2)), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (53) = 106\).
time = 0.44, size = 534, normalized size = 7.32 \begin {gather*} \frac {2}{45} \, c {\left (\frac {15 \, {\left (\frac {3 \, {\left (c \sqrt {x} + 1\right )}^{5} b c^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {10 \, {\left (c \sqrt {x} + 1\right )}^{3} b c^{5}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {3 \, {\left (c \sqrt {x} + 1\right )} b c^{5}}{c \sqrt {x} - 1}\right )} \log \left (-\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )}{\frac {{\left (c \sqrt {x} + 1\right )}^{6}}{{\left (c \sqrt {x} - 1\right )}^{6}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {15 \, {\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {20 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {15 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1} + \frac {\frac {90 \, {\left (c \sqrt {x} + 1\right )}^{5} a c^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {300 \, {\left (c \sqrt {x} + 1\right )}^{3} a c^{5}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {90 \, {\left (c \sqrt {x} + 1\right )} a c^{5}}{c \sqrt {x} - 1} + \frac {45 \, {\left (c \sqrt {x} + 1\right )}^{5} b c^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {135 \, {\left (c \sqrt {x} + 1\right )}^{4} b c^{5}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {230 \, {\left (c \sqrt {x} + 1\right )}^{3} b c^{5}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {210 \, {\left (c \sqrt {x} + 1\right )}^{2} b c^{5}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {93 \, {\left (c \sqrt {x} + 1\right )} b c^{5}}{c \sqrt {x} - 1} + 23 \, b c^{5}}{\frac {{\left (c \sqrt {x} + 1\right )}^{6}}{{\left (c \sqrt {x} - 1\right )}^{6}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{5}}{{\left (c \sqrt {x} - 1\right )}^{5}} + \frac {15 \, {\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {20 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {15 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^4,x, algorithm="giac")

[Out]

2/45*c*(15*(3*(c*sqrt(x) + 1)^5*b*c^5/(c*sqrt(x) - 1)^5 + 10*(c*sqrt(x) + 1)^3*b*c^5/(c*sqrt(x) - 1)^3 + 3*(c*
sqrt(x) + 1)*b*c^5/(c*sqrt(x) - 1))*log(-(c*sqrt(x) + 1)/(c*sqrt(x) - 1))/((c*sqrt(x) + 1)^6/(c*sqrt(x) - 1)^6
+ 6*(c*sqrt(x) + 1)^5/(c*sqrt(x) - 1)^5 + 15*(c*sqrt(x) + 1)^4/(c*sqrt(x) - 1)^4 + 20*(c*sqrt(x) + 1)^3/(c*sq
rt(x) - 1)^3 + 15*(c*sqrt(x) + 1)^2/(c*sqrt(x) - 1)^2 + 6*(c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 1) + (90*(c*sqrt(x
) + 1)^5*a*c^5/(c*sqrt(x) - 1)^5 + 300*(c*sqrt(x) + 1)^3*a*c^5/(c*sqrt(x) - 1)^3 + 90*(c*sqrt(x) + 1)*a*c^5/(c
*sqrt(x) - 1) + 45*(c*sqrt(x) + 1)^5*b*c^5/(c*sqrt(x) - 1)^5 + 135*(c*sqrt(x) + 1)^4*b*c^5/(c*sqrt(x) - 1)^4 +
230*(c*sqrt(x) + 1)^3*b*c^5/(c*sqrt(x) - 1)^3 + 210*(c*sqrt(x) + 1)^2*b*c^5/(c*sqrt(x) - 1)^2 + 93*(c*sqrt(x)
+ 1)*b*c^5/(c*sqrt(x) - 1) + 23*b*c^5)/((c*sqrt(x) + 1)^6/(c*sqrt(x) - 1)^6 + 6*(c*sqrt(x) + 1)^5/(c*sqrt(x)
- 1)^5 + 15*(c*sqrt(x) + 1)^4/(c*sqrt(x) - 1)^4 + 20*(c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 + 15*(c*sqrt(x) + 1)^
2/(c*sqrt(x) - 1)^2 + 6*(c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 1))

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Mupad [B]
time = 1.39, size = 69, normalized size = 0.95 \begin {gather*} \frac {b\,c^6\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{3}-\frac {b\,\left (15\,\ln \left (c\,\sqrt {x}+1\right )-15\,\ln \left (1-c\,\sqrt {x}\right )+6\,c\,\sqrt {x}+10\,c^3\,x^{3/2}+30\,c^5\,x^{5/2}\right )}{90\,x^3}-\frac {a}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))/x^4,x)

[Out]

(b*c^6*atanh(c*x^(1/2)))/3 - (b*(15*log(c*x^(1/2) + 1) - 15*log(1 - c*x^(1/2)) + 6*c*x^(1/2) + 10*c^3*x^(3/2)
+ 30*c^5*x^(5/2)))/(90*x^3) - a/(3*x^3)

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