3.43 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{x^4 (1-c^2 x)} \, dx\)

Optimal. Leaf size=192 \[ \frac {c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^6 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {c^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{x}-\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{2 x^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-b c^6 \text {Li}_2\left (\frac {2}{\sqrt {x} c+1}-1\right )+\frac {11}{6} b c^6 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {11 b c^5}{6 \sqrt {x}}-\frac {5 b c^3}{18 x^{3/2}}-\frac {b c}{15 x^{5/2}} \]

[Out]

-1/15*b*c/x^(5/2)-5/18*b*c^3/x^(3/2)+11/6*b*c^6*arctanh(c*x^(1/2))+1/3*(-a-b*arctanh(c*x^(1/2)))/x^3-1/2*c^2*(
a+b*arctanh(c*x^(1/2)))/x^2-c^4*(a+b*arctanh(c*x^(1/2)))/x+c^6*(a+b*arctanh(c*x^(1/2)))^2/b+2*c^6*(a+b*arctanh
(c*x^(1/2)))*ln(2-2/(1+c*x^(1/2)))-b*c^6*polylog(2,-1+2/(1+c*x^(1/2)))-11/6*b*c^5/x^(1/2)

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Rubi [A]  time = 0.57, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {44, 1593, 5982, 5916, 325, 206, 5988, 5932, 2447} \[ -b c^6 \text {PolyLog}\left (2,\frac {2}{c \sqrt {x}+1}-1\right )-\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{2 x^2}+\frac {c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}-\frac {c^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{x}+2 c^6 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-\frac {5 b c^3}{18 x^{3/2}}-\frac {11 b c^5}{6 \sqrt {x}}+\frac {11}{6} b c^6 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/(x^4*(1 - c^2*x)),x]

[Out]

-(b*c)/(15*x^(5/2)) - (5*b*c^3)/(18*x^(3/2)) - (11*b*c^5)/(6*Sqrt[x]) + (11*b*c^6*ArcTanh[c*Sqrt[x]])/6 - (a +
b*ArcTanh[c*Sqrt[x]])/(3*x^3) - (c^2*(a + b*ArcTanh[c*Sqrt[x]]))/(2*x^2) - (c^4*(a + b*ArcTanh[c*Sqrt[x]]))/x
+ (c^6*(a + b*ArcTanh[c*Sqrt[x]])^2)/b + 2*c^6*(a + b*ArcTanh[c*Sqrt[x]])*Log[2 - 2/(1 + c*Sqrt[x])] - b*c^6*
PolyLog[2, -1 + 2/(1 + c*Sqrt[x])]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 325

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5916

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

Rule 5932

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

Rule 5982

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

Rule 5988

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

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x^4 \left (1-c^2 x\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^7-c^2 x^9} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^7 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^7} \, dx,x,\sqrt {x}\right )+\left (2 c^2\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^5 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}+\frac {1}{3} (b c) \operatorname {Subst}\left (\int \frac {1}{x^6 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\left (2 c^2\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^5} \, dx,x,\sqrt {x}\right )+\left (2 c^4\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{15 x^{5/2}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{2 x^2}+\frac {1}{3} \left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\frac {1}{2} \left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\left (2 c^4\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx,x,\sqrt {x}\right )+\left (2 c^6\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{15 x^{5/2}}-\frac {5 b c^3}{18 x^{3/2}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{2 x^2}-\frac {c^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{x}+\frac {c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+\frac {1}{3} \left (b c^5\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\frac {1}{2} \left (b c^5\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\left (b c^5\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )+\left (2 c^6\right ) \operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{15 x^{5/2}}-\frac {5 b c^3}{18 x^{3/2}}-\frac {11 b c^5}{6 \sqrt {x}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{2 x^2}-\frac {c^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{x}+\frac {c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )+\frac {1}{3} \left (b c^7\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )+\frac {1}{2} \left (b c^7\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )+\left (b c^7\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-\left (2 b c^7\right ) \operatorname {Subst}\left (\int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {b c}{15 x^{5/2}}-\frac {5 b c^3}{18 x^{3/2}}-\frac {11 b c^5}{6 \sqrt {x}}+\frac {11}{6} b c^6 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{3 x^3}-\frac {c^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{2 x^2}-\frac {c^4 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{x}+\frac {c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 c^6 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )-b c^6 \text {Li}_2\left (-1+\frac {2}{1+c \sqrt {x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.81, size = 187, normalized size = 0.97 \[ -\frac {-90 a c^6 x^3 \log (x)+90 a c^4 x^2+45 a c^2 x+90 a c^6 x^3 \log \left (1-c^2 x\right )+30 a-90 b c^6 x^3 \tanh ^{-1}\left (c \sqrt {x}\right )^2+165 b c^5 x^{5/2}+25 b c^3 x^{3/2}-15 b \tanh ^{-1}\left (c \sqrt {x}\right ) \left (11 c^6 x^3+12 c^6 x^3 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )-6 c^4 x^2-3 c^2 x-2\right )+6 b c \sqrt {x}}{90 x^3}-b c^6 \text {Li}_2\left (e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/(x^4*(1 - c^2*x)),x]

[Out]

-1/90*(30*a + 6*b*c*Sqrt[x] + 45*a*c^2*x + 25*b*c^3*x^(3/2) + 90*a*c^4*x^2 + 165*b*c^5*x^(5/2) - 90*b*c^6*x^3*
ArcTanh[c*Sqrt[x]]^2 - 15*b*ArcTanh[c*Sqrt[x]]*(-2 - 3*c^2*x - 6*c^4*x^2 + 11*c^6*x^3 + 12*c^6*x^3*Log[1 - E^(
-2*ArcTanh[c*Sqrt[x]])]) - 90*a*c^6*x^3*Log[x] + 90*a*c^6*x^3*Log[1 - c^2*x])/x^3 - b*c^6*PolyLog[2, E^(-2*Arc
Tanh[c*Sqrt[x]])]

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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{c^{2} x^{5} - x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arctanh(c*sqrt(x)) + a)/(c^2*x^5 - x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{{\left (c^{2} x - 1\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(b*arctanh(c*sqrt(x)) + a)/((c^2*x - 1)*x^4), x)

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maple [B]  time = 0.07, size = 381, normalized size = 1.98 \[ -\frac {c^{2} b \arctanh \left (c \sqrt {x}\right )}{2 x^{2}}-c^{6} b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )+\frac {c^{6} b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{2}-\frac {c^{6} b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{2}-\frac {c^{4} b \arctanh \left (c \sqrt {x}\right )}{x}-c^{6} b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )+2 c^{6} b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )-c^{6} b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )+\frac {c^{6} b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )}{2}-\frac {11 b \,c^{5}}{6 \sqrt {x}}-\frac {5 b \,c^{3}}{18 x^{\frac {3}{2}}}-\frac {b c}{15 x^{\frac {5}{2}}}-\frac {c^{4} a}{x}-\frac {c^{2} a}{2 x^{2}}-c^{6} a \ln \left (c \sqrt {x}-1\right )-c^{6} a \ln \left (1+c \sqrt {x}\right )-\frac {c^{6} b \ln \left (c \sqrt {x}-1\right )^{2}}{4}+\frac {c^{6} b \ln \left (1+c \sqrt {x}\right )^{2}}{4}-\frac {11 c^{6} b \ln \left (c \sqrt {x}-1\right )}{12}+2 c^{6} a \ln \left (c \sqrt {x}\right )+\frac {11 c^{6} b \ln \left (1+c \sqrt {x}\right )}{12}-c^{6} b \dilog \left (c \sqrt {x}\right )+c^{6} b \dilog \left (\frac {1}{2}+\frac {c \sqrt {x}}{2}\right )-\frac {b \arctanh \left (c \sqrt {x}\right )}{3 x^{3}}-c^{6} b \dilog \left (1+c \sqrt {x}\right )-\frac {a}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^4*b*arctanh(c*x^(1/2))/x-c^6*b*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))-c^6*b*ln(c*x^(1/2))*ln(1+c*x^(1/2))+2*c^6
*b*arctanh(c*x^(1/2))*ln(c*x^(1/2))+1/2*c^6*b*ln(c*x^(1/2)-1)*ln(1/2+1/2*c*x^(1/2))-1/2*c^6*b*ln(-1/2*c*x^(1/2
)+1/2)*ln(1+c*x^(1/2))+1/2*c^6*b*ln(-1/2*c*x^(1/2)+1/2)*ln(1/2+1/2*c*x^(1/2))-c^6*b*arctanh(c*x^(1/2))*ln(c*x^
(1/2)-1)-1/2*c^2*b*arctanh(c*x^(1/2))/x^2-11/6*b*c^5/x^(1/2)-5/18*b*c^3/x^(3/2)-1/15*b*c/x^(5/2)-c^6*b*dilog(c
*x^(1/2))-c^4*a/x-1/2*c^2*a/x^2-c^6*a*ln(c*x^(1/2)-1)-c^6*a*ln(1+c*x^(1/2))-1/4*c^6*b*ln(c*x^(1/2)-1)^2+c^6*b*
dilog(1/2+1/2*c*x^(1/2))+1/4*c^6*b*ln(1+c*x^(1/2))^2-1/3*b*arctanh(c*x^(1/2))/x^3-11/12*c^6*b*ln(c*x^(1/2)-1)-
c^6*b*dilog(1+c*x^(1/2))+2*c^6*a*ln(c*x^(1/2))+11/12*c^6*b*ln(1+c*x^(1/2))-1/3*a/x^3

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maxima [B]  time = 0.56, size = 330, normalized size = 1.72 \[ -{\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b c^{6} - {\left (\log \left (c \sqrt {x}\right ) \log \left (-c \sqrt {x} + 1\right ) + {\rm Li}_2\left (-c \sqrt {x} + 1\right )\right )} b c^{6} + {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x}\right ) + {\rm Li}_2\left (c \sqrt {x} + 1\right )\right )} b c^{6} + \frac {11}{12} \, b c^{6} \log \left (c \sqrt {x} + 1\right ) - \frac {11}{12} \, b c^{6} \log \left (c \sqrt {x} - 1\right ) - \frac {1}{6} \, {\left (6 \, c^{6} \log \left (c \sqrt {x} + 1\right ) + 6 \, c^{6} \log \left (c \sqrt {x} - 1\right ) - 6 \, c^{6} \log \relax (x) + \frac {6 \, c^{4} x^{2} + 3 \, c^{2} x + 2}{x^{3}}\right )} a - \frac {45 \, b c^{6} x^{3} \log \left (c \sqrt {x} + 1\right )^{2} - 45 \, b c^{6} x^{3} \log \left (-c \sqrt {x} + 1\right )^{2} + 330 \, b c^{5} x^{\frac {5}{2}} + 50 \, b c^{3} x^{\frac {3}{2}} + 12 \, b c \sqrt {x} + 15 \, {\left (6 \, b c^{4} x^{2} + 3 \, b c^{2} x + 2 \, b\right )} \log \left (c \sqrt {x} + 1\right ) - 15 \, {\left (6 \, b c^{6} x^{3} \log \left (c \sqrt {x} + 1\right ) + 6 \, b c^{4} x^{2} + 3 \, b c^{2} x + 2 \, b\right )} \log \left (-c \sqrt {x} + 1\right )}{180 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1/2))*b*c^6 - (log(c*sqrt(x))*log(-c*sq
rt(x) + 1) + dilog(-c*sqrt(x) + 1))*b*c^6 + (log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(c*sqrt(x) + 1))*b*c^6
+ 11/12*b*c^6*log(c*sqrt(x) + 1) - 11/12*b*c^6*log(c*sqrt(x) - 1) - 1/6*(6*c^6*log(c*sqrt(x) + 1) + 6*c^6*log(
c*sqrt(x) - 1) - 6*c^6*log(x) + (6*c^4*x^2 + 3*c^2*x + 2)/x^3)*a - 1/180*(45*b*c^6*x^3*log(c*sqrt(x) + 1)^2 -
45*b*c^6*x^3*log(-c*sqrt(x) + 1)^2 + 330*b*c^5*x^(5/2) + 50*b*c^3*x^(3/2) + 12*b*c*sqrt(x) + 15*(6*b*c^4*x^2 +
3*b*c^2*x + 2*b)*log(c*sqrt(x) + 1) - 15*(6*b*c^6*x^3*log(c*sqrt(x) + 1) + 6*b*c^4*x^2 + 3*b*c^2*x + 2*b)*log
(-c*sqrt(x) + 1))/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{x^4\,\left (c^2\,x-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*atanh(c*x^(1/2)))/(x^4*(c^2*x - 1)),x)

[Out]

-int((a + b*atanh(c*x^(1/2)))/(x^4*(c^2*x - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a}{c^{2} x^{5} - x^{4}}\, dx - \int \frac {b \operatorname {atanh}{\left (c \sqrt {x} \right )}}{c^{2} x^{5} - x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a/(c**2*x**5 - x**4), x) - Integral(b*atanh(c*sqrt(x))/(c**2*x**5 - x**4), x)

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