3.46 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{d+e x} \, dx\)

Optimal. Leaf size=318 \[ \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{e}-\frac {2 \log \left (\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (\sqrt {x} c+1\right )}\right )}{2 e}-\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (\sqrt {-d} c+\sqrt {e}\right ) \left (\sqrt {x} c+1\right )}\right )}{2 e}+\frac {b \text {Li}_2\left (1-\frac {2}{\sqrt {x} c+1}\right )}{e} \]

[Out]

-2*(a+b*arctanh(c*x^(1/2)))*ln(2/(1+c*x^(1/2)))/e+(a+b*arctanh(c*x^(1/2)))*ln(2*c*((-d)^(1/2)-e^(1/2)*x^(1/2))
/(c*(-d)^(1/2)-e^(1/2))/(1+c*x^(1/2)))/e+(a+b*arctanh(c*x^(1/2)))*ln(2*c*((-d)^(1/2)+e^(1/2)*x^(1/2))/(c*(-d)^
(1/2)+e^(1/2))/(1+c*x^(1/2)))/e+b*polylog(2,1-2/(1+c*x^(1/2)))/e-1/2*b*polylog(2,1-2*c*((-d)^(1/2)-e^(1/2)*x^(
1/2))/(c*(-d)^(1/2)-e^(1/2))/(1+c*x^(1/2)))/e-1/2*b*polylog(2,1-2*c*((-d)^(1/2)+e^(1/2)*x^(1/2))/(c*(-d)^(1/2)
+e^(1/2))/(1+c*x^(1/2)))/e

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Rubi [A]  time = 0.32, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6044, 5920, 2402, 2315, 2447} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{2 e}-\frac {b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{2 e}+\frac {b \text {PolyLog}\left (2,1-\frac {2}{c \sqrt {x}+1}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{e}-\frac {2 \log \left (\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/(d + e*x),x]

[Out]

(-2*(a + b*ArcTanh[c*Sqrt[x]])*Log[2/(1 + c*Sqrt[x])])/e + ((a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] - Sq
rt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/e + ((a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d]
+ Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/e + (b*PolyLog[2, 1 - 2/(1 + c*Sqrt[x])])/e - (
b*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e) - (b*Poly
Log[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e)

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5920

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

Rule 6044

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && (GtQ[q, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{d+e x} \, dx &=2 \operatorname {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{d+e x^2} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (-\frac {a+b \tanh ^{-1}(c x)}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {a+b \tanh ^{-1}(c x)}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx,x,\sqrt {x}\right )}{\sqrt {e}}+\frac {\operatorname {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx,x,\sqrt {x}\right )}{\sqrt {e}}\\ &=-\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e}+2 \frac {(b c) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e}-\frac {(b c) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) (1+c x)}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e}-\frac {(b c) \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) (1+c x)}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{e}\\ &=-\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e}-\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e}+2 \frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c \sqrt {x}}\right )}{e}\\ &=-\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e}+\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{e}+\frac {b \text {Li}_2\left (1-\frac {2}{1+c \sqrt {x}}\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e}-\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 e}\\ \end {align*}

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Mathematica [C]  time = 1.62, size = 432, normalized size = 1.36 \[ \frac {a \log (d+e x)}{e}-\frac {b \left (\text {Li}_2\left (\frac {\left (-d c^2+e-2 \sqrt {-c^2 d e}\right ) e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{d c^2+e}\right )+\text {Li}_2\left (\frac {\left (-d c^2+e+2 \sqrt {-c^2 d e}\right ) e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{d c^2+e}\right )+4 i \sin ^{-1}\left (\sqrt {\frac {c^2 d}{c^2 d+e}}\right ) \tanh ^{-1}\left (\frac {c e \sqrt {x}}{\sqrt {-c^2 d e}}\right )-2 \log \left (\frac {e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )} \left (-2 \sqrt {-c^2 d e}+c^2 d \left (e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}+1\right )+e \left (e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}-1\right )\right )}{c^2 d+e}\right ) \left (\tanh ^{-1}\left (c \sqrt {x}\right )-i \sin ^{-1}\left (\sqrt {\frac {c^2 d}{c^2 d+e}}\right )\right )-2 \log \left (\frac {e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )} \left (2 \sqrt {-c^2 d e}+c^2 d \left (e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}+1\right )+e \left (e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}-1\right )\right )}{c^2 d+e}\right ) \left (\tanh ^{-1}\left (c \sqrt {x}\right )+i \sin ^{-1}\left (\sqrt {\frac {c^2 d}{c^2 d+e}}\right )\right )-2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )-2 \tanh ^{-1}\left (c \sqrt {x}\right )^2+2 \tanh ^{-1}\left (c \sqrt {x}\right ) \left (\tanh ^{-1}\left (c \sqrt {x}\right )+2 \log \left (e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}+1\right )\right )\right )}{2 e} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/(d + e*x),x]

[Out]

(a*Log[d + e*x])/e - (b*(-2*ArcTanh[c*Sqrt[x]]^2 + (4*I)*ArcSin[Sqrt[(c^2*d)/(c^2*d + e)]]*ArcTanh[(c*e*Sqrt[x
])/Sqrt[-(c^2*d*e)]] + 2*ArcTanh[c*Sqrt[x]]*(ArcTanh[c*Sqrt[x]] + 2*Log[1 + E^(-2*ArcTanh[c*Sqrt[x]])]) - 2*((
-I)*ArcSin[Sqrt[(c^2*d)/(c^2*d + e)]] + ArcTanh[c*Sqrt[x]])*Log[(-2*Sqrt[-(c^2*d*e)] + e*(-1 + E^(2*ArcTanh[c*
Sqrt[x]])) + c^2*d*(1 + E^(2*ArcTanh[c*Sqrt[x]])))/((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))] - 2*(I*ArcSin[Sqrt[
(c^2*d)/(c^2*d + e)]] + ArcTanh[c*Sqrt[x]])*Log[(2*Sqrt[-(c^2*d*e)] + e*(-1 + E^(2*ArcTanh[c*Sqrt[x]])) + c^2*
d*(1 + E^(2*ArcTanh[c*Sqrt[x]])))/((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))] - 2*PolyLog[2, -E^(-2*ArcTanh[c*Sqrt
[x]])] + PolyLog[2, (-(c^2*d) + e - 2*Sqrt[-(c^2*d*e)])/((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))] + PolyLog[2, (
-(c^2*d) + e + 2*Sqrt[-(c^2*d*e)])/((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))]))/(2*e)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctanh(c*sqrt(x)) + a)/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)/(e*x + d), x)

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maple [A]  time = 0.06, size = 462, normalized size = 1.45 \[ \frac {a \ln \left (c^{2} e x +c^{2} d \right )}{e}+\frac {b \ln \left (c^{2} e x +c^{2} d \right ) \arctanh \left (c \sqrt {x}\right )}{e}+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 e}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 e}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 e}-\frac {b \dilog \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 e}-\frac {b \dilog \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 e}-\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 e}+\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 e}+\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 e}+\frac {b \dilog \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 e}+\frac {b \dilog \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/(e*x+d),x)

[Out]

a/e*ln(c^2*e*x+c^2*d)+b/e*ln(c^2*e*x+c^2*d)*arctanh(c*x^(1/2))+1/2*b/e*ln(c*x^(1/2)-1)*ln(c^2*e*x+c^2*d)-1/2*b
*ln(c*x^(1/2)-1)/e*ln((c*(-d*e)^(1/2)-e*(c*x^(1/2)-1)-e)/(c*(-d*e)^(1/2)-e))-1/2*b*ln(c*x^(1/2)-1)/e*ln((c*(-d
*e)^(1/2)+e*(c*x^(1/2)-1)+e)/(c*(-d*e)^(1/2)+e))-1/2*b/e*dilog((c*(-d*e)^(1/2)-e*(c*x^(1/2)-1)-e)/(c*(-d*e)^(1
/2)-e))-1/2*b/e*dilog((c*(-d*e)^(1/2)+e*(c*x^(1/2)-1)+e)/(c*(-d*e)^(1/2)+e))-1/2*b/e*ln(1+c*x^(1/2))*ln(c^2*e*
x+c^2*d)+1/2*b*ln(1+c*x^(1/2))/e*ln((c*(-d*e)^(1/2)-e*(1+c*x^(1/2))+e)/(c*(-d*e)^(1/2)+e))+1/2*b*ln(1+c*x^(1/2
))/e*ln((c*(-d*e)^(1/2)+e*(1+c*x^(1/2))-e)/(c*(-d*e)^(1/2)-e))+1/2*b/e*dilog((c*(-d*e)^(1/2)-e*(1+c*x^(1/2))+e
)/(c*(-d*e)^(1/2)+e))+1/2*b/e*dilog((c*(-d*e)^(1/2)+e*(1+c*x^(1/2))-e)/(c*(-d*e)^(1/2)-e))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\log \left (c \sqrt {x} + 1\right )}{2 \, {\left (e x + d\right )}}\,{d x} - b \int \frac {\log \left (-c \sqrt {x} + 1\right )}{2 \, {\left (e x + d\right )}}\,{d x} + \frac {a \log \left (e x + d\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(1/2*log(c*sqrt(x) + 1)/(e*x + d), x) - b*integrate(1/2*log(-c*sqrt(x) + 1)/(e*x + d), x) + a*log(e
*x + d)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))/(d + e*x),x)

[Out]

int((a + b*atanh(c*x^(1/2)))/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*sqrt(x)))/(d + e*x), x)

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