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

Optimal. Leaf size=85 \[ \frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {2 b \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}} \]

[Out]

2*b*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(1/2)/d^(1/2)-2*b*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(1/2)/d^(1/
2)+2*(a+b*arctanh(c*x))*(d*x)^(1/2)/d

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5916, 329, 298, 205, 208} \[ \frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {2 b \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/Sqrt[d*x],x]

[Out]

(2*b*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(Sqrt[c]*Sqrt[d]) + (2*Sqrt[d*x]*(a + b*ArcTanh[c*x]))/d - (2*b*ArcT
anh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(Sqrt[c]*Sqrt[d])

Rule 205

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

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, 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]

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {d x}} \, dx &=\frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {(2 b c) \int \frac {\sqrt {d x}}{1-c^2 x^2} \, dx}{d}\\ &=\frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {(4 b c) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {c^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{d^2}\\ &=\frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}(c x)\right )}{d}-(2 b) \operatorname {Subst}\left (\int \frac {1}{d-c x^2} \, dx,x,\sqrt {d x}\right )+(2 b) \operatorname {Subst}\left (\int \frac {1}{d+c x^2} \, dx,x,\sqrt {d x}\right )\\ &=\frac {2 b \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}}+\frac {2 \sqrt {d x} \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 98, normalized size = 1.15 \[ \frac {\sqrt {x} \left (2 a \sqrt {c} \sqrt {x}+b \log \left (1-\sqrt {c} \sqrt {x}\right )-b \log \left (\sqrt {c} \sqrt {x}+1\right )+2 b \tan ^{-1}\left (\sqrt {c} \sqrt {x}\right )+2 b \sqrt {c} \sqrt {x} \tanh ^{-1}(c x)\right )}{\sqrt {c} \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(2*a*Sqrt[c]*Sqrt[x] + 2*b*ArcTan[Sqrt[c]*Sqrt[x]] + 2*b*Sqrt[c]*Sqrt[x]*ArcTanh[c*x] + b*Log[1 - Sqr
t[c]*Sqrt[x]] - b*Log[1 + Sqrt[c]*Sqrt[x]]))/(Sqrt[c]*Sqrt[d*x])

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fricas [A]  time = 0.55, size = 211, normalized size = 2.48 \[ \left [-\frac {2 \, \sqrt {c d} b \arctan \left (\frac {\sqrt {c d} \sqrt {d x}}{c d x}\right ) - \sqrt {c d} b \log \left (\frac {c d x - 2 \, \sqrt {c d} \sqrt {d x} + d}{c x - 1}\right ) - {\left (b c \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c\right )} \sqrt {d x}}{c d}, \frac {2 \, \sqrt {-c d} b \arctan \left (\frac {\sqrt {-c d} \sqrt {d x}}{c d x}\right ) - \sqrt {-c d} b \log \left (\frac {c d x - 2 \, \sqrt {-c d} \sqrt {d x} - d}{c x + 1}\right ) + {\left (b c \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c\right )} \sqrt {d x}}{c d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*sqrt(c*d)*b*arctan(sqrt(c*d)*sqrt(d*x)/(c*d*x)) - sqrt(c*d)*b*log((c*d*x - 2*sqrt(c*d)*sqrt(d*x) + d)/(c*
x - 1)) - (b*c*log(-(c*x + 1)/(c*x - 1)) + 2*a*c)*sqrt(d*x))/(c*d), (2*sqrt(-c*d)*b*arctan(sqrt(-c*d)*sqrt(d*x
)/(c*d*x)) - sqrt(-c*d)*b*log((c*d*x - 2*sqrt(-c*d)*sqrt(d*x) - d)/(c*x + 1)) + (b*c*log(-(c*x + 1)/(c*x - 1))
+ 2*a*c)*sqrt(d*x))/(c*d)]

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giac [A]  time = 0.15, size = 88, normalized size = 1.04 \[ \frac {{\left (2 \, c d {\left (\frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c} + \frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} c}\right )} + \sqrt {d x} \log \left (-\frac {c x + 1}{c x - 1}\right )\right )} b + 2 \, \sqrt {d x} a}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*c*d*(arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c) + arctan(sqrt(d*x)*c/sqrt(-c*d))/(sqrt(-c*d)*c)) + sqrt(d
*x)*log(-(c*x + 1)/(c*x - 1)))*b + 2*sqrt(d*x)*a)/d

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maple [A]  time = 0.04, size = 70, normalized size = 0.82 \[ \frac {2 a \sqrt {d x}}{d}+\frac {2 b \sqrt {d x}\, \arctanh \left (c x \right )}{d}+\frac {2 b \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}}-\frac {2 b \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{\sqrt {c d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a*(d*x)^(1/2)+2/d*b*(d*x)^(1/2)*arctanh(c*x)+2*b/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2))-2*b/(c*d)^(
1/2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))

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maxima [A]  time = 0.43, size = 103, normalized size = 1.21 \[ \frac {{\left (2 \, \sqrt {d x} \operatorname {artanh}\left (c x\right ) + \frac {{\left (\frac {2 \, d^{2} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c} + \frac {d^{2} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c}\right )} c}{d}\right )} b + 2 \, \sqrt {d x} a}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*sqrt(d*x)*arctanh(c*x) + (2*d^2*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c) + d^2*log((sqrt(d*x)*c - sqrt(
c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c))*c/d)*b + 2*sqrt(d*x)*a)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d*x)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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