3.1.39 \(\int \frac {a+b \tanh ^{-1}(c x)}{(d x)^{3/2}} \, dx\) [39]

Optimal. Leaf size=85 \[ \frac {2 b \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{d \sqrt {d x}}+\frac {2 b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}} \]

[Out]

2*b*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))*c^(1/2)/d^(3/2)+2*b*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))*c^(1/2)/d^(3/
2)-2*(a+b*arctanh(c*x))/d/(d*x)^(1/2)

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Rubi [A]
time = 0.04, 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 = {6049, 335, 218, 214, 211} \begin {gather*} -\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{d \sqrt {d x}}+\frac {2 b \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {2 b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d*x)^(3/2),x]

[Out]

(2*b*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/d^(3/2) - (2*(a + b*ArcTanh[c*x]))/(d*Sqrt[d*x]) + (2*b*Sqrt
[c]*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/d^(3/2)

Rule 211

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

Rule 214

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

Rule 218

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

Rule 335

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 6049

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 99, normalized size = 1.16 \begin {gather*} \frac {x \left (-2 a+2 b \sqrt {c} \sqrt {x} \text {ArcTan}\left (\sqrt {c} \sqrt {x}\right )-2 b \tanh ^{-1}(c x)-b \sqrt {c} \sqrt {x} \log \left (1-\sqrt {c} \sqrt {x}\right )+b \sqrt {c} \sqrt {x} \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{(d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(d*x)^(3/2),x]

[Out]

(x*(-2*a + 2*b*Sqrt[c]*Sqrt[x]*ArcTan[Sqrt[c]*Sqrt[x]] - 2*b*ArcTanh[c*x] - b*Sqrt[c]*Sqrt[x]*Log[1 - Sqrt[c]*
Sqrt[x]] + b*Sqrt[c]*Sqrt[x]*Log[1 + Sqrt[c]*Sqrt[x]]))/(d*x)^(3/2)

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Maple [A]
time = 0.04, size = 69, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{\sqrt {d x}}-\frac {2 b \arctanh \left (c x \right )}{\sqrt {d x}}+\frac {2 b c \arctan \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{\sqrt {d c}}+\frac {2 b c \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{\sqrt {d c}}}{d}\)	\(69\)
default	\(\frac {-\frac {2 a}{\sqrt {d x}}-\frac {2 b \arctanh \left (c x \right )}{\sqrt {d x}}+\frac {2 b c \arctan \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{\sqrt {d c}}+\frac {2 b c \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{\sqrt {d c}}}{d}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(d*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.46, size = 94, normalized size = 1.11 \begin {gather*} \frac {b {\left (\frac {{\left (\frac {2 \, d \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d}} - \frac {d \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d}}\right )} c}{d} - \frac {2 \, \operatorname {artanh}\left (c x\right )}{\sqrt {d x}}\right )} - \frac {2 \, a}{\sqrt {d x}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 221, normalized size = 2.60 \begin {gather*} \left [-\frac {2 \, b d x \sqrt {\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {\frac {c}{d}}}{c x}\right ) - b d x \sqrt {\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {\frac {c}{d}} + 1}{c x - 1}\right ) + \sqrt {d x} {\left (b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a\right )}}{d^{2} x}, -\frac {2 \, b d x \sqrt {-\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {c}{d}}}{c x}\right ) - b d x \sqrt {-\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {-\frac {c}{d}} - 1}{c x + 1}\right ) + \sqrt {d x} {\left (b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a\right )}}{d^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\left (d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(d*x)**(3/2),x)

[Out]

Integral((a + b*atanh(c*x))/(d*x)**(3/2), x)

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Giac [A]
time = 0.44, size = 93, normalized size = 1.09 \begin {gather*} \frac {2 \, b c d {\left (\frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d} - \frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} d}\right )} - \frac {b \log \left (-\frac {c d x + d}{c d x - d}\right )}{\sqrt {d x}} - \frac {2 \, a}{\sqrt {d x}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{{\left (d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d*x)^(3/2),x)

[Out]

int((a + b*atanh(c*x))/(d*x)^(3/2), x)

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