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

Optimal. Leaf size=106 \[ \frac {2 (d x)^{3/2} \left (a+b \tanh ^{-1}(c x)\right )}{3 d}-\frac {2 b \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 c^{3/2}}-\frac {2 b \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 c^{3/2}}+\frac {4 b \sqrt {d x}}{3 c} \]

[Out]

2/3*(d*x)^(3/2)*(a+b*arctanh(c*x))/d-2/3*b*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))*d^(1/2)/c^(3/2)-2/3*b*arctanh(c
^(1/2)*(d*x)^(1/2)/d^(1/2))*d^(1/2)/c^(3/2)+4/3*b*(d*x)^(1/2)/c

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Rubi [A]  time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5916, 321, 329, 212, 208, 205} \[ \frac {2 (d x)^{3/2} \left (a+b \tanh ^{-1}(c x)\right )}{3 d}-\frac {2 b \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 c^{3/2}}-\frac {2 b \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{3 c^{3/2}}+\frac {4 b \sqrt {d x}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]*(a + b*ArcTanh[c*x]),x]

[Out]

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

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 212

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] &&
!GtQ[a/b, 0]

Rule 321

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

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

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Mathematica [A]  time = 0.05, size = 114, normalized size = 1.08 \[ \frac {\sqrt {d x} \left (2 a c^{3/2} x^{3/2}+2 b c^{3/2} x^{3/2} \tanh ^{-1}(c x)+4 b \sqrt {c} \sqrt {x}+b \log \left (1-\sqrt {c} \sqrt {x}\right )-b \log \left (\sqrt {c} \sqrt {x}+1\right )-2 b \tan ^{-1}\left (\sqrt {c} \sqrt {x}\right )\right )}{3 c^{3/2} \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*x]*(4*b*Sqrt[c]*Sqrt[x] + 2*a*c^(3/2)*x^(3/2) - 2*b*ArcTan[Sqrt[c]*Sqrt[x]] + 2*b*c^(3/2)*x^(3/2)*ArcT
anh[c*x] + b*Log[1 - Sqrt[c]*Sqrt[x]] - b*Log[1 + Sqrt[c]*Sqrt[x]]))/(3*c^(3/2)*Sqrt[x])

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fricas [A]  time = 0.75, size = 223, normalized size = 2.10 \[ \left [-\frac {2 \, b \sqrt {\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {\frac {d}{c}}}{d}\right ) - b \sqrt {\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {\frac {d}{c}} + d}{c x - 1}\right ) - {\left (b c x \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c x + 4 \, b\right )} \sqrt {d x}}{3 \, c}, \frac {2 \, b \sqrt {-\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {-\frac {d}{c}}}{d}\right ) + b \sqrt {-\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {-\frac {d}{c}} - d}{c x + 1}\right ) + {\left (b c x \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c x + 4 \, b\right )} \sqrt {d x}}{3 \, c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(2*b*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) - b*sqrt(d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(d/c) + d)/(
c*x - 1)) - (b*c*x*log(-(c*x + 1)/(c*x - 1)) + 2*a*c*x + 4*b)*sqrt(d*x))/c, 1/3*(2*b*sqrt(-d/c)*arctan(sqrt(d*
x)*c*sqrt(-d/c)/d) + b*sqrt(-d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(-d/c) - d)/(c*x + 1)) + (b*c*x*log(-(c*x + 1
)/(c*x - 1)) + 2*a*c*x + 4*b)*sqrt(d*x))/c]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 89, normalized size = 0.84 \[ \frac {2 \left (d x \right )^{\frac {3}{2}} a}{3 d}+\frac {2 b \left (d x \right )^{\frac {3}{2}} \arctanh \left (c x \right )}{3 d}+\frac {4 b \sqrt {d x}}{3 c}-\frac {2 d b \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}}-\frac {2 d b \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{3 c \sqrt {c d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*(d*x)^(3/2)*a+2/3/d*b*(d*x)^(3/2)*arctanh(c*x)+4/3*b*(d*x)^(1/2)/c-2/3*d*b/c/(c*d)^(1/2)*arctan(c*(d*x)^
(1/2)/(c*d)^(1/2))-2/3*d*b/c/(c*d)^(1/2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))

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maxima [A]  time = 0.43, size = 119, normalized size = 1.12 \[ \frac {2 \, \left (d x\right )^{\frac {3}{2}} a + {\left (2 \, \left (d x\right )^{\frac {3}{2}} \operatorname {artanh}\left (c x\right ) - \frac {{\left (\frac {2 \, d^{3} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c^{2}} - \frac {d^{3} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c^{2}} - \frac {4 \, \sqrt {d x} d^{2}}{c^{2}}\right )} c}{d}\right )} b}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*(d*x)^(3/2)*a + (2*(d*x)^(3/2)*arctanh(c*x) - (2*d^3*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c^2) - d^
3*log((sqrt(d*x)*c - sqrt(c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c^2) - 4*sqrt(d*x)*d^2/c^2)*c/d)*b)/d

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

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 [C]  time = 11.73, size = 685, normalized size = 6.46 \[ \frac {2 a \left (d x\right )^{\frac {3}{2}}}{3 d} + \frac {2 b \left (\begin {cases} \frac {4 c^{2} \sqrt {d} \left (d x\right )^{\frac {3}{2}} \sqrt {\frac {1}{c}} \operatorname {atanh}{\left (c x \right )}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {4 i c^{2} \sqrt {d} \left (d x\right )^{\frac {3}{2}} \sqrt {\frac {1}{c}} \operatorname {atanh}{\left (c x \right )}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {2 i c^{2} d^{2} \log {\left (i \sqrt {d} \sqrt {\frac {1}{c}} + \sqrt {d x} \right )}}{12 c^{4} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{4} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {8 c d^{\frac {3}{2}} \sqrt {d x} \sqrt {\frac {1}{c}}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {8 i c d^{\frac {3}{2}} \sqrt {d x} \sqrt {\frac {1}{c}}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {4 i c d^{2} \log {\left (- \sqrt {d} \sqrt {\frac {1}{c}} + \sqrt {d x} \right )}}{12 c^{3} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{3} \sqrt {d} \sqrt {\frac {1}{c}}} - \frac {6 i c d^{2} \log {\left (i \sqrt {d} \sqrt {\frac {1}{c}} + \sqrt {d x} \right )}}{12 c^{3} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{3} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {4 i c d^{2} \operatorname {atanh}{\left (c x \right )}}{12 c^{3} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{3} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {4 d^{2} \log {\left (- \sqrt {d} \sqrt {\frac {1}{c}} + \sqrt {d x} \right )}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} - \frac {4 d^{2} \log {\left (- i \sqrt {d} \sqrt {\frac {1}{c}} + \sqrt {d x} \right )}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} + \frac {4 d^{2} \operatorname {atanh}{\left (c x \right )}}{12 c^{2} \sqrt {d} \sqrt {\frac {1}{c}} + 12 i c^{2} \sqrt {d} \sqrt {\frac {1}{c}}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(d*x)**(3/2)/(3*d) + 2*b*Piecewise((4*c**2*sqrt(d)*(d*x)**(3/2)*sqrt(1/c)*atanh(c*x)/(12*c**2*sqrt(d)*sqrt
(1/c) + 12*I*c**2*sqrt(d)*sqrt(1/c)) + 4*I*c**2*sqrt(d)*(d*x)**(3/2)*sqrt(1/c)*atanh(c*x)/(12*c**2*sqrt(d)*sqr
t(1/c) + 12*I*c**2*sqrt(d)*sqrt(1/c)) + 2*I*c**2*d**2*log(I*sqrt(d)*sqrt(1/c) + sqrt(d*x))/(12*c**4*sqrt(d)*sq
rt(1/c) + 12*I*c**4*sqrt(d)*sqrt(1/c)) + 8*c*d**(3/2)*sqrt(d*x)*sqrt(1/c)/(12*c**2*sqrt(d)*sqrt(1/c) + 12*I*c*
*2*sqrt(d)*sqrt(1/c)) + 8*I*c*d**(3/2)*sqrt(d*x)*sqrt(1/c)/(12*c**2*sqrt(d)*sqrt(1/c) + 12*I*c**2*sqrt(d)*sqrt
(1/c)) + 4*I*c*d**2*log(-sqrt(d)*sqrt(1/c) + sqrt(d*x))/(12*c**3*sqrt(d)*sqrt(1/c) + 12*I*c**3*sqrt(d)*sqrt(1/
c)) - 6*I*c*d**2*log(I*sqrt(d)*sqrt(1/c) + sqrt(d*x))/(12*c**3*sqrt(d)*sqrt(1/c) + 12*I*c**3*sqrt(d)*sqrt(1/c)
) + 4*I*c*d**2*atanh(c*x)/(12*c**3*sqrt(d)*sqrt(1/c) + 12*I*c**3*sqrt(d)*sqrt(1/c)) + 4*d**2*log(-sqrt(d)*sqrt
(1/c) + sqrt(d*x))/(12*c**2*sqrt(d)*sqrt(1/c) + 12*I*c**2*sqrt(d)*sqrt(1/c)) - 4*d**2*log(-I*sqrt(d)*sqrt(1/c)
+ sqrt(d*x))/(12*c**2*sqrt(d)*sqrt(1/c) + 12*I*c**2*sqrt(d)*sqrt(1/c)) + 4*d**2*atanh(c*x)/(12*c**2*sqrt(d)*s
qrt(1/c) + 12*I*c**2*sqrt(d)*sqrt(1/c)), Ne(c, 0)), (0, True))/d

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