3.1.36 \(\int (d x)^{3/2} (a+b \tanh ^{-1}(c x)) \, dx\) [36]

Optimal. Leaf size=106 \[ \frac {4 b (d x)^{3/2}}{15 c}+\frac {2 b d^{3/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}} \]

[Out]

4/15*b*(d*x)^(3/2)/c+2/5*b*d^(3/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(5/2)+2/5*(d*x)^(5/2)*(a+b*arctanh(c*
x))/d-2/5*b*d^(3/2)*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(5/2)

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Rubi [A]
time = 0.05, 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 = {6049, 327, 335, 304, 211, 214} \begin {gather*} \frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}+\frac {2 b d^{3/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}}+\frac {4 b (d x)^{3/2}}{15 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(d*x)^(3/2))/(15*c) + (2*b*d^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/2)) + (2*(d*x)^(5/2)*(a +
b*ArcTanh[c*x]))/(5*d) - (2*b*d^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/2))

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 304

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 327

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 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 (d x)^{3/2} \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac {(2 b c) \int \frac {(d x)^{5/2}}{1-c^2 x^2} \, dx}{5 d}\\ &=\frac {4 b (d x)^{3/2}}{15 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac {(2 b d) \int \frac {\sqrt {d x}}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac {4 b (d x)^{3/2}}{15 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac {(4 b) \text {Subst}\left (\int \frac {x^2}{1-\frac {c^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{5 c}\\ &=\frac {4 b (d x)^{3/2}}{15 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac {\left (2 b d^2\right ) \text {Subst}\left (\int \frac {1}{d-c x^2} \, dx,x,\sqrt {d x}\right )}{5 c^2}+\frac {\left (2 b d^2\right ) \text {Subst}\left (\int \frac {1}{d+c x^2} \, dx,x,\sqrt {d x}\right )}{5 c^2}\\ &=\frac {4 b (d x)^{3/2}}{15 c}+\frac {2 b d^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right )}{5 d}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 115, normalized size = 1.08 \begin {gather*} \frac {(d x)^{3/2} \left (4 b c^{3/2} x^{3/2}+6 a c^{5/2} x^{5/2}+6 b \text {ArcTan}\left (\sqrt {c} \sqrt {x}\right )+6 b c^{5/2} x^{5/2} \tanh ^{-1}(c x)+3 b \log \left (1-\sqrt {c} \sqrt {x}\right )-3 b \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{15 c^{5/2} x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(3/2)*(4*b*c^(3/2)*x^(3/2) + 6*a*c^(5/2)*x^(5/2) + 6*b*ArcTan[Sqrt[c]*Sqrt[x]] + 6*b*c^(5/2)*x^(5/2)*Ar
cTanh[c*x] + 3*b*Log[1 - Sqrt[c]*Sqrt[x]] - 3*b*Log[1 + Sqrt[c]*Sqrt[x]]))/(15*c^(5/2)*x^(3/2))

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Maple [A]
time = 0.04, size = 93, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+\frac {2 b \left (d x \right )^{\frac {5}{2}} \arctanh \left (c x \right )}{5}+\frac {4 b d \left (d x \right )^{\frac {3}{2}}}{15 c}+\frac {2 b \,d^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{5 c^{2} \sqrt {d c}}-\frac {2 b \,d^{3} \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{5 c^{2} \sqrt {d c}}}{d}\)	\(93\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+\frac {2 b \left (d x \right )^{\frac {5}{2}} \arctanh \left (c x \right )}{5}+\frac {4 b d \left (d x \right )^{\frac {3}{2}}}{15 c}+\frac {2 b \,d^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{5 c^{2} \sqrt {d c}}-\frac {2 b \,d^{3} \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{5 c^{2} \sqrt {d c}}}{d}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/5*(d*x)^(5/2)*a+1/5*b*(d*x)^(5/2)*arctanh(c*x)+2/15*b*d*(d*x)^(3/2)/c+1/5*b*d^3/c^2/(d*c)^(1/2)*arctan(
c*(d*x)^(1/2)/(d*c)^(1/2))-1/5*b*d^3/c^2/(d*c)^(1/2)*arctanh(c*(d*x)^(1/2)/(d*c)^(1/2)))

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Maxima [A]
time = 0.47, size = 118, normalized size = 1.11 \begin {gather*} \frac {6 \, \left (d x\right )^{\frac {5}{2}} a + {\left (6 \, \left (d x\right )^{\frac {5}{2}} \operatorname {artanh}\left (c x\right ) + \frac {{\left (\frac {4 \, \left (d x\right )^{\frac {3}{2}} d^{2}}{c^{2}} + \frac {6 \, d^{4} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c^{3}} + \frac {3 \, d^{4} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c^{3}}\right )} c}{d}\right )} b}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(6*(d*x)^(5/2)*a + (6*(d*x)^(5/2)*arctanh(c*x) + (4*(d*x)^(3/2)*d^2/c^2 + 6*d^4*arctan(sqrt(d*x)*c/sqrt(c
*d))/(sqrt(c*d)*c^3) + 3*d^4*log((sqrt(d*x)*c - sqrt(c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c^3))*c/d)*b)
/d

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Fricas [A]
time = 0.39, size = 255, normalized size = 2.41 \begin {gather*} \left [\frac {6 \, b d \sqrt {\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {\frac {d}{c}}}{d}\right ) + 3 \, b d \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 (3 \, b c^{2} d x^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a c^{2} d x^{2} + 4 \, b c d x\right )} \sqrt {d x}}{15 \, c^{2}}, \frac {6 \, b d \sqrt {-\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {-\frac {d}{c}}}{d}\right ) + 3 \, b d \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 (3 \, b c^{2} d x^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a c^{2} d x^{2} + 4 \, b c d x\right )} \sqrt {d x}}{15 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(6*b*d*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) + 3*b*d*sqrt(d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt(d/c)
+ d)/(c*x - 1)) + (3*b*c^2*d*x^2*log(-(c*x + 1)/(c*x - 1)) + 6*a*c^2*d*x^2 + 4*b*c*d*x)*sqrt(d*x))/c^2, 1/15*(
6*b*d*sqrt(-d/c)*arctan(sqrt(d*x)*c*sqrt(-d/c)/d) + 3*b*d*sqrt(-d/c)*log((c*d*x + 2*sqrt(d*x)*c*sqrt(-d/c) - d
)/(c*x + 1)) + (3*b*c^2*d*x^2*log(-(c*x + 1)/(c*x - 1)) + 6*a*c^2*d*x^2 + 4*b*c*d*x)*sqrt(d*x))/c^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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