3.1.83 \(\int (d x)^{3/2} (a+b \tanh ^{-1}(c x^2)) \, dx\) [83]

Optimal. Leaf size=317 \[ \frac {8 b d \sqrt {d x}}{5 c}-\frac {2 b d^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {\sqrt {2} b d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}-\frac {\sqrt {2} b d^{3/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {b d^{3/2} \log \left (\sqrt {d}+\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}-\frac {b d^{3/2} \log \left (\sqrt {d}+\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}} \]

[Out]

-2/5*b*d^(3/2)*arctan(c^(1/4)*(d*x)^(1/2)/d^(1/2))/c^(5/4)+2/5*(d*x)^(5/2)*(a+b*arctanh(c*x^2))/d-2/5*b*d^(3/2
)*arctanh(c^(1/4)*(d*x)^(1/2)/d^(1/2))/c^(5/4)+1/10*b*d^(3/2)*ln(d^(1/2)+x*c^(1/2)*d^(1/2)-c^(1/4)*2^(1/2)*(d*
x)^(1/2))/c^(5/4)*2^(1/2)-1/10*b*d^(3/2)*ln(d^(1/2)+x*c^(1/2)*d^(1/2)+c^(1/4)*2^(1/2)*(d*x)^(1/2))/c^(5/4)*2^(
1/2)-1/5*b*d^(3/2)*arctan(-1+c^(1/4)*2^(1/2)*(d*x)^(1/2)/d^(1/2))*2^(1/2)/c^(5/4)-1/5*b*d^(3/2)*arctan(1+c^(1/
4)*2^(1/2)*(d*x)^(1/2)/d^(1/2))*2^(1/2)/c^(5/4)+8/5*b*d*(d*x)^(1/2)/c

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Rubi [A]
time = 0.19, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {6049, 327, 335, 220, 218, 214, 211, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {2 b d^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {\sqrt {2} b d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}-\frac {\sqrt {2} b d^{3/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}+1\right )}{5 c^{5/4}}+\frac {b d^{3/2} \log \left (\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{5 \sqrt {2} c^{5/4}}-\frac {b d^{3/2} \log \left (\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}+\sqrt {d}\right )}{5 \sqrt {2} c^{5/4}}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {8 b d \sqrt {d x}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d*Sqrt[d*x])/(5*c) - (2*b*d^(3/2)*ArcTan[(c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (Sqrt[2]*b*d^(3/2)*A
rcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) - (Sqrt[2]*b*d^(3/2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*S
qrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (2*(d*x)^(5/2)*(a + b*ArcTanh[c*x^2]))/(5*d) - (2*b*d^(3/2)*ArcTanh[(c^(1/4)
*Sqrt[d*x])/Sqrt[d]])/(5*c^(5/4)) + (b*d^(3/2)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x - Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(
5*Sqrt[2]*c^(5/4)) - (b*d^(3/2)*Log[Sqrt[d] + Sqrt[c]*Sqrt[d]*x + Sqrt[2]*c^(1/4)*Sqrt[d*x]])/(5*Sqrt[2]*c^(5/
4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, 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 220

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), x], x] + Dist[r/(2*a), Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a, b},
x] && IGtQ[n/4, 1] &&  !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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

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

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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}\left (c x^2\right )\right ) \, dx &=\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {(4 b c) \int \frac {x (d x)^{5/2}}{1-c^2 x^4} \, dx}{5 d}\\ &=\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {(4 b c) \int \frac {(d x)^{7/2}}{1-c^2 x^4} \, dx}{5 d^2}\\ &=\frac {8 b d \sqrt {d x}}{5 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {\left (4 b d^2\right ) \int \frac {1}{\sqrt {d x} \left (1-c^2 x^4\right )} \, dx}{5 c}\\ &=\frac {8 b d \sqrt {d x}}{5 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {(8 b d) \text {Subst}\left (\int \frac {1}{1-\frac {c^2 x^8}{d^4}} \, dx,x,\sqrt {d x}\right )}{5 c}\\ &=\frac {8 b d \sqrt {d x}}{5 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {\left (4 b d^3\right ) \text {Subst}\left (\int \frac {1}{d^2-c x^4} \, dx,x,\sqrt {d x}\right )}{5 c}-\frac {\left (4 b d^3\right ) \text {Subst}\left (\int \frac {1}{d^2+c x^4} \, dx,x,\sqrt {d x}\right )}{5 c}\\ &=\frac {8 b d \sqrt {d x}}{5 c}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {\left (2 b d^2\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {c} x^2} \, dx,x,\sqrt {d x}\right )}{5 c}-\frac {\left (2 b d^2\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {c} x^2} \, dx,x,\sqrt {d x}\right )}{5 c}-\frac {\left (2 b d^2\right ) \text {Subst}\left (\int \frac {d-\sqrt {c} x^2}{d^2+c x^4} \, dx,x,\sqrt {d x}\right )}{5 c}-\frac {\left (2 b d^2\right ) \text {Subst}\left (\int \frac {d+\sqrt {c} x^2}{d^2+c x^4} \, dx,x,\sqrt {d x}\right )}{5 c}\\ &=\frac {8 b d \sqrt {d x}}{5 c}-\frac {2 b d^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {\left (b d^{3/2}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {d}}{\sqrt [4]{c}}+2 x}{-\frac {d}{\sqrt {c}}-\frac {\sqrt {2} \sqrt {d} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}+\frac {\left (b d^{3/2}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {d}}{\sqrt [4]{c}}-2 x}{-\frac {d}{\sqrt {c}}+\frac {\sqrt {2} \sqrt {d} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}-\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {1}{\frac {d}{\sqrt {c}}-\frac {\sqrt {2} \sqrt {d} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d x}\right )}{5 c^{3/2}}-\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {1}{\frac {d}{\sqrt {c}}+\frac {\sqrt {2} \sqrt {d} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {d x}\right )}{5 c^{3/2}}\\ &=\frac {8 b d \sqrt {d x}}{5 c}-\frac {2 b d^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {b d^{3/2} \log \left (\sqrt {d}+\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}-\frac {b d^{3/2} \log \left (\sqrt {d}+\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}-\frac {\left (\sqrt {2} b d^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {\left (\sqrt {2} b d^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}\\ &=\frac {8 b d \sqrt {d x}}{5 c}-\frac {2 b d^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {\sqrt {2} b d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}-\frac {\sqrt {2} b d^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {2 (d x)^{5/2} \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{5 d}-\frac {2 b d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d x}}{\sqrt {d}}\right )}{5 c^{5/4}}+\frac {b d^{3/2} \log \left (\sqrt {d}+\sqrt {c} \sqrt {d} x-\sqrt {2} \sqrt [4]{c} \sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}-\frac {b d^{3/2} \log \left (\sqrt {d}+\sqrt {c} \sqrt {d} x+\sqrt {2} \sqrt [4]{c} \sqrt {d x}\right )}{5 \sqrt {2} c^{5/4}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 240, normalized size = 0.76 \begin {gather*} \frac {(d x)^{3/2} \left (16 b \sqrt [4]{c} \sqrt {x}+4 a c^{5/4} x^{5/2}+2 \sqrt {2} b \text {ArcTan}\left (1-\sqrt {2} \sqrt [4]{c} \sqrt {x}\right )-2 \sqrt {2} b \text {ArcTan}\left (1+\sqrt {2} \sqrt [4]{c} \sqrt {x}\right )-4 b \text {ArcTan}\left (\sqrt [4]{c} \sqrt {x}\right )+4 b c^{5/4} x^{5/2} \tanh ^{-1}\left (c x^2\right )+2 b \log \left (1-\sqrt [4]{c} \sqrt {x}\right )-2 b \log \left (1+\sqrt [4]{c} \sqrt {x}\right )+\sqrt {2} b \log \left (1-\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )-\sqrt {2} b \log \left (1+\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )\right )}{10 c^{5/4} x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(3/2)*(16*b*c^(1/4)*Sqrt[x] + 4*a*c^(5/4)*x^(5/2) + 2*Sqrt[2]*b*ArcTan[1 - Sqrt[2]*c^(1/4)*Sqrt[x]] - 2
*Sqrt[2]*b*ArcTan[1 + Sqrt[2]*c^(1/4)*Sqrt[x]] - 4*b*ArcTan[c^(1/4)*Sqrt[x]] + 4*b*c^(5/4)*x^(5/2)*ArcTanh[c*x
^2] + 2*b*Log[1 - c^(1/4)*Sqrt[x]] - 2*b*Log[1 + c^(1/4)*Sqrt[x]] + Sqrt[2]*b*Log[1 - Sqrt[2]*c^(1/4)*Sqrt[x]
+ Sqrt[c]*x] - Sqrt[2]*b*Log[1 + Sqrt[2]*c^(1/4)*Sqrt[x] + Sqrt[c]*x]))/(10*c^(5/4)*x^(3/2))

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Maple [A]
time = 0.04, size = 303, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+\frac {2 b \left (d x \right )^{\frac {5}{2}} \arctanh \left (c \,x^{2}\right )}{5}+\frac {8 b \,d^{2} \sqrt {d x}}{5 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}{d x -\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}\right )}{10 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+1\right )}{5 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}-1\right )}{5 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{5 c}-\frac {2 b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{5 c}}{d}\)	\(303\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+\frac {2 b \left (d x \right )^{\frac {5}{2}} \arctanh \left (c \,x^{2}\right )}{5}+\frac {8 b \,d^{2} \sqrt {d x}}{5 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d x +\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}{d x -\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {d x}\, \sqrt {2}+\sqrt {\frac {d^{2}}{c}}}\right )}{10 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}+1\right )}{5 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}-1\right )}{5 c}-\frac {b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{5 c}-\frac {2 b \,d^{2} \left (\frac {d^{2}}{c}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{c}\right )^{\frac {1}{4}}}\right )}{5 c}}{d}\)	\(303\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/5*(d*x)^(5/2)*a+1/5*b*(d*x)^(5/2)*arctanh(c*x^2)+4/5*b/c*d^2*(d*x)^(1/2)-1/20*b/c*d^2*(d^2/c)^(1/4)*2^(
1/2)*ln((d*x+(d^2/c)^(1/4)*(d*x)^(1/2)*2^(1/2)+(d^2/c)^(1/2))/(d*x-(d^2/c)^(1/4)*(d*x)^(1/2)*2^(1/2)+(d^2/c)^(
1/2)))-1/10*b/c*d^2*(d^2/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2/c)^(1/4)*(d*x)^(1/2)+1)-1/10*b/c*d^2*(d^2/c)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(d^2/c)^(1/4)*(d*x)^(1/2)-1)-1/10*b/c*d^2*(d^2/c)^(1/4)*ln(((d*x)^(1/2)+(d^2/c)^(1/4
))/((d*x)^(1/2)-(d^2/c)^(1/4)))-1/5*b/c*d^2*(d^2/c)^(1/4)*arctan((d*x)^(1/2)/(d^2/c)^(1/4)))

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Maxima [A]
time = 0.47, size = 310, normalized size = 0.98 \begin {gather*} \frac {4 \, \left (d x\right )^{\frac {5}{2}} a + {\left (4 \, \left (d x\right )^{\frac {5}{2}} \operatorname {artanh}\left (c x^{2}\right ) + \frac {{\left (\frac {16 \, \sqrt {d x} d^{4}}{c^{2}} - \frac {\frac {2 \, \sqrt {2} d^{5} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} \sqrt {d} + 2 \, \sqrt {d x} \sqrt {c}\right )}}{2 \, \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}} + \frac {2 \, \sqrt {2} d^{5} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} \sqrt {d} - 2 \, \sqrt {d x} \sqrt {c}\right )}}{2 \, \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}} + \frac {\sqrt {2} d^{\frac {9}{2}} \log \left (\sqrt {c} d x + \sqrt {2} \sqrt {d x} c^{\frac {1}{4}} \sqrt {d} + d\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} d^{\frac {9}{2}} \log \left (\sqrt {c} d x - \sqrt {2} \sqrt {d x} c^{\frac {1}{4}} \sqrt {d} + d\right )}{c^{\frac {1}{4}}}}{c^{2}} - \frac {2 \, {\left (\frac {2 \, d^{5} \arctan \left (\frac {\sqrt {d x} \sqrt {c}}{\sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}} - \frac {d^{5} \log \left (\frac {\sqrt {d x} \sqrt {c} - \sqrt {\sqrt {c} d}}{\sqrt {d x} \sqrt {c} + \sqrt {\sqrt {c} d}}\right )}{\sqrt {\sqrt {c} d}}\right )}}{c^{2}}\right )} c}{d^{2}}\right )} b}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(4*(d*x)^(5/2)*a + (4*(d*x)^(5/2)*arctanh(c*x^2) + (16*sqrt(d*x)*d^4/c^2 - (2*sqrt(2)*d^5*arctan(1/2*sqrt
(2)*(sqrt(2)*c^(1/4)*sqrt(d) + 2*sqrt(d*x)*sqrt(c))/sqrt(sqrt(c)*d))/sqrt(sqrt(c)*d) + 2*sqrt(2)*d^5*arctan(-1
/2*sqrt(2)*(sqrt(2)*c^(1/4)*sqrt(d) - 2*sqrt(d*x)*sqrt(c))/sqrt(sqrt(c)*d))/sqrt(sqrt(c)*d) + sqrt(2)*d^(9/2)*
log(sqrt(c)*d*x + sqrt(2)*sqrt(d*x)*c^(1/4)*sqrt(d) + d)/c^(1/4) - sqrt(2)*d^(9/2)*log(sqrt(c)*d*x - sqrt(2)*s
qrt(d*x)*c^(1/4)*sqrt(d) + d)/c^(1/4))/c^2 - 2*(2*d^5*arctan(sqrt(d*x)*sqrt(c)/sqrt(sqrt(c)*d))/sqrt(sqrt(c)*d
) - d^5*log((sqrt(d*x)*sqrt(c) - sqrt(sqrt(c)*d))/(sqrt(d*x)*sqrt(c) + sqrt(sqrt(c)*d)))/sqrt(sqrt(c)*d))/c^2)
*c/d^2)*b)/d

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Fricas [A]
time = 0.38, size = 399, normalized size = 1.26 \begin {gather*} \frac {4 \, \left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c \arctan \left (-\frac {\left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {3}{4}} \sqrt {d x} b c^{4} d - \sqrt {b^{2} d^{3} x + \sqrt {\frac {b^{4} d^{6}}{c^{5}}} c^{2}} \left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {3}{4}} c^{4}}{b^{4} d^{6}}\right ) - 4 \, \left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c \arctan \left (-\frac {\left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {3}{4}} \sqrt {d x} b c^{4} d - \sqrt {b^{2} d^{3} x + \sqrt {-\frac {b^{4} d^{6}}{c^{5}}} c^{2}} \left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {3}{4}} c^{4}}{b^{4} d^{6}}\right ) - \left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c \log \left (\sqrt {d x} b d + \left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c\right ) + \left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c \log \left (\sqrt {d x} b d - \left (\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c\right ) - \left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c \log \left (\sqrt {d x} b d + \left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c\right ) + \left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c \log \left (\sqrt {d x} b d - \left (-\frac {b^{4} d^{6}}{c^{5}}\right )^{\frac {1}{4}} c\right ) + {\left (b c d x^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a c d x^{2} + 8 \, b d\right )} \sqrt {d x}}{5 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(4*(b^4*d^6/c^5)^(1/4)*c*arctan(-((b^4*d^6/c^5)^(3/4)*sqrt(d*x)*b*c^4*d - sqrt(b^2*d^3*x + sqrt(b^4*d^6/c^
5)*c^2)*(b^4*d^6/c^5)^(3/4)*c^4)/(b^4*d^6)) - 4*(-b^4*d^6/c^5)^(1/4)*c*arctan(-((-b^4*d^6/c^5)^(3/4)*sqrt(d*x)
*b*c^4*d - sqrt(b^2*d^3*x + sqrt(-b^4*d^6/c^5)*c^2)*(-b^4*d^6/c^5)^(3/4)*c^4)/(b^4*d^6)) - (b^4*d^6/c^5)^(1/4)
*c*log(sqrt(d*x)*b*d + (b^4*d^6/c^5)^(1/4)*c) + (b^4*d^6/c^5)^(1/4)*c*log(sqrt(d*x)*b*d - (b^4*d^6/c^5)^(1/4)*
c) - (-b^4*d^6/c^5)^(1/4)*c*log(sqrt(d*x)*b*d + (-b^4*d^6/c^5)^(1/4)*c) + (-b^4*d^6/c^5)^(1/4)*c*log(sqrt(d*x)
*b*d - (-b^4*d^6/c^5)^(1/4)*c) + (b*c*d*x^2*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 2*a*c*d*x^2 + 8*b*d)*sqrt(d*x))/c

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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^{2} \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**2)),x)

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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