3.35 \(\int (d x)^{5/2} (a+b \tanh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=124 \[ \frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {2 b d^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}-\frac {2 b d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}+\frac {4 b d^2 \sqrt {d x}}{7 c^3}+\frac {4 b (d x)^{5/2}}{35 c} \]

[Out]

4/35*b*(d*x)^(5/2)/c-2/7*b*d^(5/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(7/2)+2/7*(d*x)^(7/2)*(a+b*arctanh(c*
x))/d-2/7*b*d^(5/2)*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))/c^(7/2)+4/7*b*d^2*(d*x)^(1/2)/c^3

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Rubi [A]  time = 0.09, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}+\frac {4 b d^2 \sqrt {d x}}{7 c^3}-\frac {2 b d^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}-\frac {2 b d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}+\frac {4 b (d x)^{5/2}}{35 c} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(5/2)*(a + b*ArcTanh[c*x]),x]

[Out]

(4*b*d^2*Sqrt[d*x])/(7*c^3) + (4*b*(d*x)^(5/2))/(35*c) - (2*b*d^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(7*
c^(7/2)) + (2*(d*x)^(7/2)*(a + b*ArcTanh[c*x]))/(7*d) - (2*b*d^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(7*
c^(7/2))

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 (d x)^{5/2} \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {(2 b c) \int \frac {(d x)^{7/2}}{1-c^2 x^2} \, dx}{7 d}\\ &=\frac {4 b (d x)^{5/2}}{35 c}+\frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {(2 b d) \int \frac {(d x)^{3/2}}{1-c^2 x^2} \, dx}{7 c}\\ &=\frac {4 b d^2 \sqrt {d x}}{7 c^3}+\frac {4 b (d x)^{5/2}}{35 c}+\frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {\left (2 b d^3\right ) \int \frac {1}{\sqrt {d x} \left (1-c^2 x^2\right )} \, dx}{7 c^3}\\ &=\frac {4 b d^2 \sqrt {d x}}{7 c^3}+\frac {4 b (d x)^{5/2}}{35 c}+\frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {\left (4 b d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {c^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{7 c^3}\\ &=\frac {4 b d^2 \sqrt {d x}}{7 c^3}+\frac {4 b (d x)^{5/2}}{35 c}+\frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {\left (2 b d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d-c x^2} \, dx,x,\sqrt {d x}\right )}{7 c^3}-\frac {\left (2 b d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d+c x^2} \, dx,x,\sqrt {d x}\right )}{7 c^3}\\ &=\frac {4 b d^2 \sqrt {d x}}{7 c^3}+\frac {4 b (d x)^{5/2}}{35 c}-\frac {2 b d^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}+\frac {2 (d x)^{7/2} \left (a+b \tanh ^{-1}(c x)\right )}{7 d}-\frac {2 b d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 128, normalized size = 1.03 \[ \frac {(d x)^{5/2} \left (10 a c^{7/2} x^{7/2}+4 b c^{5/2} x^{5/2}+10 b c^{7/2} x^{7/2} \tanh ^{-1}(c x)+20 b \sqrt {c} \sqrt {x}+5 b \log \left (1-\sqrt {c} \sqrt {x}\right )-5 b \log \left (\sqrt {c} \sqrt {x}+1\right )-10 b \tan ^{-1}\left (\sqrt {c} \sqrt {x}\right )\right )}{35 c^{7/2} x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(5/2)*(a + b*ArcTanh[c*x]),x]

[Out]

((d*x)^(5/2)*(20*b*Sqrt[c]*Sqrt[x] + 4*b*c^(5/2)*x^(5/2) + 10*a*c^(7/2)*x^(7/2) - 10*b*ArcTan[Sqrt[c]*Sqrt[x]]
+ 10*b*c^(7/2)*x^(7/2)*ArcTanh[c*x] + 5*b*Log[1 - Sqrt[c]*Sqrt[x]] - 5*b*Log[1 + Sqrt[c]*Sqrt[x]]))/(35*c^(7/
2)*x^(5/2))

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fricas [A]  time = 0.58, size = 296, normalized size = 2.39 \[ \left [-\frac {10 \, b d^{2} \sqrt {\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {\frac {d}{c}}}{d}\right ) - 5 \, b d^{2} \sqrt {\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {\frac {d}{c}} + d}{c x - 1}\right ) - {\left (5 \, b c^{3} d^{2} x^{3} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a c^{3} d^{2} x^{3} + 4 \, b c^{2} d^{2} x^{2} + 20 \, b d^{2}\right )} \sqrt {d x}}{35 \, c^{3}}, \frac {10 \, b d^{2} \sqrt {-\frac {d}{c}} \arctan \left (\frac {\sqrt {d x} c \sqrt {-\frac {d}{c}}}{d}\right ) + 5 \, b d^{2} \sqrt {-\frac {d}{c}} \log \left (\frac {c d x - 2 \, \sqrt {d x} c \sqrt {-\frac {d}{c}} - d}{c x + 1}\right ) + {\left (5 \, b c^{3} d^{2} x^{3} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a c^{3} d^{2} x^{3} + 4 \, b c^{2} d^{2} x^{2} + 20 \, b d^{2}\right )} \sqrt {d x}}{35 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arctanh(c*x)),x, algorithm="fricas")

[Out]

[-1/35*(10*b*d^2*sqrt(d/c)*arctan(sqrt(d*x)*c*sqrt(d/c)/d) - 5*b*d^2*sqrt(d/c)*log((c*d*x - 2*sqrt(d*x)*c*sqrt
(d/c) + d)/(c*x - 1)) - (5*b*c^3*d^2*x^3*log(-(c*x + 1)/(c*x - 1)) + 10*a*c^3*d^2*x^3 + 4*b*c^2*d^2*x^2 + 20*b
*d^2)*sqrt(d*x))/c^3, 1/35*(10*b*d^2*sqrt(-d/c)*arctan(sqrt(d*x)*c*sqrt(-d/c)/d) + 5*b*d^2*sqrt(-d/c)*log((c*d
*x - 2*sqrt(d*x)*c*sqrt(-d/c) - d)/(c*x + 1)) + (5*b*c^3*d^2*x^3*log(-(c*x + 1)/(c*x - 1)) + 10*a*c^3*d^2*x^3
+ 4*b*c^2*d^2*x^2 + 20*b*d^2)*sqrt(d*x))/c^3]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {5}{2}} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arctanh(c*x)),x, algorithm="giac")

[Out]

integrate((d*x)^(5/2)*(b*arctanh(c*x) + a), x)

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maple [A]  time = 0.04, size = 107, normalized size = 0.86 \[ \frac {2 \left (d x \right )^{\frac {7}{2}} a}{7 d}+\frac {2 b \left (d x \right )^{\frac {7}{2}} \arctanh \left (c x \right )}{7 d}+\frac {4 b \left (d x \right )^{\frac {5}{2}}}{35 c}+\frac {4 b \,d^{2} \sqrt {d x}}{7 c^{3}}-\frac {2 d^{3} b \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}}-\frac {2 d^{3} b \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{7 c^{3} \sqrt {c d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(5/2)*(a+b*arctanh(c*x)),x)

[Out]

2/7/d*(d*x)^(7/2)*a+2/7/d*b*(d*x)^(7/2)*arctanh(c*x)+4/35*b*(d*x)^(5/2)/c+4/7*b*d^2*(d*x)^(1/2)/c^3-2/7*d^3*b/
c^3/(c*d)^(1/2)*arctan(c*(d*x)^(1/2)/(c*d)^(1/2))-2/7*d^3*b/c^3/(c*d)^(1/2)*arctanh(c*(d*x)^(1/2)/(c*d)^(1/2))

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maxima [A]  time = 0.42, size = 134, normalized size = 1.08 \[ \frac {10 \, \left (d x\right )^{\frac {7}{2}} a + {\left (10 \, \left (d x\right )^{\frac {7}{2}} \operatorname {artanh}\left (c x\right ) - \frac {{\left (\frac {10 \, d^{5} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} c^{4}} - \frac {5 \, d^{5} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} c^{4}} - \frac {4 \, {\left (\left (d x\right )^{\frac {5}{2}} c^{2} d^{2} + 5 \, \sqrt {d x} d^{4}\right )}}{c^{4}}\right )} c}{d}\right )} b}{35 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arctanh(c*x)),x, algorithm="maxima")

[Out]

1/35*(10*(d*x)^(7/2)*a + (10*(d*x)^(7/2)*arctanh(c*x) - (10*d^5*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*c^4)
- 5*d^5*log((sqrt(d*x)*c - sqrt(c*d))/(sqrt(d*x)*c + sqrt(c*d)))/(sqrt(c*d)*c^4) - 4*((d*x)^(5/2)*c^2*d^2 + 5*
sqrt(d*x)*d^4)/c^4)*c/d)*b)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))*(d*x)^(5/2),x)

[Out]

int((a + b*atanh(c*x))*(d*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {5}{2}} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(5/2)*(a+b*atanh(c*x)),x)

[Out]

Integral((d*x)**(5/2)*(a + b*atanh(c*x)), x)

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