3.41 \(\int \frac {a+b \tanh ^{-1}(c x)}{(d x)^{7/2}} \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {2 b c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}+\frac {2 b c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}-\frac {4 b c}{15 d^2 (d x)^{3/2}} \]

[Out]

-4/15*b*c/d^2/(d*x)^(3/2)+2/5*b*c^(5/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/d^(7/2)-2/5*(a+b*arctanh(c*x))/d/(
d*x)^(5/2)+2/5*b*c^(5/2)*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))/d^(7/2)

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Rubi [A]  time = 0.07, antiderivative size = 107, 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, 325, 329, 212, 208, 205} \[ -\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {2 b c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}+\frac {2 b c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}-\frac {4 b c}{15 d^2 (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d*x)^(7/2),x]

[Out]

(-4*b*c)/(15*d^2*(d*x)^(3/2)) + (2*b*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*d^(7/2)) - (2*(a + b*ArcT
anh[c*x]))/(5*d*(d*x)^(5/2)) + (2*b*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]])/(5*d^(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 \frac {a+b \tanh ^{-1}(c x)}{(d x)^{7/2}} \, dx &=-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {(2 b c) \int \frac {1}{(d x)^{5/2} \left (1-c^2 x^2\right )} \, dx}{5 d}\\ &=-\frac {4 b c}{15 d^2 (d x)^{3/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {\left (2 b c^3\right ) \int \frac {1}{\sqrt {d x} \left (1-c^2 x^2\right )} \, dx}{5 d^3}\\ &=-\frac {4 b c}{15 d^2 (d x)^{3/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {\left (4 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {c^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{5 d^4}\\ &=-\frac {4 b c}{15 d^2 (d x)^{3/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {\left (2 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{d-c x^2} \, dx,x,\sqrt {d x}\right )}{5 d^3}+\frac {\left (2 b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{d+c x^2} \, dx,x,\sqrt {d x}\right )}{5 d^3}\\ &=-\frac {4 b c}{15 d^2 (d x)^{3/2}}+\frac {2 b c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{5 d (d x)^{5/2}}+\frac {2 b c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{5 d^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 108, normalized size = 1.01 \[ \frac {x \left (-6 a-3 b c^{5/2} x^{5/2} \log \left (1-\sqrt {c} \sqrt {x}\right )+3 b c^{5/2} x^{5/2} \log \left (\sqrt {c} \sqrt {x}+1\right )+6 b c^{5/2} x^{5/2} \tan ^{-1}\left (\sqrt {c} \sqrt {x}\right )-4 b c x-6 b \tanh ^{-1}(c x)\right )}{15 (d x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(d*x)^(7/2),x]

[Out]

(x*(-6*a - 4*b*c*x + 6*b*c^(5/2)*x^(5/2)*ArcTan[Sqrt[c]*Sqrt[x]] - 6*b*ArcTanh[c*x] - 3*b*c^(5/2)*x^(5/2)*Log[
1 - Sqrt[c]*Sqrt[x]] + 3*b*c^(5/2)*x^(5/2)*Log[1 + Sqrt[c]*Sqrt[x]]))/(15*(d*x)^(7/2))

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fricas [A]  time = 0.65, size = 253, normalized size = 2.36 \[ \left [-\frac {6 \, b c^{2} d x^{3} \sqrt {\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {\frac {c}{d}}}{c x}\right ) - 3 \, b c^{2} d x^{3} \sqrt {\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {\frac {c}{d}} + 1}{c x - 1}\right ) + {\left (4 \, b c x + 3 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a\right )} \sqrt {d x}}{15 \, d^{4} x^{3}}, -\frac {6 \, b c^{2} d x^{3} \sqrt {-\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {c}{d}}}{c x}\right ) - 3 \, b c^{2} d x^{3} \sqrt {-\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {-\frac {c}{d}} - 1}{c x + 1}\right ) + {\left (4 \, b c x + 3 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a\right )} \sqrt {d x}}{15 \, d^{4} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(7/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 117, normalized size = 1.09 \[ \frac {6 \, b c^{3} {\left (\frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d^{2}} - \frac {\arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} d^{2}}\right )} - \frac {3 \, b \log \left (-\frac {c d x + d}{c d x - d}\right )}{\sqrt {d x} d^{2} x^{2}} - \frac {2 \, {\left (2 \, b c d x + 3 \, a d\right )}}{\sqrt {d x} d^{3} x^{2}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 94, normalized size = 0.88 \[ -\frac {2 a}{5 d \left (d x \right )^{\frac {5}{2}}}-\frac {2 b \arctanh \left (c x \right )}{5 d \left (d x \right )^{\frac {5}{2}}}-\frac {4 b c}{15 d^{2} \left (d x \right )^{\frac {3}{2}}}+\frac {2 b \,c^{3} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{3} \sqrt {c d}}+\frac {2 b \,c^{3} \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {c d}}\right )}{5 d^{3} \sqrt {c d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(d*x)^(7/2),x)

[Out]

-2/5/d*a/(d*x)^(5/2)-2/5/d*b/(d*x)^(5/2)*arctanh(c*x)-4/15*b*c/d^2/(d*x)^(3/2)+2/5/d^3*b*c^3/(c*d)^(1/2)*arcta
n(c*(d*x)^(1/2)/(c*d)^(1/2))+2/5/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 = 0.43, size = 112, normalized size = 1.05 \[ \frac {b {\left (\frac {{\left (\frac {6 \, c^{2} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d} - \frac {3 \, c^{2} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} d} - \frac {4}{\left (d x\right )^{\frac {3}{2}}}\right )} c}{d} - \frac {6 \, \operatorname {artanh}\left (c x\right )}{\left (d x\right )^{\frac {5}{2}}}\right )} - \frac {6 \, a}{\left (d x\right )^{\frac {5}{2}}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{{\left (d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d*x)^(7/2),x)

[Out]

int((a + b*atanh(c*x))/(d*x)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\left (d x\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(d*x)**(7/2),x)

[Out]

Integral((a + b*atanh(c*x))/(d*x)**(7/2), x)

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