3.1.42 \(\int \frac {a+b \tanh ^{-1}(c x)}{(d x)^{9/2}} \, dx\) [42]

Optimal. Leaf size=125 \[ -\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {2 b c^{7/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {2 b c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}} \]

[Out]

-4/35*b*c/d^2/(d*x)^(5/2)-2/7*b*c^(7/2)*arctan(c^(1/2)*(d*x)^(1/2)/d^(1/2))/d^(9/2)-2/7*(a+b*arctanh(c*x))/d/(
d*x)^(7/2)+2/7*b*c^(7/2)*arctanh(c^(1/2)*(d*x)^(1/2)/d^(1/2))/d^(9/2)-4/7*b*c^3/d^4/(d*x)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 125, 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 = {6049, 331, 335, 304, 211, 214} \begin {gather*} -\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}-\frac {2 b c^{7/2} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}}+\frac {2 b c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {4 b c}{35 d^2 (d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d*x)^(9/2),x]

[Out]

(-4*b*c)/(35*d^2*(d*x)^(5/2)) - (4*b*c^3)/(7*d^4*Sqrt[d*x]) - (2*b*c^(7/2)*ArcTan[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]
)/(7*d^(9/2)) - (2*(a + b*ArcTanh[c*x]))/(7*d*(d*x)^(7/2)) + (2*b*c^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]]
)/(7*d^(9/2))

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 331

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 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 \frac {a+b \tanh ^{-1}(c x)}{(d x)^{9/2}} \, dx &=-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {(2 b c) \int \frac {1}{(d x)^{7/2} \left (1-c^2 x^2\right )} \, dx}{7 d}\\ &=-\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {\left (2 b c^3\right ) \int \frac {1}{(d x)^{3/2} \left (1-c^2 x^2\right )} \, dx}{7 d^3}\\ &=-\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {\left (2 b c^5\right ) \int \frac {\sqrt {d x}}{1-c^2 x^2} \, dx}{7 d^5}\\ &=-\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {\left (4 b c^5\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {c^2 x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{7 d^6}\\ &=-\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {\left (2 b c^4\right ) \text {Subst}\left (\int \frac {1}{d-c x^2} \, dx,x,\sqrt {d x}\right )}{7 d^4}-\frac {\left (2 b c^4\right ) \text {Subst}\left (\int \frac {1}{d+c x^2} \, dx,x,\sqrt {d x}\right )}{7 d^4}\\ &=-\frac {4 b c}{35 d^2 (d x)^{5/2}}-\frac {4 b c^3}{7 d^4 \sqrt {d x}}-\frac {2 b c^{7/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )}{7 d (d x)^{7/2}}+\frac {2 b c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )}{7 d^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 122, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {d x} \left (10 a+4 b c x+20 b c^3 x^3+10 b c^{7/2} x^{7/2} \text {ArcTan}\left (\sqrt {c} \sqrt {x}\right )+10 b \tanh ^{-1}(c x)+5 b c^{7/2} x^{7/2} \log \left (1-\sqrt {c} \sqrt {x}\right )-5 b c^{7/2} x^{7/2} \log \left (1+\sqrt {c} \sqrt {x}\right )\right )}{35 d^5 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(d*x)^(9/2),x]

[Out]

-1/35*(Sqrt[d*x]*(10*a + 4*b*c*x + 20*b*c^3*x^3 + 10*b*c^(7/2)*x^(7/2)*ArcTan[Sqrt[c]*Sqrt[x]] + 10*b*ArcTanh[
c*x] + 5*b*c^(7/2)*x^(7/2)*Log[1 - Sqrt[c]*Sqrt[x]] - 5*b*c^(7/2)*x^(7/2)*Log[1 + Sqrt[c]*Sqrt[x]]))/(d^5*x^4)

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Maple [A]
time = 0.05, size = 107, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \arctanh \left (c x \right )}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \,c^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{7 d^{3} \sqrt {d c}}+\frac {2 b \,c^{4} \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{7 d^{3} \sqrt {d c}}-\frac {4 b c}{35 d \left (d x \right )^{\frac {5}{2}}}-\frac {4 b \,c^{3}}{7 d^{3} \sqrt {d x}}}{d}\)	\(107\)
default	\(\frac {-\frac {2 a}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \arctanh \left (c x \right )}{7 \left (d x \right )^{\frac {7}{2}}}-\frac {2 b \,c^{4} \arctan \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{7 d^{3} \sqrt {d c}}+\frac {2 b \,c^{4} \arctanh \left (\frac {c \sqrt {d x}}{\sqrt {d c}}\right )}{7 d^{3} \sqrt {d c}}-\frac {4 b c}{35 d \left (d x \right )^{\frac {5}{2}}}-\frac {4 b \,c^{3}}{7 d^{3} \sqrt {d x}}}{d}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/(d*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 130, normalized size = 1.04 \begin {gather*} -\frac {b {\left (\frac {{\left (\frac {10 \, c^{3} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d^{2}} + \frac {5 \, c^{3} \log \left (\frac {\sqrt {d x} c - \sqrt {c d}}{\sqrt {d x} c + \sqrt {c d}}\right )}{\sqrt {c d} d^{2}} + \frac {4 \, {\left (5 \, c^{2} d^{2} x^{2} + d^{2}\right )}}{\left (d x\right )^{\frac {5}{2}} d^{2}}\right )} c}{d} + \frac {10 \, \operatorname {artanh}\left (c x\right )}{\left (d x\right )^{\frac {7}{2}}}\right )} + \frac {10 \, a}{\left (d x\right )^{\frac {7}{2}}}}{35 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(9/2),x, algorithm="maxima")

[Out]

-1/35*(b*((10*c^3*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*d^2) + 5*c^3*log((sqrt(d*x)*c - sqrt(c*d))/(sqrt(d*
x)*c + sqrt(c*d)))/(sqrt(c*d)*d^2) + 4*(5*c^2*d^2*x^2 + d^2)/((d*x)^(5/2)*d^2))*c/d + 10*arctanh(c*x)/(d*x)^(7
/2)) + 10*a/(d*x)^(7/2))/d

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Fricas [A]
time = 0.36, size = 272, normalized size = 2.18 \begin {gather*} \left [\frac {10 \, b c^{3} d x^{4} \sqrt {\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {\frac {c}{d}}}{c x}\right ) + 5 \, b c^{3} d x^{4} \sqrt {\frac {c}{d}} \log \left (\frac {c x + 2 \, \sqrt {d x} \sqrt {\frac {c}{d}} + 1}{c x - 1}\right ) - {\left (20 \, b c^{3} x^{3} + 4 \, b c x + 5 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a\right )} \sqrt {d x}}{35 \, d^{5} x^{4}}, -\frac {10 \, b c^{3} d x^{4} \sqrt {-\frac {c}{d}} \arctan \left (\frac {\sqrt {d x} \sqrt {-\frac {c}{d}}}{c x}\right ) - 5 \, b c^{3} d x^{4} \sqrt {-\frac {c}{d}} \log \left (\frac {c x - 2 \, \sqrt {d x} \sqrt {-\frac {c}{d}} - 1}{c x + 1}\right ) + {\left (20 \, b c^{3} x^{3} + 4 \, b c x + 5 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 10 \, a\right )} \sqrt {d x}}{35 \, d^{5} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(9/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{\left (d x\right )^{\frac {9}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/(d*x)**(9/2),x)

[Out]

Integral((a + b*atanh(c*x))/(d*x)**(9/2), x)

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Giac [A]
time = 0.51, size = 135, normalized size = 1.08 \begin {gather*} -\frac {\frac {10 \, b c^{4} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {c d}}\right )}{\sqrt {c d} d^{3}} + \frac {10 \, b c^{4} \arctan \left (\frac {\sqrt {d x} c}{\sqrt {-c d}}\right )}{\sqrt {-c d} d^{3}} + \frac {5 \, b \log \left (-\frac {c d x + d}{c d x - d}\right )}{\sqrt {d x} d^{3} x^{3}} + \frac {2 \, {\left (10 \, b c^{3} d^{3} x^{3} + 2 \, b c d^{3} x + 5 \, a d^{3}\right )}}{\sqrt {d x} d^{6} x^{3}}}{35 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/(d*x)^(9/2),x, algorithm="giac")

[Out]

-1/35*(10*b*c^4*arctan(sqrt(d*x)*c/sqrt(c*d))/(sqrt(c*d)*d^3) + 10*b*c^4*arctan(sqrt(d*x)*c/sqrt(-c*d))/(sqrt(
-c*d)*d^3) + 5*b*log(-(c*d*x + d)/(c*d*x - d))/(sqrt(d*x)*d^3*x^3) + 2*(10*b*c^3*d^3*x^3 + 2*b*c*d^3*x + 5*a*d
^3)/(sqrt(d*x)*d^6*x^3))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{{\left (d\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d*x)^(9/2),x)

[Out]

int((a + b*atanh(c*x))/(d*x)^(9/2), x)

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