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

Optimal. Leaf size=75 \[ \frac {(d x)^{m+1} \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{d (m+1)}-\frac {2 b c d (d x)^{m-1} \, _2F_1\left (1,\frac {1-m}{4};\frac {5-m}{4};\frac {c^2}{x^4}\right )}{1-m^2} \]

[Out]

(d*x)^(1+m)*(a+b*arctanh(c/x^2))/d/(1+m)-2*b*c*d*(d*x)^(-1+m)*hypergeom([1, 1/4-1/4*m],[5/4-1/4*m],c^2/x^4)/(-
m^2+1)

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Rubi [A]  time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6097, 16, 339, 364} \[ \frac {(d x)^{m+1} \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{d (m+1)}-\frac {2 b c d (d x)^{m-1} \, _2F_1\left (1,\frac {1-m}{4};\frac {5-m}{4};\frac {c^2}{x^4}\right )}{1-m^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a + b*ArcTanh[c/x^2]),x]

[Out]

((d*x)^(1 + m)*(a + b*ArcTanh[c/x^2]))/(d*(1 + m)) - (2*b*c*d*(d*x)^(-1 + m)*Hypergeometric2F1[1, (1 - m)/4, (
5 - m)/4, c^2/x^4])/(1 - m^2)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 339

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Dist[((c*x)^(m + 1)*(1/x)^(m + 1))/c, Subst
[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, b, c, m, p}, x] && ILtQ[n, 0] &&  !RationalQ[m]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int (d x)^m \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right ) \, dx &=\frac {(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{d (1+m)}+\frac {(2 b c) \int \frac {(d x)^{1+m}}{\left (1-\frac {c^2}{x^4}\right ) x^3} \, dx}{d (1+m)}\\ &=\frac {(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{d (1+m)}+\frac {\left (2 b c d^2\right ) \int \frac {(d x)^{-2+m}}{1-\frac {c^2}{x^4}} \, dx}{1+m}\\ &=\frac {(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{d (1+m)}-\frac {\left (2 b c d \left (\frac {1}{x}\right )^{-1+m} (d x)^{-1+m}\right ) \operatorname {Subst}\left (\int \frac {x^{-m}}{1-c^2 x^4} \, dx,x,\frac {1}{x}\right )}{1+m}\\ &=\frac {(d x)^{1+m} \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{d (1+m)}-\frac {2 b c d (d x)^{-1+m} \, _2F_1\left (1,\frac {1-m}{4};\frac {5-m}{4};\frac {c^2}{x^4}\right )}{1-m^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 68, normalized size = 0.91 \[ \frac {(d x)^m \left ((m-1) x^2 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+2 b c \, _2F_1\left (1,\frac {1}{4}-\frac {m}{4};\frac {5}{4}-\frac {m}{4};\frac {c^2}{x^4}\right )\right )}{(m-1) (m+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^m*((-1 + m)*x^2*(a + b*ArcTanh[c/x^2]) + 2*b*c*Hypergeometric2F1[1, 1/4 - m/4, 5/4 - m/4, c^2/x^4]))/((
-1 + m)*(1 + m)*x)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctanh(c/x^2) + a)*(d*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{m} \left (a +b \arctanh \left (\frac {c}{x^{2}}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (4 \, c d^{m} \int \frac {x^{2} x^{m}}{{\left (m + 1\right )} x^{4} - c^{2} {\left (m + 1\right )}}\,{d x} + \frac {d^{m} x x^{m} \log \left (x^{2} + c\right ) - d^{m} x x^{m} \log \left (x^{2} - c\right )}{m + 1}\right )} b + \frac {\left (d x\right )^{m + 1} a}{d {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*c*d^m*integrate(x^2*x^m/((m + 1)*x^4 - c^2*(m + 1)), x) + (d^m*x*x^m*log(x^2 + c) - d^m*x*x^m*log(x^2 -
c))/(m + 1))*b + (d*x)^(m + 1)*a/(d*(m + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a + b*atanh(c/x^2)),x)

[Out]

int((d*x)^m*(a + b*atanh(c/x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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