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

Optimal. Leaf size=166 \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{2 b c d e \log (d+e x)}{c^2 d^4-e^4}-\frac{b c d \log \left (1-c x^2\right )}{2 e \left (c d^2-e^2\right )}+\frac{b c d \log \left (c x^2+1\right )}{2 e \left (c d^2+e^2\right )}+\frac{b \sqrt{c} \tan ^{-1}\left (\sqrt{c} x\right )}{c d^2+e^2}-\frac{b \sqrt{c} \tanh ^{-1}\left (\sqrt{c} x\right )}{c d^2-e^2} \]

[Out]

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

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Rubi [A]  time = 0.278595, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {6273, 12, 6725, 635, 207, 260, 203} \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{2 b c d e \log (d+e x)}{c^2 d^4-e^4}-\frac{b c d \log \left (1-c x^2\right )}{2 e \left (c d^2-e^2\right )}+\frac{b c d \log \left (c x^2+1\right )}{2 e \left (c d^2+e^2\right )}+\frac{b \sqrt{c} \tan ^{-1}\left (\sqrt{c} x\right )}{c d^2+e^2}-\frac{b \sqrt{c} \tanh ^{-1}\left (\sqrt{c} x\right )}{c d^2-e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6273

Int[((a_.) + ArcTanh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 - u^2), x],
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m
+ 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 207

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}\left (c x^2\right )}{(d+e x)^2} \, dx &=-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{b \int \frac{2 c x}{(d+e x) \left (1-c^2 x^4\right )} \, dx}{e}\\ &=-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{(2 b c) \int \frac{x}{(d+e x) \left (1-c^2 x^4\right )} \, dx}{e}\\ &=-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{(2 b c) \int \left (-\frac{d e^3}{\left (-c^2 d^4+e^4\right ) (d+e x)}+\frac{e-c d x}{2 \left (c d^2-e^2\right ) \left (-1+c x^2\right )}+\frac{e+c d x}{2 \left (c d^2+e^2\right ) \left (1+c x^2\right )}\right ) \, dx}{e}\\ &=-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{2 b c d e \log (d+e x)}{c^2 d^4-e^4}+\frac{(b c) \int \frac{e-c d x}{-1+c x^2} \, dx}{e \left (c d^2-e^2\right )}+\frac{(b c) \int \frac{e+c d x}{1+c x^2} \, dx}{e \left (c d^2+e^2\right )}\\ &=-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{2 b c d e \log (d+e x)}{c^2 d^4-e^4}+\frac{(b c) \int \frac{1}{-1+c x^2} \, dx}{c d^2-e^2}-\frac{\left (b c^2 d\right ) \int \frac{x}{-1+c x^2} \, dx}{e \left (c d^2-e^2\right )}+\frac{(b c) \int \frac{1}{1+c x^2} \, dx}{c d^2+e^2}+\frac{\left (b c^2 d\right ) \int \frac{x}{1+c x^2} \, dx}{e \left (c d^2+e^2\right )}\\ &=\frac{b \sqrt{c} \tan ^{-1}\left (\sqrt{c} x\right )}{c d^2+e^2}-\frac{b \sqrt{c} \tanh ^{-1}\left (\sqrt{c} x\right )}{c d^2-e^2}-\frac{a+b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}+\frac{2 b c d e \log (d+e x)}{c^2 d^4-e^4}-\frac{b c d \log \left (1-c x^2\right )}{2 e \left (c d^2-e^2\right )}+\frac{b c d \log \left (1+c x^2\right )}{2 e \left (c d^2+e^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.356395, size = 261, normalized size = 1.57 \[ \frac{1}{2} \left (-\frac{2 a}{e (d+e x)}+\frac{b c^2 d^3 \log \left (c x^2+1\right )}{c^2 d^4 e-e^5}-\frac{b c d e \log \left (1-c^2 x^4\right )}{c^2 d^4-e^4}+\frac{4 b c d e \log (d+e x)}{c^2 d^4-e^4}+\frac{b \sqrt{c} \left (c^{3/2} d^3-c d^2 e-e^3\right ) \log \left (1-\sqrt{c} x\right )}{e^5-c^2 d^4 e}+\frac{b \sqrt{c} \left (c^{3/2} d^3+c d^2 e+e^3\right ) \log \left (\sqrt{c} x+1\right )}{e^5-c^2 d^4 e}+\frac{2 b \sqrt{c} \tan ^{-1}\left (\sqrt{c} x\right )}{c d^2+e^2}-\frac{2 b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.034, size = 181, normalized size = 1.1 \begin{align*} -{\frac{a}{ \left ( ex+d \right ) e}}-{\frac{b{\it Artanh} \left ( c{x}^{2} \right ) }{ \left ( ex+d \right ) e}}+{\frac{bcd\ln \left ( c{x}^{2}+1 \right ) }{e \left ( 2\,c{d}^{2}+2\,{e}^{2} \right ) }}+2\,{\frac{b\sqrt{c}\arctan \left ( x\sqrt{c} \right ) }{2\,c{d}^{2}+2\,{e}^{2}}}-{\frac{bcd\ln \left ( c{x}^{2}-1 \right ) }{e \left ( 2\,c{d}^{2}-2\,{e}^{2} \right ) }}-2\,{\frac{b\sqrt{c}{\it Artanh} \left ( x\sqrt{c} \right ) }{2\,c{d}^{2}-2\,{e}^{2}}}+2\,{\frac{becd\ln \left ( ex+d \right ) }{ \left ( c{d}^{2}-{e}^{2} \right ) \left ( c{d}^{2}+{e}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(e*x+d)/e-b/(e*x+d)/e*arctanh(c*x^2)+b/e*c/(2*c*d^2+2*e^2)*d*ln(c*x^2+1)+2*b*c^(1/2)/(2*c*d^2+2*e^2)*arctan
(x*c^(1/2))-b/e*c/(2*c*d^2-2*e^2)*d*ln(c*x^2-1)-2*b*c^(1/2)/(2*c*d^2-2*e^2)*arctanh(x*c^(1/2))+2*b*e*c*d/(c*d^
2-e^2)/(c*d^2+e^2)*ln(e*x+d)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 26.8824, size = 1350, normalized size = 8.13 \begin{align*} \left [-\frac{2 \, a c^{2} d^{4} - 2 \, a e^{4} - 2 \,{\left (b c d^{3} e - b d e^{3} +{\left (b c d^{2} e^{2} - b e^{4}\right )} x\right )} \sqrt{c} \arctan \left (\sqrt{c} x\right ) +{\left (b c d^{3} e + b d e^{3} +{\left (b c d^{2} e^{2} + b e^{4}\right )} x\right )} \sqrt{c} \log \left (\frac{c x^{2} + 2 \, \sqrt{c} x + 1}{c x^{2} - 1}\right ) -{\left (b c^{2} d^{4} - b c d^{2} e^{2} +{\left (b c^{2} d^{3} e - b c d e^{3}\right )} x\right )} \log \left (c x^{2} + 1\right ) +{\left (b c^{2} d^{4} + b c d^{2} e^{2} +{\left (b c^{2} d^{3} e + b c d e^{3}\right )} x\right )} \log \left (c x^{2} - 1\right ) - 4 \,{\left (b c d e^{3} x + b c d^{2} e^{2}\right )} \log \left (e x + d\right ) +{\left (b c^{2} d^{4} - b e^{4}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{2 \,{\left (c^{2} d^{5} e - d e^{5} +{\left (c^{2} d^{4} e^{2} - e^{6}\right )} x\right )}}, -\frac{2 \, a c^{2} d^{4} - 2 \, a e^{4} - 2 \,{\left (b c d^{3} e + b d e^{3} +{\left (b c d^{2} e^{2} + b e^{4}\right )} x\right )} \sqrt{-c} \arctan \left (\sqrt{-c} x\right ) -{\left (b c d^{3} e - b d e^{3} +{\left (b c d^{2} e^{2} - b e^{4}\right )} x\right )} \sqrt{-c} \log \left (\frac{c x^{2} + 2 \, \sqrt{-c} x - 1}{c x^{2} + 1}\right ) -{\left (b c^{2} d^{4} - b c d^{2} e^{2} +{\left (b c^{2} d^{3} e - b c d e^{3}\right )} x\right )} \log \left (c x^{2} + 1\right ) +{\left (b c^{2} d^{4} + b c d^{2} e^{2} +{\left (b c^{2} d^{3} e + b c d e^{3}\right )} x\right )} \log \left (c x^{2} - 1\right ) - 4 \,{\left (b c d e^{3} x + b c d^{2} e^{2}\right )} \log \left (e x + d\right ) +{\left (b c^{2} d^{4} - b e^{4}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{2 \,{\left (c^{2} d^{5} e - d e^{5} +{\left (c^{2} d^{4} e^{2} - e^{6}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.17415, size = 366, normalized size = 2.2 \begin{align*} \frac{1}{2} \,{\left (\frac{c d \log \left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{e^{2}}{{\left (x e + d\right )}^{2}}\right )}{c d^{2} e + e^{3}} - \frac{c d \log \left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} - \frac{e^{2}}{{\left (x e + d\right )}^{2}}\right )}{c d^{2} e - e^{3}} - \frac{e^{\left (-1\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{x e + d} + \frac{2 \, c \arctan \left (\frac{{\left (c d - \frac{c d^{2}}{x e + d} + \frac{e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{-c}}\right )}{{\left (c d^{2} - e^{2}\right )} \sqrt{-c}} + \frac{2 \, \sqrt{c} \arctan \left (\frac{{\left (c d - \frac{c d^{2}}{x e + d} - \frac{e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{c}}\right )}{c d^{2} + e^{2}}\right )} b - \frac{a e^{\left (-1\right )}}{x e + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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