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]

(-a-b*arctanh(c*x^2))/e/(e*x+d)+2*b*c*d*e*ln(e*x+d)/(c^2*d^4-e^4)-1/2*b*c*d*ln(-c*x^2+1)/e/(c*d^2-e^2)+1/2*b*c
*d*ln(c*x^2+1)/e/(c*d^2+e^2)+b*arctan(x*c^(1/2))*c^(1/2)/(c*d^2+e^2)-b*arctanh(x*c^(1/2))*c^(1/2)/(c*d^2-e^2)

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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 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]

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 = 0.36, size = 261, normalized size = 1.57 \[ \frac {1}{2} \left (-\frac {2 a}{e (d+e x)}-\frac {b c d e \log \left (1-c^2 x^4\right )}{c^2 d^4-e^4}+\frac {4 b c d e \log (d+e x)}{c^2 d^4-e^4}+\frac {b c^2 d^3 \log \left (c x^2+1\right )}{c^2 d^4 e-e^5}+\frac {b \sqrt {c} \left (c^{3/2} d^3-c d^2 e-e^3\right ) \log \left (1-\sqrt {c} x\right )}{e^5-c^2 d^4 e}+\frac {b \sqrt {c} \left (c^{3/2} d^3+c d^2 e+e^3\right ) \log \left (\sqrt {c} x+1\right )}{e^5-c^2 d^4 e}+\frac {2 b \sqrt {c} \tan ^{-1}\left (\sqrt {c} x\right )}{c d^2+e^2}-\frac {2 b \tanh ^{-1}\left (c x^2\right )}{e (d+e x)}\right ) \]

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

________________________________________________________________________________________

fricas [B]  time = 8.20, size = 634, normalized size = 3.82 \[ \left [-\frac {2 \, a c^{2} d^{4} - 2 \, a e^{4} - 2 \, {\left (b c d^{3} e - b d e^{3} + {\left (b c d^{2} e^{2} - b e^{4}\right )} x\right )} \sqrt {c} \arctan \left (\sqrt {c} x\right ) + {\left (b c d^{3} e + b d e^{3} + {\left (b c d^{2} e^{2} + b e^{4}\right )} x\right )} \sqrt {c} \log \left (\frac {c x^{2} + 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right ) - {\left (b c^{2} d^{4} - b c d^{2} e^{2} + {\left (b c^{2} d^{3} e - b c d e^{3}\right )} x\right )} \log \left (c x^{2} + 1\right ) + {\left (b c^{2} d^{4} + b c d^{2} e^{2} + {\left (b c^{2} d^{3} e + b c d e^{3}\right )} x\right )} \log \left (c x^{2} - 1\right ) - 4 \, {\left (b c d e^{3} x + b c d^{2} e^{2}\right )} \log \left (e x + d\right ) + {\left (b c^{2} d^{4} - b e^{4}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{2 \, {\left (c^{2} d^{5} e - d e^{5} + {\left (c^{2} d^{4} e^{2} - e^{6}\right )} x\right )}}, -\frac {2 \, a c^{2} d^{4} - 2 \, a e^{4} - 2 \, {\left (b c d^{3} e + b d e^{3} + {\left (b c d^{2} e^{2} + b e^{4}\right )} x\right )} \sqrt {-c} \arctan \left (\sqrt {-c} x\right ) - {\left (b c d^{3} e - b d e^{3} + {\left (b c d^{2} e^{2} - b e^{4}\right )} x\right )} \sqrt {-c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right ) - {\left (b c^{2} d^{4} - b c d^{2} e^{2} + {\left (b c^{2} d^{3} e - b c d e^{3}\right )} x\right )} \log \left (c x^{2} + 1\right ) + {\left (b c^{2} d^{4} + b c d^{2} e^{2} + {\left (b c^{2} d^{3} e + b c d e^{3}\right )} x\right )} \log \left (c x^{2} - 1\right ) - 4 \, {\left (b c d e^{3} x + b c d^{2} e^{2}\right )} \log \left (e x + d\right ) + {\left (b c^{2} d^{4} - b e^{4}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{2 \, {\left (c^{2} d^{5} e - d e^{5} + {\left (c^{2} d^{4} e^{2} - e^{6}\right )} x\right )}}\right ] \]

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)]

________________________________________________________________________________________

giac [A]  time = 0.23, size = 271, normalized size = 1.63 \[ \frac {1}{2} \, {\left (\frac {c d \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {e^{2}}{{\left (x e + d\right )}^{2}}\right )}{c d^{2} e + e^{3}} - \frac {c d \log \left (c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} - \frac {e^{2}}{{\left (x e + d\right )}^{2}}\right )}{c d^{2} e - e^{3}} - \frac {e^{\left (-1\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{x e + d} + \frac {2 \, c \arctan \left (\frac {{\left (c d - \frac {c d^{2}}{x e + d} + \frac {e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {-c}}\right )}{{\left (c d^{2} - e^{2}\right )} \sqrt {-c}} + \frac {2 \, \sqrt {c} \arctan \left (\frac {{\left (c d - \frac {c d^{2}}{x e + d} - \frac {e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt {c}}\right )}{c d^{2} + e^{2}}\right )} b - \frac {a e^{\left (-1\right )}}{x e + d} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 181, normalized size = 1.09 \[ -\frac {a}{\left (e x +d \right ) e}-\frac {b \arctanh \left (c \,x^{2}\right )}{\left (e x +d \right ) e}+\frac {b c d \ln \left (c \,x^{2}+1\right )}{e \left (2 c \,d^{2}+2 e^{2}\right )}+\frac {2 b \sqrt {c}\, \arctan \left (x \sqrt {c}\right )}{2 c \,d^{2}+2 e^{2}}+\frac {2 b e c d \ln \left (e x +d \right )}{\left (c \,d^{2}-e^{2}\right ) \left (c \,d^{2}+e^{2}\right )}-\frac {b c d \ln \left (c \,x^{2}-1\right )}{e \left (2 c \,d^{2}-2 e^{2}\right )}-\frac {2 b \sqrt {c}\, \arctanh \left (x \sqrt {c}\right )}{2 c \,d^{2}-2 e^{2}} \]

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))+2*b*e*c*d/(c*d^2-e^2)/(c*d^2+e^2)*ln(e*x+d)-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))

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 172, normalized size = 1.04 \[ \frac {1}{2} \, {\left ({\left (\frac {4 \, d e \log \left (e x + d\right )}{c^{2} d^{4} - e^{4}} + \frac {d \log \left (c x^{2} + 1\right )}{c d^{2} e + e^{3}} - \frac {d \log \left (c x^{2} - 1\right )}{c d^{2} e - e^{3}} + \frac {2 \, \arctan \left (\sqrt {c} x\right )}{{\left (c d^{2} + e^{2}\right )} \sqrt {c}} + \frac {\log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{{\left (c d^{2} - e^{2}\right )} \sqrt {c}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{e^{2} x + d e}\right )} b - \frac {a}{e^{2} x + d e} \]

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]

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

________________________________________________________________________________________

mupad [B]  time = 2.60, size = 727, normalized size = 4.38 \[ \frac {\ln \left (\frac {16\,b^4\,c^{10}\,x}{e}-\frac {\left (b\,\sqrt {-c}\,e+b\,c\,d\right )\,\left (8\,b^3\,c^9\,e-\frac {\left (b\,\sqrt {-c}\,e+b\,c\,d\right )\,\left (\frac {\left (b\,\sqrt {-c}\,e+b\,c\,d\right )\,\left (\frac {8\,x\,\left (4\,b\,c^{11}\,d^5\,e^2+52\,b\,c^9\,d\,e^6\right )}{e^2}-\frac {\left (b\,\sqrt {-c}\,e+b\,c\,d\right )\,\left (384\,c^8\,d\,e^6+\frac {8\,x\,\left (24\,c^{10}\,d^4\,e^5+40\,c^8\,e^9\right )}{e^2}+128\,c^{10}\,d^5\,e^2\right )}{2\,\left (c\,d^2\,e+e^3\right )}+320\,b\,c^9\,d^2\,e^3\right )}{2\,\left (c\,d^2\,e+e^3\right )}-64\,b^2\,c^{10}\,d^2\,e\,x\right )}{2\,\left (c\,d^2\,e+e^3\right )}+\frac {8\,b^3\,c^{11}\,d^3\,x}{e^2}\right )}{2\,\left (c\,d^2\,e+e^3\right )}\right )\,\left (b\,\sqrt {-c}\,e+b\,c\,d\right )}{2\,\left (c\,d^2\,e+e^3\right )}-\frac {a}{x\,e^2+d\,e}-\frac {\ln \left (\frac {\left (b\,\sqrt {-c}\,e-b\,c\,d\right )\,\left (8\,b^3\,c^9\,e-\frac {\left (b\,\sqrt {-c}\,e-b\,c\,d\right )\,\left (\frac {\left (b\,\sqrt {-c}\,e-b\,c\,d\right )\,\left (\frac {8\,x\,\left (4\,b\,c^{11}\,d^5\,e^2+52\,b\,c^9\,d\,e^6\right )}{e^2}+\frac {\left (b\,\sqrt {-c}\,e-b\,c\,d\right )\,\left (384\,c^8\,d\,e^6+\frac {8\,x\,\left (24\,c^{10}\,d^4\,e^5+40\,c^8\,e^9\right )}{e^2}+128\,c^{10}\,d^5\,e^2\right )}{2\,\left (c\,d^2\,e+e^3\right )}+320\,b\,c^9\,d^2\,e^3\right )}{2\,\left (c\,d^2\,e+e^3\right )}+64\,b^2\,c^{10}\,d^2\,e\,x\right )}{2\,\left (c\,d^2\,e+e^3\right )}+\frac {8\,b^3\,c^{11}\,d^3\,x}{e^2}\right )}{2\,\left (c\,d^2\,e+e^3\right )}+\frac {16\,b^4\,c^{10}\,x}{e}\right )\,\left (b\,\sqrt {-c}\,e-b\,c\,d\right )}{2\,\left (c\,d^2\,e+e^3\right )}-\frac {b\,\ln \left (c\,x^2+1\right )}{2\,e\,\left (d+e\,x\right )}-\frac {b\,\sqrt {c}\,\ln \left (\sqrt {c}\,x-1\right )}{2\,\left (e^2+\sqrt {c}\,d\,e\right )}+\frac {b\,\sqrt {c}\,\ln \left (\sqrt {c}\,x+1\right )}{2\,\left (e^2-\sqrt {c}\,d\,e\right )}+\frac {b\,\ln \left (1-c\,x^2\right )}{e\,\left (2\,d+2\,e\,x\right )}-\frac {2\,b\,c\,d\,e\,\ln \left (d+e\,x\right )}{e^4-c^2\,d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((16*b^4*c^10*x)/e - ((b*(-c)^(1/2)*e + b*c*d)*(8*b^3*c^9*e - ((b*(-c)^(1/2)*e + b*c*d)*(((b*(-c)^(1/2)*e
+ b*c*d)*((8*x*(4*b*c^11*d^5*e^2 + 52*b*c^9*d*e^6))/e^2 - ((b*(-c)^(1/2)*e + b*c*d)*(384*c^8*d*e^6 + (8*x*(40*
c^8*e^9 + 24*c^10*d^4*e^5))/e^2 + 128*c^10*d^5*e^2))/(2*(e^3 + c*d^2*e)) + 320*b*c^9*d^2*e^3))/(2*(e^3 + c*d^2
*e)) - 64*b^2*c^10*d^2*e*x))/(2*(e^3 + c*d^2*e)) + (8*b^3*c^11*d^3*x)/e^2))/(2*(e^3 + c*d^2*e)))*(b*(-c)^(1/2)
*e + b*c*d))/(2*(e^3 + c*d^2*e)) - a/(d*e + e^2*x) - (log(((b*(-c)^(1/2)*e - b*c*d)*(8*b^3*c^9*e - ((b*(-c)^(1
/2)*e - b*c*d)*(((b*(-c)^(1/2)*e - b*c*d)*((8*x*(4*b*c^11*d^5*e^2 + 52*b*c^9*d*e^6))/e^2 + ((b*(-c)^(1/2)*e -
b*c*d)*(384*c^8*d*e^6 + (8*x*(40*c^8*e^9 + 24*c^10*d^4*e^5))/e^2 + 128*c^10*d^5*e^2))/(2*(e^3 + c*d^2*e)) + 32
0*b*c^9*d^2*e^3))/(2*(e^3 + c*d^2*e)) + 64*b^2*c^10*d^2*e*x))/(2*(e^3 + c*d^2*e)) + (8*b^3*c^11*d^3*x)/e^2))/(
2*(e^3 + c*d^2*e)) + (16*b^4*c^10*x)/e)*(b*(-c)^(1/2)*e - b*c*d))/(2*(e^3 + c*d^2*e)) - (b*log(c*x^2 + 1))/(2*
e*(d + e*x)) - (b*c^(1/2)*log(c^(1/2)*x - 1))/(2*(e^2 + c^(1/2)*d*e)) + (b*c^(1/2)*log(c^(1/2)*x + 1))/(2*(e^2
- c^(1/2)*d*e)) + (b*log(1 - c*x^2))/(e*(2*d + 2*e*x)) - (2*b*c*d*e*log(d + e*x))/(e^4 - c^2*d^4)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________