3.9.25 \(\int \frac {e^{\tanh ^{-1}(a+b x)}}{x^4} \, dx\) [825]

Optimal. Leaf size=213 \[ -\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}-\frac {(3+2 a) b \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^2 (1+a) x^2}-\frac {(4+a) (1+2 a) b^2 \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^3 (1+a)^2 x}-\frac {\left (1+2 a+2 a^2\right ) b^3 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1-a) \left (1-a^2\right )^{5/2}} \]

[Out]

-(2*a^2+2*a+1)*b^3*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b*x-a+1)^(1/2))/(1-a)/(-a^2+1)^(5/2)-1/3*
(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/(1-a)/x^3-1/6*(3+2*a)*b*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/(1-a)^2/(1+a)/x^2-1/
6*(4+a)*(1+2*a)*b^2*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/(1-a)^3/(1+a)^2/x

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Rubi [A]
time = 0.14, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6298, 101, 156, 12, 95, 214} \begin {gather*} -\frac {\left (2 a^2+2 a+1\right ) b^3 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \left (1-a^2\right )^{5/2}}-\frac {(a+4) (2 a+1) b^2 \sqrt {-a-b x+1} \sqrt {a+b x+1}}{6 (1-a)^3 (a+1)^2 x}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{3 (1-a) x^3}-\frac {(2 a+3) b \sqrt {-a-b x+1} \sqrt {a+b x+1}}{6 (1-a)^2 (a+1) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]/x^4,x]

[Out]

-1/3*(Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/((1 - a)*x^3) - ((3 + 2*a)*b*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(
6*(1 - a)^2*(1 + a)*x^2) - ((4 + a)*(1 + 2*a)*b^2*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(6*(1 - a)^3*(1 + a)^2*
x) - ((1 + 2*a + 2*a^2)*b^3*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/((1 - a)
*(1 - a^2)^(5/2))

Rule 12

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

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 6298

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a+b x)}}{x^4} \, dx &=\int \frac {\sqrt {1+a+b x}}{x^4 \sqrt {1-a-b x}} \, dx\\ &=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}+\frac {\int \frac {(3+2 a) b+2 b^2 x}{x^3 \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{3 (1-a)}\\ &=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}-\frac {(3+2 a) b \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^2 (1+a) x^2}-\frac {\int \frac {-(4+a) (1+2 a) b^2-(3+2 a) b^3 x}{x^2 \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{6 (1-a)^2 (1+a)}\\ &=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}-\frac {(3+2 a) b \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^2 (1+a) x^2}-\frac {(4+a) (1+2 a) b^2 \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^3 (1+a)^2 x}+\frac {\int \frac {3 \left (1+2 a+2 a^2\right ) b^3}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{6 (1-a)^3 (1+a)^2}\\ &=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}-\frac {(3+2 a) b \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^2 (1+a) x^2}-\frac {(4+a) (1+2 a) b^2 \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^3 (1+a)^2 x}+\frac {\left (\left (1+2 a+2 a^2\right ) b^3\right ) \int \frac {1}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{2 (1-a)^3 (1+a)^2}\\ &=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}-\frac {(3+2 a) b \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^2 (1+a) x^2}-\frac {(4+a) (1+2 a) b^2 \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^3 (1+a)^2 x}+\frac {\left (\left (1+2 a+2 a^2\right ) b^3\right ) \text {Subst}\left (\int \frac {1}{-1-a-(-1+a) x^2} \, dx,x,\frac {\sqrt {1+a+b x}}{\sqrt {1-a-b x}}\right )}{(1-a)^3 (1+a)^2}\\ &=-\frac {\sqrt {1-a-b x} \sqrt {1+a+b x}}{3 (1-a) x^3}-\frac {(3+2 a) b \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^2 (1+a) x^2}-\frac {(4+a) (1+2 a) b^2 \sqrt {1-a-b x} \sqrt {1+a+b x}}{6 (1-a)^3 (1+a)^2 x}-\frac {\left (1+2 a+2 a^2\right ) b^3 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1-a)^3 (1+a)^2 \sqrt {1-a^2}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 193, normalized size = 0.91 \begin {gather*} \frac {2 (-1+a) (1+a) \sqrt {1-a-b x} (1+a+b x)^{3/2}-(1+4 a) b x \sqrt {1-a-b x} (1+a+b x)^{3/2}+\frac {3 \left (1+2 a+2 a^2\right ) b^2 x^2 \left (\sqrt {-1-a} \sqrt {-1+a} \sqrt {-((-1+a+b x) (1+a+b x))}-2 b x \tanh ^{-1}\left (\frac {\sqrt {-1-a} \sqrt {1-a-b x}}{\sqrt {-1+a} \sqrt {1+a+b x}}\right )\right )}{\sqrt {-1-a} (-1+a)^{3/2}}}{6 \left (-1+a^2\right )^2 x^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a + b*x]/x^4,x]

[Out]

(2*(-1 + a)*(1 + a)*Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2) - (1 + 4*a)*b*x*Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2
) + (3*(1 + 2*a + 2*a^2)*b^2*x^2*(Sqrt[-1 - a]*Sqrt[-1 + a]*Sqrt[-((-1 + a + b*x)*(1 + a + b*x))] - 2*b*x*ArcT
anh[(Sqrt[-1 - a]*Sqrt[1 - a - b*x])/(Sqrt[-1 + a]*Sqrt[1 + a + b*x])]))/(Sqrt[-1 - a]*(-1 + a)^(3/2)))/(6*(-1
+ a^2)^2*x^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(599\) vs. \(2(185)=370\).
time = 0.07, size = 600, normalized size = 2.82

method	result	size
risch	\(-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}-1\right ) \left (2 a^{2} b^{2} x^{2}-2 a^{3} b x +9 a \,b^{2} x^{2}+2 a^{4}-3 a^{2} b x +4 b^{2} x^{2}+2 a b x -4 a^{2}+3 b x +2\right )}{6 \left (-1+a \right ) x^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \left (a^{2}-1\right )^{2}}+\frac {b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right ) a^{2}}{\left (a^{2}-1\right )^{2} \left (-1+a \right ) \sqrt {-a^{2}+1}}+\frac {b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right ) a}{\left (a^{2}-1\right )^{2} \left (-1+a \right ) \sqrt {-a^{2}+1}}+\frac {b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (a^{2}-1\right )^{2} \left (-1+a \right ) \sqrt {-a^{2}+1}}\)	\(351\)
default	\(\left (1+a \right ) \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{3 \left (-a^{2}+1\right ) x^{3}}+\frac {5 b a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 \left (-a^{2}+1\right ) x^{2}}+\frac {3 b a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) x}-\frac {b a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{2 \left (-a^{2}+1\right )}-\frac {b^{2} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{3 \left (-a^{2}+1\right )}+\frac {2 b^{2} \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) x}-\frac {b a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{3 \left (-a^{2}+1\right )}\right )+b \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 \left (-a^{2}+1\right ) x^{2}}+\frac {3 b a \left (-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) x}-\frac {b a \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{2 \left (-a^{2}+1\right )}-\frac {b^{2} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (-a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(600\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

(1+a)*(-1/3/(-a^2+1)/x^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+5/3*b*a/(-a^2+1)*(-1/2/(-a^2+1)/x^2*(-b^2*x^2-2*a*b*x-
a^2+1)^(1/2)+3/2*b*a/(-a^2+1)*(-1/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-b*a/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*
a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x))-1/2*b^2/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2
+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x))+2/3*b^2/(-a^2+1)*(-1/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-b
*a/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)))+b*(-1/2/(-a^2+1)/
x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+3/2*b*a/(-a^2+1)*(-1/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-b*a/(-a^2+1)
^(3/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x))-1/2*b^2/(-a^2+1)^(3/2)*ln((-2
*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is

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Fricas [A]
time = 0.44, size = 488, normalized size = 2.29 \begin {gather*} \left [-\frac {3 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \sqrt {-a^{2} + 1} b^{3} x^{3} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, {\left (2 \, a^{6} + {\left (2 \, a^{4} + 9 \, a^{3} + 2 \, a^{2} - 9 \, a - 4\right )} b^{2} x^{2} - 6 \, a^{4} - {\left (2 \, a^{5} + 3 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} + 2 \, a + 3\right )} b x + 6 \, a^{2} - 2\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{12 \, {\left (a^{7} - a^{6} - 3 \, a^{5} + 3 \, a^{4} + 3 \, a^{3} - 3 \, a^{2} - a + 1\right )} x^{3}}, -\frac {3 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \sqrt {a^{2} - 1} b^{3} x^{3} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - {\left (2 \, a^{6} + {\left (2 \, a^{4} + 9 \, a^{3} + 2 \, a^{2} - 9 \, a - 4\right )} b^{2} x^{2} - 6 \, a^{4} - {\left (2 \, a^{5} + 3 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} + 2 \, a + 3\right )} b x + 6 \, a^{2} - 2\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \, {\left (a^{7} - a^{6} - 3 \, a^{5} + 3 \, a^{4} + 3 \, a^{3} - 3 \, a^{2} - a + 1\right )} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[-1/12*(3*(2*a^2 + 2*a + 1)*sqrt(-a^2 + 1)*b^3*x^3*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 2*sqrt
(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) - 2*(2*a^6 + (2*a^4 + 9*a^3
+ 2*a^2 - 9*a - 4)*b^2*x^2 - 6*a^4 - (2*a^5 + 3*a^4 - 4*a^3 - 6*a^2 + 2*a + 3)*b*x + 6*a^2 - 2)*sqrt(-b^2*x^2
- 2*a*b*x - a^2 + 1))/((a^7 - a^6 - 3*a^5 + 3*a^4 + 3*a^3 - 3*a^2 - a + 1)*x^3), -1/6*(3*(2*a^2 + 2*a + 1)*sqr
t(a^2 - 1)*b^3*x^3*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b^2*x^
2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) - (2*a^6 + (2*a^4 + 9*a^3 + 2*a^2 - 9*a - 4)*b^2*x^2 - 6*a^4 - (2*a^5
+ 3*a^4 - 4*a^3 - 6*a^2 + 2*a + 3)*b*x + 6*a^2 - 2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a^7 - a^6 - 3*a^5 +
3*a^4 + 3*a^3 - 3*a^2 - a + 1)*x^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b x + 1}{x^{4} \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)/(1-(b*x+a)**2)**(1/2)/x**4,x)

[Out]

Integral((a + b*x + 1)/(x**4*sqrt(-(a + b*x - 1)*(a + b*x + 1))), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1664 vs. \(2 (176) = 352\).
time = 0.43, size = 1664, normalized size = 7.81 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

-(2*a^2*b^4 + 2*a*b^4 + b^4)*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt
(a^2 - 1))/((a^5*abs(b) - a^4*abs(b) - 2*a^3*abs(b) + 2*a^2*abs(b) + a*abs(b) - abs(b))*sqrt(a^2 - 1)) + 1/3*(
12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^7*b^4/(b^2*x + a*b)^2 + 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^
2 + 1)*abs(b) + b)^4*a^7*b^4/(b^2*x + a*b)^4 + 6*a^7*b^4 - 24*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*
a^6*b^4/(b^2*x + a*b) + 24*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^6*b^4/(b^2*x + a*b)^2 - 36*(sqr
t(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^6*b^4/(b^2*x + a*b)^3 + 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)
*abs(b) + b)^4*a^6*b^4/(b^2*x + a*b)^4 - 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^5*a^6*b^4/(b^2*x +
a*b)^5 + 12*a^6*b^4 - 57*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^5*b^4/(b^2*x + a*b) + 36*(sqrt(-b^
2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^5*b^4/(b^2*x + a*b)^2 - 72*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(
b) + b)^3*a^5*b^4/(b^2*x + a*b)^3 + 30*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^5*b^4/(b^2*x + a*b)
^4 - 15*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^5*a^5*b^4/(b^2*x + a*b)^5 - 2*a^5*b^4 + 84*(sqrt(-b^2*
x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^4*b^4/(b^2*x + a*b)^2 - 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b)
+ b)^3*a^4*b^4/(b^2*x + a*b)^3 + 51*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^4*b^4/(b^2*x + a*b)^4
+ 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^5*a^4*b^4/(b^2*x + a*b)^5 - 3*a^4*b^4 + 12*(sqrt(-b^2*x^
2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^3*b^4/(b^2*x + a*b) - 30*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^
3*a^3*b^4/(b^2*x + a*b)^3 - 18*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^3*b^4/(b^2*x + a*b)^4 + 6*(
sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^5*a^3*b^4/(b^2*x + a*b)^5 + 2*a^3*b^4 - 6*(sqrt(-b^2*x^2 - 2*a*
b*x - a^2 + 1)*abs(b) + b)*a^2*b^4/(b^2*x + a*b) - 18*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^2*b^
4/(b^2*x + a*b)^2 - 4*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^2*b^4/(b^2*x + a*b)^3 - 18*(sqrt(-b^
2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^4*a^2*b^4/(b^2*x + a*b)^4 - 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b
) + b)^5*a^2*b^4/(b^2*x + a*b)^5 + 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a*b^4/(b^2*x + a*b)^2
+ 12*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a*b^4/(b^2*x + a*b)^3 + 12*(sqrt(-b^2*x^2 - 2*a*b*x - a
^2 + 1)*abs(b) + b)^4*a*b^4/(b^2*x + a*b)^4 - 8*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*b^4/(b^2*x +
a*b)^3)/((a^8*abs(b) - a^7*abs(b) - 2*a^6*abs(b) + 2*a^5*abs(b) + a^4*abs(b) - a^3*abs(b))*((sqrt(-b^2*x^2 -
2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 + a - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^
2*x + a*b))^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,x+1}{x^4\,\sqrt {1-{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + 1)/(x^4*(1 - (a + b*x)^2)^(1/2)),x)

[Out]

int((a + b*x + 1)/(x^4*(1 - (a + b*x)^2)^(1/2)), x)

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