3.825 \(\int \frac {e^{\tanh ^{-1}(a+b x)}}{x^4} \, dx\)

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

[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 = 0.18, 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 = {6163, 99, 151, 12, 93, 208} \[ -\frac {\left (2 a^2+2 a+1\right ) b^3 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \left (1-a^2\right )^{5/2}}-\frac {(a+4) (2 a+1) b^2 \sqrt {-a-b x+1} \sqrt {a+b x+1}}{6 (1-a)^3 (a+1)^2 x}-\frac {(2 a+3) b \sqrt {-a-b x+1} \sqrt {a+b x+1}}{6 (1-a)^2 (a+1) x^2}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{3 (1-a) x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]/x^4,x]

[Out]

-(Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(3*(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 93

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 99

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 151

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] && LtQ[m, -1] && IntegerQ[m]

Rule 208

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

Rule 6163

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 ) \operatorname {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 = 0.27, size = 193, normalized size = 0.91 \[ \frac {\frac {3 \left (2 a^2+2 a+1\right ) b^2 x^2 \left (\sqrt {-a-1} \sqrt {a-1} \sqrt {-((a+b x-1) (a+b x+1))}-2 b x \tanh ^{-1}\left (\frac {\sqrt {-a-1} \sqrt {-a-b x+1}}{\sqrt {a-1} \sqrt {a+b x+1}}\right )\right )}{\sqrt {-a-1} (a-1)^{3/2}}-(4 a+1) b x \sqrt {-a-b x+1} (a+b x+1)^{3/2}+2 (a-1) (a+1) \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{6 \left (a^2-1\right )^2 x^3} \]

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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fricas [A]  time = 1.76, size = 488, normalized size = 2.29 \[ \left [-\frac {3 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \sqrt {-a^{2} + 1} b^{3} x^{3} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, {\left (2 \, a^{6} + {\left (2 \, a^{4} + 9 \, a^{3} + 2 \, a^{2} - 9 \, a - 4\right )} b^{2} x^{2} - 6 \, a^{4} - {\left (2 \, a^{5} + 3 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} + 2 \, a + 3\right )} b x + 6 \, a^{2} - 2\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{12 \, {\left (a^{7} - a^{6} - 3 \, a^{5} + 3 \, a^{4} + 3 \, a^{3} - 3 \, a^{2} - a + 1\right )} x^{3}}, -\frac {3 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \sqrt {a^{2} - 1} b^{3} x^{3} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - {\left (2 \, a^{6} + {\left (2 \, a^{4} + 9 \, a^{3} + 2 \, a^{2} - 9 \, a - 4\right )} b^{2} x^{2} - 6 \, a^{4} - {\left (2 \, a^{5} + 3 \, a^{4} - 4 \, a^{3} - 6 \, a^{2} + 2 \, a + 3\right )} b x + 6 \, a^{2} - 2\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \, {\left (a^{7} - a^{6} - 3 \, a^{5} + 3 \, a^{4} + 3 \, a^{3} - 3 \, a^{2} - a + 1\right )} x^{3}}\right ] \]

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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giac [B]  time = 1.05, size = 1376, normalized size = 6.46 \[ \text {result too large to display} \]

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

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maple [B]  time = 0.04, size = 683, normalized size = 3.21 \[ -\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{3 \left (-a^{2}+1\right ) x^{3}}-\frac {5 a b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 \left (-a^{2}+1\right )^{2} x^{2}}-\frac {5 a^{2} b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 \left (-a^{2}+1\right )^{3} x}-\frac {5 a^{3} b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (-a^{2}+1\right )^{\frac {7}{2}}}-\frac {3 a \,b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (-a^{2}+1\right )^{\frac {5}{2}}}-\frac {2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{3 \left (-a^{2}+1\right )^{2} x}-\frac {a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{3 \left (-a^{2}+1\right ) x^{3}}-\frac {5 a^{2} b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 \left (-a^{2}+1\right )^{2} x^{2}}-\frac {5 a^{3} b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 \left (-a^{2}+1\right )^{3} x}-\frac {5 a^{4} b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (-a^{2}+1\right )^{\frac {7}{2}}}-\frac {3 a^{2} b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {5}{2}}}-\frac {13 a \,b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 \left (-a^{2}+1\right )^{2} x}-\frac {b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{2 \left (-a^{2}+1\right ) x^{2}}-\frac {b^{3} \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{2 \left (-a^{2}+1\right )^{\frac {3}{2}}} \]

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)

[Out]

-1/3/(-a^2+1)/x^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-5/6*a*b/(-a^2+1)^2/x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-5/2*a^2
*b^2/(-a^2+1)^3/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-5/2*a^3*b^3/(-a^2+1)^(7/2)*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)-3/2*a*b^3/(-a^2+1)^(5/2)*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)^2/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/3*a/(-a^2+1)/x^3*(-b^2*x^2-2*
a*b*x-a^2+1)^(1/2)-5/6*a^2*b/(-a^2+1)^2/x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-5/2*a^3*b^2/(-a^2+1)^3/x*(-b^2*x^2-
2*a*b*x-a^2+1)^(1/2)-5/2*a^4*b^3/(-a^2+1)^(7/2)*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)-3*a^2*b^3/(-a^2+1)^(5/2)*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
3/6*a*b^2/(-a^2+1)^2/x*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2*b/(-a^2+1)/x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2*b^
3/(-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 \[ \text {Exception raised: ValueError} \]

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 a-1 positive, negative or zero?

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

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

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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