3.83 \(\int \frac {1}{x^2 (c+(a+b x)^2)} \, dx\)

Optimal. Leaf size=79 \[ -\frac {2 a b \log (x)}{\left (a^2+c\right )^2}+\frac {a b \log \left ((a+b x)^2+c\right )}{\left (a^2+c\right )^2}+\frac {b \left (a^2-c\right ) \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2+c\right )^2}-\frac {1}{x \left (a^2+c\right )} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {371, 710, 801, 635, 203, 260} \[ -\frac {2 a b \log (x)}{\left (a^2+c\right )^2}+\frac {a b \log \left ((a+b x)^2+c\right )}{\left (a^2+c\right )^2}+\frac {b \left (a^2-c\right ) \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )}{\sqrt {c} \left (a^2+c\right )^2}-\frac {1}{x \left (a^2+c\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(c + (a + b*x)^2)),x]

[Out]

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

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 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 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

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 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 81, normalized size = 1.03 \[ \frac {b x \left (a^2-c\right ) \tan ^{-1}\left (\frac {a+b x}{\sqrt {c}}\right )-\sqrt {c} \left (-a b x \log \left (a^2+2 a b x+b^2 x^2+c\right )+a^2+2 a b x \log (x)+c\right )}{\sqrt {c} x \left (a^2+c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(c + (a + b*x)^2)),x]

[Out]

(b*(a^2 - c)*x*ArcTan[(a + b*x)/Sqrt[c]] - Sqrt[c]*(a^2 + c + 2*a*b*x*Log[x] - a*b*x*Log[a^2 + c + 2*a*b*x + b
^2*x^2]))/(Sqrt[c]*(a^2 + c)^2*x)

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fricas [A]  time = 0.92, size = 229, normalized size = 2.90 \[ \left [\frac {2 \, a b c x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right ) - 4 \, a b c x \log \relax (x) + {\left (a^{2} b - b c\right )} \sqrt {-c} x \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 2 \, {\left (b x + a\right )} \sqrt {-c} - c}{b^{2} x^{2} + 2 \, a b x + a^{2} + c}\right ) - 2 \, a^{2} c - 2 \, c^{2}}{2 \, {\left (a^{4} c + 2 \, a^{2} c^{2} + c^{3}\right )} x}, \frac {a b c x \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right ) - 2 \, a b c x \log \relax (x) + {\left (a^{2} b - b c\right )} \sqrt {c} x \arctan \left (\frac {b x + a}{\sqrt {c}}\right ) - a^{2} c - c^{2}}{{\left (a^{4} c + 2 \, a^{2} c^{2} + c^{3}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 117, normalized size = 1.48 \[ \frac {a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{a^{4} + 2 \, a^{2} c + c^{2}} - \frac {2 \, a b \log \left ({\left | x \right |}\right )}{a^{4} + 2 \, a^{2} c + c^{2}} + \frac {{\left (a^{2} b^{2} - b^{2} c\right )} \arctan \left (\frac {b x + a}{\sqrt {c}}\right )}{{\left (a^{4} + 2 \, a^{2} c + c^{2}\right )} b \sqrt {c}} - \frac {1}{{\left (a^{2} + c\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b*log(b^2*x^2 + 2*a*b*x + a^2 + c)/(a^4 + 2*a^2*c + c^2) - 2*a*b*log(abs(x))/(a^4 + 2*a^2*c + c^2) + (a^2*b^
2 - b^2*c)*arctan((b*x + a)/sqrt(c))/((a^4 + 2*a^2*c + c^2)*b*sqrt(c)) - 1/((a^2 + c)*x)

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maple [A]  time = 0.01, size = 123, normalized size = 1.56 \[ \frac {a^{2} b \arctan \left (\frac {2 b^{2} x +2 b a}{2 b \sqrt {c}}\right )}{\left (a^{2}+c \right )^{2} \sqrt {c}}-\frac {2 a b \ln \relax (x )}{\left (a^{2}+c \right )^{2}}+\frac {a b \ln \left (b^{2} x^{2}+2 a b x +a^{2}+c \right )}{\left (a^{2}+c \right )^{2}}-\frac {b \sqrt {c}\, \arctan \left (\frac {2 b^{2} x +2 b a}{2 b \sqrt {c}}\right )}{\left (a^{2}+c \right )^{2}}-\frac {1}{\left (a^{2}+c \right ) x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.59, size = 123, normalized size = 1.56 \[ \frac {a b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + c\right )}{a^{4} + 2 \, a^{2} c + c^{2}} - \frac {2 \, a b \log \relax (x)}{a^{4} + 2 \, a^{2} c + c^{2}} + \frac {{\left (a^{2} b^{2} - b^{2} c\right )} \arctan \left (\frac {b^{2} x + a b}{b \sqrt {c}}\right )}{{\left (a^{4} + 2 \, a^{2} c + c^{2}\right )} b \sqrt {c}} - \frac {1}{{\left (a^{2} + c\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.58, size = 425, normalized size = 5.38 \[ \frac {\ln \left ({\left (-c\right )}^{13/2}-35\,a^2\,{\left (-c\right )}^{11/2}+34\,a^4\,{\left (-c\right )}^{9/2}+34\,a^6\,{\left (-c\right )}^{7/2}-35\,a^8\,{\left (-c\right )}^{5/2}+a^{10}\,{\left (-c\right )}^{3/2}+a\,c^6-a^{11}\,c+35\,a^3\,c^5+34\,a^5\,c^4-34\,a^7\,c^3-35\,a^9\,c^2+b\,c^6\,x-a^{10}\,b\,c\,x+35\,a^2\,b\,c^5\,x+34\,a^4\,b\,c^4\,x-34\,a^6\,b\,c^3\,x-35\,a^8\,b\,c^2\,x\right )\,\left (b\,{\left (-c\right )}^{3/2}+2\,a\,b\,c+a^2\,b\,\sqrt {-c}\right )}{2\,\left (a^4\,c+2\,a^2\,c^2+c^3\right )}-\frac {1}{x\,\left (a^2+c\right )}-\frac {\ln \left ({\left (-c\right )}^{13/2}-35\,a^2\,{\left (-c\right )}^{11/2}+34\,a^4\,{\left (-c\right )}^{9/2}+34\,a^6\,{\left (-c\right )}^{7/2}-35\,a^8\,{\left (-c\right )}^{5/2}+a^{10}\,{\left (-c\right )}^{3/2}-a\,c^6+a^{11}\,c-35\,a^3\,c^5-34\,a^5\,c^4+34\,a^7\,c^3+35\,a^9\,c^2-b\,c^6\,x+a^{10}\,b\,c\,x-35\,a^2\,b\,c^5\,x-34\,a^4\,b\,c^4\,x+34\,a^6\,b\,c^3\,x+35\,a^8\,b\,c^2\,x\right )\,\left (b\,{\left (-c\right )}^{3/2}-2\,a\,b\,c+a^2\,b\,\sqrt {-c}\right )}{2\,\left (a^4\,c+2\,a^2\,c^2+c^3\right )}-\frac {2\,a\,b\,\ln \relax (x)}{{\left (a^2+c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(c + (a + b*x)^2)),x)

[Out]

(log((-c)^(13/2) - 35*a^2*(-c)^(11/2) + 34*a^4*(-c)^(9/2) + 34*a^6*(-c)^(7/2) - 35*a^8*(-c)^(5/2) + a^10*(-c)^
(3/2) + a*c^6 - a^11*c + 35*a^3*c^5 + 34*a^5*c^4 - 34*a^7*c^3 - 35*a^9*c^2 + b*c^6*x - a^10*b*c*x + 35*a^2*b*c
^5*x + 34*a^4*b*c^4*x - 34*a^6*b*c^3*x - 35*a^8*b*c^2*x)*(b*(-c)^(3/2) + 2*a*b*c + a^2*b*(-c)^(1/2)))/(2*(a^4*
c + c^3 + 2*a^2*c^2)) - 1/(x*(c + a^2)) - (log((-c)^(13/2) - 35*a^2*(-c)^(11/2) + 34*a^4*(-c)^(9/2) + 34*a^6*(
-c)^(7/2) - 35*a^8*(-c)^(5/2) + a^10*(-c)^(3/2) - a*c^6 + a^11*c - 35*a^3*c^5 - 34*a^5*c^4 + 34*a^7*c^3 + 35*a
^9*c^2 - b*c^6*x + a^10*b*c*x - 35*a^2*b*c^5*x - 34*a^4*b*c^4*x + 34*a^6*b*c^3*x + 35*a^8*b*c^2*x)*(b*(-c)^(3/
2) - 2*a*b*c + a^2*b*(-c)^(1/2)))/(2*(a^4*c + c^3 + 2*a^2*c^2)) - (2*a*b*log(x))/(c + a^2)^2

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sympy [B]  time = 11.12, size = 1620, normalized size = 20.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*log(x + (-16*a**13*b**2*c/(a**2 + c)**4 + 48*a**11*b**2*c**2/(a**2 + c)**4 + 352*a**9*b**2*c**3/(a**2 +
c)**4 - 20*a**9*b**2*c/(a**2 + c)**2 + 608*a**7*b**2*c**4/(a**2 + c)**4 - 64*a**7*b**2*c**2/(a**2 + c)**2 + 4
32*a**5*b**2*c**5/(a**2 + c)**4 - 72*a**5*b**2*c**3/(a**2 + c)**2 + 36*a**5*b**2*c + 112*a**3*b**2*c**6/(a**2
+ c)**4 - 32*a**3*b**2*c**4/(a**2 + c)**2 - 88*a**3*b**2*c**2 - 4*a*b**2*c**5/(a**2 + c)**2 + 4*a*b**2*c**3)/(
a**6*b**3 + 33*a**4*b**3*c - 33*a**2*b**3*c**2 - b**3*c**3))/(a**2 + c)**2 + (a*b/(a**2 + c)**2 - b*sqrt(-c)*(
a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))*log(x + (-4*a**11*c*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*
(a**4 + 2*a**2*c + c**2)))**2 + 12*a**9*c**2*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c
+ c**2)))**2 + 10*a**8*b*c*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 88*a**
7*c**3*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 32*a**6*b*c**2*(a*b/(a*
*2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 36*a**5*b**2*c + 152*a**5*c**4*(a*b/(a**2
+ c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 36*a**4*b*c**3*(a*b/(a**2 + c)**2 - b*sq
rt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) - 88*a**3*b**2*c**2 + 108*a**3*c**5*(a*b/(a**2 + c)**2 - b*s
qrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 16*a**2*b*c**4*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 -
c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 4*a*b**2*c**3 + 28*a*c**6*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*
c*(a**4 + 2*a**2*c + c**2)))**2 + 2*b*c**5*(a*b/(a**2 + c)**2 - b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c +
c**2))))/(a**6*b**3 + 33*a**4*b**3*c - 33*a**2*b**3*c**2 - b**3*c**3)) + (a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2
- c)/(2*c*(a**4 + 2*a**2*c + c**2)))*log(x + (-4*a**11*c*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**
4 + 2*a**2*c + c**2)))**2 + 12*a**9*c**2*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c*
*2)))**2 + 10*a**8*b*c*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 88*a**7*c*
*3*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 32*a**6*b*c**2*(a*b/(a**2 +
c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) + 36*a**5*b**2*c + 152*a**5*c**4*(a*b/(a**2 + c
)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 36*a**4*b*c**3*(a*b/(a**2 + c)**2 + b*sqrt(-
c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2))) - 88*a**3*b**2*c**2 + 108*a**3*c**5*(a*b/(a**2 + c)**2 + b*sqrt(
-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2)))**2 + 16*a**2*b*c**4*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/
(2*c*(a**4 + 2*a**2*c + c**2))) + 4*a*b**2*c**3 + 28*a*c**6*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a
**4 + 2*a**2*c + c**2)))**2 + 2*b*c**5*(a*b/(a**2 + c)**2 + b*sqrt(-c)*(a**2 - c)/(2*c*(a**4 + 2*a**2*c + c**2
))))/(a**6*b**3 + 33*a**4*b**3*c - 33*a**2*b**3*c**2 - b**3*c**3)) - 1/(x*(a**2 + c))

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