3.2.74 \(\int \frac {1}{x^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac {b}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b \log (x) (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 44} \begin {gather*} -\frac {b}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b \log (x) (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(-2*b)/(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(2*a^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a + b*x)/(a^
3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*(a + b*x)*Log[x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*(a + b*
x)*Log[a + b*x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 81, normalized size = 0.49 \begin {gather*} \frac {-a \left (2 a^2+9 a b x+6 b^2 x^2\right )-6 b x \log (x) (a+b x)^2+6 b x (a+b x)^2 \log (a+b x)}{2 a^4 x (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(-(a*(2*a^2 + 9*a*b*x + 6*b^2*x^2)) - 6*b*x*(a + b*x)^2*Log[x] + 6*b*x*(a + b*x)^2*Log[a + b*x])/(2*a^4*x*(a +
b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 3.35, size = 778, normalized size = 4.72 \begin {gather*} \frac {-a^{16} b+2 a^{15} b^2 x+61 a^{14} b^3 x^2+864 a^{13} b^4 x^3+7540 a^{12} b^5 x^4+45344 a^{11} b^6 x^5+199056 a^{10} b^7 x^6+658944 a^9 b^8 x^7+1674816 a^8 b^9 x^8+3294720 a^7 b^{10} x^9+5015296 a^6 b^{11} x^{10}+5857280 a^5 b^{12} x^{11}+5151744 a^4 b^{13} x^{12}+3301376 a^3 b^{14} x^{13}+1454080 a^2 b^{15} x^{14}+393216 a b^{16} x^{15}+\sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (-a^{15}-a^{14} b x-60 a^{13} b^2 x^2-804 a^{12} b^3 x^3-6736 a^{11} b^4 x^4-38608 a^{10} b^5 x^5-160448 a^9 b^6 x^6-498496 a^8 b^7 x^7-1176320 a^7 b^8 x^8-2118400 a^6 b^9 x^9-2896896 a^5 b^{10} x^{10}-2960384 a^4 b^{11} x^{11}-2191360 a^3 b^{12} x^{12}-1110016 a^2 b^{13} x^{13}-344064 a b^{14} x^{14}-49152 b^{15} x^{15}\right )+49152 b^{17} x^{16}}{a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (-2 a^{14} b^2-54 a^{13} b^3 x-676 a^{12} b^4 x^2-5200 a^{11} b^5 x^3-27456 a^{10} b^6 x^4-105248 a^9 b^7 x^5-302016 a^8 b^8 x^6-658944 a^7 b^9 x^7-1098240 a^6 b^{10} x^8-1391104 a^5 b^{11} x^9-1317888 a^4 b^{12} x^{10}-905216 a^3 b^{13} x^{11}-425984 a^2 b^{14} x^{12}-122880 a b^{15} x^{13}-16384 b^{16} x^{14}\right )+a^3 \sqrt {b^2} x^2 \left (2 a^{15} b+56 a^{14} b^2 x+730 a^{13} b^3 x^2+5876 a^{12} b^4 x^3+32656 a^{11} b^5 x^4+132704 a^{10} b^6 x^5+407264 a^9 b^7 x^6+960960 a^8 b^8 x^7+1757184 a^7 b^9 x^8+2489344 a^6 b^{10} x^9+2708992 a^5 b^{11} x^{10}+2223104 a^4 b^{12} x^{11}+1331200 a^3 b^{13} x^{12}+548864 a^2 b^{14} x^{13}+139264 a b^{15} x^{14}+16384 b^{16} x^{15}\right )}-\frac {6 b \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )}{a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(-(a^16*b) + 2*a^15*b^2*x + 61*a^14*b^3*x^2 + 864*a^13*b^4*x^3 + 7540*a^12*b^5*x^4 + 45344*a^11*b^6*x^5 + 1990
56*a^10*b^7*x^6 + 658944*a^9*b^8*x^7 + 1674816*a^8*b^9*x^8 + 3294720*a^7*b^10*x^9 + 5015296*a^6*b^11*x^10 + 58
57280*a^5*b^12*x^11 + 5151744*a^4*b^13*x^12 + 3301376*a^3*b^14*x^13 + 1454080*a^2*b^15*x^14 + 393216*a*b^16*x^
15 + 49152*b^17*x^16 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-a^15 - a^14*b*x - 60*a^13*b^2*x^2 - 804*a^12*
b^3*x^3 - 6736*a^11*b^4*x^4 - 38608*a^10*b^5*x^5 - 160448*a^9*b^6*x^6 - 498496*a^8*b^7*x^7 - 1176320*a^7*b^8*x
^8 - 2118400*a^6*b^9*x^9 - 2896896*a^5*b^10*x^10 - 2960384*a^4*b^11*x^11 - 2191360*a^3*b^12*x^12 - 1110016*a^2
*b^13*x^13 - 344064*a*b^14*x^14 - 49152*b^15*x^15))/(a^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-2*a^14*b^2 - 54*a
^13*b^3*x - 676*a^12*b^4*x^2 - 5200*a^11*b^5*x^3 - 27456*a^10*b^6*x^4 - 105248*a^9*b^7*x^5 - 302016*a^8*b^8*x^
6 - 658944*a^7*b^9*x^7 - 1098240*a^6*b^10*x^8 - 1391104*a^5*b^11*x^9 - 1317888*a^4*b^12*x^10 - 905216*a^3*b^13
*x^11 - 425984*a^2*b^14*x^12 - 122880*a*b^15*x^13 - 16384*b^16*x^14) + a^3*Sqrt[b^2]*x^2*(2*a^15*b + 56*a^14*b
^2*x + 730*a^13*b^3*x^2 + 5876*a^12*b^4*x^3 + 32656*a^11*b^5*x^4 + 132704*a^10*b^6*x^5 + 407264*a^9*b^7*x^6 +
960960*a^8*b^8*x^7 + 1757184*a^7*b^9*x^8 + 2489344*a^6*b^10*x^9 + 2708992*a^5*b^11*x^10 + 2223104*a^4*b^12*x^1
1 + 1331200*a^3*b^13*x^12 + 548864*a^2*b^14*x^13 + 139264*a*b^15*x^14 + 16384*b^16*x^15)) - (6*b*ArcTanh[(Sqrt
[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/a^4

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fricas [A]  time = 0.41, size = 109, normalized size = 0.66 \begin {gather*} -\frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \relax (x)}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^
2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.06, size = 117, normalized size = 0.71 \begin {gather*} \frac {\left (-6 b^{3} x^{3} \ln \relax (x )+6 b^{3} x^{3} \ln \left (b x +a \right )-12 a \,b^{2} x^{2} \ln \relax (x )+12 a \,b^{2} x^{2} \ln \left (b x +a \right )-6 a^{2} b x \ln \relax (x )+6 a^{2} b x \ln \left (b x +a \right )-6 a \,b^{2} x^{2}-9 a^{2} b x -2 a^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(6*b^3*x^3*ln(b*x+a)-6*b^3*x^3*ln(x)+12*a*b^2*x^2*ln(b*x+a)-12*a*b^2*x^2*ln(x)+6*a^2*b*x*ln(b*x+a)-6*a^2*b
*x*ln(x)-6*a*b^2*x^2-9*a^2*b*x-2*a^3)*(b*x+a)/x/a^4/((b*x+a)^2)^(3/2)

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maxima [A]  time = 1.36, size = 107, normalized size = 0.65 \begin {gather*} \frac {3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} - \frac {3 \, b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}} - \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x} - \frac {1}{2 \, a^{2} b {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

3*(-1)^(2*a*b*x + 2*a^2)*b*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^4 - 3*b/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3) -
1/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x) - 1/2/(a^2*b*(x + a/b)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

int(1/(x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/(x**2*((a + b*x)**2)**(3/2)), x)

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