3.5.66 \(\int \frac {1}{x (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac {1}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\log (x) \left (a+b x^2\right )}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.08, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1112, 266, 44} \begin {gather*} \frac {1}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\log (x) \left (a+b x^2\right )}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 74, normalized size = 0.50 \begin {gather*} \frac {a \left (3 a+2 b x^2\right )+4 \log (x) \left (a+b x^2\right )^2-2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^3 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [B]  time = 2.25, size = 755, normalized size = 5.14 \begin {gather*} \frac {\left (\sqrt {a^2+2 a b x^2+b^2 x^4}-\sqrt {b^2} x^2\right )^4 \tanh ^{-1}\left (\frac {\sqrt {b^2} x^2}{a}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{a}\right )}{a^3 \left (a^4+4 a^3 b x^2+12 a^2 b^2 x^4-8 a b \sqrt {b^2} x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}-4 a^2 \sqrt {b^2} x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}-8 \left (b^2\right )^{3/2} x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}+16 a b^3 x^6+8 b^4 x^8\right )}+\frac {\sqrt {b^2} \left (-a^{11}-3 a^9 b^2 x^4-50 a^8 b^3 x^6-368 a^7 b^4 x^8-1568 a^6 b^5 x^{10}-4256 a^5 b^6 x^{12}-7616 a^4 b^7 x^{14}-8960 a^3 b^8 x^{16}-6656 a^2 b^9 x^{18}-2816 a b^{10} x^{20}-512 b^{11} x^{22}\right )+\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-a^{10} b+a^9 b^2 x^2+2 a^8 b^3 x^4+48 a^7 b^4 x^6+320 a^6 b^5 x^8+1248 a^5 b^6 x^{10}+3008 a^4 b^7 x^{12}+4608 a^3 b^8 x^{14}+4352 a^2 b^9 x^{16}+2304 a b^{10} x^{18}+512 b^{11} x^{20}\right )}{2 a^2 b \sqrt {b^2} x^4 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-2 a^9 b-34 a^8 b^2 x^2-256 a^7 b^3 x^4-1120 a^6 b^4 x^6-3136 a^5 b^5 x^8-5824 a^4 b^6 x^{10}-7168 a^3 b^7 x^{12}-5632 a^2 b^8 x^{14}-2560 a b^9 x^{16}-512 b^{10} x^{18}\right )+2 a^2 b x^4 \left (2 a^{10} b^2+36 a^9 b^3 x^2+290 a^8 b^4 x^4+1376 a^7 b^5 x^6+4256 a^6 b^6 x^8+8960 a^5 b^7 x^{10}+12992 a^4 b^8 x^{12}+12800 a^3 b^9 x^{14}+8192 a^2 b^{10} x^{16}+3072 a b^{11} x^{18}+512 b^{12} x^{20}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-(a^10*b) + a^9*b^2*x^2 + 2*a^8*b^3*x^4 + 48*a^7*b^4*x^6 + 320*a^6*b^5*x^8 +
1248*a^5*b^6*x^10 + 3008*a^4*b^7*x^12 + 4608*a^3*b^8*x^14 + 4352*a^2*b^9*x^16 + 2304*a*b^10*x^18 + 512*b^11*x
^20) + Sqrt[b^2]*(-a^11 - 3*a^9*b^2*x^4 - 50*a^8*b^3*x^6 - 368*a^7*b^4*x^8 - 1568*a^6*b^5*x^10 - 4256*a^5*b^6*
x^12 - 7616*a^4*b^7*x^14 - 8960*a^3*b^8*x^16 - 6656*a^2*b^9*x^18 - 2816*a*b^10*x^20 - 512*b^11*x^22))/(2*a^2*b
*Sqrt[b^2]*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-2*a^9*b - 34*a^8*b^2*x^2 - 256*a^7*b^3*x^4 - 1120*a^6*b^4*x^6
- 3136*a^5*b^5*x^8 - 5824*a^4*b^6*x^10 - 7168*a^3*b^7*x^12 - 5632*a^2*b^8*x^14 - 2560*a*b^9*x^16 - 512*b^10*x
^18) + 2*a^2*b*x^4*(2*a^10*b^2 + 36*a^9*b^3*x^2 + 290*a^8*b^4*x^4 + 1376*a^7*b^5*x^6 + 4256*a^6*b^6*x^8 + 8960
*a^5*b^7*x^10 + 12992*a^4*b^8*x^12 + 12800*a^3*b^9*x^14 + 8192*a^2*b^10*x^16 + 3072*a*b^11*x^18 + 512*b^12*x^2
0)) + ((-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*ArcTanh[(Sqrt[b^2]*x^2)/a - Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]/a])/(a^3*(a^4 + 4*a^3*b*x^2 + 12*a^2*b^2*x^4 + 16*a*b^3*x^6 + 8*b^4*x^8 - 4*a^2*Sqrt[b^2]*x^2*Sqrt
[a^2 + 2*a*b*x^2 + b^2*x^4] - 8*a*b*Sqrt[b^2]*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4] - 8*(b^2)^(3/2)*x^6*Sqrt[a^2
+ 2*a*b*x^2 + b^2*x^4]))

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fricas [A]  time = 2.92, size = 90, normalized size = 0.61 \begin {gather*} \frac {2 \, a b x^{2} + 3 \, a^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \relax (x)}{4 \, {\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 79, normalized size = 0.54 \begin {gather*} -\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {2 \, a b x^{2} + 3 \, a^{2}}{4 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 107, normalized size = 0.73 \begin {gather*} \frac {\left (4 b^{2} x^{4} \ln \relax (x )-2 b^{2} x^{4} \ln \left (b \,x^{2}+a \right )+8 a b \,x^{2} \ln \relax (x )-4 a b \,x^{2} \ln \left (b \,x^{2}+a \right )+2 a b \,x^{2}+4 a^{2} \ln \relax (x )-2 a^{2} \ln \left (b \,x^{2}+a \right )+3 a^{2}\right ) \left (b \,x^{2}+a \right )}{4 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/4*(4*ln(x)*x^4*b^2-2*b^2*x^4*ln(b*x^2+a)+8*ln(x)*x^2*a*b-4*a*b*x^2*ln(b*x^2+a)+2*a*b*x^2+4*a^2*ln(x)-2*a^2*l
n(b*x^2+a)+3*a^2)*(b*x^2+a)/a^3/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 1.44, size = 57, normalized size = 0.39 \begin {gather*} \frac {2 \, b x^{2} + 3 \, a}{4 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {\log \relax (x)}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

int(1/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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