3.1.23 \(\int \frac {1}{(a x+b x^3)^2} \, dx\) [23]

Optimal. Leaf size=57 \[ -\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a+b x^2\right )}-\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1607, 296, 331, 211} \begin {gather*} -\frac {3 \sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^3)^(-2),x]

[Out]

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

Rule 211

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

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a x+b x^3\right )^2} \, dx &=\int \frac {1}{x^2 \left (a+b x^2\right )^2} \, dx\\ &=\frac {1}{2 a x \left (a+b x^2\right )}+\frac {3 \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{2 a}\\ &=-\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a+b x^2\right )}-\frac {(3 b) \int \frac {1}{a+b x^2} \, dx}{2 a^2}\\ &=-\frac {3}{2 a^2 x}+\frac {1}{2 a x \left (a+b x^2\right )}-\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 54, normalized size = 0.95 \begin {gather*} -\frac {1}{a^2 x}-\frac {b x}{2 a^2 \left (a+b x^2\right )}-\frac {3 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^3)^(-2),x]

[Out]

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

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Maple [A]
time = 0.35, size = 45, normalized size = 0.79

method	result	size
default	\(-\frac {b \left (\frac {x}{2 b \,x^{2}+2 a}+\frac {3 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}-\frac {1}{a^{2} x}\)	\(45\)
risch	\(\frac {-\frac {3 b \,x^{2}}{2 a^{2}}-\frac {1}{a}}{x \left (b \,x^{2}+a \right )}+\frac {3 \left (\munderset {\textit {\_R} =\RootOf \left (a^{5} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (3 a^{5} \textit {\_R}^{2}+2 b \right ) x +a^{3} \textit {\_R} \right )\right )}{4}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 49, normalized size = 0.86 \begin {gather*} -\frac {3 \, b x^{2} + 2 \, a}{2 \, {\left (a^{2} b x^{3} + a^{3} x\right )}} - \frac {3 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*b*x^2 + 2*a)/(a^2*b*x^3 + a^3*x) - 3/2*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2)

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Fricas [A]
time = 1.82, size = 136, normalized size = 2.39 \begin {gather*} \left [-\frac {6 \, b x^{2} - 3 \, {\left (b x^{3} + a x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 4 \, a}{4 \, {\left (a^{2} b x^{3} + a^{3} x\right )}}, -\frac {3 \, b x^{2} + 3 \, {\left (b x^{3} + a x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 2 \, a}{2 \, {\left (a^{2} b x^{3} + a^{3} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*b*x^2 - 3*(b*x^3 + a*x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 4*a)/(a^2*b*x^3
+ a^3*x), -1/2*(3*b*x^2 + 3*(b*x^3 + a*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 2*a)/(a^2*b*x^3 + a^3*x)]

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Sympy [A]
time = 0.14, size = 92, normalized size = 1.61 \begin {gather*} \frac {3 \sqrt {- \frac {b}{a^{5}}} \log {\left (- \frac {a^{3} \sqrt {- \frac {b}{a^{5}}}}{b} + x \right )}}{4} - \frac {3 \sqrt {- \frac {b}{a^{5}}} \log {\left (\frac {a^{3} \sqrt {- \frac {b}{a^{5}}}}{b} + x \right )}}{4} + \frac {- 2 a - 3 b x^{2}}{2 a^{3} x + 2 a^{2} b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.17, size = 47, normalized size = 0.82 \begin {gather*} -\frac {3 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} - \frac {3 \, b x^{2} + 2 \, a}{2 \, {\left (b x^{3} + a x\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) - 1/2*(3*b*x^2 + 2*a)/((b*x^3 + a*x)*a^2)

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Mupad [B]
time = 4.98, size = 44, normalized size = 0.77 \begin {gather*} -\frac {\frac {1}{a}+\frac {3\,b\,x^2}{2\,a^2}}{b\,x^3+a\,x}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b*x^3)^2,x)

[Out]

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

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