3.251 \(\int \frac {1}{x^2 (a-b x^2)^5} \, dx\)

Optimal. Leaf size=118 \[ \frac {315 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{11/2}}-\frac {315}{128 a^5 x}+\frac {105}{128 a^4 x \left (a-b x^2\right )}+\frac {21}{64 a^3 x \left (a-b x^2\right )^2}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {1}{8 a x \left (a-b x^2\right )^4} \]

[Out]

-315/128/a^5/x+1/8/a/x/(-b*x^2+a)^4+3/16/a^2/x/(-b*x^2+a)^3+21/64/a^3/x/(-b*x^2+a)^2+105/128/a^4/x/(-b*x^2+a)+
315/128*arctanh(x*b^(1/2)/a^(1/2))*b^(1/2)/a^(11/2)

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Rubi [A]  time = 0.05, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {290, 325, 208} \[ \frac {105}{128 a^4 x \left (a-b x^2\right )}+\frac {21}{64 a^3 x \left (a-b x^2\right )^2}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {315 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{11/2}}-\frac {315}{128 a^5 x}+\frac {1}{8 a x \left (a-b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a - b*x^2)^5),x]

[Out]

-315/(128*a^5*x) + 1/(8*a*x*(a - b*x^2)^4) + 3/(16*a^2*x*(a - b*x^2)^3) + 21/(64*a^3*x*(a - b*x^2)^2) + 105/(1
28*a^4*x*(a - b*x^2)) + (315*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(11/2))

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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

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]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (a-b x^2\right )^5} \, dx &=\frac {1}{8 a x \left (a-b x^2\right )^4}+\frac {9 \int \frac {1}{x^2 \left (a-b x^2\right )^4} \, dx}{8 a}\\ &=\frac {1}{8 a x \left (a-b x^2\right )^4}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {21 \int \frac {1}{x^2 \left (a-b x^2\right )^3} \, dx}{16 a^2}\\ &=\frac {1}{8 a x \left (a-b x^2\right )^4}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {21}{64 a^3 x \left (a-b x^2\right )^2}+\frac {105 \int \frac {1}{x^2 \left (a-b x^2\right )^2} \, dx}{64 a^3}\\ &=\frac {1}{8 a x \left (a-b x^2\right )^4}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {21}{64 a^3 x \left (a-b x^2\right )^2}+\frac {105}{128 a^4 x \left (a-b x^2\right )}+\frac {315 \int \frac {1}{x^2 \left (a-b x^2\right )} \, dx}{128 a^4}\\ &=-\frac {315}{128 a^5 x}+\frac {1}{8 a x \left (a-b x^2\right )^4}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {21}{64 a^3 x \left (a-b x^2\right )^2}+\frac {105}{128 a^4 x \left (a-b x^2\right )}+\frac {(315 b) \int \frac {1}{a-b x^2} \, dx}{128 a^5}\\ &=-\frac {315}{128 a^5 x}+\frac {1}{8 a x \left (a-b x^2\right )^4}+\frac {3}{16 a^2 x \left (a-b x^2\right )^3}+\frac {21}{64 a^3 x \left (a-b x^2\right )^2}+\frac {105}{128 a^4 x \left (a-b x^2\right )}+\frac {315 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 92, normalized size = 0.78 \[ \frac {\frac {\sqrt {a} \left (-128 a^4+837 a^3 b x^2-1533 a^2 b^2 x^4+1155 a b^3 x^6-315 b^4 x^8\right )}{x \left (a-b x^2\right )^4}+315 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a - b*x^2)^5),x]

[Out]

((Sqrt[a]*(-128*a^4 + 837*a^3*b*x^2 - 1533*a^2*b^2*x^4 + 1155*a*b^3*x^6 - 315*b^4*x^8))/(x*(a - b*x^2)^4) + 31
5*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(11/2))

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fricas [A]  time = 0.82, size = 334, normalized size = 2.83 \[ \left [-\frac {630 \, b^{4} x^{8} - 2310 \, a b^{3} x^{6} + 3066 \, a^{2} b^{2} x^{4} - 1674 \, a^{3} b x^{2} + 256 \, a^{4} - 315 \, {\left (b^{4} x^{9} - 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} - 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt {\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {\frac {b}{a}} + a}{b x^{2} - a}\right )}{256 \, {\left (a^{5} b^{4} x^{9} - 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} - 4 \, a^{8} b x^{3} + a^{9} x\right )}}, -\frac {315 \, b^{4} x^{8} - 1155 \, a b^{3} x^{6} + 1533 \, a^{2} b^{2} x^{4} - 837 \, a^{3} b x^{2} + 128 \, a^{4} + 315 \, {\left (b^{4} x^{9} - 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} - 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt {-\frac {b}{a}} \arctan \left (x \sqrt {-\frac {b}{a}}\right )}{128 \, {\left (a^{5} b^{4} x^{9} - 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} - 4 \, a^{8} b x^{3} + a^{9} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^5,x, algorithm="fricas")

[Out]

[-1/256*(630*b^4*x^8 - 2310*a*b^3*x^6 + 3066*a^2*b^2*x^4 - 1674*a^3*b*x^2 + 256*a^4 - 315*(b^4*x^9 - 4*a*b^3*x
^7 + 6*a^2*b^2*x^5 - 4*a^3*b*x^3 + a^4*x)*sqrt(b/a)*log((b*x^2 + 2*a*x*sqrt(b/a) + a)/(b*x^2 - a)))/(a^5*b^4*x
^9 - 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 - 4*a^8*b*x^3 + a^9*x), -1/128*(315*b^4*x^8 - 1155*a*b^3*x^6 + 1533*a^2*b^2
*x^4 - 837*a^3*b*x^2 + 128*a^4 + 315*(b^4*x^9 - 4*a*b^3*x^7 + 6*a^2*b^2*x^5 - 4*a^3*b*x^3 + a^4*x)*sqrt(-b/a)*
arctan(x*sqrt(-b/a)))/(a^5*b^4*x^9 - 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 - 4*a^8*b*x^3 + a^9*x)]

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giac [A]  time = 0.63, size = 83, normalized size = 0.70 \[ -\frac {315 \, b \arctan \left (\frac {b x}{\sqrt {-a b}}\right )}{128 \, \sqrt {-a b} a^{5}} - \frac {1}{a^{5} x} - \frac {187 \, b^{4} x^{7} - 643 \, a b^{3} x^{5} + 765 \, a^{2} b^{2} x^{3} - 325 \, a^{3} b x}{128 \, {\left (b x^{2} - a\right )}^{4} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^5,x, algorithm="giac")

[Out]

-315/128*b*arctan(b*x/sqrt(-a*b))/(sqrt(-a*b)*a^5) - 1/(a^5*x) - 1/128*(187*b^4*x^7 - 643*a*b^3*x^5 + 765*a^2*
b^2*x^3 - 325*a^3*b*x)/((b*x^2 - a)^4*a^5)

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maple [A]  time = 0.02, size = 78, normalized size = 0.66 \[ -\frac {\left (-\frac {315 \arctanh \left (\frac {b x}{\sqrt {a b}}\right )}{128 \sqrt {a b}}+\frac {\frac {187}{128} b^{3} x^{7}-\frac {643}{128} a \,b^{2} x^{5}+\frac {765}{128} a^{2} b \,x^{3}-\frac {325}{128} a^{3} x}{\left (b \,x^{2}-a \right )^{4}}\right ) b}{a^{5}}-\frac {1}{a^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(-b*x^2+a)^5,x)

[Out]

-1/a^5/x-1/a^5*b*((187/128*b^3*x^7-643/128*a*b^2*x^5+765/128*a^2*b*x^3-325/128*a^3*x)/(b*x^2-a)^4-315/128/(a*b
)^(1/2)*arctanh(1/(a*b)^(1/2)*b*x))

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maxima [A]  time = 2.90, size = 130, normalized size = 1.10 \[ -\frac {315 \, b^{4} x^{8} - 1155 \, a b^{3} x^{6} + 1533 \, a^{2} b^{2} x^{4} - 837 \, a^{3} b x^{2} + 128 \, a^{4}}{128 \, {\left (a^{5} b^{4} x^{9} - 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} - 4 \, a^{8} b x^{3} + a^{9} x\right )}} - \frac {315 \, b \log \left (\frac {b x - \sqrt {a b}}{b x + \sqrt {a b}}\right )}{256 \, \sqrt {a b} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-b*x^2+a)^5,x, algorithm="maxima")

[Out]

-1/128*(315*b^4*x^8 - 1155*a*b^3*x^6 + 1533*a^2*b^2*x^4 - 837*a^3*b*x^2 + 128*a^4)/(a^5*b^4*x^9 - 4*a^6*b^3*x^
7 + 6*a^7*b^2*x^5 - 4*a^8*b*x^3 + a^9*x) - 315/256*b*log((b*x - sqrt(a*b))/(b*x + sqrt(a*b)))/(sqrt(a*b)*a^5)

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mupad [B]  time = 5.17, size = 110, normalized size = 0.93 \[ \frac {315\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{128\,a^{11/2}}-\frac {\frac {1}{a}-\frac {837\,b\,x^2}{128\,a^2}+\frac {1533\,b^2\,x^4}{128\,a^3}-\frac {1155\,b^3\,x^6}{128\,a^4}+\frac {315\,b^4\,x^8}{128\,a^5}}{a^4\,x-4\,a^3\,b\,x^3+6\,a^2\,b^2\,x^5-4\,a\,b^3\,x^7+b^4\,x^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a - b*x^2)^5),x)

[Out]

(315*b^(1/2)*atanh((b^(1/2)*x)/a^(1/2)))/(128*a^(11/2)) - (1/a - (837*b*x^2)/(128*a^2) + (1533*b^2*x^4)/(128*a
^3) - (1155*b^3*x^6)/(128*a^4) + (315*b^4*x^8)/(128*a^5))/(a^4*x + b^4*x^9 - 4*a^3*b*x^3 - 4*a*b^3*x^7 + 6*a^2
*b^2*x^5)

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sympy [A]  time = 0.66, size = 155, normalized size = 1.31 \[ - \frac {315 \sqrt {\frac {b}{a^{11}}} \log {\left (- \frac {a^{6} \sqrt {\frac {b}{a^{11}}}}{b} + x \right )}}{256} + \frac {315 \sqrt {\frac {b}{a^{11}}} \log {\left (\frac {a^{6} \sqrt {\frac {b}{a^{11}}}}{b} + x \right )}}{256} - \frac {128 a^{4} - 837 a^{3} b x^{2} + 1533 a^{2} b^{2} x^{4} - 1155 a b^{3} x^{6} + 315 b^{4} x^{8}}{128 a^{9} x - 512 a^{8} b x^{3} + 768 a^{7} b^{2} x^{5} - 512 a^{6} b^{3} x^{7} + 128 a^{5} b^{4} x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(-b*x**2+a)**5,x)

[Out]

-315*sqrt(b/a**11)*log(-a**6*sqrt(b/a**11)/b + x)/256 + 315*sqrt(b/a**11)*log(a**6*sqrt(b/a**11)/b + x)/256 -
(128*a**4 - 837*a**3*b*x**2 + 1533*a**2*b**2*x**4 - 1155*a*b**3*x**6 + 315*b**4*x**8)/(128*a**9*x - 512*a**8*b
*x**3 + 768*a**7*b**2*x**5 - 512*a**6*b**3*x**7 + 128*a**5*b**4*x**9)

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