3.3.19 \(\int \frac {1}{(b x^2+c x^4)^3} \, dx\) [219]

Optimal. Leaf size=100 \[ -\frac {63}{40 b^3 x^5}+\frac {21 c}{8 b^4 x^3}-\frac {63 c^2}{8 b^5 x}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac {63 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}} \]

[Out]

-63/40/b^3/x^5+21/8*c/b^4/x^3-63/8*c^2/b^5/x+1/4/b/x^5/(c*x^2+b)^2+9/8/b^2/x^5/(c*x^2+b)-63/8*c^(5/2)*arctan(x
*c^(1/2)/b^(1/2))/b^(11/2)

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Rubi [A]
time = 0.03, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1607, 296, 331, 211} \begin {gather*} -\frac {63 c^{5/2} \text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}}-\frac {63 c^2}{8 b^5 x}+\frac {21 c}{8 b^4 x^3}-\frac {63}{40 b^3 x^5}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(-3),x]

[Out]

-63/(40*b^3*x^5) + (21*c)/(8*b^4*x^3) - (63*c^2)/(8*b^5*x) + 1/(4*b*x^5*(b + c*x^2)^2) + 9/(8*b^2*x^5*(b + c*x
^2)) - (63*c^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(11/2))

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 (b x^2+c x^4\right )^3} \, dx &=\int \frac {1}{x^6 \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9 \int \frac {1}{x^6 \left (b+c x^2\right )^2} \, dx}{4 b}\\ &=\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}+\frac {63 \int \frac {1}{x^6 \left (b+c x^2\right )} \, dx}{8 b^2}\\ &=-\frac {63}{40 b^3 x^5}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac {(63 c) \int \frac {1}{x^4 \left (b+c x^2\right )} \, dx}{8 b^3}\\ &=-\frac {63}{40 b^3 x^5}+\frac {21 c}{8 b^4 x^3}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}+\frac {\left (63 c^2\right ) \int \frac {1}{x^2 \left (b+c x^2\right )} \, dx}{8 b^4}\\ &=-\frac {63}{40 b^3 x^5}+\frac {21 c}{8 b^4 x^3}-\frac {63 c^2}{8 b^5 x}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac {\left (63 c^3\right ) \int \frac {1}{b+c x^2} \, dx}{8 b^5}\\ &=-\frac {63}{40 b^3 x^5}+\frac {21 c}{8 b^4 x^3}-\frac {63 c^2}{8 b^5 x}+\frac {1}{4 b x^5 \left (b+c x^2\right )^2}+\frac {9}{8 b^2 x^5 \left (b+c x^2\right )}-\frac {63 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 90, normalized size = 0.90 \begin {gather*} -\frac {8 b^4-24 b^3 c x^2+168 b^2 c^2 x^4+525 b c^3 x^6+315 c^4 x^8}{40 b^5 x^5 \left (b+c x^2\right )^2}-\frac {63 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(-3),x]

[Out]

-1/40*(8*b^4 - 24*b^3*c*x^2 + 168*b^2*c^2*x^4 + 525*b*c^3*x^6 + 315*c^4*x^8)/(b^5*x^5*(b + c*x^2)^2) - (63*c^(
5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(11/2))

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Maple [A]
time = 0.10, size = 75, normalized size = 0.75

method	result	size
default	\(-\frac {c^{3} \left (\frac {\frac {15}{8} c \,x^{3}+\frac {17}{8} b x}{\left (c \,x^{2}+b \right )^{2}}+\frac {63 \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{5}}-\frac {1}{5 b^{3} x^{5}}+\frac {c}{b^{4} x^{3}}-\frac {6 c^{2}}{b^{5} x}\)	\(75\)
risch	\(\frac {-\frac {63 c^{4} x^{8}}{8 b^{5}}-\frac {105 c^{3} x^{6}}{8 b^{4}}-\frac {21 c^{2} x^{4}}{5 b^{3}}+\frac {3 c \,x^{2}}{5 b^{2}}-\frac {1}{5 b}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {63 \left (\munderset {\textit {\_R} =\RootOf \left (b^{11} \textit {\_Z}^{2}+c^{5}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{11}+2 c^{5}\right ) x +b^{6} c^{2} \textit {\_R} \right )\right )}{16}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

-1/b^5*c^3*((15/8*c*x^3+17/8*b*x)/(c*x^2+b)^2+63/8/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))-1/5/b^3/x^5+c/b^4/x^3-
6*c^2/b^5/x

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Maxima [A]
time = 0.49, size = 97, normalized size = 0.97 \begin {gather*} -\frac {315 \, c^{4} x^{8} + 525 \, b c^{3} x^{6} + 168 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} + 8 \, b^{4}}{40 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}} - \frac {63 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/40*(315*c^4*x^8 + 525*b*c^3*x^6 + 168*b^2*c^2*x^4 - 24*b^3*c*x^2 + 8*b^4)/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*
x^5) - 63/8*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^5)

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Fricas [A]
time = 0.43, size = 264, normalized size = 2.64 \begin {gather*} \left [-\frac {630 \, c^{4} x^{8} + 1050 \, b c^{3} x^{6} + 336 \, b^{2} c^{2} x^{4} - 48 \, b^{3} c x^{2} + 16 \, b^{4} - 315 \, {\left (c^{4} x^{9} + 2 \, b c^{3} x^{7} + b^{2} c^{2} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{80 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}, -\frac {315 \, c^{4} x^{8} + 525 \, b c^{3} x^{6} + 168 \, b^{2} c^{2} x^{4} - 24 \, b^{3} c x^{2} + 8 \, b^{4} + 315 \, {\left (c^{4} x^{9} + 2 \, b c^{3} x^{7} + b^{2} c^{2} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{40 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/80*(630*c^4*x^8 + 1050*b*c^3*x^6 + 336*b^2*c^2*x^4 - 48*b^3*c*x^2 + 16*b^4 - 315*(c^4*x^9 + 2*b*c^3*x^7 +
b^2*c^2*x^5)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)
, -1/40*(315*c^4*x^8 + 525*b*c^3*x^6 + 168*b^2*c^2*x^4 - 24*b^3*c*x^2 + 8*b^4 + 315*(c^4*x^9 + 2*b*c^3*x^7 + b
^2*c^2*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5)]

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Sympy [A]
time = 0.27, size = 150, normalized size = 1.50 \begin {gather*} \frac {63 \sqrt {- \frac {c^{5}}{b^{11}}} \log {\left (- \frac {b^{6} \sqrt {- \frac {c^{5}}{b^{11}}}}{c^{3}} + x \right )}}{16} - \frac {63 \sqrt {- \frac {c^{5}}{b^{11}}} \log {\left (\frac {b^{6} \sqrt {- \frac {c^{5}}{b^{11}}}}{c^{3}} + x \right )}}{16} + \frac {- 8 b^{4} + 24 b^{3} c x^{2} - 168 b^{2} c^{2} x^{4} - 525 b c^{3} x^{6} - 315 c^{4} x^{8}}{40 b^{7} x^{5} + 80 b^{6} c x^{7} + 40 b^{5} c^{2} x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

63*sqrt(-c**5/b**11)*log(-b**6*sqrt(-c**5/b**11)/c**3 + x)/16 - 63*sqrt(-c**5/b**11)*log(b**6*sqrt(-c**5/b**11
)/c**3 + x)/16 + (-8*b**4 + 24*b**3*c*x**2 - 168*b**2*c**2*x**4 - 525*b*c**3*x**6 - 315*c**4*x**8)/(40*b**7*x*
*5 + 80*b**6*c*x**7 + 40*b**5*c**2*x**9)

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Giac [A]
time = 5.10, size = 80, normalized size = 0.80 \begin {gather*} -\frac {63 \, c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{5}} - \frac {15 \, c^{4} x^{3} + 17 \, b c^{3} x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{5}} - \frac {30 \, c^{2} x^{4} - 5 \, b c x^{2} + b^{2}}{5 \, b^{5} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-63/8*c^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^5) - 1/8*(15*c^4*x^3 + 17*b*c^3*x)/((c*x^2 + b)^2*b^5) - 1/5*(30*
c^2*x^4 - 5*b*c*x^2 + b^2)/(b^5*x^5)

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Mupad [B]
time = 4.24, size = 92, normalized size = 0.92 \begin {gather*} -\frac {\frac {1}{5\,b}-\frac {3\,c\,x^2}{5\,b^2}+\frac {21\,c^2\,x^4}{5\,b^3}+\frac {105\,c^3\,x^6}{8\,b^4}+\frac {63\,c^4\,x^8}{8\,b^5}}{b^2\,x^5+2\,b\,c\,x^7+c^2\,x^9}-\frac {63\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2 + c*x^4)^3,x)

[Out]

- (1/(5*b) - (3*c*x^2)/(5*b^2) + (21*c^2*x^4)/(5*b^3) + (105*c^3*x^6)/(8*b^4) + (63*c^4*x^8)/(8*b^5))/(b^2*x^5
+ c^2*x^9 + 2*b*c*x^7) - (63*c^(5/2)*atan((c^(1/2)*x)/b^(1/2)))/(8*b^(11/2))

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