3.215 \(\int \frac {x^4}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=76 \[ -\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2}}-\frac {15}{8 b^3 x}+\frac {5}{8 b^2 x \left (b+c x^2\right )}+\frac {1}{4 b x \left (b+c x^2\right )^2} \]

[Out]

-15/8/b^3/x+1/4/b/x/(c*x^2+b)^2+5/8/b^2/x/(c*x^2+b)-15/8*arctan(x*c^(1/2)/b^(1/2))*c^(1/2)/b^(7/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1584, 290, 325, 205} \[ \frac {5}{8 b^2 x \left (b+c x^2\right )}-\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2}}-\frac {15}{8 b^3 x}+\frac {1}{4 b x \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(b*x^2 + c*x^4)^3,x]

[Out]

-15/(8*b^3*x) + 1/(4*b*x*(b + c*x^2)^2) + 5/(8*b^2*x*(b + c*x^2)) - (15*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(
8*b^(7/2))

Rule 205

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

Rule 1584

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

Rubi steps

\begin {align*} \int \frac {x^4}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {1}{x^2 \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{4 b x \left (b+c x^2\right )^2}+\frac {5 \int \frac {1}{x^2 \left (b+c x^2\right )^2} \, dx}{4 b}\\ &=\frac {1}{4 b x \left (b+c x^2\right )^2}+\frac {5}{8 b^2 x \left (b+c x^2\right )}+\frac {15 \int \frac {1}{x^2 \left (b+c x^2\right )} \, dx}{8 b^2}\\ &=-\frac {15}{8 b^3 x}+\frac {1}{4 b x \left (b+c x^2\right )^2}+\frac {5}{8 b^2 x \left (b+c x^2\right )}-\frac {(15 c) \int \frac {1}{b+c x^2} \, dx}{8 b^3}\\ &=-\frac {15}{8 b^3 x}+\frac {1}{4 b x \left (b+c x^2\right )^2}+\frac {5}{8 b^2 x \left (b+c x^2\right )}-\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 68, normalized size = 0.89 \[ -\frac {15 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{7/2}}-\frac {8 b^2+25 b c x^2+15 c^2 x^4}{8 b^3 x \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(b*x^2 + c*x^4)^3,x]

[Out]

-1/8*(8*b^2 + 25*b*c*x^2 + 15*c^2*x^4)/(b^3*x*(b + c*x^2)^2) - (15*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(
7/2))

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 202, normalized size = 2.66 \[ \left [-\frac {30 \, c^{2} x^{4} + 50 \, b c x^{2} - 15 \, {\left (c^{2} x^{5} + 2 \, b c x^{3} + b^{2} x\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right ) + 16 \, b^{2}}{16 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}}, -\frac {15 \, c^{2} x^{4} + 25 \, b c x^{2} + 15 \, {\left (c^{2} x^{5} + 2 \, b c x^{3} + b^{2} x\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right ) + 8 \, b^{2}}{8 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(30*c^2*x^4 + 50*b*c*x^2 - 15*(c^2*x^5 + 2*b*c*x^3 + b^2*x)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) -
b)/(c*x^2 + b)) + 16*b^2)/(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x), -1/8*(15*c^2*x^4 + 25*b*c*x^2 + 15*(c^2*x^5 + 2
*b*c*x^3 + b^2*x)*sqrt(c/b)*arctan(x*sqrt(c/b)) + 8*b^2)/(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 57, normalized size = 0.75 \[ -\frac {15 \, c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} - \frac {7 \, c^{2} x^{3} + 9 \, b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{3}} - \frac {1}{b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8*c*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) - 1/8*(7*c^2*x^3 + 9*b*c*x)/((c*x^2 + b)^2*b^3) - 1/(b^3*x)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 66, normalized size = 0.87 \[ -\frac {7 c^{2} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {9 c x}{8 \left (c \,x^{2}+b \right )^{2} b^{2}}-\frac {15 c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{3}}-\frac {1}{b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/b^3*c^2/(c*x^2+b)^2*x^3-9/8/b^2*c/(c*x^2+b)^2*x-15/8/b^3*c/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)-1/b^3/x

________________________________________________________________________________________

maxima [A]  time = 3.00, size = 71, normalized size = 0.93 \[ -\frac {15 \, c^{2} x^{4} + 25 \, b c x^{2} + 8 \, b^{2}}{8 \, {\left (b^{3} c^{2} x^{5} + 2 \, b^{4} c x^{3} + b^{5} x\right )}} - \frac {15 \, c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(15*c^2*x^4 + 25*b*c*x^2 + 8*b^2)/(b^3*c^2*x^5 + 2*b^4*c*x^3 + b^5*x) - 15/8*c*arctan(c*x/sqrt(b*c))/(sqr
t(b*c)*b^3)

________________________________________________________________________________________

mupad [B]  time = 4.26, size = 66, normalized size = 0.87 \[ -\frac {\frac {1}{b}+\frac {25\,c\,x^2}{8\,b^2}+\frac {15\,c^2\,x^4}{8\,b^3}}{b^2\,x+2\,b\,c\,x^3+c^2\,x^5}-\frac {15\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )}{8\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2 + c*x^4)^3,x)

[Out]

- (1/b + (25*c*x^2)/(8*b^2) + (15*c^2*x^4)/(8*b^3))/(b^2*x + c^2*x^5 + 2*b*c*x^3) - (15*c^(1/2)*atan((c^(1/2)*
x)/b^(1/2)))/(8*b^(7/2))

________________________________________________________________________________________

sympy [A]  time = 0.45, size = 116, normalized size = 1.53 \[ \frac {15 \sqrt {- \frac {c}{b^{7}}} \log {\left (- \frac {b^{4} \sqrt {- \frac {c}{b^{7}}}}{c} + x \right )}}{16} - \frac {15 \sqrt {- \frac {c}{b^{7}}} \log {\left (\frac {b^{4} \sqrt {- \frac {c}{b^{7}}}}{c} + x \right )}}{16} + \frac {- 8 b^{2} - 25 b c x^{2} - 15 c^{2} x^{4}}{8 b^{5} x + 16 b^{4} c x^{3} + 8 b^{3} c^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(-c/b**7)*log(-b**4*sqrt(-c/b**7)/c + x)/16 - 15*sqrt(-c/b**7)*log(b**4*sqrt(-c/b**7)/c + x)/16 + (-8*b
**2 - 25*b*c*x**2 - 15*c**2*x**4)/(8*b**5*x + 16*b**4*c*x**3 + 8*b**3*c**2*x**5)

________________________________________________________________________________________