3.256 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=81 \[ -\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}}-\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^7}-\frac {3 c \sqrt {b x^2+c x^4}}{8 x^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2020, 2008, 206} \[ -\frac {3 c^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}}-\frac {3 c \sqrt {b x^2+c x^4}}{8 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^8,x]

[Out]

(-3*c*Sqrt[b*x^2 + c*x^4])/(8*x^3) - (b*x^2 + c*x^4)^(3/2)/(4*x^7) - (3*c^2*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c
*x^4]])/(8*Sqrt[b])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 80, normalized size = 0.99 \[ -\frac {2 b^2+3 c^2 x^4 \sqrt {\frac {c x^2}{b}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^2}{b}+1}\right )+7 b c x^2+5 c^2 x^4}{8 x^3 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^8,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 164, normalized size = 2.02 \[ \left [\frac {3 \, \sqrt {b} c^{2} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (5 \, b c x^{2} + 2 \, b^{2}\right )}}{16 \, b x^{5}}, \frac {3 \, \sqrt {-b} c^{2} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (5 \, b c x^{2} + 2 \, b^{2}\right )}}{8 \, b x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.24, size = 76, normalized size = 0.94 \[ \frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} - \frac {5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} c^{3} \mathrm {sgn}\relax (x) - 3 \, \sqrt {c x^{2} + b} b c^{3} \mathrm {sgn}\relax (x)}{c^{2} x^{4}}}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 125, normalized size = 1.54 \[ -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} c^{2} x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{2} x^{4}+\left (c \,x^{2}+b \right )^{\frac {5}{2}} c \,x^{2}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \right )}{8 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^8,x)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^8, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^8,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^8, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**8,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**8, x)

________________________________________________________________________________________