\(\int \frac {(b x^2+c x^4)^{3/2}}{x^{10}} \, dx\) [257]

Optimal result

Integrand size = 19, antiderivative size = 109 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=-\frac {c \sqrt {b x^2+c x^4}}{8 x^5}-\frac {c^2 \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}+\frac {c^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}} \]

[Out]

-1/6*(c*x^4+b*x^2)^(3/2)/x^9+1/16*c^3*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(3/2)-1/8*c*(c*x^4+b*x^2)^(1/2)
/x^5-1/16*c^2*(c*x^4+b*x^2)^(1/2)/b/x^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2045, 2050, 2033, 212} \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\frac {c^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}}-\frac {c^2 \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}-\frac {c \sqrt {b x^2+c x^4}}{8 x^5} \]

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^10,x]

[Out]

-1/8*(c*Sqrt[b*x^2 + c*x^4])/x^5 - (c^2*Sqrt[b*x^2 + c*x^4])/(16*b*x^3) - (b*x^2 + c*x^4)^(3/2)/(6*x^9) + (c^3
*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(16*b^(3/2))

Rule 212

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

Rule 2033

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 2045

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]

Rule 2050

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}+\frac {1}{2} c \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx \\ & = -\frac {c \sqrt {b x^2+c x^4}}{8 x^5}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}+\frac {1}{8} c^2 \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx \\ & = -\frac {c \sqrt {b x^2+c x^4}}{8 x^5}-\frac {c^2 \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}-\frac {c^3 \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{16 b} \\ & = -\frac {c \sqrt {b x^2+c x^4}}{8 x^5}-\frac {c^2 \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}+\frac {c^3 \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{16 b} \\ & = -\frac {c \sqrt {b x^2+c x^4}}{8 x^5}-\frac {c^2 \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}+\frac {c^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-\sqrt {b} \sqrt {b+c x^2} \left (8 b^2+14 b c x^2+3 c^2 x^4\right )+3 c^3 x^6 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{48 b^{3/2} x^7 \sqrt {b+c x^2}} \]

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^10,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-(Sqrt[b]*Sqrt[b + c*x^2]*(8*b^2 + 14*b*c*x^2 + 3*c^2*x^4)) + 3*c^3*x^6*ArcTanh[Sqrt[b
+ c*x^2]/Sqrt[b]]))/(48*b^(3/2)*x^7*Sqrt[b + c*x^2])

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {\left (3 c^{2} x^{4}+14 b c \,x^{2}+8 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 x^{7} b}+\frac {c^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 b^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\)	\(100\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{3} x^{6}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{3} x^{6}+\left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{2} x^{4}-3 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b c \,x^{2}-8 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2}\right )}{48 x^{9} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3}}\)	\(145\)

[In]

int((c*x^4+b*x^2)^(3/2)/x^10,x,method=_RETURNVERBOSE)

[Out]

-1/48*(3*c^2*x^4+14*b*c*x^2+8*b^2)/x^7/b*(x^2*(c*x^2+b))^(1/2)+1/16*c^3/b^(3/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^(1
/2))/x)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.70 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\left [\frac {3 \, \sqrt {b} c^{3} x^{7} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, {\left (3 \, b c^{2} x^{4} + 14 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, b^{2} x^{7}}, -\frac {3 \, \sqrt {-b} c^{3} x^{7} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (3 \, b c^{2} x^{4} + 14 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, b^{2} x^{7}}\right ] \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^10,x, algorithm="fricas")

[Out]

[1/96*(3*sqrt(b)*c^3*x^7*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*(3*b*c^2*x^4 + 14*b^2*c
*x^2 + 8*b^3)*sqrt(c*x^4 + b*x^2))/(b^2*x^7), -1/48*(3*sqrt(-b)*c^3*x^7*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c
*x^3 + b*x)) + (3*b*c^2*x^4 + 14*b^2*c*x^2 + 8*b^3)*sqrt(c*x^4 + b*x^2))/(b^2*x^7)]

Sympy [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{10}}\, dx \]

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**10,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**10, x)

Maxima [F]

\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{10}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^10,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^10, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=-\frac {\frac {3 \, c^{4} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b} b} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} c^{4} \mathrm {sgn}\left (x\right ) + 8 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} b c^{4} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {c x^{2} + b} b^{2} c^{4} \mathrm {sgn}\left (x\right )}{b c^{3} x^{6}}}{48 \, c} \]

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^10,x, algorithm="giac")

[Out]

-1/48*(3*c^4*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b) + (3*(c*x^2 + b)^(5/2)*c^4*sgn(x) + 8*(c*x^2
+ b)^(3/2)*b*c^4*sgn(x) - 3*sqrt(c*x^2 + b)*b^2*c^4*sgn(x))/(b*c^3*x^6))/c

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{10}} \,d x \]

[In]

int((b*x^2 + c*x^4)^(3/2)/x^10,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^10, x)