3.3.56 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^8} \, dx\) [256]

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

[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

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Rubi [A]
time = 0.06, 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 = {2045, 2033, 212} \begin {gather*} -\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} \end {gather*}

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 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]

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 ) \text {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*}

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Mathematica [A]
time = 0.10, size = 92, normalized size = 1.14 \begin {gather*} -\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b} \sqrt {b+c x^2} \left (2 b+5 c x^2\right )+3 c^2 x^4 \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{8 \sqrt {b} x^5 \sqrt {b+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 125, normalized size = 1.54

method	result	size
risch	\(-\frac {\left (5 c \,x^{2}+2 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 x^{5}}-\frac {3 c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 \sqrt {b}\, x \sqrt {c \,x^{2}+b}}\)	\(86\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{2} x^{4}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{2} x^{4}+\left (c \,x^{2}+b \right )^{\frac {5}{2}} c \,x^{2}-3 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \right )}{8 x^{7} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [A]
time = 0.39, size = 164, normalized size = 2.02 \begin {gather*} \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 ] \end {gather*}

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)]

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

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)

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Giac [A]
time = 4.28, size = 76, normalized size = 0.94 \begin {gather*} \frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} - \frac {5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} c^{3} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {c x^{2} + b} b c^{3} \mathrm {sgn}\left (x\right )}{c^{2} x^{4}}}{8 \, c} \end {gather*}

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

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

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)

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