3.255 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=79 \[ \frac {3 c \sqrt {b x^2+c x^4}}{2 x}-\frac {3}{2} \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2020, 2021, 2008, 206} \[ -\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5}+\frac {3 c \sqrt {b x^2+c x^4}}{2 x}-\frac {3}{2} \sqrt {b} c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^6,x]

[Out]

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

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]

Rule 2021

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 + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 44, normalized size = 0.56 \[ \frac {c \left (x^2 \left (b+c x^2\right )\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x^2}{b}+1\right )}{5 b^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^6,x]

[Out]

(c*(x^2*(b + c*x^2))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/b])/(5*b^2*x^5)

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fricas [A]  time = 0.69, size = 147, normalized size = 1.86 \[ \left [\frac {3 \, \sqrt {b} c x^{3} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} - b\right )}}{4 \, x^{3}}, \frac {3 \, \sqrt {-b} c x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} - b\right )}}{2 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 69, normalized size = 0.87 \[ \frac {\frac {3 \, b c^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} + 2 \, \sqrt {c x^{2} + b} c^{2} \mathrm {sgn}\relax (x) - \frac {\sqrt {c x^{2} + b} b c \mathrm {sgn}\relax (x)}{x^{2}}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 102, normalized size = 1.29 \[ -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} c \,x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {c \,x^{2}+b}\, b c \,x^{2}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} c \,x^{2}+\left (c \,x^{2}+b \right )^{\frac {5}{2}}\right )}{2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^6,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^6,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**6,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**6, x)

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