3.242 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^5} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2018, 662, 664, 620, 206} \[ -\frac {\left (b x^2+c x^4\right )^{3/2}}{x^4}+\frac {3}{2} c \sqrt {b x^2+c x^4}+\frac {3}{2} b \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^5,x]

[Out]

(3*c*Sqrt[b*x^2 + c*x^4])/2 - (b*x^2 + c*x^4)^(3/2)/x^4 + (3*b*Sqrt[c]*ArcTanh[(Sqrt[c]*x^2)/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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.71 \[ -\frac {b \sqrt {x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {c x^2}{b}\right )}{x^2 \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^5,x]

[Out]

-((b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((c*x^2)/b)])/(x^2*Sqrt[1 + (c*x^2)/b]))

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fricas [A]  time = 0.55, size = 139, normalized size = 1.83 \[ \left [\frac {3 \, b \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 2 \, b\right )}}{4 \, x^{2}}, -\frac {3 \, b \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 2 \, b\right )}}{2 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^5,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 79, normalized size = 1.04 \[ \frac {1}{2} \, \sqrt {c x^{2} + b} c x \mathrm {sgn}\relax (x) - \frac {3}{4} \, b \sqrt {c} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\relax (x) + \frac {2 \, b^{2} \sqrt {c} \mathrm {sgn}\relax (x)}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 107, normalized size = 1.41 \[ \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{2} c x \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 \sqrt {c \,x^{2}+b}\, b \,c^{\frac {3}{2}} x^{2}+2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} x^{2}-2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\right )}{2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \sqrt {c}\, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^5,x)

[Out]

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

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maxima [A]  time = 1.44, size = 71, normalized size = 0.93 \[ \frac {3}{4} \, b \sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{2 \, x^{2}} + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{2 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^5,x, algorithm="maxima")

[Out]

3/4*b*sqrt(c)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 3/2*sqrt(c*x^4 + b*x^2)*b/x^2 + 1/2*(c*x^4 +
b*x^2)^(3/2)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^5,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**5,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**5, x)

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