\(\int \frac {(b x^2+c x^4)^{3/2}}{x^3} \, dx\) [241]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3}{8} b \sqrt {b x^2+c x^4}+\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^2}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, 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 = {2043, 678, 634, 212} \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {c}}+\frac {3}{8} b \sqrt {b x^2+c x^4}+\frac {\left (b x^2+c x^4\right )^{3/2}}{4 x^2} \]

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^3,x]

[Out]

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

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 634

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 678

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 2043

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {1}{8} \sqrt {x^2 \left (b+c x^2\right )} \left (5 b+2 c x^2-\frac {3 b^2 \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )}{\sqrt {c} x \sqrt {b+c x^2}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {\left (2 c \,x^{2}+5 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8}+\frac {3 b^{2} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 \sqrt {c}\, x \sqrt {c \,x^{2}+b}}\)	\(76\)
default	\(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 x \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}+3 \sqrt {c}\, \sqrt {c \,x^{2}+b}\, b x +3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{2}\right )}{8 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}}\)	\(84\)
pseudoelliptic	\(\frac {4 c^{\frac {3}{2}} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+3 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}-3 \ln \left (2\right ) b^{2}+10 b \sqrt {c}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 \sqrt {c}}\)	\(90\)

[In]

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

[Out]

1/8*(2*c*x^2+5*b)*(x^2*(c*x^2+b))^(1/2)+3/8*b^2*ln(x*c^(1/2)+(c*x^2+b)^(1/2))/c^(1/2)*(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 = 145, normalized size of antiderivative = 1.81 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, b^{2} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} + 5 \, b c\right )}}{16 \, c}, -\frac {3 \, b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (2 \, c^{2} x^{2} + 5 \, b c\right )}}{8 \, c}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=\frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, \sqrt {c}} + \frac {3}{8} \, \sqrt {c x^{4} + b x^{2}} b + \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{4 \, x^{2}} \]

[In]

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

[Out]

3/16*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) + 3/8*sqrt(c*x^4 + b*x^2)*b + 1/4*(c*x^4 + b
*x^2)^(3/2)/x^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^3} \, dx=-\frac {3 \, b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, \sqrt {c}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, \sqrt {c}} + \frac {1}{8} \, {\left (2 \, c x^{2} \mathrm {sgn}\left (x\right ) + 5 \, b \mathrm {sgn}\left (x\right )\right )} \sqrt {c x^{2} + b} x \]

[In]

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

[Out]

-3/8*b^2*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))*sgn(x)/sqrt(c) + 3/16*b^2*log(abs(b))*sgn(x)/sqrt(c) + 1/8*(2*
c*x^2*sgn(x) + 5*b*sgn(x))*sqrt(c*x^2 + b)*x

Mupad [F(-1)]

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

[In]

int((b*x^2 + c*x^4)^(3/2)/x^3,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^3, x)