\(\int \frac {(b x^2+c x^4)^{3/2}}{x^4} \, dx\) [254]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {b \sqrt {b x^2+c x^4}}{x}+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}-b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2046, 2033, 212} \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=b^{3/2} \left (-\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\right )+\frac {b \sqrt {b x^2+c x^4}}{x}+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3} \]

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^4,x]

[Out]

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

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 2046

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {x \left (4 b^2+5 b c x^2+c^2 x^4-3 b^{3/2} \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{3 \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )-\left (c \,x^{2}+b \right )^{\frac {3}{2}}-3 \sqrt {c \,x^{2}+b}\, b \right )}{3 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}}}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.92 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, b^{\frac {3}{2}} x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} + 4 \, b\right )}}{6 \, x}, \frac {3 \, \sqrt {-b} b x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} + 4 \, b\right )}}{3 \, x}\right ] \]

[In]

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

[Out]

[1/6*(3*b^(3/2)*x*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)*(c*x^2 + 4
*b))/x, 1/3*(3*sqrt(-b)*b*x*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^4 + b*x^2)*(c*x^2 +
4*b))/x]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {b^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + \frac {1}{3} \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + \sqrt {c x^{2} + b} b \mathrm {sgn}\left (x\right ) - \frac {{\left (3 \, b^{2} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 4 \, \sqrt {-b} b^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-b}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((b*x^2 + c*x^4)^(3/2)/x^4,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^4, x)