Integrand size = 19, antiderivative size = 73 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {b \sqrt {b x^2+c x^4}}{x}+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}-b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \] Output:
b*(c*x^4+b*x^2)^(1/2)/x+1/3*(c*x^4+b*x^2)^(3/2)/x^3-b^(3/2)*arctanh(b^(1/2 )*x/(c*x^4+b*x^2)^(1/2))
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {x \left (4 b^2+5 b c x^2+c^2 x^4-3 b^{3/2} \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[(b*x^2 + c*x^4)^(3/2)/x^4,x]
Output:
(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)])
Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1426, 1426, 1400, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 1426 |
\(\displaystyle b \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 1426 |
\(\displaystyle b \left (b \int \frac {1}{\sqrt {c x^4+b x^2}}dx+\frac {\sqrt {b x^2+c x^4}}{x}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle b \left (\frac {\sqrt {b x^2+c x^4}}{x}-b \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle b \left (\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\) |
Input:
Int[(b*x^2 + c*x^4)^(3/2)/x^4,x]
Output:
(b*x^2 + c*x^4)^(3/2)/(3*x^3) + b*(Sqrt[b*x^2 + c*x^4]/x - Sqrt[b]*ArcTanh [(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
Time = 0.65 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )-\left (c \,x^{2}+b \right )^{\frac {3}{2}}-3 \sqrt {c \,x^{2}+b}\, b \right )}{3 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}}}\) | \(78\) |
Input:
int((c*x^4+b*x^2)^(3/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-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)
Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.85 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\left [\frac {3 \, b^{\frac {3}{2}} x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} + 4 \, b\right )}}{6 \, x}, \frac {3 \, \sqrt {-b} b x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (c x^{2} + 4 \, b\right )}}{3 \, x}\right ] \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^4,x, algorithm="fricas")
Output:
[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(sqr t(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b*x^2)*(c*x^2 + 4*b))/x]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(3/2)/x**4,x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)/x**4, x)
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^4,x, algorithm="maxima")
Output:
integrate((c*x^4 + b*x^2)^(3/2)/x^4, x)
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {b^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + \frac {1}{3} \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + \sqrt {c x^{2} + b} b \mathrm {sgn}\left (x\right ) - \frac {{\left (3 \, b^{2} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 4 \, \sqrt {-b} b^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-b}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^4,x, algorithm="giac")
Output:
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)
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^4} \,d x \] Input:
int((b*x^2 + c*x^4)^(3/2)/x^4,x)
Output:
int((b*x^2 + c*x^4)^(3/2)/x^4, x)
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx=\frac {4 \sqrt {c \,x^{2}+b}\, b}{3}+\frac {\sqrt {c \,x^{2}+b}\, c \,x^{2}}{3}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b -\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b \] Input:
int((c*x^4+b*x^2)^(3/2)/x^4,x)
Output:
(4*sqrt(b + c*x**2)*b + sqrt(b + c*x**2)*c*x**2 + 3*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*b - 3*sqrt(b)*log((sqrt(b + c*x**2 ) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b)/3