Integrand size = 26, antiderivative size = 100 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\frac {(2 b B+A c) \sqrt {b x^2+c x^4}}{2 b x}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5}-\frac {(2 b B+A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {b}} \] Output:
1/2*(A*c+2*B*b)*(c*x^4+b*x^2)^(1/2)/b/x-1/2*A*(c*x^4+b*x^2)^(3/2)/b/x^5-1/ 2*(A*c+2*B*b)*arctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))/b^(1/2)
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b} \left (A-2 B x^2\right ) \sqrt {b+c x^2}+(2 b B+A c) x^2 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{2 \sqrt {b} x^3 \sqrt {b+c x^2}} \] Input:
Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^4,x]
Output:
-1/2*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b]*(A - 2*B*x^2)*Sqrt[b + c*x^2] + (2*b* B + A*c)*x^2*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(Sqrt[b]*x^3*Sqrt[b + c*x^ 2])
Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1944, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle \frac {(A c+2 b B) \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5}\) |
\(\Big \downarrow \) 1426 |
\(\displaystyle \frac {(A c+2 b B) \left (b \int \frac {1}{\sqrt {c x^4+b x^2}}dx+\frac {\sqrt {b x^2+c x^4}}{x}\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle \frac {(A c+2 b B) \left (\frac {\sqrt {b x^2+c x^4}}{x}-b \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(A c+2 b B) \left (\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{2 b x^5}\) |
Input:
Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^4,x]
Output:
-1/2*(A*(b*x^2 + c*x^4)^(3/2))/(b*x^5) + ((2*b*B + A*c)*(Sqrt[b*x^2 + c*x^ 4]/x - Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]]))/(2*b)
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]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 0.76 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95
method | result | size |
risch | \(-\frac {A \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x^{3}}+\frac {\left (-\frac {\left (A c +2 B b \right ) \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{2 \sqrt {b}}+B \sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\) | \(95\) |
default | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (A \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c \,x^{2}+2 B \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x^{2}-A \sqrt {c \,x^{2}+b}\, c \,x^{2}-2 B \sqrt {c \,x^{2}+b}\, b \,x^{2}+A \left (c \,x^{2}+b \right )^{\frac {3}{2}}\right )}{2 x^{3} \sqrt {c \,x^{2}+b}\, b}\) | \(135\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/2*A/x^3*(x^2*(c*x^2+b))^(1/2)+(-1/2*(A*c+2*B*b)/b^(1/2)*ln((2*b+2*b^(1/ 2)*(c*x^2+b)^(1/2))/x)+B*(c*x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b )^(1/2)
Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.64 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\left [\frac {{\left (2 \, B b + A c\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, B b x^{2} - A b\right )}}{4 \, b x^{3}}, \frac {{\left (2 \, B b + A c\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (2 \, B b x^{2} - A b\right )}}{2 \, b x^{3}}\right ] \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^4,x, algorithm="fricas")
Output:
[1/4*((2*B*b + A*c)*sqrt(b)*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*B*b*x^2 - A*b))/(b*x^3), 1/2*(( 2*B*b + A*c)*sqrt(-b)*x^3*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqr t(c*x^4 + b*x^2)*(2*B*b*x^2 - A*b))/(b*x^3)]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{4}}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**4,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**4, x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{4}} \,d x } \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^4,x, algorithm="maxima")
Output:
integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^4, x)
Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\frac {1}{2} \, {\left (\frac {2 \, \sqrt {c x^{2} + b} B \mathrm {sgn}\left (x\right )}{c} + \frac {{\left (2 \, B b \mathrm {sgn}\left (x\right ) + A c \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} c} - \frac {\sqrt {c x^{2} + b} A \mathrm {sgn}\left (x\right )}{c x^{2}}\right )} c \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^4,x, algorithm="giac")
Output:
1/2*(2*sqrt(c*x^2 + b)*B*sgn(x)/c + (2*B*b*sgn(x) + A*c*sgn(x))*arctan(sqr t(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*c) - sqrt(c*x^2 + b)*A*sgn(x)/(c*x^2))*c
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^4} \,d x \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^4,x)
Output:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^4, x)
Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.62 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^4} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, a b +2 \sqrt {c \,x^{2}+b}\, b^{2} x^{2}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{2}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{2}-\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{2}-2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{2}}{2 b \,x^{2}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^4,x)
Output:
( - sqrt(b + c*x**2)*a*b + 2*sqrt(b + c*x**2)*b**2*x**2 + sqrt(b)*log((sqr t(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c*x**2 + 2*sqrt(b)*log((sq rt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*x**2 - sqrt(b)*log((sq rt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c*x**2 - 2*sqrt(b)*log((s qrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*x**2)/(2*b*x**2)