Integrand size = 26, antiderivative size = 80 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=-\frac {B \sqrt {b x^2+c x^4}}{x^2}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}+B \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right ) \] Output:
-B*(c*x^4+b*x^2)^(1/2)/x^2-1/3*A*(c*x^4+b*x^2)^(3/2)/b/x^6+B*c^(1/2)*arcta nh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.24 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b+c x^2} \left (3 b B x^2+A \left (b+c x^2\right )\right )+3 b B \sqrt {c} x^3 \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{3 b x^4 \sqrt {b+c x^2}} \] Input:
Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^5,x]
Output:
-1/3*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b + c*x^2]*(3*b*B*x^2 + A*(b + c*x^2)) + 3*b*B*Sqrt[c]*x^3*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]]))/(b*x^4*Sqrt[b + c *x^2])
Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1940, 1220, 1125, 25, 27, 1091, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \sqrt {c x^4+b x^2}}{x^6}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (B \int \frac {\sqrt {c x^4+b x^2}}{x^4}dx^2-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}\right )\) |
\(\Big \downarrow \) 1125 |
\(\displaystyle \frac {1}{2} \left (B \left (-\int -\frac {c}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (B \left (\int \frac {c}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (B \left (c \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (B \left (2 c \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (B \left (2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{3 b x^6}\right )\) |
Input:
Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^5,x]
Output:
((-2*A*(b*x^2 + c*x^4)^(3/2))/(3*b*x^6) + B*((-2*Sqrt[b*x^2 + c*x^4])/x^2 + 2*Sqrt[c]*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]]))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 2)*(d + e*x))), x] - Simp[e^(2*m + 2) Int[(1/Sqrt[a + b*x + c*x^2])*Expan dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x ), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[m, 0] && EqQ[m + p, -3/2]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-\frac {\left (A c \,x^{2}+3 B b \,x^{2}+A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3 x^{4} b}+\frac {B \sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\) | \(86\) |
pseudoelliptic | \(\frac {3 x^{4} B b \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) \sqrt {c}-2 \left (\left (A c +3 B b \right ) x^{2}+A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{6 b \,x^{4}}\) | \(87\) |
default | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (-3 B \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x^{4}+3 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, x^{2}-3 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c \,x^{3}+A \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\right )}{3 x^{4} \sqrt {c \,x^{2}+b}\, b \sqrt {c}}\) | \(109\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/3*(A*c*x^2+3*B*b*x^2+A*b)/x^4/b*(x^2*(c*x^2+b))^(1/2)+B*c^(1/2)*ln(c^(1 /2)*x+(c*x^2+b)^(1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.00 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=\left [\frac {3 \, B b \sqrt {c} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left ({\left (3 \, B b + A c\right )} x^{2} + A b\right )}}{6 \, b x^{4}}, -\frac {3 \, B b \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left ({\left (3 \, B b + A c\right )} x^{2} + A b\right )}}{3 \, b x^{4}}\right ] \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^5,x, algorithm="fricas")
Output:
[1/6*(3*B*b*sqrt(c)*x^4*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*sqrt(c*x^4 + b*x^2)*((3*B*b + A*c)*x^2 + A*b))/(b*x^4), -1/3*(3*B*b*sq rt(-c)*x^4*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + sqrt(c*x^4 + b*x^2)*((3*B*b + A*c)*x^2 + A*b))/(b*x^4)]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{5}}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**5,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**5, x)
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.20 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=\frac {1}{2} \, {\left (\sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {2 \, \sqrt {c x^{4} + b x^{2}}}{x^{2}}\right )} B - \frac {1}{3} \, A {\left (\frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{2}} + \frac {\sqrt {c x^{4} + b x^{2}}}{x^{4}}\right )} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^5,x, algorithm="maxima")
Output:
1/2*(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)/x^2)*B - 1/3*A*(sqrt(c*x^4 + b*x^2)*c/(b*x^2) + sqrt(c*x^4 + b *x^2)/x^4)
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (68) = 136\).
Time = 0.34 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.04 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=-\frac {1}{2} \, B \sqrt {c} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b \sqrt {c} \mathrm {sgn}\left (x\right ) + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{2} \sqrt {c} \mathrm {sgn}\left (x\right ) + 3 \, B b^{3} \sqrt {c} \mathrm {sgn}\left (x\right ) + A b^{2} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^5,x, algorithm="giac")
Output:
-1/2*B*sqrt(c)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 2/3*(3*(sqrt( c)*x - sqrt(c*x^2 + b))^4*B*b*sqrt(c)*sgn(x) + 3*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*c^(3/2)*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^2*sqrt(c)* sgn(x) + 3*B*b^3*sqrt(c)*sgn(x) + A*b^2*c^(3/2)*sgn(x))/((sqrt(c)*x - sqrt (c*x^2 + b))^2 - b)^3
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^5} \,d x \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^5,x)
Output:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^5, x)
Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.24 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^5} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, a b -\sqrt {c \,x^{2}+b}\, a c \,x^{2}-3 \sqrt {c \,x^{2}+b}\, b^{2} x^{2}+3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{3}-\sqrt {c}\, a c \,x^{3}+\sqrt {c}\, b^{2} x^{3}}{3 b \,x^{3}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^5,x)
Output:
( - sqrt(b + c*x**2)*a*b - sqrt(b + c*x**2)*a*c*x**2 - 3*sqrt(b + c*x**2)* b**2*x**2 + 3*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**2*x** 3 - sqrt(c)*a*c*x**3 + sqrt(c)*b**2*x**3)/(3*b*x**3)