Integrand size = 26, antiderivative size = 97 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\frac {(b B+2 A c) \sqrt {b x^2+c x^4}}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4}+\frac {(b B+2 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {c}} \] Output:
1/2*(2*A*c+B*b)*(c*x^4+b*x^2)^(1/2)/b-A*(c*x^4+b*x^2)^(3/2)/b/x^4+1/2*(2*A *c+B*b)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(1/2)
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-2 A+B x^2+\frac {2 (b B+2 A c) x \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )}{\sqrt {c} \sqrt {b+c x^2}}\right )}{2 x^2} \] Input:
Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^3,x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-2*A + B*x^2 + (2*(b*B + 2*A*c)*x*ArcTanh[(Sqrt[c] *x)/(-Sqrt[b] + Sqrt[b + c*x^2])])/(Sqrt[c]*Sqrt[b + c*x^2])))/(2*x^2)
Time = 0.45 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1940, 1220, 1131, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \sqrt {c x^4+b x^2}}{x^4}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+b B) \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx^2}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^4}\right )\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+b B) \left (\frac {1}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2+\sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^4}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+b B) \left (b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}+\sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {(2 A c+b B) \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}+\sqrt {b x^2+c x^4}\right )}{b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{b x^4}\right )\) |
Input:
Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^3,x]
Output:
((-2*A*(b*x^2 + c*x^4)^(3/2))/(b*x^4) + ((b*B + 2*A*c)*(Sqrt[b*x^2 + c*x^4 ] + (b*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c]))/b)/2
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_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[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] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
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 = 0.46 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86
method | result | size |
risch | \(-\frac {\left (-B \,x^{2}+2 A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x^{2}}+\frac {\left (A c +\frac {B b}{2}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{\sqrt {c}\, x \sqrt {c \,x^{2}+b}}\) | \(83\) |
pseudoelliptic | \(\frac {-2 \sqrt {c}\, \left (-\frac {B \,x^{2}}{2}+A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}+x^{2} \left (A c +\frac {B b}{2}\right ) \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right )}{2 \sqrt {c}\, x^{2}}\) | \(84\) |
default | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (-2 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x^{2}-B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b \,x^{2}+2 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}-2 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c x -B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} x \right )}{2 x^{2} \sqrt {c \,x^{2}+b}\, b \sqrt {c}}\) | \(132\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2*(-B*x^2+2*A)/x^2*(x^2*(c*x^2+b))^(1/2)+(A*c+1/2*B*b)*ln(c^(1/2)*x+(c* x^2+b)^(1/2))/c^(1/2)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.66 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\left [\frac {{\left (B b + 2 \, A c\right )} \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} - 2 \, A c\right )}}{4 \, c x^{2}}, -\frac {{\left (B b + 2 \, A c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} - 2 \, A c\right )}}{2 \, c x^{2}}\right ] \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="fricas")
Output:
[1/4*((B*b + 2*A*c)*sqrt(c)*x^2*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*s qrt(c)) + 2*sqrt(c*x^4 + b*x^2)*(B*c*x^2 - 2*A*c))/(c*x^2), -1/2*((B*b + 2 *A*c)*sqrt(-c)*x^2*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - sqrt (c*x^4 + b*x^2)*(B*c*x^2 - 2*A*c))/(c*x^2)]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{3}}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**3,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**3, x)
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\frac {1}{2} \, {\left (\sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {2 \, \sqrt {c x^{4} + b x^{2}}}{x^{2}}\right )} A + \frac {1}{4} \, {\left (\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{\sqrt {c}} + 2 \, \sqrt {c x^{4} + b x^{2}}\right )} B \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,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)*A + 1/4*(b*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c ))/sqrt(c) + 2*sqrt(c*x^4 + b*x^2))*B
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\frac {1}{2} \, \sqrt {c x^{2} + b} B x \mathrm {sgn}\left (x\right ) + \frac {2 \, A b \sqrt {c} \mathrm {sgn}\left (x\right )}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b} - \frac {{\left (B b \mathrm {sgn}\left (x\right ) + 2 \, A c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right )}{4 \, \sqrt {c}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="giac")
Output:
1/2*sqrt(c*x^2 + b)*B*x*sgn(x) + 2*A*b*sqrt(c)*sgn(x)/((sqrt(c)*x - sqrt(c *x^2 + b))^2 - b) - 1/4*(B*b*sgn(x) + 2*A*c*sgn(x))*log((sqrt(c)*x - sqrt( c*x^2 + b))^2)/sqrt(c)
Timed out. \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^3} \,d x \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^3,x)
Output:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^3, x)
Time = 0.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx=\frac {-8 \sqrt {c \,x^{2}+b}\, a c +4 \sqrt {c \,x^{2}+b}\, b c \,x^{2}+8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c x +4 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x -8 \sqrt {c}\, a c x -\sqrt {c}\, b^{2} x}{8 c x} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x)
Output:
( - 8*sqrt(b + c*x**2)*a*c + 4*sqrt(b + c*x**2)*b*c*x**2 + 8*sqrt(c)*log(( sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*a*c*x + 4*sqrt(c)*log((sqrt(b + c*x **2) + sqrt(c)*x)/sqrt(b))*b**2*x - 8*sqrt(c)*a*c*x - sqrt(c)*b**2*x)/(8*c *x)