Integrand size = 26, antiderivative size = 100 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=-\frac {(b B-4 A c) \sqrt {b x^2+c x^4}}{8 c}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2}-\frac {b (b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{3/2}} \] Output:
-1/8*(-4*A*c+B*b)*(c*x^4+b*x^2)^(1/2)/c+1/4*B*(c*x^4+b*x^2)^(3/2)/c/x^2-1/ 8*b*(-4*A*c+B*b)*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(3/2)
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (b B+4 A c+2 B c x^2\right )+b (b B-4 A c) \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{8 c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x,x]
Output:
(x*(Sqrt[c]*x*(b + c*x^2)*(b*B + 4*A*c + 2*B*c*x^2) + b*(b*B - 4*A*c)*Sqrt [b + c*x^2]*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]]))/(8*c^(3/2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.45 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1940, 1221, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \sqrt {c x^4+b x^2}}{x^2}dx^2\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{2 c x^2}-\frac {(b B-4 A c) \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx^2}{4 c}\right )\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{2 c x^2}-\frac {(b B-4 A c) \left (\frac {1}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2+\sqrt {b x^2+c x^4}\right )}{4 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{2 c x^2}-\frac {(b B-4 A c) \left (b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}+\sqrt {b x^2+c x^4}\right )}{4 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {B \left (b x^2+c x^4\right )^{3/2}}{2 c x^2}-\frac {(b B-4 A c) \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}+\sqrt {b x^2+c x^4}\right )}{4 c}\right )\) |
Input:
Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x,x]
Output:
((B*(b*x^2 + c*x^4)^(3/2))/(2*c*x^2) - ((b*B - 4*A*c)*(Sqrt[b*x^2 + c*x^4] + (b*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c]))/(4*c))/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[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 )/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c *f - b*g))/(c*e*(m + 2*p + 2)) Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 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 = 0.44 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\left (2 B c \,x^{2}+4 A c +B b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 c}+\frac {b \left (4 A c -B b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 c^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\) | \(91\) |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (2 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, x +4 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x -B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b x +4 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c -B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2}\right )}{8 \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x}\) | \(124\) |
pseudoelliptic | \(\frac {4 B \,c^{\frac {3}{2}} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}-4 A \ln \left (2\right ) b c +4 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b c +8 A \,c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}+B \ln \left (2\right ) b^{2}-B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2}+2 B b \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}}{16 c^{\frac {3}{2}}}\) | \(155\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x,method=_RETURNVERBOSE)
Output:
1/8*(2*B*c*x^2+4*A*c+B*b)/c*(x^2*(c*x^2+b))^(1/2)+1/8*b*(4*A*c-B*b)/c^(3/2 )*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 = 0.11 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.72 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\left [-\frac {{\left (B b^{2} - 4 \, A b c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{2} x^{2} + B b c + 4 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, c^{2}}, \frac {{\left (B b^{2} - 4 \, A b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (2 \, B c^{2} x^{2} + B b c + 4 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, c^{2}}\right ] \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x, algorithm="fricas")
Output:
[-1/16*((B*b^2 - 4*A*b*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2) *sqrt(c)) - 2*(2*B*c^2*x^2 + B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^2, 1/ 8*((B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (2*B*c^2*x^2 + B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^2]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x, x)
Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.28 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\frac {1}{4} \, {\left (\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{\sqrt {c}} + 2 \, \sqrt {c x^{4} + b x^{2}}\right )} A + \frac {1}{16} \, {\left (4 \, \sqrt {c x^{4} + b x^{2}} x^{2} - \frac {b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {c x^{4} + b x^{2}} b}{c}\right )} B \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x, algorithm="maxima")
Output:
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))*A + 1/16*(4*sqrt(c*x^4 + b*x^2)*x^2 - b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(3/2) + 2*sqrt(c*x^4 + b*x^2)*b/c)*B
Time = 0.21 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\frac {1}{8} \, {\left (2 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c \mathrm {sgn}\left (x\right ) + 4 \, A c^{2} \mathrm {sgn}\left (x\right )}{c^{2}}\right )} \sqrt {c x^{2} + b} x + \frac {{\left (B b^{2} \mathrm {sgn}\left (x\right ) - 4 \, A b c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {3}{2}}} - \frac {{\left (B b^{2} \log \left ({\left | b \right |}\right ) - 4 \, A b c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {3}{2}}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x, algorithm="giac")
Output:
1/8*(2*B*x^2*sgn(x) + (B*b*c*sgn(x) + 4*A*c^2*sgn(x))/c^2)*sqrt(c*x^2 + b) *x + 1/8*(B*b^2*sgn(x) - 4*A*b*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(3/2) - 1/16*(B*b^2*log(abs(b)) - 4*A*b*c*log(abs(b)))*sgn(x)/c^(3 /2)
Time = 9.97 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.17 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\frac {A\,\sqrt {c\,x^4+b\,x^2}}{2}+\frac {B\,\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}}{2}+\frac {A\,b\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{4\,\sqrt {c}}-\frac {B\,b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{3/2}} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x,x)
Output:
(A*(b*x^2 + c*x^4)^(1/2))/2 + (B*(b/(4*c) + x^2/2)*(b*x^2 + c*x^4)^(1/2))/ 2 + (A*b*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(4*c^(1/2)) - (B*b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(16*c^(3/2))
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x} \, dx=\frac {4 \sqrt {c \,x^{2}+b}\, a \,c^{2} x +\sqrt {c \,x^{2}+b}\, b^{2} c x +2 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{3}+4 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b c -\sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}}{8 c^{2}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x)
Output:
(4*sqrt(b + c*x**2)*a*c**2*x + sqrt(b + c*x**2)*b**2*c*x + 2*sqrt(b + c*x* *2)*b*c**2*x**3 + 4*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*a* b*c - sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**3)/(8*c**2)