Integrand size = 26, antiderivative size = 78 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {A \sqrt {b x^2+c x^4}}{x}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3}-A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \] Output:
A*(c*x^4+b*x^2)^(1/2)/x+1/3*B*(c*x^4+b*x^2)^(3/2)/c/x^3-A*b^(1/2)*arctanh( b^(1/2)*x/(c*x^4+b*x^2)^(1/2))
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {x \left (\left (b+c x^2\right ) \left (b B+3 A c+B c x^2\right )-3 A \sqrt {b} c \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{3 c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^2,x]
Output:
(x*((b + c*x^2)*(b*B + 3*A*c + B*c*x^2) - 3*A*Sqrt[b]*c*Sqrt[b + c*x^2]*Ar cTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(3*c*Sqrt[x^2*(b + c*x^2)])
Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1945, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle A \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx+\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 1426 |
\(\displaystyle A \left (b \int \frac {1}{\sqrt {c x^4+b x^2}}dx+\frac {\sqrt {b x^2+c x^4}}{x}\right )+\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle A \left (\frac {\sqrt {b x^2+c x^4}}{x}-b \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}\right )+\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle A \left (\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\right )+\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 c x^3}\) |
Input:
Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^2,x]
Output:
(B*(b*x^2 + c*x^4)^(3/2))/(3*c*x^3) + A*(Sqrt[b*x^2 + c*x^4]/x - Sqrt[b]*A rcTanh[(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]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.68 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (-3 A \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {b}\, c +B \left (c \,x^{2}+b \right )^{\frac {3}{2}}+3 A \sqrt {c \,x^{2}+b}\, c \right )}{3 x \sqrt {c \,x^{2}+b}\, c}\) | \(84\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
1/3*(c*x^4+b*x^2)^(1/2)*(-3*A*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b^(1/2)* c+B*(c*x^2+b)^(3/2)+3*A*(c*x^2+b)^(1/2)*c)/x/(c*x^2+b)^(1/2)/c
Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.97 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\left [\frac {3 \, A \sqrt {b} c x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} + B b + 3 \, A c\right )}}{6 \, c x}, \frac {3 \, A \sqrt {-b} c x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} + B b + 3 \, A c\right )}}{3 \, c x}\right ] \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^2,x, algorithm="fricas")
Output:
[1/6*(3*A*sqrt(b)*c*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)*(B*c*x^2 + B*b + 3*A*c))/(c*x), 1/3*(3*A*sqr t(-b)*c*x*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b*x^2) *(B*c*x^2 + B*b + 3*A*c))/(c*x)]
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{2}}\, dx \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**2,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**2, x)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{2}} \,d x } \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^2,x, algorithm="maxima")
Output:
A*integrate(sqrt(c*x^2 + b)/x, x) + 1/3*(c*x^2 + b)^(3/2)*B/c
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {A b \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} - \frac {{\left (3 \, A b c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + B \sqrt {-b} b^{\frac {3}{2}} + 3 \, A \sqrt {-b} \sqrt {b} c\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-b} c} + \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}} B c^{2} \mathrm {sgn}\left (x\right ) + 3 \, \sqrt {c x^{2} + b} A c^{3} \mathrm {sgn}\left (x\right )}{3 \, c^{3}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^2,x, algorithm="giac")
Output:
A*b*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - 1/3*(3*A*b*c*arctan (sqrt(b)/sqrt(-b)) + B*sqrt(-b)*b^(3/2) + 3*A*sqrt(-b)*sqrt(b)*c)*sgn(x)/( sqrt(-b)*c) + 1/3*((c*x^2 + b)^(3/2)*B*c^2*sgn(x) + 3*sqrt(c*x^2 + b)*A*c^ 3*sgn(x))/c^3
Time = 9.64 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {A\,\sqrt {c\,x^4+b\,x^2}}{x}+\frac {B\,\left (c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{3\,c\,x}+\frac {A\,\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,\sqrt {c\,x^4+b\,x^2}\,1{}\mathrm {i}}{\sqrt {c}\,x^2\,\sqrt {\frac {b}{c\,x^2}+1}} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^2,x)
Output:
(A*(b*x^2 + c*x^4)^(1/2))/x + (B*(b + c*x^2)*(b*x^2 + c*x^4)^(1/2))/(3*c*x ) + (A*b^(1/2)*asin((b^(1/2)*1i)/(c^(1/2)*x))*(b*x^2 + c*x^4)^(1/2)*1i)/(c ^(1/2)*x^2*(b/(c*x^2) + 1)^(1/2))
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.28 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^2} \, dx=\frac {3 \sqrt {c \,x^{2}+b}\, a c +\sqrt {c \,x^{2}+b}\, b^{2}+\sqrt {c \,x^{2}+b}\, b c \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c -3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c}{3 c} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^2,x)
Output:
(3*sqrt(b + c*x**2)*a*c + sqrt(b + c*x**2)*b**2 + sqrt(b + c*x**2)*b*c*x** 2 + 3*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c - 3*sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c)/(3*c)