Integrand size = 22, antiderivative size = 90 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {(4 b c-3 a d) \sqrt {c+\frac {d}{x^2}} x^2}{8 c^2}+\frac {a \sqrt {c+\frac {d}{x^2}} x^4}{4 c}-\frac {d (4 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8 c^{5/2}} \] Output:
1/8*(-3*a*d+4*b*c)*(c+d/x^2)^(1/2)*x^2/c^2+1/4*a*(c+d/x^2)^(1/2)*x^4/c-1/8 *d*(-3*a*d+4*b*c)*arctanh((c+d/x^2)^(1/2)/c^(1/2))/c^(5/2)
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {\sqrt {c} \left (d+c x^2\right ) \left (4 b c-3 a d+2 a c x^2\right )+\frac {2 d (-4 b c+3 a d) \sqrt {d+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {d}+\sqrt {d+c x^2}}\right )}{x}}{8 c^{5/2} \sqrt {c+\frac {d}{x^2}}} \] Input:
Integrate[((a + b/x^2)*x^3)/Sqrt[c + d/x^2],x]
Output:
(Sqrt[c]*(d + c*x^2)*(4*b*c - 3*a*d + 2*a*c*x^2) + (2*d*(-4*b*c + 3*a*d)*S qrt[d + c*x^2]*ArcTanh[(Sqrt[c]*x)/(-Sqrt[d] + Sqrt[d + c*x^2])])/x)/(8*c^ (5/2)*Sqrt[c + d/x^2])
Time = 0.34 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {948, 87, 52, 73, 221}
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.
\(\displaystyle \int \frac {x^3 \left (a+\frac {b}{x^2}\right )}{\sqrt {c+\frac {d}{x^2}}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \frac {\left (a+\frac {b}{x^2}\right ) x^6}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (\frac {a x^4 \sqrt {c+\frac {d}{x^2}}}{2 c}-\frac {(4 b c-3 a d) \int \frac {x^4}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}}{4 c}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {a x^4 \sqrt {c+\frac {d}{x^2}}}{2 c}-\frac {(4 b c-3 a d) \left (-\frac {d \int \frac {x^2}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}}{2 c}-\frac {x^2 \sqrt {c+\frac {d}{x^2}}}{c}\right )}{4 c}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {a x^4 \sqrt {c+\frac {d}{x^2}}}{2 c}-\frac {(4 b c-3 a d) \left (-\frac {\int \frac {1}{\frac {1}{d x^4}-\frac {c}{d}}d\sqrt {c+\frac {d}{x^2}}}{c}-\frac {x^2 \sqrt {c+\frac {d}{x^2}}}{c}\right )}{4 c}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {a x^4 \sqrt {c+\frac {d}{x^2}}}{2 c}-\frac {(4 b c-3 a d) \left (\frac {d \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {x^2 \sqrt {c+\frac {d}{x^2}}}{c}\right )}{4 c}\right )\) |
Input:
Int[((a + b/x^2)*x^3)/Sqrt[c + d/x^2],x]
Output:
((a*Sqrt[c + d/x^2]*x^4)/(2*c) - ((4*b*c - 3*a*d)*(-((Sqrt[c + d/x^2]*x^2) /c) + (d*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/c^(3/2)))/(4*c))/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10
method | result | size |
risch | \(\frac {\left (2 a c \,x^{2}-3 a d +4 c b \right ) \left (c \,x^{2}+d \right )}{8 c^{2} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}+\frac {d \left (3 a d -4 c b \right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) \sqrt {c \,x^{2}+d}}{8 c^{\frac {5}{2}} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x}\) | \(99\) |
default | \(\frac {\sqrt {c \,x^{2}+d}\, \left (2 c^{\frac {5}{2}} \sqrt {c \,x^{2}+d}\, a \,x^{3}-3 c^{\frac {3}{2}} \sqrt {c \,x^{2}+d}\, a d x +4 c^{\frac {5}{2}} \sqrt {c \,x^{2}+d}\, b x +3 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) a c \,d^{2}-4 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) b \,c^{2} d \right )}{8 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \,c^{\frac {7}{2}}}\) | \(129\) |
Input:
int((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8*(2*a*c*x^2-3*a*d+4*b*c)*(c*x^2+d)/c^2/((c*x^2+d)/x^2)^(1/2)+1/8*d*(3*a *d-4*b*c)/c^(5/2)*ln(c^(1/2)*x+(c*x^2+d)^(1/2))/((c*x^2+d)/x^2)^(1/2)/x*(c *x^2+d)^(1/2)
Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.13 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\left [-\frac {{\left (4 \, b c d - 3 \, a d^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left (2 \, a c^{2} x^{4} + {\left (4 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, c^{3}}, \frac {{\left (4 \, b c d - 3 \, a d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (2 \, a c^{2} x^{4} + {\left (4 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, c^{3}}\right ] \] Input:
integrate((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x, algorithm="fricas")
Output:
[-1/16*((4*b*c*d - 3*a*d^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x ^2 + d)/x^2) - d) - 2*(2*a*c^2*x^4 + (4*b*c^2 - 3*a*c*d)*x^2)*sqrt((c*x^2 + d)/x^2))/c^3, 1/8*((4*b*c*d - 3*a*d^2)*sqrt(-c)*arctan(sqrt(-c)*x^2*sqrt ((c*x^2 + d)/x^2)/(c*x^2 + d)) + (2*a*c^2*x^4 + (4*b*c^2 - 3*a*c*d)*x^2)*s qrt((c*x^2 + d)/x^2))/c^3]
Time = 24.73 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.67 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {a x^{5}}{4 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {a \sqrt {d} x^{3}}{8 c \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {3 a d^{\frac {3}{2}} x}{8 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 a d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{8 c^{\frac {5}{2}}} + \frac {b \sqrt {d} x \sqrt {\frac {c x^{2}}{d} + 1}}{2 c} - \frac {b d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2 c^{\frac {3}{2}}} \] Input:
integrate((a+b/x**2)*x**3/(c+d/x**2)**(1/2),x)
Output:
a*x**5/(4*sqrt(d)*sqrt(c*x**2/d + 1)) - a*sqrt(d)*x**3/(8*c*sqrt(c*x**2/d + 1)) - 3*a*d**(3/2)*x/(8*c**2*sqrt(c*x**2/d + 1)) + 3*a*d**2*asinh(sqrt(c )*x/sqrt(d))/(8*c**(5/2)) + b*sqrt(d)*x*sqrt(c*x**2/d + 1)/(2*c) - b*d*asi nh(sqrt(c)*x/sqrt(d))/(2*c**(3/2))
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (74) = 148\).
Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {1}{4} \, b {\left (\frac {2 \, \sqrt {c + \frac {d}{x^{2}}} d}{{\left (c + \frac {d}{x^{2}}\right )} c - c^{2}} + \frac {d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {3}{2}}}\right )} - \frac {1}{16} \, a {\left (\frac {3 \, d^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} - 5 \, \sqrt {c + \frac {d}{x^{2}}} c d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} c^{2} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} c^{3} + c^{4}}\right )} \] Input:
integrate((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x, algorithm="maxima")
Output:
1/4*b*(2*sqrt(c + d/x^2)*d/((c + d/x^2)*c - c^2) + d*log((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/c^(3/2)) - 1/16*a*(3*d^2*log((sqrt (c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c)))/c^(5/2) + 2*(3*(c + d/ x^2)^(3/2)*d^2 - 5*sqrt(c + d/x^2)*c*d^2)/((c + d/x^2)^2*c^2 - 2*(c + d/x^ 2)*c^3 + c^4))
Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {1}{8} \, \sqrt {c x^{2} + d} x {\left (\frac {2 \, a x^{2}}{c \mathrm {sgn}\left (x\right )} + \frac {4 \, b c^{2} \mathrm {sgn}\left (x\right ) - 3 \, a c d \mathrm {sgn}\left (x\right )}{c^{3}}\right )} - \frac {{\left (4 \, b c d \log \left ({\left | d \right |}\right ) - 3 \, a d^{2} \log \left ({\left | d \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {5}{2}}} + \frac {{\left (4 \, b c d - 3 \, a d^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + d} \right |}\right )}{8 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \] Input:
integrate((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x, algorithm="giac")
Output:
1/8*sqrt(c*x^2 + d)*x*(2*a*x^2/(c*sgn(x)) + (4*b*c^2*sgn(x) - 3*a*c*d*sgn( x))/c^3) - 1/16*(4*b*c*d*log(abs(d)) - 3*a*d^2*log(abs(d)))*sgn(x)/c^(5/2) + 1/8*(4*b*c*d - 3*a*d^2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + d)))/(c^(5/2) *sgn(x))
Time = 5.00 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {5\,a\,x^4\,\sqrt {c+\frac {d}{x^2}}}{8\,c}-\frac {3\,a\,x^4\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{8\,c^2}+\frac {b\,x^2\,\sqrt {c+\frac {d}{x^2}}}{2\,c}-\frac {b\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2\,c^{3/2}}+\frac {3\,a\,d^2\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{8\,c^{5/2}} \] Input:
int((x^3*(a + b/x^2))/(c + d/x^2)^(1/2),x)
Output:
(5*a*x^4*(c + d/x^2)^(1/2))/(8*c) - (3*a*x^4*(c + d/x^2)^(3/2))/(8*c^2) + (b*x^2*(c + d/x^2)^(1/2))/(2*c) - (b*d*atanh((c + d/x^2)^(1/2)/c^(1/2)))/( 2*c^(3/2)) + (3*a*d^2*atanh((c + d/x^2)^(1/2)/c^(1/2)))/(8*c^(5/2))
Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) x^3}{\sqrt {c+\frac {d}{x^2}}} \, dx=\frac {2 \sqrt {c \,x^{2}+d}\, a \,c^{2} x^{3}-3 \sqrt {c \,x^{2}+d}\, a c d x +4 \sqrt {c \,x^{2}+d}\, b \,c^{2} x +3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) a \,d^{2}-4 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) b c d}{8 c^{3}} \] Input:
int((a+b/x^2)*x^3/(c+d/x^2)^(1/2),x)
Output:
(2*sqrt(c*x**2 + d)*a*c**2*x**3 - 3*sqrt(c*x**2 + d)*a*c*d*x + 4*sqrt(c*x* *2 + d)*b*c**2*x + 3*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*a *d**2 - 4*sqrt(c)*log((sqrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*b*c*d)/(8*c* *3)