Integrand size = 22, antiderivative size = 90 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\frac {4}{3} b d \sqrt {c+\frac {d}{x^2}} x+\frac {1}{3} b c \sqrt {c+\frac {d}{x^2}} x^3+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^5}{5 c}-b d^{3/2} \text {arctanh}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right ) \] Output:
4/3*b*d*(c+d/x^2)^(1/2)*x+1/3*b*c*(c+d/x^2)^(1/2)*x^3+1/5*a*(c+d/x^2)^(5/2 )*x^5/c-b*d^(3/2)*arctanh(d^(1/2)/(c+d/x^2)^(1/2)/x)
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\frac {1}{15} \sqrt {c+\frac {d}{x^2}} x \left (\frac {3 a \left (d+c x^2\right )^2}{c}+5 b \left (4 d+c x^2\right )-\frac {15 b d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+c x^2}}{\sqrt {d}}\right )}{\sqrt {d+c x^2}}\right ) \] Input:
Integrate[(a + b/x^2)*(c + d/x^2)^(3/2)*x^4,x]
Output:
(Sqrt[c + d/x^2]*x*((3*a*(d + c*x^2)^2)/c + 5*b*(4*d + c*x^2) - (15*b*d^(3 /2)*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]])/Sqrt[d + c*x^2]))/15
Time = 0.37 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {953, 858, 247, 247, 224, 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.
\(\displaystyle \int x^4 \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle b \int \left (c+\frac {d}{x^2}\right )^{3/2} x^2dx+\frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4d\frac {1}{x}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b \left (d \int \sqrt {c+\frac {d}{x^2}} x^2d\frac {1}{x}-\frac {1}{3} x^3 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b \left (d \left (d \int \frac {1}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x}-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{3} x^3 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b \left (d \left (d \int \frac {1}{1-\frac {d}{x^2}}d\frac {1}{\sqrt {c+\frac {d}{x^2}} x}-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{3} x^3 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}-b \left (d \left (\sqrt {d} \text {arctanh}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{3} x^3 \left (c+\frac {d}{x^2}\right )^{3/2}\right )\) |
Input:
Int[(a + b/x^2)*(c + d/x^2)^(3/2)*x^4,x]
Output:
(a*(c + d/x^2)^(5/2)*x^5)/(5*c) - b*(-1/3*((c + d/x^2)^(3/2)*x^3) + d*(-(S qrt[c + d/x^2]*x) + Sqrt[d]*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*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[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[d/e^n Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G tQ[m + n, -1]))
Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10
method | result | size |
default | \(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (-3 a \left (c \,x^{2}+d \right )^{\frac {5}{2}}+15 d^{\frac {3}{2}} \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c \,x^{2}+d}}{x}\right ) b c -5 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b c -15 \sqrt {c \,x^{2}+d}\, b c d \right )}{15 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c}\) | \(99\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)*x^4,x,method=_RETURNVERBOSE)
Output:
-1/15*((c*x^2+d)/x^2)^(3/2)*x^3*(-3*a*(c*x^2+d)^(5/2)+15*d^(3/2)*ln(2*(d^( 1/2)*(c*x^2+d)^(1/2)+d)/x)*b*c-5*(c*x^2+d)^(3/2)*b*c-15*(c*x^2+d)^(1/2)*b* c*d)/(c*x^2+d)^(3/2)/c
Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.19 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\left [\frac {15 \, b c d^{\frac {3}{2}} \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (3 \, a c^{2} x^{5} + {\left (5 \, b c^{2} + 6 \, a c d\right )} x^{3} + {\left (20 \, b c d + 3 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{30 \, c}, \frac {15 \, b c \sqrt {-d} d \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{d}\right ) + {\left (3 \, a c^{2} x^{5} + {\left (5 \, b c^{2} + 6 \, a c d\right )} x^{3} + {\left (20 \, b c d + 3 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, c}\right ] \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)*x^4,x, algorithm="fricas")
Output:
[1/30*(15*b*c*d^(3/2)*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2* d)/x^2) + 2*(3*a*c^2*x^5 + (5*b*c^2 + 6*a*c*d)*x^3 + (20*b*c*d + 3*a*d^2)* x)*sqrt((c*x^2 + d)/x^2))/c, 1/15*(15*b*c*sqrt(-d)*d*arctan(sqrt(-d)*x*sqr t((c*x^2 + d)/x^2)/d) + (3*a*c^2*x^5 + (5*b*c^2 + 6*a*c*d)*x^3 + (20*b*c*d + 3*a*d^2)*x)*sqrt((c*x^2 + d)/x^2))/c]
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (80) = 160\).
Time = 3.06 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.04 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\frac {a c \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {2 a d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {a d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{5 c} + \frac {b \sqrt {c} d x}{\sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c \sqrt {d} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{3} + \frac {b d^{\frac {3}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{3} - b d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )} + \frac {b d^{2}}{\sqrt {c} x \sqrt {1 + \frac {d}{c x^{2}}}} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(3/2)*x**4,x)
Output:
a*c*sqrt(d)*x**4*sqrt(c*x**2/d + 1)/5 + 2*a*d**(3/2)*x**2*sqrt(c*x**2/d + 1)/5 + a*d**(5/2)*sqrt(c*x**2/d + 1)/(5*c) + b*sqrt(c)*d*x/sqrt(1 + d/(c*x **2)) + b*c*sqrt(d)*x**2*sqrt(c*x**2/d + 1)/3 + b*d**(3/2)*sqrt(c*x**2/d + 1)/3 - b*d**(3/2)*asinh(sqrt(d)/(sqrt(c)*x)) + b*d**2/(sqrt(c)*x*sqrt(1 + d/(c*x**2)))
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5}}{5 \, c} + \frac {1}{6} \, {\left (2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} x^{3} + 6 \, \sqrt {c + \frac {d}{x^{2}}} d x + 3 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )\right )} b \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)*x^4,x, algorithm="maxima")
Output:
1/5*a*(c + d/x^2)^(5/2)*x^5/c + 1/6*(2*(c + d/x^2)^(3/2)*x^3 + 6*sqrt(c + d/x^2)*d*x + 3*d^(3/2)*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)* x + sqrt(d))))*b
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.56 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\frac {b d^{2} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-d}} - \frac {{\left (15 \, b c d^{2} \arctan \left (\frac {\sqrt {d}}{\sqrt {-d}}\right ) + 20 \, b c \sqrt {-d} d^{\frac {3}{2}} + 3 \, a \sqrt {-d} d^{\frac {5}{2}}\right )} \mathrm {sgn}\left (x\right )}{15 \, c \sqrt {-d}} + \frac {3 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a c^{4} \mathrm {sgn}\left (x\right ) + 5 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{5} \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {c x^{2} + d} b c^{5} d \mathrm {sgn}\left (x\right )}{15 \, c^{5}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)*x^4,x, algorithm="giac")
Output:
b*d^2*arctan(sqrt(c*x^2 + d)/sqrt(-d))*sgn(x)/sqrt(-d) - 1/15*(15*b*c*d^2* arctan(sqrt(d)/sqrt(-d)) + 20*b*c*sqrt(-d)*d^(3/2) + 3*a*sqrt(-d)*d^(5/2)) *sgn(x)/(c*sqrt(-d)) + 1/15*(3*(c*x^2 + d)^(5/2)*a*c^4*sgn(x) + 5*(c*x^2 + d)^(3/2)*b*c^5*sgn(x) + 15*sqrt(c*x^2 + d)*b*c^5*d*sgn(x))/c^5
Timed out. \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\int x^4\,\left (a+\frac {b}{x^2}\right )\,{\left (c+\frac {d}{x^2}\right )}^{3/2} \,d x \] Input:
int(x^4*(a + b/x^2)*(c + d/x^2)^(3/2),x)
Output:
int(x^4*(a + b/x^2)*(c + d/x^2)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.57 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^4 \, dx=\frac {3 \sqrt {c \,x^{2}+d}\, a \,c^{2} x^{4}+6 \sqrt {c \,x^{2}+d}\, a c d \,x^{2}+3 \sqrt {c \,x^{2}+d}\, a \,d^{2}+5 \sqrt {c \,x^{2}+d}\, b \,c^{2} x^{2}+20 \sqrt {c \,x^{2}+d}\, b c d +15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x -\sqrt {d}}{\sqrt {d}}\right ) b c d -15 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x +\sqrt {d}}{\sqrt {d}}\right ) b c d}{15 c} \] Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)*x^4,x)
Output:
(3*sqrt(c*x**2 + d)*a*c**2*x**4 + 6*sqrt(c*x**2 + d)*a*c*d*x**2 + 3*sqrt(c *x**2 + d)*a*d**2 + 5*sqrt(c*x**2 + d)*b*c**2*x**2 + 20*sqrt(c*x**2 + d)*b *c*d + 15*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x - sqrt(d))/sqrt(d))*b* c*d - 15*sqrt(d)*log((sqrt(c*x**2 + d) + sqrt(c)*x + sqrt(d))/sqrt(d))*b*c *d)/(15*c)