Integrand size = 29, antiderivative size = 134 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=-\frac {\left (2 a c^2-d^2\right ) x^2}{2 b^2 c^3}+\frac {d \left (2 a c^2-d^2\right ) \sqrt {a+b x^2}}{b^3 c^4}-\frac {d \left (a+b x^2\right )^{3/2}}{3 b^3 c^2}+\frac {\left (a+b x^2\right )^2}{4 b^3 c}+\frac {\left (a c^2-d^2\right )^2 \log \left (d+c \sqrt {a+b x^2}\right )}{b^3 c^5} \]
-1/2*(2*a*c^2-d^2)*x^2/b^2/c^3-1/3*d*(b*x^2+a)^(3/2)/b^3/c^2+1/4*(b*x^2+a) ^2/b^3/c+(a*c^2-d^2)^2*ln(d+c*(b*x^2+a)^(1/2))/b^3/c^5+d*(2*a*c^2-d^2)*(b* x^2+a)^(1/2)/b^3/c^4
Time = 0.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {3 c^2 \left (a+b x^2\right ) \left (-3 a c^2+2 d^2+b c^2 x^2\right )-4 c d \sqrt {a+b x^2} \left (-5 a c^2+3 d^2+b c^2 x^2\right )+12 \left (-a c^2+d^2\right )^2 \log \left (d+c \sqrt {a+b x^2}\right )}{12 b^3 c^5} \]
(3*c^2*(a + b*x^2)*(-3*a*c^2 + 2*d^2 + b*c^2*x^2) - 4*c*d*Sqrt[a + b*x^2]* (-5*a*c^2 + 3*d^2 + b*c^2*x^2) + 12*(-(a*c^2) + d^2)^2*Log[d + c*Sqrt[a + b*x^2]])/(12*b^3*c^5)
Time = 0.50 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2586, 7267, 476, 2009}
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.
\(\displaystyle \int \frac {x^5}{d \sqrt {a+b x^2}+a c+b c x^2} \, dx\) |
\(\Big \downarrow \) 2586 |
\(\displaystyle \frac {1}{2} \int \frac {x^4}{b c x^2+a c+d \sqrt {b x^2+a}}dx^2\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {\int \frac {\left (a-x^4\right )^2}{\sqrt {b x^2+a} c+d}d\sqrt {b x^2+a}}{b^3}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (\frac {x^6}{c}-\frac {d x^4}{c^2}+\frac {2 a c^2 d-d^3}{c^4}-\frac {\left (2 a c^2-d^2\right ) \sqrt {b x^2+a}}{c^3}+\frac {\left (a c^2-d^2\right )^2}{c^4 \left (\sqrt {b x^2+a} c+d\right )}\right )d\sqrt {b x^2+a}}{b^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\left (a c^2-d^2\right )^2 \log \left (c \sqrt {a+b x^2}+d\right )}{c^5}+\frac {d \sqrt {a+b x^2} \left (2 a c^2-d^2\right )}{c^4}-\frac {x^4 \left (2 a c^2-d^2\right )}{2 c^3}-\frac {d x^6}{3 c^2}+\frac {x^8}{4 c}}{b^3}\) |
(-1/2*((2*a*c^2 - d^2)*x^4)/c^3 - (d*x^6)/(3*c^2) + x^8/(4*c) + (d*(2*a*c^ 2 - d^2)*Sqrt[a + b*x^2])/c^4 + ((a*c^2 - d^2)^2*Log[d + c*Sqrt[a + b*x^2] ])/c^5)/b^3
3.6.44.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] ), x_Symbol] :> Simp[1/n Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* d, 0] && IntegerQ[(m + 1)/n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(1811\) vs. \(2(122)=244\).
Time = 0.14 (sec) , antiderivative size = 1812, normalized size of antiderivative = 13.52
d*(-1/3/b^3/c^2*(b*x^2+a)^(3/2)-1/2/b^2*c^2*a^2/((-a*b)^(1/2)*c^2+(-(a*c^2 -d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*((b*(x-1 /b*(-a*b)^(1/2))^2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2)+(-a*b)^(1/2) *ln(((x-1/b*(-a*b)^(1/2))*b+(-a*b)^(1/2))/b^(1/2)+(b*(x-1/b*(-a*b)^(1/2))^ 2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2))/b^(1/2))-1/2/b^2*c^2*a^2/((- a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2) *b*c^2)^(1/2))*((b*(x+1/b*(-a*b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/ 2)))^(1/2)-(-a*b)^(1/2)*ln(((x+1/b*(-a*b)^(1/2))*b-(-a*b)^(1/2))/b^(1/2)+( b*(x+1/b*(-a*b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/2)))^(1/2))/b^(1/ 2))-1/2*(-a^2*c^4+2*a*c^2*d^2-d^4)/b^2/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c ^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))/c^2*((b*(x+(-(a*c ^2-d^2)*b*c^2)^(1/2)/b/c^2)^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x+(-(a*c^2 -d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2)-(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*ln ((-(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2+(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)*b)/ b^(1/2)+(b*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2-2*(-(a*c^2-d^2)*b*c^2)^( 1/2)/c^2*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2))/b^(1/2)-d^2/ c^2/(d^2/c^2)^(1/2)*ln((2*d^2/c^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x+(-(a *c^2-d^2)*b*c^2)^(1/2)/b/c^2)+2*(d^2/c^2)^(1/2)*(b*(x+(-(a*c^2-d^2)*b*c^2) ^(1/2)/b/c^2)^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x+(-(a*c^2-d^2)*b*c^2)^( 1/2)/b/c^2)+d^2/c^2)^(1/2))/(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)))-1/2*...
Time = 0.29 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.74 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {3 \, b^{2} c^{4} x^{4} - 6 \, {\left (a b c^{4} - b c^{2} d^{2}\right )} x^{2} + 6 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) + 3 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 3 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 4 \, {\left (b c^{3} d x^{2} - 5 \, a c^{3} d + 3 \, c d^{3}\right )} \sqrt {b x^{2} + a}}{12 \, b^{3} c^{5}} \]
1/12*(3*b^2*c^4*x^4 - 6*(a*b*c^4 - b*c^2*d^2)*x^2 + 6*(a^2*c^4 - 2*a*c^2*d ^2 + d^4)*log(b*c^2*x^2 + a*c^2 - d^2) + 3*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*l og(-(b*c^2*x^2 + a*c^2 + 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) - 3*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(-(b*c^2*x^2 + a*c^2 - 2*sqrt(b*x^2 + a)*c*d + d^2)/ x^2) - 4*(b*c^3*d*x^2 - 5*a*c^3*d + 3*c*d^3)*sqrt(b*x^2 + a))/(b^3*c^5)
Time = 1.91 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\begin {cases} \frac {2 \left (\frac {\left (a + b x^{2}\right )^{2}}{8 c} - \frac {d \left (a + b x^{2}\right )^{\frac {3}{2}}}{6 c^{2}} + \frac {\left (a + b x^{2}\right ) \left (- 2 a c^{2} + d^{2}\right )}{4 c^{3}} + \frac {\sqrt {a + b x^{2}} \cdot \left (2 a c^{2} d - d^{3}\right )}{2 c^{4}} + \frac {\left (a c^{2} - d^{2}\right )^{2} \left (\begin {cases} \frac {\sqrt {a + b x^{2}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{2}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{2 c^{4}}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{3 \cdot \left (2 \sqrt {a} d + 2 a c\right )} & \text {otherwise} \end {cases} \]
Piecewise((2*((a + b*x**2)**2/(8*c) - d*(a + b*x**2)**(3/2)/(6*c**2) + (a + b*x**2)*(-2*a*c**2 + d**2)/(4*c**3) + sqrt(a + b*x**2)*(2*a*c**2*d - d** 3)/(2*c**4) + (a*c**2 - d**2)**2*Piecewise((sqrt(a + b*x**2)/d, Eq(c, 0)), (log(c*sqrt(a + b*x**2) + d)/c, True))/(2*c**4))/b**3, Ne(b, 0)), (x**6/( 3*(2*sqrt(a)*d + 2*a*c)), True))
Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {\frac {3 \, {\left (b x^{2} + a\right )}^{2} c^{3} - 4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{2} d - 6 \, {\left (2 \, a c^{3} - c d^{2}\right )} {\left (b x^{2} + a\right )} + 12 \, {\left (2 \, a c^{2} d - d^{3}\right )} \sqrt {b x^{2} + a}}{c^{4}} + \frac {12 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt {b x^{2} + a} c + d\right )}{c^{5}}}{12 \, b^{3}} \]
1/12*((3*(b*x^2 + a)^2*c^3 - 4*(b*x^2 + a)^(3/2)*c^2*d - 6*(2*a*c^3 - c*d^ 2)*(b*x^2 + a) + 12*(2*a*c^2*d - d^3)*sqrt(b*x^2 + a))/c^4 + 12*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^2 + a)*c + d)/c^5)/b^3
Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.16 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{b^{3} c^{5}} + \frac {3 \, {\left (b x^{2} + a\right )}^{2} b^{9} c^{3} - 12 \, {\left (b x^{2} + a\right )} a b^{9} c^{3} - 4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{9} c^{2} d + 24 \, \sqrt {b x^{2} + a} a b^{9} c^{2} d + 6 \, {\left (b x^{2} + a\right )} b^{9} c d^{2} - 12 \, \sqrt {b x^{2} + a} b^{9} d^{3}}{12 \, b^{12} c^{4}} \]
(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(abs(sqrt(b*x^2 + a)*c + d))/(b^3*c^5) + 1/12*(3*(b*x^2 + a)^2*b^9*c^3 - 12*(b*x^2 + a)*a*b^9*c^3 - 4*(b*x^2 + a)^( 3/2)*b^9*c^2*d + 24*sqrt(b*x^2 + a)*a*b^9*c^2*d + 6*(b*x^2 + a)*b^9*c*d^2 - 12*sqrt(b*x^2 + a)*b^9*d^3)/(b^12*c^4)
Time = 17.48 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.25 \[ \int \frac {x^5}{a c+b c x^2+d \sqrt {a+b x^2}} \, dx=\frac {x^4}{4\,b\,c}-\sqrt {b\,x^2+a}\,\left (\frac {d^3}{b^3\,c^4}-\frac {2\,a\,d}{b^3\,c^2}\right )-\frac {d\,{\left (b\,x^2+a\right )}^{3/2}}{3\,b^3\,c^2}-\frac {x^2\,\left (a\,c^2-d^2\right )}{2\,b^2\,c^3}+\frac {\mathrm {atanh}\left (\frac {c\,\sqrt {b\,x^2+a}}{d}\right )\,{\left (a\,c^2-d^2\right )}^2}{b^3\,c^5}+\frac {\ln \left (b\,c^2\,x^2+a\,c^2-d^2\right )\,\left (a^2\,c^4-2\,a\,c^2\,d^2+d^4\right )}{2\,b^3\,c^5} \]