Integrand size = 63, antiderivative size = 36 \[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\frac {e \arctan \left (\frac {\sqrt {d} x \left (a+b x^2\right )^{3/2}}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}} \] Output:
e*arctan(d^(1/2)*x*(b*x^2+a)^(3/2)/c^(1/2))/c^(1/2)/d^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 6.27 (sec) , antiderivative size = 22291, normalized size of antiderivative = 619.19 \[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\text {Result too large to show} \] Input:
Integrate[(Sqrt[a + b*x^2]*(a*e + 4*b*e*x^2))/(c + a^3*d*x^2 + 3*a^2*b*d*x ^4 + 3*a*b^2*d*x^6 + b^3*d*x^8),x]
Output:
Result too large to show
Time = 1.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {7239, 27, 7260, 218}
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.
\(\displaystyle \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8+c} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e \sqrt {a+b x^2} \left (a+4 b x^2\right )}{d x^2 \left (a+b x^2\right )^3+c}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e \int \frac {\sqrt {b x^2+a} \left (4 b x^2+a\right )}{d x^2 \left (b x^2+a\right )^3+c}dx\) |
\(\Big \downarrow \) 7260 |
\(\displaystyle e \int \frac {1}{d x^2 \left (b x^2+a\right )^3+c}d\left (x \left (b x^2+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {e \arctan \left (\frac {\sqrt {d} x \left (a+b x^2\right )^{3/2}}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}}\) |
Input:
Int[(Sqrt[a + b*x^2]*(a*e + 4*b*e*x^2))/(c + a^3*d*x^2 + 3*a^2*b*d*x^4 + 3 *a*b^2*d*x^6 + b^3*d*x^8),x]
Output:
(e*ArcTan[(Sqrt[d]*x*(a + b*x^2)^(3/2))/Sqrt[c]])/(Sqrt[c]*Sqrt[d])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*(v_)^(r_.)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(q_.))^(m_.), x_Symbol] : > With[{c = Simplify[u/(p*w*D[v, x] + q*v*D[w, x])]}, Simp[c*(p/(r + 1)) Subst[Int[(a + b*x^(p/(r + 1)))^m, x], x, v^(r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q, r}, x] && EqQ[p, q*(r + 1)] && NeQ[r, -1] && Integ erQ[p/(r + 1)]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 3.61
method | result | size |
default | \(-\frac {e \,a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\left (a^{4} d -4 b c \right ) \textit {\_Z}^{6}+6 c \,b^{2} \textit {\_Z}^{4}-4 c \,b^{3} \textit {\_Z}^{2}+b^{4} c \right )}{\sum }\frac {\textit {\_R} \left (\textit {\_R}^{4}+2 \textit {\_R}^{2} b -3 b^{2}\right ) \ln \left (\frac {-\textit {\_R} x +\sqrt {b \,x^{2}+a}}{x}\right )}{3 \textit {\_R}^{4} a^{4} d +4 \textit {\_R}^{6} c -12 \textit {\_R}^{4} b c +12 \textit {\_R}^{2} b^{2} c -4 b^{3} c}\right )}{2}\) | \(130\) |
pseudoelliptic | \(-\frac {e \,a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\left (a^{4} d -4 b c \right ) \textit {\_Z}^{6}+6 c \,b^{2} \textit {\_Z}^{4}-4 c \,b^{3} \textit {\_Z}^{2}+b^{4} c \right )}{\sum }\frac {\textit {\_R} \left (\textit {\_R}^{4}+2 \textit {\_R}^{2} b -3 b^{2}\right ) \ln \left (\frac {-\textit {\_R} x +\sqrt {b \,x^{2}+a}}{x}\right )}{3 \textit {\_R}^{4} a^{4} d +4 \textit {\_R}^{6} c -12 \textit {\_R}^{4} b c +12 \textit {\_R}^{2} b^{2} c -4 b^{3} c}\right )}{2}\) | \(130\) |
Input:
int((b*x^2+a)^(1/2)*(4*b*e*x^2+a*e)/(b^3*d*x^8+3*a*b^2*d*x^6+3*a^2*b*d*x^4 +a^3*d*x^2+c),x,method=_RETURNVERBOSE)
Output:
-1/2*e*a^2*sum(_R*(_R^4+2*_R^2*b-3*b^2)*ln((-_R*x+(b*x^2+a)^(1/2))/x)/(3*_ R^4*a^4*d+4*_R^6*c-12*_R^4*b*c+12*_R^2*b^2*c-4*b^3*c),_R=RootOf(c*_Z^8+(a^ 4*d-4*b*c)*_Z^6+6*c*b^2*_Z^4-4*c*b^3*_Z^2+b^4*c))
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (26) = 52\).
Time = 0.80 (sec) , antiderivative size = 454, normalized size of antiderivative = 12.61 \[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\left [-\frac {\sqrt {-c d} e \log \left (\frac {b^{6} d^{2} x^{16} + 6 \, a b^{5} d^{2} x^{14} + 15 \, a^{2} b^{4} d^{2} x^{12} + 20 \, a^{3} b^{3} d^{2} x^{10} + 3 \, {\left (5 \, a^{4} b^{2} d^{2} - 2 \, b^{3} c d\right )} x^{8} - 6 \, a^{3} c d x^{2} + 6 \, {\left (a^{5} b d^{2} - 3 \, a b^{2} c d\right )} x^{6} + {\left (a^{6} d^{2} - 18 \, a^{2} b c d\right )} x^{4} - 4 \, {\left (b^{4} d x^{11} + 4 \, a b^{3} d x^{9} + 6 \, a^{2} b^{2} d x^{7} + 4 \, a^{3} b d x^{5} + {\left (a^{4} d - b c\right )} x^{3} - a c x\right )} \sqrt {b x^{2} + a} \sqrt {-c d} + c^{2}}{b^{6} d^{2} x^{16} + 6 \, a b^{5} d^{2} x^{14} + 15 \, a^{2} b^{4} d^{2} x^{12} + 20 \, a^{3} b^{3} d^{2} x^{10} + {\left (15 \, a^{4} b^{2} d^{2} + 2 \, b^{3} c d\right )} x^{8} + 2 \, a^{3} c d x^{2} + 6 \, {\left (a^{5} b d^{2} + a b^{2} c d\right )} x^{6} + {\left (a^{6} d^{2} + 6 \, a^{2} b c d\right )} x^{4} + c^{2}}\right )}{4 \, c d}, \frac {\sqrt {c d} e \arctan \left (\frac {{\left (b^{3} d x^{8} + 3 \, a b^{2} d x^{6} + 3 \, a^{2} b d x^{4} + a^{3} d x^{2} - c\right )} \sqrt {b x^{2} + a} \sqrt {c d}}{2 \, {\left (b^{2} c d x^{5} + 2 \, a b c d x^{3} + a^{2} c d x\right )}}\right )}{2 \, c d}\right ] \] Input:
integrate((b*x^2+a)^(1/2)*(4*b*e*x^2+a*e)/(b^3*d*x^8+3*a*b^2*d*x^6+3*a^2*b *d*x^4+a^3*d*x^2+c),x, algorithm="fricas")
Output:
[-1/4*sqrt(-c*d)*e*log((b^6*d^2*x^16 + 6*a*b^5*d^2*x^14 + 15*a^2*b^4*d^2*x ^12 + 20*a^3*b^3*d^2*x^10 + 3*(5*a^4*b^2*d^2 - 2*b^3*c*d)*x^8 - 6*a^3*c*d* x^2 + 6*(a^5*b*d^2 - 3*a*b^2*c*d)*x^6 + (a^6*d^2 - 18*a^2*b*c*d)*x^4 - 4*( b^4*d*x^11 + 4*a*b^3*d*x^9 + 6*a^2*b^2*d*x^7 + 4*a^3*b*d*x^5 + (a^4*d - b* c)*x^3 - a*c*x)*sqrt(b*x^2 + a)*sqrt(-c*d) + c^2)/(b^6*d^2*x^16 + 6*a*b^5* d^2*x^14 + 15*a^2*b^4*d^2*x^12 + 20*a^3*b^3*d^2*x^10 + (15*a^4*b^2*d^2 + 2 *b^3*c*d)*x^8 + 2*a^3*c*d*x^2 + 6*(a^5*b*d^2 + a*b^2*c*d)*x^6 + (a^6*d^2 + 6*a^2*b*c*d)*x^4 + c^2))/(c*d), 1/2*sqrt(c*d)*e*arctan(1/2*(b^3*d*x^8 + 3 *a*b^2*d*x^6 + 3*a^2*b*d*x^4 + a^3*d*x^2 - c)*sqrt(b*x^2 + a)*sqrt(c*d)/(b ^2*c*d*x^5 + 2*a*b*c*d*x^3 + a^2*c*d*x))/(c*d)]
Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\text {Timed out} \] Input:
integrate((b*x**2+a)**(1/2)*(4*b*e*x**2+a*e)/(b**3*d*x**8+3*a*b**2*d*x**6+ 3*a**2*b*d*x**4+a**3*d*x**2+c),x)
Output:
Timed out
\[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\int { \frac {{\left (4 \, b e x^{2} + a e\right )} \sqrt {b x^{2} + a}}{b^{3} d x^{8} + 3 \, a b^{2} d x^{6} + 3 \, a^{2} b d x^{4} + a^{3} d x^{2} + c} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)*(4*b*e*x^2+a*e)/(b^3*d*x^8+3*a*b^2*d*x^6+3*a^2*b *d*x^4+a^3*d*x^2+c),x, algorithm="maxima")
Output:
integrate((4*b*e*x^2 + a*e)*sqrt(b*x^2 + a)/(b^3*d*x^8 + 3*a*b^2*d*x^6 + 3 *a^2*b*d*x^4 + a^3*d*x^2 + c), x)
\[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\int { \frac {{\left (4 \, b e x^{2} + a e\right )} \sqrt {b x^{2} + a}}{b^{3} d x^{8} + 3 \, a b^{2} d x^{6} + 3 \, a^{2} b d x^{4} + a^{3} d x^{2} + c} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)*(4*b*e*x^2+a*e)/(b^3*d*x^8+3*a*b^2*d*x^6+3*a^2*b *d*x^4+a^3*d*x^2+c),x, algorithm="giac")
Output:
sage0*x
Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=\int \frac {\sqrt {b\,x^2+a}\,\left (4\,b\,e\,x^2+a\,e\right )}{d\,a^3\,x^2+3\,d\,a^2\,b\,x^4+3\,d\,a\,b^2\,x^6+d\,b^3\,x^8+c} \,d x \] Input:
int(((a + b*x^2)^(1/2)*(a*e + 4*b*e*x^2))/(c + a^3*d*x^2 + b^3*d*x^8 + 3*a ^2*b*d*x^4 + 3*a*b^2*d*x^6),x)
Output:
int(((a + b*x^2)^(1/2)*(a*e + 4*b*e*x^2))/(c + a^3*d*x^2 + b^3*d*x^8 + 3*a ^2*b*d*x^4 + 3*a*b^2*d*x^6), x)
\[ \int \frac {\sqrt {a+b x^2} \left (a e+4 b e x^2\right )}{c+a^3 d x^2+3 a^2 b d x^4+3 a b^2 d x^6+b^3 d x^8} \, dx=e \left (\left (\int \frac {\sqrt {b \,x^{2}+a}}{b^{3} d \,x^{8}+3 a \,b^{2} d \,x^{6}+3 a^{2} b d \,x^{4}+a^{3} d \,x^{2}+c}d x \right ) a +4 \left (\int \frac {\sqrt {b \,x^{2}+a}\, x^{2}}{b^{3} d \,x^{8}+3 a \,b^{2} d \,x^{6}+3 a^{2} b d \,x^{4}+a^{3} d \,x^{2}+c}d x \right ) b \right ) \] Input:
int((b*x^2+a)^(1/2)*(4*b*e*x^2+a*e)/(b^3*d*x^8+3*a*b^2*d*x^6+3*a^2*b*d*x^4 +a^3*d*x^2+c),x)
Output:
e*(int(sqrt(a + b*x**2)/(a**3*d*x**2 + 3*a**2*b*d*x**4 + 3*a*b**2*d*x**6 + b**3*d*x**8 + c),x)*a + 4*int((sqrt(a + b*x**2)*x**2)/(a**3*d*x**2 + 3*a* *2*b*d*x**4 + 3*a*b**2*d*x**6 + b**3*d*x**8 + c),x)*b)