Integrand size = 37, antiderivative size = 89 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {x \left (d+e x^2\right )^{3/2}}{3 a \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}}+\frac {2 x \sqrt {d+e x^2}}{3 a^2 \sqrt {a d+(b d+a e) x^2+b e x^4}} \] Output:
1/3*x*(e*x^2+d)^(3/2)/a/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2)+2/3*x*(e*x^2+d)^ (1/2)/a^2/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2)
Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.54 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {\left (d+e x^2\right )^{3/2} \left (3 a x+2 b x^3\right )}{3 a^2 \left (\left (a+b x^2\right ) \left (d+e x^2\right )\right )^{3/2}} \] Input:
Integrate[(d + e*x^2)^(5/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(5/2),x]
Output:
((d + e*x^2)^(3/2)*(3*a*x + 2*b*x^3))/(3*a^2*((a + b*x^2)*(d + e*x^2))^(3/ 2))
Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {1395, 209, 208}
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.
\(\displaystyle \int \frac {\left (d+e x^2\right )^{5/2}}{\left (x^2 (a e+b d)+a d+b e x^4\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1395 |
\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\sqrt {a+b x^2} \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right ) \sqrt {d+e x^2}}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\) |
Input:
Int[(d + e*x^2)^(5/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(5/2),x]
Output:
(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2 *Sqrt[a + b*x^2])))/Sqrt[a*d + (b*d + a*e)*x^2 + b*e*x^4]
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, x \left (2 b \,x^{2}+3 a \right )}{3 \sqrt {e \,x^{2}+d}\, \left (b \,x^{2}+a \right )^{2} a^{2}}\) | \(52\) |
orering | \(\frac {x \left (2 b \,x^{2}+3 a \right ) \left (b \,x^{2}+a \right ) \left (e \,x^{2}+d \right )^{\frac {5}{2}}}{3 a^{2} \left (a d +\left (a e +b d \right ) x^{2}+b e \,x^{4}\right )^{\frac {5}{2}}}\) | \(56\) |
gosper | \(\frac {\left (b \,x^{2}+a \right ) x \left (2 b \,x^{2}+3 a \right ) \left (e \,x^{2}+d \right )^{\frac {5}{2}}}{3 a^{2} \left (b e \,x^{4}+a e \,x^{2}+b d \,x^{2}+a d \right )^{\frac {5}{2}}}\) | \(57\) |
Input:
int((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x,method=_RETURNVERB OSE)
Output:
1/3/(e*x^2+d)^(1/2)*((e*x^2+d)*(b*x^2+a))^(1/2)*x*(2*b*x^2+3*a)/(b*x^2+a)^ 2/a^2
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} {\left (2 \, b x^{3} + 3 \, a x\right )} \sqrt {e x^{2} + d}}{3 \, {\left (a^{2} b^{2} e x^{6} + a^{4} d + {\left (a^{2} b^{2} d + 2 \, a^{3} b e\right )} x^{4} + {\left (2 \, a^{3} b d + a^{4} e\right )} x^{2}\right )}} \] Input:
integrate((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm=" fricas")
Output:
1/3*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*(2*b*x^3 + 3*a*x)*sqrt(e*x^2 + d )/(a^2*b^2*e*x^6 + a^4*d + (a^2*b^2*d + 2*a^3*b*e)*x^4 + (2*a^3*b*d + a^4* e)*x^2)
Timed out. \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((e*x**2+d)**(5/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(5/2),x)
Output:
Timed out
\[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {5}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm=" maxima")
Output:
integrate((e*x^2 + d)^(5/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(5/2), x)
\[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {5}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm=" giac")
Output:
integrate((e*x^2 + d)^(5/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(5/2), x)
Time = 17.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.45 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {\left (\frac {2\,x^3\,\sqrt {e\,x^2+d}}{3\,a^2\,b\,e}+\frac {x\,\sqrt {e\,x^2+d}}{a\,b^2\,e}\right )\,\sqrt {b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d}}{x^6+\frac {a^2\,d}{b^2\,e}+\frac {x^4\,\left (2\,a\,e+b\,d\right )}{b\,e}+\frac {x^2\,\left (e\,a^4+2\,b\,d\,a^3\right )}{a^2\,b^2\,e}} \] Input:
int((d + e*x^2)^(5/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(5/2),x)
Output:
(((2*x^3*(d + e*x^2)^(1/2))/(3*a^2*b*e) + (x*(d + e*x^2)^(1/2))/(a*b^2*e)) *(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2))/(x^6 + (a^2*d)/(b^2*e) + (x^4*(2 *a*e + b*d))/(b*e) + (x^2*(a^4*e + 2*a^3*b*d))/(a^2*b^2*e))
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {3 \sqrt {b \,x^{2}+a}\, a b x +2 \sqrt {b \,x^{2}+a}\, b^{2} x^{3}-2 \sqrt {b}\, a^{2}-4 \sqrt {b}\, a b \,x^{2}-2 \sqrt {b}\, b^{2} x^{4}}{3 a^{2} b \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x)
Output:
(3*sqrt(a + b*x**2)*a*b*x + 2*sqrt(a + b*x**2)*b**2*x**3 - 2*sqrt(b)*a**2 - 4*sqrt(b)*a*b*x**2 - 2*sqrt(b)*b**2*x**4)/(3*a**2*b*(a**2 + 2*a*b*x**2 + b**2*x**4))