Integrand size = 24, antiderivative size = 126 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=\frac {a x}{d \left (d+e x^2\right )^{7/2}}+\frac {(b d+6 a e) x^3}{3 d^2 \left (d+e x^2\right )^{7/2}}+\frac {\left (3 c d^2+4 e (b d+6 a e)\right ) x^5}{15 d^3 \left (d+e x^2\right )^{7/2}}+\frac {2 e \left (3 c d^2+4 e (b d+6 a e)\right ) x^7}{105 d^4 \left (d+e x^2\right )^{7/2}} \]
a*x/d/(e*x^2+d)^(7/2)+1/3*(6*a*e+b*d)*x^3/d^2/(e*x^2+d)^(7/2)+1/15*(3*c*d^ 2+4*e*(6*a*e+b*d))*x^5/d^3/(e*x^2+d)^(7/2)+2/105*e*(3*c*d^2+4*e*(6*a*e+b*d ))*x^7/d^4/(e*x^2+d)^(7/2)
Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.82 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=\frac {105 a d^3 x+35 b d^3 x^3+210 a d^2 e x^3+21 c d^3 x^5+28 b d^2 e x^5+168 a d e^2 x^5+6 c d^2 e x^7+8 b d e^2 x^7+48 a e^3 x^7}{105 d^4 \left (d+e x^2\right )^{7/2}} \]
(105*a*d^3*x + 35*b*d^3*x^3 + 210*a*d^2*e*x^3 + 21*c*d^3*x^5 + 28*b*d^2*e* x^5 + 168*a*d*e^2*x^5 + 6*c*d^2*e*x^7 + 8*b*d*e^2*x^7 + 48*a*e^3*x^7)/(105 *d^4*(d + e*x^2)^(7/2))
Time = 0.32 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1469, 2075, 362, 245, 242}
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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 1469 |
\(\displaystyle \frac {\int \frac {x^2 \left (6 a e+d \left (c x^2+b\right )\right )}{\left (e x^2+d\right )^{9/2}}dx}{d}+\frac {a x}{d \left (d+e x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {x^2 \left (c d x^2+b d+6 a e\right )}{\left (e x^2+d\right )^{9/2}}dx}{d}+\frac {a x}{d \left (d+e x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 362 |
\(\displaystyle \frac {\frac {1}{7} \left (\frac {4 (6 a e+b d)}{d}+\frac {3 c d}{e}\right ) \int \frac {x^2}{\left (e x^2+d\right )^{7/2}}dx+\frac {x^3 \left (6 a e+b d-\frac {c d^2}{e}\right )}{7 d \left (d+e x^2\right )^{7/2}}}{d}+\frac {a x}{d \left (d+e x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {\frac {1}{7} \left (\frac {4 (6 a e+b d)}{d}+\frac {3 c d}{e}\right ) \left (\frac {2 e \int \frac {x^4}{\left (e x^2+d\right )^{7/2}}dx}{3 d}+\frac {x^3}{3 d \left (d+e x^2\right )^{5/2}}\right )+\frac {x^3 \left (6 a e+b d-\frac {c d^2}{e}\right )}{7 d \left (d+e x^2\right )^{7/2}}}{d}+\frac {a x}{d \left (d+e x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {\frac {x^3 \left (6 a e+b d-\frac {c d^2}{e}\right )}{7 d \left (d+e x^2\right )^{7/2}}+\frac {1}{7} \left (\frac {2 e x^5}{15 d^2 \left (d+e x^2\right )^{5/2}}+\frac {x^3}{3 d \left (d+e x^2\right )^{5/2}}\right ) \left (\frac {4 (6 a e+b d)}{d}+\frac {3 c d}{e}\right )}{d}+\frac {a x}{d \left (d+e x^2\right )^{7/2}}\) |
(a*x)/(d*(d + e*x^2)^(7/2)) + (((b*d - (c*d^2)/e + 6*a*e)*x^3)/(7*d*(d + e *x^2)^(7/2)) + (((3*c*d)/e + (4*(b*d + 6*a*e))/d)*(x^3/(3*d*(d + e*x^2)^(5 /2)) + (2*e*x^5)/(15*d^2*(d + e*x^2)^(5/2))))/7)/d
3.3.83.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e *(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1)) I nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[a^p*x*((d + e*x^2)^(q + 1)/d), x] + Simp[1/d Int[x^2*(d + e*x^2)^q*(d*PolynomialQuotient[(a + b*x^2 + c*x^4)^p - a^p, x^2, x] - e* a^p*(2*q + 3)), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && ILtQ[q + 1/2, 0] && LtQ[4 *p + 2*q + 1, 0]
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {\left (\left (\frac {1}{5} c \,x^{4}+\frac {1}{3} b \,x^{2}+a \right ) d^{3}+2 e \left (\frac {1}{35} c \,x^{4}+\frac {2}{15} b \,x^{2}+a \right ) x^{2} d^{2}+\frac {8 \left (\frac {b \,x^{2}}{21}+a \right ) e^{2} x^{4} d}{5}+\frac {16 a \,e^{3} x^{6}}{35}\right ) x}{\left (e \,x^{2}+d \right )^{\frac {7}{2}} d^{4}}\) | \(83\) |
gosper | \(\frac {x \left (48 a \,e^{3} x^{6}+8 b d \,e^{2} x^{6}+6 c \,d^{2} e \,x^{6}+168 a d \,e^{2} x^{4}+28 b \,d^{2} e \,x^{4}+21 c \,d^{3} x^{4}+210 a \,d^{2} e \,x^{2}+35 b \,d^{3} x^{2}+105 d^{3} a \right )}{105 \left (e \,x^{2}+d \right )^{\frac {7}{2}} d^{4}}\) | \(100\) |
trager | \(\frac {x \left (48 a \,e^{3} x^{6}+8 b d \,e^{2} x^{6}+6 c \,d^{2} e \,x^{6}+168 a d \,e^{2} x^{4}+28 b \,d^{2} e \,x^{4}+21 c \,d^{3} x^{4}+210 a \,d^{2} e \,x^{2}+35 b \,d^{3} x^{2}+105 d^{3} a \right )}{105 \left (e \,x^{2}+d \right )^{\frac {7}{2}} d^{4}}\) | \(100\) |
default | \(a \left (\frac {x}{7 d \left (e \,x^{2}+d \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}\right )}{7 d}}{d}\right )+c \left (-\frac {x^{3}}{4 e \left (e \,x^{2}+d \right )^{\frac {7}{2}}}+\frac {3 d \left (-\frac {x}{6 e \left (e \,x^{2}+d \right )^{\frac {7}{2}}}+\frac {d \left (\frac {x}{7 d \left (e \,x^{2}+d \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}\right )}{7 d}}{d}\right )}{6 e}\right )}{4 e}\right )+b \left (-\frac {x}{6 e \left (e \,x^{2}+d \right )^{\frac {7}{2}}}+\frac {d \left (\frac {x}{7 d \left (e \,x^{2}+d \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}\right )}{7 d}}{d}\right )}{6 e}\right )\) | \(295\) |
((1/5*c*x^4+1/3*b*x^2+a)*d^3+2*e*(1/35*c*x^4+2/15*b*x^2+a)*x^2*d^2+8/5*(1/ 21*b*x^2+a)*e^2*x^4*d+16/35*a*e^3*x^6)/(e*x^2+d)^(7/2)*x/d^4
Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=\frac {{\left (2 \, {\left (3 \, c d^{2} e + 4 \, b d e^{2} + 24 \, a e^{3}\right )} x^{7} + 7 \, {\left (3 \, c d^{3} + 4 \, b d^{2} e + 24 \, a d e^{2}\right )} x^{5} + 105 \, a d^{3} x + 35 \, {\left (b d^{3} + 6 \, a d^{2} e\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{105 \, {\left (d^{4} e^{4} x^{8} + 4 \, d^{5} e^{3} x^{6} + 6 \, d^{6} e^{2} x^{4} + 4 \, d^{7} e x^{2} + d^{8}\right )}} \]
1/105*(2*(3*c*d^2*e + 4*b*d*e^2 + 24*a*e^3)*x^7 + 7*(3*c*d^3 + 4*b*d^2*e + 24*a*d*e^2)*x^5 + 105*a*d^3*x + 35*(b*d^3 + 6*a*d^2*e)*x^3)*sqrt(e*x^2 + d)/(d^4*e^4*x^8 + 4*d^5*e^3*x^6 + 6*d^6*e^2*x^4 + 4*d^7*e*x^2 + d^8)
Leaf count of result is larger than twice the leaf count of optimal. 1989 vs. \(2 (119) = 238\).
Time = 35.55 (sec) , antiderivative size = 1989, normalized size of antiderivative = 15.79 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=\text {Too large to display} \]
a*(35*d**14*x/(35*d**(37/2)*sqrt(1 + e*x**2/d) + 210*d**(35/2)*e*x**2*sqrt (1 + e*x**2/d) + 525*d**(33/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 700*d**(31/2 )*e**3*x**6*sqrt(1 + e*x**2/d) + 525*d**(29/2)*e**4*x**8*sqrt(1 + e*x**2/d ) + 210*d**(27/2)*e**5*x**10*sqrt(1 + e*x**2/d) + 35*d**(25/2)*e**6*x**12* sqrt(1 + e*x**2/d)) + 175*d**13*e*x**3/(35*d**(37/2)*sqrt(1 + e*x**2/d) + 210*d**(35/2)*e*x**2*sqrt(1 + e*x**2/d) + 525*d**(33/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 700*d**(31/2)*e**3*x**6*sqrt(1 + e*x**2/d) + 525*d**(29/2)*e* *4*x**8*sqrt(1 + e*x**2/d) + 210*d**(27/2)*e**5*x**10*sqrt(1 + e*x**2/d) + 35*d**(25/2)*e**6*x**12*sqrt(1 + e*x**2/d)) + 371*d**12*e**2*x**5/(35*d** (37/2)*sqrt(1 + e*x**2/d) + 210*d**(35/2)*e*x**2*sqrt(1 + e*x**2/d) + 525* d**(33/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 700*d**(31/2)*e**3*x**6*sqrt(1 + e*x**2/d) + 525*d**(29/2)*e**4*x**8*sqrt(1 + e*x**2/d) + 210*d**(27/2)*e** 5*x**10*sqrt(1 + e*x**2/d) + 35*d**(25/2)*e**6*x**12*sqrt(1 + e*x**2/d)) + 429*d**11*e**3*x**7/(35*d**(37/2)*sqrt(1 + e*x**2/d) + 210*d**(35/2)*e*x* *2*sqrt(1 + e*x**2/d) + 525*d**(33/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 700*d **(31/2)*e**3*x**6*sqrt(1 + e*x**2/d) + 525*d**(29/2)*e**4*x**8*sqrt(1 + e *x**2/d) + 210*d**(27/2)*e**5*x**10*sqrt(1 + e*x**2/d) + 35*d**(25/2)*e**6 *x**12*sqrt(1 + e*x**2/d)) + 286*d**10*e**4*x**9/(35*d**(37/2)*sqrt(1 + e* x**2/d) + 210*d**(35/2)*e*x**2*sqrt(1 + e*x**2/d) + 525*d**(33/2)*e**2*x** 4*sqrt(1 + e*x**2/d) + 700*d**(31/2)*e**3*x**6*sqrt(1 + e*x**2/d) + 525...
Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (112) = 224\).
Time = 0.19 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.80 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=-\frac {c x^{3}}{4 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} e} + \frac {16 \, a x}{35 \, \sqrt {e x^{2} + d} d^{4}} + \frac {8 \, a x}{35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{3}} + \frac {6 \, a x}{35 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d^{2}} + \frac {a x}{7 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} d} + \frac {3 \, c x}{140 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e^{2}} + \frac {2 \, c x}{35 \, \sqrt {e x^{2} + d} d^{2} e^{2}} + \frac {c x}{35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d e^{2}} - \frac {3 \, c d x}{28 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} e^{2}} - \frac {b x}{7 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} e} + \frac {8 \, b x}{105 \, \sqrt {e x^{2} + d} d^{3} e} + \frac {4 \, b x}{105 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} e} + \frac {b x}{35 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d e} \]
-1/4*c*x^3/((e*x^2 + d)^(7/2)*e) + 16/35*a*x/(sqrt(e*x^2 + d)*d^4) + 8/35* a*x/((e*x^2 + d)^(3/2)*d^3) + 6/35*a*x/((e*x^2 + d)^(5/2)*d^2) + 1/7*a*x/( (e*x^2 + d)^(7/2)*d) + 3/140*c*x/((e*x^2 + d)^(5/2)*e^2) + 2/35*c*x/(sqrt( e*x^2 + d)*d^2*e^2) + 1/35*c*x/((e*x^2 + d)^(3/2)*d*e^2) - 3/28*c*d*x/((e* x^2 + d)^(7/2)*e^2) - 1/7*b*x/((e*x^2 + d)^(7/2)*e) + 8/105*b*x/(sqrt(e*x^ 2 + d)*d^3*e) + 4/105*b*x/((e*x^2 + d)^(3/2)*d^2*e) + 1/35*b*x/((e*x^2 + d )^(5/2)*d*e)
Time = 0.32 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (x^{2} {\left (\frac {2 \, {\left (3 \, c d^{2} e^{4} + 4 \, b d e^{5} + 24 \, a e^{6}\right )} x^{2}}{d^{4} e^{3}} + \frac {7 \, {\left (3 \, c d^{3} e^{3} + 4 \, b d^{2} e^{4} + 24 \, a d e^{5}\right )}}{d^{4} e^{3}}\right )} + \frac {35 \, {\left (b d^{3} e^{3} + 6 \, a d^{2} e^{4}\right )}}{d^{4} e^{3}}\right )} x^{2} + \frac {105 \, a}{d}\right )} x}{105 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}}} \]
1/105*((x^2*(2*(3*c*d^2*e^4 + 4*b*d*e^5 + 24*a*e^6)*x^2/(d^4*e^3) + 7*(3*c *d^3*e^3 + 4*b*d^2*e^4 + 24*a*d*e^5)/(d^4*e^3)) + 35*(b*d^3*e^3 + 6*a*d^2* e^4)/(d^4*e^3))*x^2 + 105*a/d)*x/(e*x^2 + d)^(7/2)
Time = 7.88 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.22 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{9/2}} \, dx=\frac {x\,\left (\frac {a}{7\,d}-\frac {d\,\left (\frac {b}{7\,d}-\frac {c}{7\,e}\right )}{e}\right )}{{\left (e\,x^2+d\right )}^{7/2}}-\frac {x\,\left (\frac {c}{5\,e^2}-\frac {-c\,d^2+b\,d\,e+6\,a\,e^2}{35\,d^2\,e^2}\right )}{{\left (e\,x^2+d\right )}^{5/2}}+\frac {x\,\left (3\,c\,d^2+4\,b\,d\,e+24\,a\,e^2\right )}{105\,d^3\,e^2\,{\left (e\,x^2+d\right )}^{3/2}}+\frac {x\,\left (6\,c\,d^2+8\,b\,d\,e+48\,a\,e^2\right )}{105\,d^4\,e^2\,\sqrt {e\,x^2+d}} \]