Integrand size = 24, antiderivative size = 86 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {a x}{d \left (d+e x^2\right )^{5/2}}+\frac {(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac {\left (3 c d^2+2 e (b d+4 a e)\right ) x^5}{15 d^3 \left (d+e x^2\right )^{5/2}} \]
a*x/d/(e*x^2+d)^(5/2)+1/3*(4*a*e+b*d)*x^3/d^2/(e*x^2+d)^(5/2)+1/15*(3*c*d^ 2+2*e*(4*a*e+b*d))*x^5/d^3/(e*x^2+d)^(5/2)
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {15 a d^2 x+5 b d^2 x^3+20 a d e x^3+3 c d^2 x^5+2 b d e x^5+8 a e^2 x^5}{15 d^3 \left (d+e x^2\right )^{5/2}} \]
(15*a*d^2*x + 5*b*d^2*x^3 + 20*a*d*e*x^3 + 3*c*d^2*x^5 + 2*b*d*e*x^5 + 8*a *e^2*x^5)/(15*d^3*(d + e*x^2)^(5/2))
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1469, 2075, 362, 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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 1469 |
\(\displaystyle \frac {\int \frac {x^2 \left (4 a e+d \left (c x^2+b\right )\right )}{\left (e x^2+d\right )^{7/2}}dx}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {x^2 \left (c d x^2+b d+4 a e\right )}{\left (e x^2+d\right )^{7/2}}dx}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 362 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {8 a e}{d}+2 b+\frac {3 c d}{e}\right ) \int \frac {x^2}{\left (e x^2+d\right )^{5/2}}dx+\frac {x^3 \left (4 a e+b d-\frac {c d^2}{e}\right )}{5 d \left (d+e x^2\right )^{5/2}}}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {\frac {x^3 \left (4 a e+b d-\frac {c d^2}{e}\right )}{5 d \left (d+e x^2\right )^{5/2}}+\frac {x^3 \left (\frac {8 a e}{d}+2 b+\frac {3 c d}{e}\right )}{15 d \left (d+e x^2\right )^{3/2}}}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
(a*x)/(d*(d + e*x^2)^(5/2)) + (((b*d - (c*d^2)/e + 4*a*e)*x^3)/(5*d*(d + e *x^2)^(5/2)) + ((2*b + (3*c*d)/e + (8*a*e)/d)*x^3)/(15*d*(d + e*x^2)^(3/2) ))/d
3.3.82.3.1 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[((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 = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(\frac {x \left (\left (\frac {1}{5} c \,x^{4}+\frac {1}{3} b \,x^{2}+a \right ) d^{2}+\frac {4 e \left (\frac {b \,x^{2}}{10}+a \right ) x^{2} d}{3}+\frac {8 a \,e^{2} x^{4}}{15}\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(58\) |
gosper | \(\frac {x \left (8 a \,e^{2} x^{4}+2 b d e \,x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+5 b \,d^{2} x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(66\) |
trager | \(\frac {x \left (8 a \,e^{2} x^{4}+2 b d e \,x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+5 b \,d^{2} x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(66\) |
default | \(a \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )+c \left (-\frac {x^{3}}{2 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {3 d \left (-\frac {x}{4 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {d \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )}{4 e}\right )}{2 e}\right )+b \left (-\frac {x}{4 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {d \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )}{4 e}\right )\) | \(232\) |
1/(e*x^2+d)^(5/2)*x*((1/5*c*x^4+1/3*b*x^2+a)*d^2+4/3*e*(1/10*b*x^2+a)*x^2* d+8/15*a*e^2*x^4)/d^3
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left ({\left (3 \, c d^{2} + 2 \, b d e + 8 \, a e^{2}\right )} x^{5} + 15 \, a d^{2} x + 5 \, {\left (b d^{2} + 4 \, a d e\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{15 \, {\left (d^{3} e^{3} x^{6} + 3 \, d^{4} e^{2} x^{4} + 3 \, d^{5} e x^{2} + d^{6}\right )}} \]
1/15*((3*c*d^2 + 2*b*d*e + 8*a*e^2)*x^5 + 15*a*d^2*x + 5*(b*d^2 + 4*a*d*e) *x^3)*sqrt(e*x^2 + d)/(d^3*e^3*x^6 + 3*d^4*e^2*x^4 + 3*d^5*e*x^2 + d^6)
Leaf count of result is larger than twice the leaf count of optimal. 639 vs. \(2 (80) = 160\).
Time = 14.75 (sec) , antiderivative size = 639, normalized size of antiderivative = 7.43 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=a \left (\frac {15 d^{5} x}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {35 d^{4} e x^{3}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {28 d^{3} e^{2} x^{5}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {8 d^{2} e^{3} x^{7}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + b \left (\frac {5 d x^{3}}{15 d^{\frac {9}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 30 d^{\frac {7}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {5}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {2 e x^{5}}{15 d^{\frac {9}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 30 d^{\frac {7}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {5}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + \frac {c x^{5}}{5 d^{\frac {7}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 10 d^{\frac {5}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 5 d^{\frac {3}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}} \]
a*(15*d**5*x/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e* *3*x**6*sqrt(1 + e*x**2/d)) + 35*d**4*e*x**3/(15*d**(17/2)*sqrt(1 + e*x**2 /d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt (1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d)) + 28*d**3*e**2 *x**5/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x* *2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6 *sqrt(1 + e*x**2/d)) + 8*d**2*e**3*x**7/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d))) + b*(5*d*x**3/(15* d**(9/2)*sqrt(1 + e*x**2/d) + 30*d**(7/2)*e*x**2*sqrt(1 + e*x**2/d) + 15*d **(5/2)*e**2*x**4*sqrt(1 + e*x**2/d)) + 2*e*x**5/(15*d**(9/2)*sqrt(1 + e*x **2/d) + 30*d**(7/2)*e*x**2*sqrt(1 + e*x**2/d) + 15*d**(5/2)*e**2*x**4*sqr t(1 + e*x**2/d))) + c*x**5/(5*d**(7/2)*sqrt(1 + e*x**2/d) + 10*d**(5/2)*e* x**2*sqrt(1 + e*x**2/d) + 5*d**(3/2)*e**2*x**4*sqrt(1 + e*x**2/d))
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (76) = 152\).
Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.01 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=-\frac {c x^{3}}{2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e} + \frac {8 \, a x}{15 \, \sqrt {e x^{2} + d} d^{3}} + \frac {4 \, a x}{15 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}} + \frac {a x}{5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d} + \frac {c x}{10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}} + \frac {c x}{5 \, \sqrt {e x^{2} + d} d e^{2}} - \frac {3 \, c d x}{10 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e^{2}} - \frac {b x}{5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e} + \frac {2 \, b x}{15 \, \sqrt {e x^{2} + d} d^{2} e} + \frac {b x}{15 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d e} \]
-1/2*c*x^3/((e*x^2 + d)^(5/2)*e) + 8/15*a*x/(sqrt(e*x^2 + d)*d^3) + 4/15*a *x/((e*x^2 + d)^(3/2)*d^2) + 1/5*a*x/((e*x^2 + d)^(5/2)*d) + 1/10*c*x/((e* x^2 + d)^(3/2)*e^2) + 1/5*c*x/(sqrt(e*x^2 + d)*d*e^2) - 3/10*c*d*x/((e*x^2 + d)^(5/2)*e^2) - 1/5*b*x/((e*x^2 + d)^(5/2)*e) + 2/15*b*x/(sqrt(e*x^2 + d)*d^2*e) + 1/15*b*x/((e*x^2 + d)^(3/2)*d*e)
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {{\left (3 \, c d^{2} e^{2} + 2 \, b d e^{3} + 8 \, a e^{4}\right )} x^{2}}{d^{3} e^{2}} + \frac {5 \, {\left (b d^{2} e^{2} + 4 \, a d e^{3}\right )}}{d^{3} e^{2}}\right )} + \frac {15 \, a}{d}\right )} x}{15 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}}} \]
1/15*(x^2*((3*c*d^2*e^2 + 2*b*d*e^3 + 8*a*e^4)*x^2/(d^3*e^2) + 5*(b*d^2*e^ 2 + 4*a*d*e^3)/(d^3*e^2)) + 15*a/d)*x/(e*x^2 + d)^(5/2)
Time = 7.87 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.55 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {3\,c\,d^4\,x-6\,c\,d^3\,x\,\left (e\,x^2+d\right )-3\,b\,d^3\,e\,x+8\,a\,e^2\,x\,{\left (e\,x^2+d\right )}^2+3\,c\,d^2\,x\,{\left (e\,x^2+d\right )}^2+3\,a\,d^2\,e^2\,x+4\,a\,d\,e^2\,x\,\left (e\,x^2+d\right )+2\,b\,d\,e\,x\,{\left (e\,x^2+d\right )}^2+b\,d^2\,e\,x\,\left (e\,x^2+d\right )}{15\,d^3\,e^2\,{\left (e\,x^2+d\right )}^{5/2}} \]