Integrand size = 19, antiderivative size = 91 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {\left (a+\frac {c d^2}{e^2}\right ) x}{5 d \left (d+e x^2\right )^{5/2}}+\frac {2 \left (\frac {2 a}{d^2}-\frac {3 c}{e^2}\right ) x}{15 \left (d+e x^2\right )^{3/2}}+\frac {\left (\frac {8 a}{d^2}+\frac {3 c}{e^2}\right ) x}{15 d \sqrt {d+e x^2}} \] Output:
1/5*(a+c*d^2/e^2)*x/d/(e*x^2+d)^(5/2)+2/15*(2*a/d^2-3*c/e^2)*x/(e*x^2+d)^( 3/2)+1/15*(8*a/d^2+3*c/e^2)*x/d/(e*x^2+d)^(1/2)
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {15 a d^2 x+20 a d e x^3+3 c d^2 x^5+8 a e^2 x^5}{15 d^3 \left (d+e x^2\right )^{5/2}} \] Input:
Integrate[(a + c*x^4)/(d + e*x^2)^(7/2),x]
Output:
(15*a*d^2*x + 20*a*d*e*x^3 + 3*c*d^2*x^5 + 8*a*e^2*x^5)/(15*d^3*(d + e*x^2 )^(5/2))
Time = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1470, 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.
\(\displaystyle \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 1470 |
\(\displaystyle \frac {\int \frac {x^2 \left (c d x^2+4 a e\right )}{\left (e x^2+d\right )^{7/2}}dx}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 362 |
\(\displaystyle \frac {\frac {1}{5} \left (\frac {8 a e}{d}+\frac {3 c d}{e}\right ) \int \frac {x^2}{\left (e x^2+d\right )^{5/2}}dx-\frac {x^3 \left (\frac {c d}{e}-\frac {4 a e}{d}\right )}{5 \left (d+e x^2\right )^{5/2}}}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {\frac {x^3 \left (\frac {8 a e}{d}+\frac {3 c d}{e}\right )}{15 d \left (d+e x^2\right )^{3/2}}-\frac {x^3 \left (\frac {c d}{e}-\frac {4 a e}{d}\right )}{5 \left (d+e x^2\right )^{5/2}}}{d}+\frac {a x}{d \left (d+e x^2\right )^{5/2}}\) |
Input:
Int[(a + c*x^4)/(d + e*x^2)^(7/2),x]
Output:
(a*x)/(d*(d + e*x^2)^(5/2)) + (-1/5*(((c*d)/e - (4*a*e)/d)*x^3)/(d + e*x^2 )^(5/2) + (((3*c*d)/e + (8*a*e)/d)*x^3)/(15*d*(d + e*x^2)^(3/2)))/d
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_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si mp[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 + c*x^4)^p - a^p, x^2, x] - e*a^p*(2*q + 3)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && ILtQ[q + 1/2, 0] && LtQ[4*p + 2*q + 1, 0]
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(\frac {x \left (8 a \,e^{2} x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(49\) |
trager | \(\frac {x \left (8 a \,e^{2} x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(49\) |
pseudoelliptic | \(\frac {x \left (8 a \,e^{2} x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(49\) |
orering | \(\frac {x \left (8 a \,e^{2} x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\) | \(49\) |
default | \(a \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )+c \left (-\frac {x^{3}}{2 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {3 d \left (-\frac {x}{4 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {d \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )}{4 e}\right )}{2 e}\right )\) | \(156\) |
Input:
int((c*x^4+a)/(e*x^2+d)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*x*(8*a*e^2*x^4+3*c*d^2*x^4+20*a*d*e*x^2+15*a*d^2)/(e*x^2+d)^(5/2)/d^3
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left (20 \, a d e x^{3} + {\left (3 \, c d^{2} + 8 \, a e^{2}\right )} x^{5} + 15 \, a d^{2} x\right )} \sqrt {e x^{2} + d}}{15 \, {\left (d^{3} e^{3} x^{6} + 3 \, d^{4} e^{2} x^{4} + 3 \, d^{5} e x^{2} + d^{6}\right )}} \] Input:
integrate((c*x^4+a)/(e*x^2+d)^(7/2),x, algorithm="fricas")
Output:
1/15*(20*a*d*e*x^3 + (3*c*d^2 + 8*a*e^2)*x^5 + 15*a*d^2*x)*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. 488 vs. \(2 (83) = 166\).
Time = 9.67 (sec) , antiderivative size = 488, normalized size of antiderivative = 5.36 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=a \left (\frac {15 d^{5} x}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {35 d^{4} e x^{3}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {28 d^{3} e^{2} x^{5}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {8 d^{2} e^{3} x^{7}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + \frac {c x^{5}}{5 d^{\frac {7}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 10 d^{\frac {5}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 5 d^{\frac {3}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}} \] Input:
integrate((c*x**4+a)/(e*x**2+d)**(7/2),x)
Output:
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))) + 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))
Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=-\frac {c x^{3}}{2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e} + \frac {8 \, a x}{15 \, \sqrt {e x^{2} + d} d^{3}} + \frac {4 \, a x}{15 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}} + \frac {a x}{5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d} + \frac {c x}{10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}} + \frac {c x}{5 \, \sqrt {e x^{2} + d} d e^{2}} - \frac {3 \, c d x}{10 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e^{2}} \] Input:
integrate((c*x^4+a)/(e*x^2+d)^(7/2),x, algorithm="maxima")
Output:
-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)
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.63 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {20 \, a e}{d^{2}} + \frac {{\left (3 \, c d^{2} e^{2} + 8 \, a e^{4}\right )} x^{2}}{d^{3} e^{2}}\right )} + \frac {15 \, a}{d}\right )} x}{15 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}}} \] Input:
integrate((c*x^4+a)/(e*x^2+d)^(7/2),x, algorithm="giac")
Output:
1/15*(x^2*(20*a*e/d^2 + (3*c*d^2*e^2 + 8*a*e^4)*x^2/(d^3*e^2)) + 15*a/d)*x /(e*x^2 + d)^(5/2)
Time = 17.41 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {3\,c\,d^4\,x-6\,c\,d^3\,x\,\left (e\,x^2+d\right )+8\,a\,e^2\,x\,{\left (e\,x^2+d\right )}^2+3\,c\,d^2\,x\,{\left (e\,x^2+d\right )}^2+3\,a\,d^2\,e^2\,x+4\,a\,d\,e^2\,x\,\left (e\,x^2+d\right )}{15\,d^3\,e^2\,{\left (e\,x^2+d\right )}^{5/2}} \] Input:
int((a + c*x^4)/(d + e*x^2)^(7/2),x)
Output:
(3*c*d^4*x - 6*c*d^3*x*(d + e*x^2) + 8*a*e^2*x*(d + e*x^2)^2 + 3*c*d^2*x*( d + e*x^2)^2 + 3*a*d^2*e^2*x + 4*a*d*e^2*x*(d + e*x^2))/(15*d^3*e^2*(d + e *x^2)^(5/2))
Time = 0.17 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.30 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {15 \sqrt {e \,x^{2}+d}\, a \,d^{2} e^{3} x +20 \sqrt {e \,x^{2}+d}\, a d \,e^{4} x^{3}+8 \sqrt {e \,x^{2}+d}\, a \,e^{5} x^{5}+3 \sqrt {e \,x^{2}+d}\, c \,d^{2} e^{3} x^{5}-8 \sqrt {e}\, a \,d^{3} e^{2}-24 \sqrt {e}\, a \,d^{2} e^{3} x^{2}-24 \sqrt {e}\, a d \,e^{4} x^{4}-8 \sqrt {e}\, a \,e^{5} x^{6}+3 \sqrt {e}\, c \,d^{5}+9 \sqrt {e}\, c \,d^{4} e \,x^{2}+9 \sqrt {e}\, c \,d^{3} e^{2} x^{4}+3 \sqrt {e}\, c \,d^{2} e^{3} x^{6}}{15 d^{3} e^{3} \left (e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}\right )} \] Input:
int((c*x^4+a)/(e*x^2+d)^(7/2),x)
Output:
(15*sqrt(d + e*x**2)*a*d**2*e**3*x + 20*sqrt(d + e*x**2)*a*d*e**4*x**3 + 8 *sqrt(d + e*x**2)*a*e**5*x**5 + 3*sqrt(d + e*x**2)*c*d**2*e**3*x**5 - 8*sq rt(e)*a*d**3*e**2 - 24*sqrt(e)*a*d**2*e**3*x**2 - 24*sqrt(e)*a*d*e**4*x**4 - 8*sqrt(e)*a*e**5*x**6 + 3*sqrt(e)*c*d**5 + 9*sqrt(e)*c*d**4*e*x**2 + 9* sqrt(e)*c*d**3*e**2*x**4 + 3*sqrt(e)*c*d**2*e**3*x**6)/(15*d**3*e**3*(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6))