Integrand size = 21, antiderivative size = 174 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=-\frac {d x \left (a+b x^2\right )^3}{7 c (b c-a d) \left (c+d x^2\right )^{7/2}}+\frac {(7 b c-6 a d) x \left (a+b x^2\right )^2}{35 c^2 (b c-a d) \left (c+d x^2\right )^{5/2}}+\frac {4 a (7 b c-6 a d) x \left (a+b x^2\right )}{105 c^3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {8 a^2 (7 b c-6 a d) x}{105 c^4 (b c-a d) \sqrt {c+d x^2}} \]
-1/7*d*x*(b*x^2+a)^3/c/(-a*d+b*c)/(d*x^2+c)^(7/2)+1/35*(-6*a*d+7*b*c)*x*(b *x^2+a)^2/c^2/(-a*d+b*c)/(d*x^2+c)^(5/2)+4/105*a*(-6*a*d+7*b*c)*x*(b*x^2+a )/c^3/(-a*d+b*c)/(d*x^2+c)^(3/2)+8/105*a^2*(-6*a*d+7*b*c)*x/c^4/(-a*d+b*c) /(d*x^2+c)^(1/2)
Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.61 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=\frac {3 b^2 c^2 x^5 \left (7 c+2 d x^2\right )+2 a b c x^3 \left (35 c^2+28 c d x^2+8 d^2 x^4\right )+3 a^2 \left (35 c^3 x+70 c^2 d x^3+56 c d^2 x^5+16 d^3 x^7\right )}{105 c^4 \left (c+d x^2\right )^{7/2}} \]
(3*b^2*c^2*x^5*(7*c + 2*d*x^2) + 2*a*b*c*x^3*(35*c^2 + 28*c*d*x^2 + 8*d^2* x^4) + 3*a^2*(35*c^3*x + 70*c^2*d*x^3 + 56*c*d^2*x^5 + 16*d^3*x^7))/(105*c ^4*(c + d*x^2)^(7/2))
Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {296, 292, 292, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx\) |
| \(\Big \downarrow \) 296 |
| \(\displaystyle \frac {(7 b c-6 a d) \int \frac {\left (b x^2+a\right )^2}{\left (d x^2+c\right )^{7/2}}dx}{7 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^3}{7 c \left (c+d x^2\right )^{7/2} (b c-a d)}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {(7 b c-6 a d) \left (\frac {4 a \int \frac {b x^2+a}{\left (d x^2+c\right )^{5/2}}dx}{5 c}+\frac {x \left (a+b x^2\right )^2}{5 c \left (c+d x^2\right )^{5/2}}\right )}{7 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^3}{7 c \left (c+d x^2\right )^{7/2} (b c-a d)}\) |
| \(\Big \downarrow \) 292 |
| \(\displaystyle \frac {(7 b c-6 a d) \left (\frac {4 a \left (\frac {2 a \int \frac {1}{\left (d x^2+c\right )^{3/2}}dx}{3 c}+\frac {x \left (a+b x^2\right )}{3 c \left (c+d x^2\right )^{3/2}}\right )}{5 c}+\frac {x \left (a+b x^2\right )^2}{5 c \left (c+d x^2\right )^{5/2}}\right )}{7 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^3}{7 c \left (c+d x^2\right )^{7/2} (b c-a d)}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {(7 b c-6 a d) \left (\frac {4 a \left (\frac {x \left (a+b x^2\right )}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 a x}{3 c^2 \sqrt {c+d x^2}}\right )}{5 c}+\frac {x \left (a+b x^2\right )^2}{5 c \left (c+d x^2\right )^{5/2}}\right )}{7 c (b c-a d)}-\frac {d x \left (a+b x^2\right )^3}{7 c \left (c+d x^2\right )^{7/2} (b c-a d)}\) |
-1/7*(d*x*(a + b*x^2)^3)/(c*(b*c - a*d)*(c + d*x^2)^(7/2)) + ((7*b*c - 6*a *d)*((x*(a + b*x^2)^2)/(5*c*(c + d*x^2)^(5/2)) + (4*a*((x*(a + b*x^2))/(3* c*(c + d*x^2)^(3/2)) + (2*a*x)/(3*c^2*Sqrt[c + d*x^2])))/(5*c)))/(7*c*(b*c - a*d))
3.1.98.3.1 Defintions of rubi rules used
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_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( a*(p + 1))) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ {a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt Q[q, 0] && NeQ[p, -1]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Time = 2.38 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.55
| method | result | size |
| pseudoelliptic | \(\frac {x \left (\left (\frac {1}{5} b^{2} x^{4}+\frac {2}{3} a b \,x^{2}+a^{2}\right ) c^{3}+2 x^{2} d \left (\frac {1}{35} b^{2} x^{4}+\frac {4}{15} a b \,x^{2}+a^{2}\right ) c^{2}+\frac {8 x^{4} d^{2} a \left (\frac {2 b \,x^{2}}{21}+a \right ) c}{5}+\frac {16 a^{2} d^{3} x^{6}}{35}\right )}{\left (d \,x^{2}+c \right )^{\frac {7}{2}} c^{4}}\) | \(96\) |
| gosper | \(\frac {x \left (48 a^{2} d^{3} x^{6}+16 a b c \,d^{2} x^{6}+6 b^{2} c^{2} d \,x^{6}+168 a^{2} c \,d^{2} x^{4}+56 a b \,c^{2} d \,x^{4}+21 b^{2} c^{3} x^{4}+210 a^{2} c^{2} d \,x^{2}+70 a b \,c^{3} x^{2}+105 a^{2} c^{3}\right )}{105 \left (d \,x^{2}+c \right )^{\frac {7}{2}} c^{4}}\) | \(115\) |
| trager | \(\frac {x \left (48 a^{2} d^{3} x^{6}+16 a b c \,d^{2} x^{6}+6 b^{2} c^{2} d \,x^{6}+168 a^{2} c \,d^{2} x^{4}+56 a b \,c^{2} d \,x^{4}+21 b^{2} c^{3} x^{4}+210 a^{2} c^{2} d \,x^{2}+70 a b \,c^{3} x^{2}+105 a^{2} c^{3}\right )}{105 \left (d \,x^{2}+c \right )^{\frac {7}{2}} c^{4}}\) | \(115\) |
| default | \(a^{2} \left (\frac {x}{7 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}\right )}{7 c}}{c}\right )+b^{2} \left (-\frac {x^{3}}{4 d \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {3 c \left (-\frac {x}{6 d \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {c \left (\frac {x}{7 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}\right )}{7 c}}{c}\right )}{6 d}\right )}{4 d}\right )+2 a b \left (-\frac {x}{6 d \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {c \left (\frac {x}{7 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}\right )}{7 c}}{c}\right )}{6 d}\right )\) | \(301\) |
x*((1/5*b^2*x^4+2/3*a*b*x^2+a^2)*c^3+2*x^2*d*(1/35*b^2*x^4+4/15*a*b*x^2+a^ 2)*c^2+8/5*x^4*d^2*a*(2/21*b*x^2+a)*c+16/35*a^2*d^3*x^6)/(d*x^2+c)^(7/2)/c ^4
Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=\frac {{\left (2 \, {\left (3 \, b^{2} c^{2} d + 8 \, a b c d^{2} + 24 \, a^{2} d^{3}\right )} x^{7} + 105 \, a^{2} c^{3} x + 7 \, {\left (3 \, b^{2} c^{3} + 8 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} x^{5} + 70 \, {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{105 \, {\left (c^{4} d^{4} x^{8} + 4 \, c^{5} d^{3} x^{6} + 6 \, c^{6} d^{2} x^{4} + 4 \, c^{7} d x^{2} + c^{8}\right )}} \]
1/105*(2*(3*b^2*c^2*d + 8*a*b*c*d^2 + 24*a^2*d^3)*x^7 + 105*a^2*c^3*x + 7* (3*b^2*c^3 + 8*a*b*c^2*d + 24*a^2*c*d^2)*x^5 + 70*(a*b*c^3 + 3*a^2*c^2*d)* x^3)*sqrt(d*x^2 + c)/(c^4*d^4*x^8 + 4*c^5*d^3*x^6 + 6*c^6*d^2*x^4 + 4*c^7* d*x^2 + c^8)
\[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {9}{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=-\frac {b^{2} x^{3}}{4 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} d} + \frac {16 \, a^{2} x}{35 \, \sqrt {d x^{2} + c} c^{4}} + \frac {8 \, a^{2} x}{35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} + \frac {6 \, a^{2} x}{35 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} c^{2}} + \frac {a^{2} x}{7 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} c} + \frac {3 \, b^{2} x}{140 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} d^{2}} + \frac {2 \, b^{2} x}{35 \, \sqrt {d x^{2} + c} c^{2} d^{2}} + \frac {b^{2} x}{35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c d^{2}} - \frac {3 \, b^{2} c x}{28 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} d^{2}} - \frac {2 \, a b x}{7 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} d} + \frac {16 \, a b x}{105 \, \sqrt {d x^{2} + c} c^{3} d} + \frac {8 \, a b x}{105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} d} + \frac {2 \, a b x}{35 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} c d} \]
-1/4*b^2*x^3/((d*x^2 + c)^(7/2)*d) + 16/35*a^2*x/(sqrt(d*x^2 + c)*c^4) + 8 /35*a^2*x/((d*x^2 + c)^(3/2)*c^3) + 6/35*a^2*x/((d*x^2 + c)^(5/2)*c^2) + 1 /7*a^2*x/((d*x^2 + c)^(7/2)*c) + 3/140*b^2*x/((d*x^2 + c)^(5/2)*d^2) + 2/3 5*b^2*x/(sqrt(d*x^2 + c)*c^2*d^2) + 1/35*b^2*x/((d*x^2 + c)^(3/2)*c*d^2) - 3/28*b^2*c*x/((d*x^2 + c)^(7/2)*d^2) - 2/7*a*b*x/((d*x^2 + c)^(7/2)*d) + 16/105*a*b*x/(sqrt(d*x^2 + c)*c^3*d) + 8/105*a*b*x/((d*x^2 + c)^(3/2)*c^2* d) + 2/35*a*b*x/((d*x^2 + c)^(5/2)*c*d)
Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (x^{2} {\left (\frac {2 \, {\left (3 \, b^{2} c^{2} d^{4} + 8 \, a b c d^{5} + 24 \, a^{2} d^{6}\right )} x^{2}}{c^{4} d^{3}} + \frac {7 \, {\left (3 \, b^{2} c^{3} d^{3} + 8 \, a b c^{2} d^{4} + 24 \, a^{2} c d^{5}\right )}}{c^{4} d^{3}}\right )} + \frac {70 \, {\left (a b c^{3} d^{3} + 3 \, a^{2} c^{2} d^{4}\right )}}{c^{4} d^{3}}\right )} x^{2} + \frac {105 \, a^{2}}{c}\right )} x}{105 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}}} \]
1/105*((x^2*(2*(3*b^2*c^2*d^4 + 8*a*b*c*d^5 + 24*a^2*d^6)*x^2/(c^4*d^3) + 7*(3*b^2*c^3*d^3 + 8*a*b*c^2*d^4 + 24*a^2*c*d^5)/(c^4*d^3)) + 70*(a*b*c^3* d^3 + 3*a^2*c^2*d^4)/(c^4*d^3))*x^2 + 105*a^2/c)*x/(d*x^2 + c)^(7/2)
Time = 5.04 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx=\frac {x\,\left (\frac {a^2}{7\,c}+\frac {c\,\left (\frac {b^2}{7\,d}-\frac {2\,a\,b}{7\,c}\right )}{d}\right )}{{\left (d\,x^2+c\right )}^{7/2}}-\frac {x\,\left (\frac {b^2}{5\,d^2}-\frac {6\,a^2\,d^2+2\,a\,b\,c\,d-b^2\,c^2}{35\,c^2\,d^2}\right )}{{\left (d\,x^2+c\right )}^{5/2}}+\frac {x\,\left (24\,a^2\,d^2+8\,a\,b\,c\,d+3\,b^2\,c^2\right )}{105\,c^3\,d^2\,{\left (d\,x^2+c\right )}^{3/2}}+\frac {x\,\left (48\,a^2\,d^2+16\,a\,b\,c\,d+6\,b^2\,c^2\right )}{105\,c^4\,d^2\,\sqrt {d\,x^2+c}} \]
(x*(a^2/(7*c) + (c*(b^2/(7*d) - (2*a*b)/(7*c)))/d))/(c + d*x^2)^(7/2) - (x *(b^2/(5*d^2) - (6*a^2*d^2 - b^2*c^2 + 2*a*b*c*d)/(35*c^2*d^2)))/(c + d*x^ 2)^(5/2) + (x*(24*a^2*d^2 + 3*b^2*c^2 + 8*a*b*c*d))/(105*c^3*d^2*(c + d*x^ 2)^(3/2)) + (x*(48*a^2*d^2 + 6*b^2*c^2 + 16*a*b*c*d))/(105*c^4*d^2*(c + d* x^2)^(1/2))