Integrand size = 21, antiderivative size = 106 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=-\frac {(b c-a d) x \left (a+b x^2\right )}{c d \sqrt {c+d x^2}}+\frac {b (3 b c-2 a d) x \sqrt {c+d x^2}}{2 c d^2}-\frac {b (3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{5/2}} \]
-1/2*b*(-4*a*d+3*b*c)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)-(-a*d+b*c )*x*(b*x^2+a)/c/d/(d*x^2+c)^(1/2)+1/2*b*(-2*a*d+3*b*c)*x*(d*x^2+c)^(1/2)/c /d^2
Time = 0.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {d} x \left (-4 a b c d+2 a^2 d^2+b^2 c \left (3 c+d x^2\right )\right )}{c \sqrt {c+d x^2}}+b (3 b c-4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 d^{5/2}} \]
((Sqrt[d]*x*(-4*a*b*c*d + 2*a^2*d^2 + b^2*c*(3*c + d*x^2)))/(c*Sqrt[c + d* x^2]) + b*(3*b*c - 4*a*d)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(2*d^(5/2))
Time = 0.23 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {315, 27, 299, 224, 219}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\int \frac {b \left ((3 b c-2 a d) x^2+a c\right )}{\sqrt {d x^2+c}}dx}{c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {(3 b c-2 a d) x^2+a c}{\sqrt {d x^2+c}}dx}{c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {b \left (\frac {x \sqrt {c+d x^2} (3 b c-2 a d)}{2 d}-\frac {c (3 b c-4 a d) \int \frac {1}{\sqrt {d x^2+c}}dx}{2 d}\right )}{c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {b \left (\frac {x \sqrt {c+d x^2} (3 b c-2 a d)}{2 d}-\frac {c (3 b c-4 a d) \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{2 d}\right )}{c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b \left (\frac {x \sqrt {c+d x^2} (3 b c-2 a d)}{2 d}-\frac {c (3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}\right )}{c d}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}}\) |
-(((b*c - a*d)*x*(a + b*x^2))/(c*d*Sqrt[c + d*x^2])) + (b*(((3*b*c - 2*a*d )*x*Sqrt[c + d*x^2])/(2*d) - (c*(3*b*c - 4*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*d^(3/2))))/(c*d)
3.7.52.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Time = 2.95 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {d \,x^{2}+c}\, \left (a d -\frac {3 b c}{4}\right ) b c \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \left (-2 \left (-\frac {b \,x^{2}}{4}+a \right ) b c \,d^{\frac {3}{2}}+\frac {3 b^{2} c^{2} \sqrt {d}}{2}+a^{2} d^{\frac {5}{2}}\right )}{\sqrt {d \,x^{2}+c}\, d^{\frac {5}{2}} c}\) | \(92\) |
risch | \(\frac {b^{2} x \sqrt {d \,x^{2}+c}}{2 d^{2}}+\frac {\frac {2 a^{2} d^{2} x}{c \sqrt {d \,x^{2}+c}}-\frac {b^{2} c x}{\sqrt {d \,x^{2}+c}}+\left (4 a b \,d^{2}-3 b^{2} c d \right ) \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d^{2}}\) | \(115\) |
default | \(\frac {a^{2} x}{c \sqrt {d \,x^{2}+c}}+b^{2} \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )+2 a b \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )\) | \(123\) |
1/(d*x^2+c)^(1/2)/d^(5/2)*(2*(d*x^2+c)^(1/2)*(a*d-3/4*b*c)*b*c*arctanh((d* x^2+c)^(1/2)/x/d^(1/2))+x*(-2*(-1/4*b*x^2+a)*b*c*d^(3/2)+3/2*b^2*c^2*d^(1/ 2)+a^2*d^(5/2)))/c
Time = 0.28 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.59 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (b^{2} c d^{2} x^{3} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, \frac {{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d^{2} x^{3} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}\right ] \]
[-1/4*((3*b^2*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*c*d^2)*x^2)*sqrt(d) *log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(b^2*c*d^2*x^3 + (3*b ^2*c^2*d - 4*a*b*c*d^2 + 2*a^2*d^3)*x)*sqrt(d*x^2 + c))/(c*d^4*x^2 + c^2*d ^3), 1/2*((3*b^2*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*c*d^2)*x^2)*sqrt (-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (b^2*c*d^2*x^3 + (3*b^2*c^2*d - 4*a*b*c*d^2 + 2*a^2*d^3)*x)*sqrt(d*x^2 + c))/(c*d^4*x^2 + c^2*d^3)]
\[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{3}}{2 \, \sqrt {d x^{2} + c} d} + \frac {a^{2} x}{\sqrt {d x^{2} + c} c} + \frac {3 \, b^{2} c x}{2 \, \sqrt {d x^{2} + c} d^{2}} - \frac {2 \, a b x}{\sqrt {d x^{2} + c} d} - \frac {3 \, b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {5}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} \]
1/2*b^2*x^3/(sqrt(d*x^2 + c)*d) + a^2*x/(sqrt(d*x^2 + c)*c) + 3/2*b^2*c*x/ (sqrt(d*x^2 + c)*d^2) - 2*a*b*x/(sqrt(d*x^2 + c)*d) - 3/2*b^2*c*arcsinh(d* x/sqrt(c*d))/d^(5/2) + 2*a*b*arcsinh(d*x/sqrt(c*d))/d^(3/2)
Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {b^{2} x^{2}}{d} + \frac {3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}}{c d^{3}}\right )} x}{2 \, \sqrt {d x^{2} + c}} + \frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{2 \, d^{\frac {5}{2}}} \]
1/2*(b^2*x^2/d + (3*b^2*c^2*d - 4*a*b*c*d^2 + 2*a^2*d^3)/(c*d^3))*x/sqrt(d *x^2 + c) + 1/2*(3*b^2*c - 4*a*b*d)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c))) /d^(5/2)
Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \]