Integrand size = 21, antiderivative size = 149 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {\left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) x \sqrt {a+b x^2}}{16 b^2}+\frac {d (8 b c-3 a d) x \left (a+b x^2\right )^{3/2}}{24 b^2}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b}+\frac {a \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \]
1/24*d*(-3*a*d+8*b*c)*x*(b*x^2+a)^(3/2)/b^2+1/6*d*x*(b*x^2+a)^(3/2)*(d*x^2 +c)/b+1/16*a*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/ 2))/b^(5/2)+1/16*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)*x*(b*x^2+a)^(1/2)/b^2
Time = 0.16 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (-3 a^2 d^2+2 a b d \left (6 c+d x^2\right )+8 b^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )\right )-3 a \left (8 b^2 c^2-4 a b c d+a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{48 b^{5/2}} \]
(Sqrt[b]*x*Sqrt[a + b*x^2]*(-3*a^2*d^2 + 2*a*b*d*(6*c + d*x^2) + 8*b^2*(3* c^2 + 3*c*d*x^2 + d^2*x^4)) - 3*a*(8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Log[-( Sqrt[b]*x) + Sqrt[a + b*x^2]])/(48*b^(5/2))
Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {318, 299, 211, 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.
| \(\displaystyle \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {\int \sqrt {b x^2+a} \left (d (8 b c-3 a d) x^2+c (6 b c-a d)\right )dx}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {3 \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \int \sqrt {b x^2+a}dx}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} (8 b c-3 a d)}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {3 \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} (8 b c-3 a d)}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {3 \left (a^2 d^2-4 a b c d+8 b^2 c^2\right ) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} (8 b c-3 a d)}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3 \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) \left (a^2 d^2-4 a b c d+8 b^2 c^2\right )}{4 b}+\frac {d x \left (a+b x^2\right )^{3/2} (8 b c-3 a d)}{4 b}}{6 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}{6 b}\) |
(d*x*(a + b*x^2)^(3/2)*(c + d*x^2))/(6*b) + ((d*(8*b*c - 3*a*d)*x*(a + b*x ^2)^(3/2))/(4*b) + (3*(8*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*((x*Sqrt[a + b*x^2 ])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4*b))/(6*b)
3.1.46.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Time = 2.36 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(\frac {a \left (a^{2} d^{2}-4 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-x \sqrt {b \,x^{2}+a}\, \left (\left (-\frac {8}{3} d^{2} x^{4}-8 c d \,x^{2}-8 c^{2}\right ) b^{\frac {5}{2}}+d \left (\left (-\frac {2 d \,x^{2}}{3}-4 c \right ) b^{\frac {3}{2}}+a d \sqrt {b}\right ) a \right )}{16 b^{\frac {5}{2}}}\) | \(110\) |
| risch | \(-\frac {x \left (-8 b^{2} d^{2} x^{4}-2 x^{2} a b \,d^{2}-24 x^{2} b^{2} c d +3 a^{2} d^{2}-12 a b c d -24 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}}{48 b^{2}}+\frac {a \left (a^{2} d^{2}-4 a b c d +8 b^{2} c^{2}\right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\) | \(115\) |
| default | \(c^{2} \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+2 c d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\) | \(187\) |
1/16*(a*(a^2*d^2-4*a*b*c*d+8*b^2*c^2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))-x *(b*x^2+a)^(1/2)*((-8/3*d^2*x^4-8*c*d*x^2-8*c^2)*b^(5/2)+d*((-2/3*d*x^2-4* c)*b^(3/2)+a*d*b^(1/2))*a))/b^(5/2)
Time = 0.29 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.77 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\left [\frac {3 \, {\left (8 \, a b^{2} c^{2} - 4 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} d^{2} x^{5} + 2 \, {\left (12 \, b^{3} c d + a b^{2} d^{2}\right )} x^{3} + 3 \, {\left (8 \, b^{3} c^{2} + 4 \, a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{3}}, -\frac {3 \, {\left (8 \, a b^{2} c^{2} - 4 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d^{2} x^{5} + 2 \, {\left (12 \, b^{3} c d + a b^{2} d^{2}\right )} x^{3} + 3 \, {\left (8 \, b^{3} c^{2} + 4 \, a b^{2} c d - a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}\right ] \]
[1/96*(3*(8*a*b^2*c^2 - 4*a^2*b*c*d + a^3*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sq rt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*b^3*d^2*x^5 + 2*(12*b^3*c*d + a*b^2*d^ 2)*x^3 + 3*(8*b^3*c^2 + 4*a*b^2*c*d - a^2*b*d^2)*x)*sqrt(b*x^2 + a))/b^3, -1/48*(3*(8*a*b^2*c^2 - 4*a^2*b*c*d + a^3*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/ sqrt(b*x^2 + a)) - (8*b^3*d^2*x^5 + 2*(12*b^3*c*d + a*b^2*d^2)*x^3 + 3*(8* b^3*c^2 + 4*a*b^2*c*d - a^2*b*d^2)*x)*sqrt(b*x^2 + a))/b^3]
Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.27 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {d^{2} x^{5}}{6} + \frac {x^{3} \left (\frac {a d^{2}}{6} + 2 b c d\right )}{4 b} + \frac {x \left (2 a c d - \frac {3 a \left (\frac {a d^{2}}{6} + 2 b c d\right )}{4 b} + b c^{2}\right )}{2 b}\right ) + \left (a c^{2} - \frac {a \left (2 a c d - \frac {3 a \left (\frac {a d^{2}}{6} + 2 b c d\right )}{4 b} + b c^{2}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x**2)*(d**2*x**5/6 + x**3*(a*d**2/6 + 2*b*c*d)/(4*b) + x*(2*a*c*d - 3*a*(a*d**2/6 + 2*b*c*d)/(4*b) + b*c**2)/(2*b)) + (a*c**2 - a*(2*a*c*d - 3*a*(a*d**2/6 + 2*b*c*d)/(4*b) + b*c**2)/(2*b))*Piecewise(( log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt (b*x**2), True)), Ne(b, 0)), (sqrt(a)*(c**2*x + 2*c*d*x**3/3 + d**2*x**5/5 ), True))
Time = 0.21 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.13 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} d^{2} x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} c^{2} x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} c d x}{2 \, b} - \frac {\sqrt {b x^{2} + a} a c d x}{4 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{2} x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} a^{2} d^{2} x}{16 \, b^{2}} + \frac {a c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} - \frac {a^{2} c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{4 \, b^{\frac {3}{2}}} + \frac {a^{3} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \]
1/6*(b*x^2 + a)^(3/2)*d^2*x^3/b + 1/2*sqrt(b*x^2 + a)*c^2*x + 1/2*(b*x^2 + a)^(3/2)*c*d*x/b - 1/4*sqrt(b*x^2 + a)*a*c*d*x/b - 1/8*(b*x^2 + a)^(3/2)* a*d^2*x/b^2 + 1/16*sqrt(b*x^2 + a)*a^2*d^2*x/b^2 + 1/2*a*c^2*arcsinh(b*x/s qrt(a*b))/sqrt(b) - 1/4*a^2*c*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 1/16*a^3* d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.87 \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, d^{2} x^{2} + \frac {12 \, b^{4} c d + a b^{3} d^{2}}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (8 \, b^{4} c^{2} + 4 \, a b^{3} c d - a^{2} b^{2} d^{2}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (8 \, a b^{2} c^{2} - 4 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \]
1/48*(2*(4*d^2*x^2 + (12*b^4*c*d + a*b^3*d^2)/b^4)*x^2 + 3*(8*b^4*c^2 + 4* a*b^3*c*d - a^2*b^2*d^2)/b^4)*sqrt(b*x^2 + a)*x - 1/16*(8*a*b^2*c^2 - 4*a^ 2*b*c*d + a^3*d^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \sqrt {a+b x^2} \left (c+d x^2\right )^2 \, dx=\int \sqrt {b\,x^2+a}\,{\left (d\,x^2+c\right )}^2 \,d x \]