Integrand size = 22, antiderivative size = 155 \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {a^2 (8 A b-3 a B) x \sqrt {a+b x^2}}{128 b^2}+\frac {a (8 A b-3 a B) x^3 \sqrt {a+b x^2}}{64 b}+\frac {(8 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a^3 (8 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \]
1/48*(8*A*b-3*B*a)*x^3*(b*x^2+a)^(3/2)/b+1/8*B*x^3*(b*x^2+a)^(5/2)/b-1/128 *a^3*(8*A*b-3*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)+1/128*a^2*(8 *A*b-3*B*a)*x*(b*x^2+a)^(1/2)/b^2+1/64*a*(8*A*b-3*B*a)*x^3*(b*x^2+a)^(1/2) /b
Time = 0.48 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (-9 a^3 B+6 a^2 b \left (4 A+B x^2\right )+16 b^3 x^4 \left (4 A+3 B x^2\right )+8 a b^2 x^2 \left (14 A+9 B x^2\right )\right )+6 a^3 (-8 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{384 b^{5/2}} \]
(Sqrt[b]*x*Sqrt[a + b*x^2]*(-9*a^3*B + 6*a^2*b*(4*A + B*x^2) + 16*b^3*x^4* (4*A + 3*B*x^2) + 8*a*b^2*x^2*(14*A + 9*B*x^2)) + 6*a^3*(-8*A*b + 3*a*B)*A rcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(384*b^(5/2))
Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {363, 248, 248, 262, 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.
| \(\displaystyle \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(8 A b-3 a B) \int x^2 \left (b x^2+a\right )^{3/2}dx}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(8 A b-3 a B) \left (\frac {1}{2} a \int x^2 \sqrt {b x^2+a}dx+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(8 A b-3 a B) \left (\frac {1}{2} a \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {b x^2+a}}dx+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(8 A b-3 a B) \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {(8 A b-3 a B) \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(8 A b-3 a B) \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )}{8 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}\) |
(B*x^3*(a + b*x^2)^(5/2))/(8*b) + ((8*A*b - 3*a*B)*((x^3*(a + b*x^2)^(3/2) )/6 + (a*((x^3*Sqrt[a + b*x^2])/4 + (a*((x*Sqrt[a + b*x^2])/(2*b) - (a*Arc Tanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/4))/2))/(8*b)
3.6.24.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Time = 2.97 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(\frac {\frac {7 \left (-\frac {3}{14} A \,a^{3} b +\frac {9}{112} B \,a^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{24}+\frac {7 x \sqrt {b \,x^{2}+a}\, \left (\frac {3 \left (\frac {x^{2} B}{4}+A \right ) a^{2} b^{\frac {3}{2}}}{14}+x^{2} a \left (\frac {9 x^{2} B}{14}+A \right ) b^{\frac {5}{2}}+\frac {4 \left (\frac {3 x^{2} B}{4}+A \right ) x^{4} b^{\frac {7}{2}}}{7}-\frac {9 B \,a^{3} \sqrt {b}}{112}\right )}{24}}{b^{\frac {5}{2}}}\) | \(108\) |
| risch | \(\frac {x \left (48 b^{3} B \,x^{6}+64 A \,b^{3} x^{4}+72 B a \,b^{2} x^{4}+112 a A \,b^{2} x^{2}+6 B \,a^{2} b \,x^{2}+24 a^{2} b A -9 a^{3} B \right ) \sqrt {b \,x^{2}+a}}{384 b^{2}}-\frac {a^{3} \left (8 A b -3 B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}\) | \(112\) |
| default | \(B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(176\) |
7/24/b^(5/2)*((-3/14*A*a^3*b+9/112*B*a^4)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2 ))+x*(b*x^2+a)^(1/2)*(3/14*(1/4*x^2*B+A)*a^2*b^(3/2)+x^2*a*(9/14*x^2*B+A)* b^(5/2)+4/7*(3/4*x^2*B+A)*x^4*b^(7/2)-9/112*B*a^3*b^(1/2)))
Time = 0.28 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.68 \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\left [-\frac {3 \, {\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, B b^{4} x^{7} + 8 \, {\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{3}}, -\frac {3 \, {\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (48 \, B b^{4} x^{7} + 8 \, {\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{3}}\right ] \]
[-1/768*(3*(3*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)* sqrt(b)*x - a) - 2*(48*B*b^4*x^7 + 8*(9*B*a*b^3 + 8*A*b^4)*x^5 + 2*(3*B*a^ 2*b^2 + 56*A*a*b^3)*x^3 - 3*(3*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/ b^3, -1/384*(3*(3*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (48*B*b^4*x^7 + 8*(9*B*a*b^3 + 8*A*b^4)*x^5 + 2*(3*B*a^2*b^2 + 56 *A*a*b^3)*x^3 - 3*(3*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^3]
Time = 0.42 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.52 \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\begin {cases} - \frac {a \left (A a^{2} - \frac {3 a \left (2 A a b + B a^{2} - \frac {5 a \left (A b^{2} + \frac {9 B a b}{8}\right )}{6 b}\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {B b x^{7}}{8} + \frac {x^{5} \left (A b^{2} + \frac {9 B a b}{8}\right )}{6 b} + \frac {x^{3} \cdot \left (2 A a b + B a^{2} - \frac {5 a \left (A b^{2} + \frac {9 B a b}{8}\right )}{6 b}\right )}{4 b} + \frac {x \left (A a^{2} - \frac {3 a \left (2 A a b + B a^{2} - \frac {5 a \left (A b^{2} + \frac {9 B a b}{8}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{3}}{3} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-a*(A*a**2 - 3*a*(2*A*a*b + B*a**2 - 5*a*(A*b**2 + 9*B*a*b/8)/( 6*b))/(4*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), N e(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(2*b) + sqrt(a + b*x**2)*(B*b*x** 7/8 + x**5*(A*b**2 + 9*B*a*b/8)/(6*b) + x**3*(2*A*a*b + B*a**2 - 5*a*(A*b* *2 + 9*B*a*b/8)/(6*b))/(4*b) + x*(A*a**2 - 3*a*(2*A*a*b + B*a**2 - 5*a*(A* b**2 + 9*B*a*b/8)/(6*b))/(4*b))/(2*b)), Ne(b, 0)), (a**(3/2)*(A*x**3/3 + B *x**5/5), True))
Time = 0.23 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05 \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{3}}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} A a^{2} x}{16 \, b} + \frac {3 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} \]
1/8*(b*x^2 + a)^(5/2)*B*x^3/b - 1/16*(b*x^2 + a)^(5/2)*B*a*x/b^2 + 1/64*(b *x^2 + a)^(3/2)*B*a^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*B*a^3*x/b^2 + 1/6*(b*x ^2 + a)^(5/2)*A*x/b - 1/24*(b*x^2 + a)^(3/2)*A*a*x/b - 1/16*sqrt(b*x^2 + a )*A*a^2*x/b + 3/128*B*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/16*A*a^3*arcs inh(b*x/sqrt(a*b))/b^(3/2)
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.86 \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B b x^{2} + \frac {9 \, B a b^{6} + 8 \, A b^{7}}{b^{6}}\right )} x^{2} + \frac {3 \, B a^{2} b^{5} + 56 \, A a b^{6}}{b^{6}}\right )} x^{2} - \frac {3 \, {\left (3 \, B a^{3} b^{4} - 8 \, A a^{2} b^{5}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} \]
1/384*(2*(4*(6*B*b*x^2 + (9*B*a*b^6 + 8*A*b^7)/b^6)*x^2 + (3*B*a^2*b^5 + 5 6*A*a*b^6)/b^6)*x^2 - 3*(3*B*a^3*b^4 - 8*A*a^2*b^5)/b^6)*sqrt(b*x^2 + a)*x - 1/128*(3*B*a^4 - 8*A*a^3*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5 /2)
Timed out. \[ \int x^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\int x^2\,\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2} \,d x \]