Integrand size = 22, antiderivative size = 155 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=-\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {a^3 (8 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}} \]
1/8*B*x^5*(b*x^2+a)^(3/2)/b+1/128*a^3*(8*A*b-5*B*a)*arctanh(x*b^(1/2)/(b*x ^2+a)^(1/2))/b^(7/2)-1/128*a^2*(8*A*b-5*B*a)*x*(b*x^2+a)^(1/2)/b^3+1/192*a *(8*A*b-5*B*a)*x^3*(b*x^2+a)^(1/2)/b^2+1/48*(8*A*b-5*B*a)*x^5*(b*x^2+a)^(1 /2)/b
Time = 0.49 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (-24 a^2 A b+15 a^3 B+16 a A b^2 x^2-10 a^2 b B x^2+64 A b^3 x^4+8 a b^2 B x^4+48 b^3 B x^6\right )}{384 b^3}-\frac {a^3 (-8 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{64 b^{7/2}} \]
(x*Sqrt[a + b*x^2]*(-24*a^2*A*b + 15*a^3*B + 16*a*A*b^2*x^2 - 10*a^2*b*B*x ^2 + 64*A*b^3*x^4 + 8*a*b^2*B*x^4 + 48*b^3*B*x^6))/(384*b^3) - (a^3*(-8*A* b + 5*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(64*b^(7/2))
Time = 0.24 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {363, 248, 262, 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.
| \(\displaystyle \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(8 A b-5 a B) \int x^4 \sqrt {b x^2+a}dx}{8 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(8 A b-5 a B) \left (\frac {1}{6} a \int \frac {x^4}{\sqrt {b x^2+a}}dx+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )}{8 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(8 A b-5 a B) \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )}{8 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(8 A b-5 a B) \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )}{8 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {(8 A b-5 a B) \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )}{8 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(8 A b-5 a B) \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )}{8 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}\) |
(B*x^5*(a + b*x^2)^(3/2))/(8*b) + ((8*A*b - 5*a*B)*((x^5*Sqrt[a + b*x^2])/ 6 + (a*((x^3*Sqrt[a + b*x^2])/(4*b) - (3*a*((x*Sqrt[a + b*x^2])/(2*b) - (a *ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/(4*b)))/6))/(8*b)
3.6.5.3.1 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 = 2.82 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(\frac {\frac {3 \left (A \,a^{3} b -\frac {5}{8} B \,a^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{2}+x \sqrt {b \,x^{2}+a}\, \left (-\frac {3 \left (\frac {5 x^{2} B}{12}+A \right ) a^{2} b^{\frac {3}{2}}}{2}+x^{2} a \left (\frac {x^{2} B}{2}+A \right ) b^{\frac {5}{2}}+\left (3 B \,x^{6}+4 A \,x^{4}\right ) b^{\frac {7}{2}}+\frac {15 B \,a^{3} \sqrt {b}}{16}\right )}{24 b^{\frac {7}{2}}}\) | \(109\) |
| risch | \(-\frac {x \left (-48 b^{3} B \,x^{6}-64 A \,b^{3} x^{4}-8 B a \,b^{2} x^{4}-16 a A \,b^{2} x^{2}+10 B \,a^{2} b \,x^{2}+24 a^{2} b A -15 a^{3} B \right ) \sqrt {b \,x^{2}+a}}{384 b^{3}}+\frac {a^{3} \left (8 A b -5 B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {7}{2}}}\) | \(112\) |
| default | \(B \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 b}-\frac {5 a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )}{8 b}\right )+A \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )\) | \(192\) |
1/24*(3/2*(A*a^3*b-5/8*B*a^4)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+x*(b*x^2+ a)^(1/2)*(-3/2*(5/12*x^2*B+A)*a^2*b^(3/2)+x^2*a*(1/2*x^2*B+A)*b^(5/2)+(3*B *x^6+4*A*x^4)*b^(7/2)+15/16*B*a^3*b^(1/2)))/b^(7/2)
Time = 0.32 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.66 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\left [-\frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, B b^{4} x^{7} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{5} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, B b^{4} x^{7} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{5} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{4}}\right ] \]
[-1/768*(3*(5*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*(B*a*b^3 + 8*A*b^4)*x^5 - 2*(5*B*a^2* b^2 - 8*A*a*b^3)*x^3 + 3*(5*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^4 , 1/384*(3*(5*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*(B*a*b^3 + 8*A*b^4)*x^5 - 2*(5*B*a^2*b^2 - 8*A*a*b^ 3)*x^3 + 3*(5*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^4]
Time = 0.41 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.13 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\begin {cases} \frac {3 a^{2} \left (A a - \frac {5 a \left (A b + \frac {B a}{8}\right )}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {B x^{7}}{8} - \frac {3 a x \left (A a - \frac {5 a \left (A b + \frac {B a}{8}\right )}{6 b}\right )}{8 b^{2}} + \frac {x^{5} \left (A b + \frac {B a}{8}\right )}{6 b} + \frac {x^{3} \left (A a - \frac {5 a \left (A b + \frac {B a}{8}\right )}{6 b}\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{5}}{5} + \frac {B x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
Piecewise((3*a**2*(A*a - 5*a*(A*b + B*a/8)/(6*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), Tru e))/(8*b**2) + sqrt(a + b*x**2)*(B*x**7/8 - 3*a*x*(A*a - 5*a*(A*b + B*a/8) /(6*b))/(8*b**2) + x**5*(A*b + B*a/8)/(6*b) + x**3*(A*a - 5*a*(A*b + B*a/8 )/(6*b))/(4*b)), Ne(b, 0)), (sqrt(a)*(A*x**5/5 + B*x**7/7), True))
Time = 0.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.07 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{5}}{8 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x^{3}}{48 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{3}}{6 \, b} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A a^{2} x}{16 \, b^{2}} - \frac {5 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \]
1/8*(b*x^2 + a)^(3/2)*B*x^5/b - 5/48*(b*x^2 + a)^(3/2)*B*a*x^3/b^2 + 1/6*( b*x^2 + a)^(3/2)*A*x^3/b + 5/64*(b*x^2 + a)^(3/2)*B*a^2*x/b^3 - 5/128*sqrt (b*x^2 + a)*B*a^3*x/b^3 - 1/8*(b*x^2 + a)^(3/2)*A*a*x/b^2 + 1/16*sqrt(b*x^ 2 + a)*A*a^2*x/b^2 - 5/128*B*a^4*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 1/16*A*a ^3*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B x^{2} + \frac {B a b^{5} + 8 \, A b^{6}}{b^{6}}\right )} x^{2} - \frac {5 \, B a^{2} b^{4} - 8 \, A a b^{5}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{3} b^{3} - 8 \, A a^{2} b^{4}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {7}{2}}} \]
1/384*(2*(4*(6*B*x^2 + (B*a*b^5 + 8*A*b^6)/b^6)*x^2 - (5*B*a^2*b^4 - 8*A*a *b^5)/b^6)*x^2 + 3*(5*B*a^3*b^3 - 8*A*a^2*b^4)/b^6)*sqrt(b*x^2 + a)*x + 1/ 128*(5*B*a^4 - 8*A*a^3*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\int x^4\,\left (B\,x^2+A\right )\,\sqrt {b\,x^2+a} \,d x \]