Integrand size = 22, antiderivative size = 188 \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=-\frac {3 a^3 (2 A b-a B) x \sqrt {a+b x^2}}{256 b^3}+\frac {a^2 (2 A b-a B) x^3 \sqrt {a+b x^2}}{128 b^2}+\frac {a (2 A b-a B) x^5 \sqrt {a+b x^2}}{32 b}+\frac {(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {3 a^4 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{7/2}} \]
1/16*(2*A*b-B*a)*x^5*(b*x^2+a)^(3/2)/b+1/10*B*x^5*(b*x^2+a)^(5/2)/b+3/256* a^4*(2*A*b-B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(7/2)-3/256*a^3*(2*A* b-B*a)*x*(b*x^2+a)^(1/2)/b^3+1/128*a^2*(2*A*b-B*a)*x^3*(b*x^2+a)^(1/2)/b^2 +1/32*a*(2*A*b-B*a)*x^5*(b*x^2+a)^(1/2)/b
Time = 0.61 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.80 \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (15 a^4 B-10 a^3 b \left (3 A+B x^2\right )+4 a^2 b^2 x^2 \left (5 A+2 B x^2\right )+32 b^4 x^6 \left (5 A+4 B x^2\right )+16 a b^3 x^4 \left (15 A+11 B x^2\right )\right )+30 a^4 (-2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a}-\sqrt {a+b x^2}}\right )}{1280 b^{7/2}} \]
(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^4*B - 10*a^3*b*(3*A + B*x^2) + 4*a^2*b^2* x^2*(5*A + 2*B*x^2) + 32*b^4*x^6*(5*A + 4*B*x^2) + 16*a*b^3*x^4*(15*A + 11 *B*x^2)) + 30*a^4*(-2*A*b + a*B)*ArcTanh[(Sqrt[b]*x)/(Sqrt[a] - Sqrt[a + b *x^2])])/(1280*b^(7/2))
Time = 0.26 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {363, 248, 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.
| \(\displaystyle \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(2 A b-a B) \int x^4 \left (b x^2+a\right )^{3/2}dx}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(2 A b-a B) \left (\frac {3}{8} a \int x^4 \sqrt {b x^2+a}dx+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(2 A b-a B) \left (\frac {3}{8} a \left (\frac {1}{6} a \int \frac {x^4}{\sqrt {b x^2+a}}dx+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(2 A b-a B) \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(2 A b-a B) \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {(2 A b-a B) \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(2 A b-a B) \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {B x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
(B*x^5*(a + b*x^2)^(5/2))/(10*b) + ((2*A*b - a*B)*((x^5*(a + b*x^2)^(3/2)) /8 + (3*a*((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))/(2*b)
3.6.22.3.1 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 = 2.86 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(\frac {\left (\frac {3}{2} a^{4} b A -\frac {3}{4} a^{5} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+x \left (-\frac {3 \left (\frac {x^{2} B}{3}+A \right ) a^{3} b^{\frac {3}{2}}}{2}+a^{2} x^{2} \left (\frac {2 x^{2} B}{5}+A \right ) b^{\frac {5}{2}}+12 x^{4} \left (\frac {11 x^{2} B}{15}+A \right ) a \,b^{\frac {7}{2}}+8 x^{6} \left (\frac {4 x^{2} B}{5}+A \right ) b^{\frac {9}{2}}+\frac {3 B \,a^{4} \sqrt {b}}{4}\right ) \sqrt {b \,x^{2}+a}}{64 b^{\frac {7}{2}}}\) | \(127\) |
| risch | \(-\frac {x \left (-128 B \,x^{8} b^{4}-160 A \,x^{6} b^{4}-176 B \,x^{6} a \,b^{3}-240 A a \,b^{3} x^{4}-8 B \,a^{2} b^{2} x^{4}-20 A \,a^{2} b^{2} x^{2}+10 B \,a^{3} b \,x^{2}+30 A \,a^{3} b -15 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{1280 b^{3}}+\frac {3 a^{4} \left (2 A b -B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {7}{2}}}\) | \(136\) |
| default | \(B \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{10 b}-\frac {a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )}{2 b}\right )+A \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )\) | \(224\) |
1/64/b^(7/2)*((3/2*a^4*b*A-3/4*a^5*B)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+x *(-3/2*(1/3*x^2*B+A)*a^3*b^(3/2)+a^2*x^2*(2/5*x^2*B+A)*b^(5/2)+12*x^4*(11/ 15*x^2*B+A)*a*b^(7/2)+8*x^6*(4/5*x^2*B+A)*b^(9/2)+3/4*B*a^4*b^(1/2))*(b*x^ 2+a)^(1/2))
Time = 0.30 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.59 \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\left [-\frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (128 \, B b^{5} x^{9} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{5} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} + 15 \, {\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{2560 \, b^{4}}, \frac {15 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (128 \, B b^{5} x^{9} + 16 \, {\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{5} - 10 \, {\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} + 15 \, {\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{1280 \, b^{4}}\right ] \]
[-1/2560*(15*(B*a^5 - 2*A*a^4*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)* sqrt(b)*x - a) - 2*(128*B*b^5*x^9 + 16*(11*B*a*b^4 + 10*A*b^5)*x^7 + 8*(B* a^2*b^3 + 30*A*a*b^4)*x^5 - 10*(B*a^3*b^2 - 2*A*a^2*b^3)*x^3 + 15*(B*a^4*b - 2*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^4, 1/1280*(15*(B*a^5 - 2*A*a^4*b)*sq rt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (128*B*b^5*x^9 + 16*(11*B*a*b^ 4 + 10*A*b^5)*x^7 + 8*(B*a^2*b^3 + 30*A*a*b^4)*x^5 - 10*(B*a^3*b^2 - 2*A*a ^2*b^3)*x^3 + 15*(B*a^4*b - 2*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^4]
Time = 0.47 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.59 \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\begin {cases} \frac {3 a^{2} \left (A a^{2} - \frac {5 a \left (2 A a b + B a^{2} - \frac {7 a \left (A b^{2} + \frac {11 B a b}{10}\right )}{8 b}\right )}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {B b x^{9}}{10} - \frac {3 a x \left (A a^{2} - \frac {5 a \left (2 A a b + B a^{2} - \frac {7 a \left (A b^{2} + \frac {11 B a b}{10}\right )}{8 b}\right )}{6 b}\right )}{8 b^{2}} + \frac {x^{7} \left (A b^{2} + \frac {11 B a b}{10}\right )}{8 b} + \frac {x^{5} \cdot \left (2 A a b + B a^{2} - \frac {7 a \left (A b^{2} + \frac {11 B a b}{10}\right )}{8 b}\right )}{6 b} + \frac {x^{3} \left (A a^{2} - \frac {5 a \left (2 A a b + B a^{2} - \frac {7 a \left (A b^{2} + \frac {11 B a b}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{5}}{5} + \frac {B x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
Piecewise((3*a**2*(A*a**2 - 5*a*(2*A*a*b + B*a**2 - 7*a*(A*b**2 + 11*B*a*b /10)/(8*b))/(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), True))/(8*b**2) + sqrt(a + b*x**2) *(B*b*x**9/10 - 3*a*x*(A*a**2 - 5*a*(2*A*a*b + B*a**2 - 7*a*(A*b**2 + 11*B *a*b/10)/(8*b))/(6*b))/(8*b**2) + x**7*(A*b**2 + 11*B*a*b/10)/(8*b) + x**5 *(2*A*a*b + B*a**2 - 7*a*(A*b**2 + 11*B*a*b/10)/(8*b))/(6*b) + x**3*(A*a** 2 - 5*a*(2*A*a*b + B*a**2 - 7*a*(A*b**2 + 11*B*a*b/10)/(8*b))/(6*b))/(4*b) ), Ne(b, 0)), (a**(3/2)*(A*x**5/5 + B*x**7/7), True))
Time = 0.20 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09 \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{5}}{10 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x^{3}}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x^{3}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{32 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} A a^{3} x}{128 \, b^{2}} - \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} \]
1/10*(b*x^2 + a)^(5/2)*B*x^5/b - 1/16*(b*x^2 + a)^(5/2)*B*a*x^3/b^2 + 1/8* (b*x^2 + a)^(5/2)*A*x^3/b + 1/32*(b*x^2 + a)^(5/2)*B*a^2*x/b^3 - 1/128*(b* x^2 + a)^(3/2)*B*a^3*x/b^3 - 3/256*sqrt(b*x^2 + a)*B*a^4*x/b^3 - 1/16*(b*x ^2 + a)^(5/2)*A*a*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*A*a^2*x/b^2 + 3/128*sqrt( b*x^2 + a)*A*a^3*x/b^2 - 3/256*B*a^5*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 3/12 8*A*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.85 \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B b x^{2} + \frac {11 \, B a b^{8} + 10 \, A b^{9}}{b^{8}}\right )} x^{2} + \frac {B a^{2} b^{7} + 30 \, A a b^{8}}{b^{8}}\right )} x^{2} - \frac {5 \, {\left (B a^{3} b^{6} - 2 \, A a^{2} b^{7}\right )}}{b^{8}}\right )} x^{2} + \frac {15 \, {\left (B a^{4} b^{5} - 2 \, A a^{3} b^{6}\right )}}{b^{8}}\right )} \sqrt {b x^{2} + a} x + \frac {3 \, {\left (B a^{5} - 2 \, A a^{4} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {7}{2}}} \]
1/1280*(2*(4*(2*(8*B*b*x^2 + (11*B*a*b^8 + 10*A*b^9)/b^8)*x^2 + (B*a^2*b^7 + 30*A*a*b^8)/b^8)*x^2 - 5*(B*a^3*b^6 - 2*A*a^2*b^7)/b^8)*x^2 + 15*(B*a^4 *b^5 - 2*A*a^3*b^6)/b^8)*sqrt(b*x^2 + a)*x + 3/256*(B*a^5 - 2*A*a^4*b)*log (abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\int x^4\,\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2} \,d x \]