Integrand size = 22, antiderivative size = 188 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {a^3 (10 A b-3 a B) x \sqrt {a+b x^2}}{256 b^2}+\frac {a^2 (10 A b-3 a B) x^3 \sqrt {a+b x^2}}{128 b}+\frac {a (10 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{96 b}+\frac {(10 A b-3 a B) x^3 \left (a+b x^2\right )^{5/2}}{80 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a^4 (10 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}} \] Output:
1/256*a^3*(10*A*b-3*B*a)*x*(b*x^2+a)^(1/2)/b^2+1/128*a^2*(10*A*b-3*B*a)*x^ 3*(b*x^2+a)^(1/2)/b+1/96*a*(10*A*b-3*B*a)*x^3*(b*x^2+a)^(3/2)/b+1/80*(10*A *b-3*B*a)*x^3*(b*x^2+a)^(5/2)/b+1/10*B*x^3*(b*x^2+a)^(7/2)/b-1/256*a^4*(10 *A*b-3*B*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
Time = 0.67 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.80 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (-45 a^4 B+30 a^3 b \left (5 A+B x^2\right )+96 b^4 x^6 \left (5 A+4 B x^2\right )+16 a b^3 x^4 \left (85 A+63 B x^2\right )+4 a^2 b^2 x^2 \left (295 A+186 B x^2\right )\right )+30 a^4 (-10 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{3840 b^{5/2}} \] Input:
Integrate[x^2*(a + b*x^2)^(5/2)*(A + B*x^2),x]
Output:
(Sqrt[b]*x*Sqrt[a + b*x^2]*(-45*a^4*B + 30*a^3*b*(5*A + B*x^2) + 96*b^4*x^ 6*(5*A + 4*B*x^2) + 16*a*b^3*x^4*(85*A + 63*B*x^2) + 4*a^2*b^2*x^2*(295*A + 186*B*x^2)) + 30*a^4*(-10*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + S qrt[a + b*x^2])])/(3840*b^(5/2))
Time = 0.26 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.85, 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, 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.
| \(\displaystyle \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {(10 A b-3 a B) \int x^2 \left (b x^2+a\right )^{5/2}dx}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(10 A b-3 a B) \left (\frac {5}{8} a \int x^2 \left (b x^2+a\right )^{3/2}dx+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}\right )}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(10 A b-3 a B) \left (\frac {5}{8} a \left (\frac {1}{2} a \int x^2 \sqrt {b x^2+a}dx+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}\right )}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {(10 A b-3 a B) \left (\frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {b x^2+a}}dx+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}\right )}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {(10 A b-3 a B) \left (\frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}\right )}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {(10 A b-3 a B) \left (\frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}\right )}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(10 A b-3 a B) \left (\frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )+\frac {1}{4} x^3 \sqrt {a+b x^2}\right )+\frac {1}{6} x^3 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}\right )}{10 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}\) |
Input:
Int[x^2*(a + b*x^2)^(5/2)*(A + B*x^2),x]
Output:
(B*x^3*(a + b*x^2)^(7/2))/(10*b) + ((10*A*b - 3*a*B)*((x^3*(a + b*x^2)^(5/ 2))/8 + (5*a*((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*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2* b^(3/2))))/4))/2))/8))/(10*b)
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 = 0.69 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(\frac {\frac {59 \left (-\frac {15}{118} a^{4} b A +\frac {9}{236} a^{5} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{192}+\frac {59 \sqrt {b \,x^{2}+a}\, x \left (\frac {15 a^{3} \left (\frac {x^{2} B}{5}+A \right ) b^{\frac {3}{2}}}{118}+a^{2} x^{2} \left (\frac {186 x^{2} B}{295}+A \right ) b^{\frac {5}{2}}+\frac {68 a \,x^{4} \left (\frac {63 x^{2} B}{85}+A \right ) b^{\frac {7}{2}}}{59}+\frac {24 \left (\frac {4 x^{2} B}{5}+A \right ) x^{6} b^{\frac {9}{2}}}{59}-\frac {9 B \,a^{4} \sqrt {b}}{236}\right )}{192}}{b^{\frac {5}{2}}}\) | \(127\) |
| risch | \(\frac {x \left (384 B \,x^{8} b^{4}+480 A \,x^{6} b^{4}+1008 B \,x^{6} a \,b^{3}+1360 A \,x^{4} a \,b^{3}+744 B \,a^{2} b^{2} x^{4}+1180 A \,a^{2} b^{2} x^{2}+30 B \,a^{3} b \,x^{2}+150 A \,a^{3} b -45 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{3840 b^{2}}-\frac {a^{4} \left (10 A b -3 B a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {5}{2}}}\) | \(136\) |
| default | \(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8 b}\right )+B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{10 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8 b}\right )}{10 b}\right )\) | \(208\) |
Input:
int(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x,method=_RETURNVERBOSE)
Output:
59/192*((-15/118*a^4*b*A+9/236*a^5*B)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+( b*x^2+a)^(1/2)*x*(15/118*a^3*(1/5*x^2*B+A)*b^(3/2)+a^2*x^2*(186/295*x^2*B+ A)*b^(5/2)+68/59*a*x^4*(63/85*x^2*B+A)*b^(7/2)+24/59*(4/5*x^2*B+A)*x^6*b^( 9/2)-9/236*B*a^4*b^(1/2)))/b^(5/2)
Time = 0.14 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.64 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\left [-\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (384 \, B b^{5} x^{9} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{5} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x^{3} - 15 \, {\left (3 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{7680 \, b^{3}}, -\frac {15 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (384 \, B b^{5} x^{9} + 48 \, {\left (21 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \, {\left (93 \, B a^{2} b^{3} + 170 \, A a b^{4}\right )} x^{5} + 10 \, {\left (3 \, B a^{3} b^{2} + 118 \, A a^{2} b^{3}\right )} x^{3} - 15 \, {\left (3 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{3840 \, b^{3}}\right ] \] Input:
integrate(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="fricas")
Output:
[-1/7680*(15*(3*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(384*B*b^5*x^9 + 48*(21*B*a*b^4 + 10*A*b^5)*x^7 + 8* (93*B*a^2*b^3 + 170*A*a*b^4)*x^5 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x^3 - 15*(3*B*a^4*b - 10*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/3840*(15*(3*B*a^ 5 - 10*A*a^4*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (384*B*b^5*x ^9 + 48*(21*B*a*b^4 + 10*A*b^5)*x^7 + 8*(93*B*a^2*b^3 + 170*A*a*b^4)*x^5 + 10*(3*B*a^3*b^2 + 118*A*a^2*b^3)*x^3 - 15*(3*B*a^4*b - 10*A*a^3*b^2)*x)*s qrt(b*x^2 + a))/b^3]
Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (173) = 346\).
Time = 0.48 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.95 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\begin {cases} - \frac {a \left (A a^{3} - \frac {3 a \left (3 A a^{2} b + B a^{3} - \frac {5 a \left (3 A a b^{2} + 3 B a^{2} b - \frac {7 a \left (A b^{3} + \frac {21 B a b^{2}}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {B b^{2} x^{9}}{10} + \frac {x^{7} \left (A b^{3} + \frac {21 B a b^{2}}{10}\right )}{8 b} + \frac {x^{5} \cdot \left (3 A a b^{2} + 3 B a^{2} b - \frac {7 a \left (A b^{3} + \frac {21 B a b^{2}}{10}\right )}{8 b}\right )}{6 b} + \frac {x^{3} \cdot \left (3 A a^{2} b + B a^{3} - \frac {5 a \left (3 A a b^{2} + 3 B a^{2} b - \frac {7 a \left (A b^{3} + \frac {21 B a b^{2}}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b} + \frac {x \left (A a^{3} - \frac {3 a \left (3 A a^{2} b + B a^{3} - \frac {5 a \left (3 A a b^{2} + 3 B a^{2} b - \frac {7 a \left (A b^{3} + \frac {21 B a b^{2}}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (\frac {A x^{3}}{3} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(b*x**2+a)**(5/2)*(B*x**2+A),x)
Output:
Piecewise((-a*(A*a**3 - 3*a*(3*A*a**2*b + B*a**3 - 5*a*(3*A*a*b**2 + 3*B*a **2*b - 7*a*(A*b**3 + 21*B*a*b**2/10)/(8*b))/(6*b))/(4*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))/(2*b) + sqrt(a + b*x**2)*(B*b**2*x**9/10 + x**7*(A*b**3 + 21* B*a*b**2/10)/(8*b) + x**5*(3*A*a*b**2 + 3*B*a**2*b - 7*a*(A*b**3 + 21*B*a* b**2/10)/(8*b))/(6*b) + x**3*(3*A*a**2*b + B*a**3 - 5*a*(3*A*a*b**2 + 3*B* a**2*b - 7*a*(A*b**3 + 21*B*a*b**2/10)/(8*b))/(6*b))/(4*b) + x*(A*a**3 - 3 *a*(3*A*a**2*b + B*a**3 - 5*a*(3*A*a*b**2 + 3*B*a**2*b - 7*a*(A*b**3 + 21* B*a*b**2/10)/(8*b))/(6*b))/(4*b))/(2*b)), Ne(b, 0)), (a**(5/2)*(A*x**3/3 + B*x**5/5), True))
Time = 0.05 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.06 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{3}}{10 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a x}{80 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{160 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A a x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} A a^{3} x}{128 \, b} + \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {5}{2}}} - \frac {5 \, A a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} \] Input:
integrate(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="maxima")
Output:
1/10*(b*x^2 + a)^(7/2)*B*x^3/b - 3/80*(b*x^2 + a)^(7/2)*B*a*x/b^2 + 1/160* (b*x^2 + a)^(5/2)*B*a^2*x/b^2 + 1/128*(b*x^2 + a)^(3/2)*B*a^3*x/b^2 + 3/25 6*sqrt(b*x^2 + a)*B*a^4*x/b^2 + 1/8*(b*x^2 + a)^(7/2)*A*x/b - 1/48*(b*x^2 + a)^(5/2)*A*a*x/b - 5/192*(b*x^2 + a)^(3/2)*A*a^2*x/b - 5/128*sqrt(b*x^2 + a)*A*a^3*x/b + 3/256*B*a^5*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 5/128*A*a^4* arcsinh(b*x/sqrt(a*b))/b^(3/2)
Time = 0.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B b^{2} x^{2} + \frac {21 \, B a b^{9} + 10 \, A b^{10}}{b^{8}}\right )} x^{2} + \frac {93 \, B a^{2} b^{8} + 170 \, A a b^{9}}{b^{8}}\right )} x^{2} + \frac {5 \, {\left (3 \, B a^{3} b^{7} + 118 \, A a^{2} b^{8}\right )}}{b^{8}}\right )} x^{2} - \frac {15 \, {\left (3 \, B a^{4} b^{6} - 10 \, A a^{3} b^{7}\right )}}{b^{8}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {5}{2}}} \] Input:
integrate(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="giac")
Output:
1/3840*(2*(4*(6*(8*B*b^2*x^2 + (21*B*a*b^9 + 10*A*b^10)/b^8)*x^2 + (93*B*a ^2*b^8 + 170*A*a*b^9)/b^8)*x^2 + 5*(3*B*a^3*b^7 + 118*A*a^2*b^8)/b^8)*x^2 - 15*(3*B*a^4*b^6 - 10*A*a^3*b^7)/b^8)*sqrt(b*x^2 + a)*x - 1/256*(3*B*a^5 - 10*A*a^4*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\int x^2\,\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2} \,d x \] Input:
int(x^2*(A + B*x^2)*(a + b*x^2)^(5/2),x)
Output:
int(x^2*(A + B*x^2)*(a + b*x^2)^(5/2), x)
Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.63 \[ \int x^2 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {105 \sqrt {b \,x^{2}+a}\, a^{4} b x +1210 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x^{3}+2104 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{5}+1488 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{7}+384 \sqrt {b \,x^{2}+a}\, b^{5} x^{9}-105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{5}}{3840 b^{2}} \] Input:
int(x^2*(b*x^2+a)^(5/2)*(B*x^2+A),x)
Output:
(105*sqrt(a + b*x**2)*a**4*b*x + 1210*sqrt(a + b*x**2)*a**3*b**2*x**3 + 21 04*sqrt(a + b*x**2)*a**2*b**3*x**5 + 1488*sqrt(a + b*x**2)*a*b**4*x**7 + 3 84*sqrt(a + b*x**2)*b**5*x**9 - 105*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b )*x)/sqrt(a))*a**5)/(3840*b**2)