\(\int \frac {x^4 (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\) [282]

Optimal result

Integrand size = 22, antiderivative size = 114 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {a (A b-a B) x}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {(4 A b-7 a B) x}{3 b^3 \sqrt {a+b x^2}}+\frac {B x \sqrt {a+b x^2}}{2 b^3}+\frac {(2 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

1/3*a*(A*b-B*a)*x/b^3/(b*x^2+a)^(3/2)-1/3*(4*A*b-7*B*a)*x/b^3/(b*x^2+a)^(1 
/2)+1/2*B*x*(b*x^2+a)^(1/2)/b^3+1/2*(2*A*b-5*B*a)*arctanh(b^(1/2)*x/(b*x^2 
+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x \left (-6 a A b+15 a^2 B-8 A b^2 x^2+20 a b B x^2+3 b^2 B x^4\right )}{6 b^3 \left (a+b x^2\right )^{3/2}}+\frac {(2 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{7/2}} \] Input:

Integrate[(x^4*(A + B*x^2))/(a + b*x^2)^(5/2),x]
 

Output:

(x*(-6*a*A*b + 15*a^2*B - 8*A*b^2*x^2 + 20*a*b*B*x^2 + 3*b^2*B*x^4))/(6*b^ 
3*(a + b*x^2)^(3/2)) + ((2*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sq 
rt[a + b*x^2])])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {360, 1471, 27, 299, 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.

\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 360

\(\displaystyle \frac {a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {-3 b^2 B x^4-3 b (A b-a B) x^2+a (A b-a B)}{\left (b x^2+a\right )^{3/2}}dx}{3 b^3}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {\frac {x (4 A b-7 a B)}{\sqrt {a+b x^2}}-\frac {\int \frac {3 a \left (b B x^2+A b-2 a B\right )}{\sqrt {b x^2+a}}dx}{a}}{3 b^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {\frac {x (4 A b-7 a B)}{\sqrt {a+b x^2}}-3 \int \frac {b B x^2+A b-2 a B}{\sqrt {b x^2+a}}dx}{3 b^3}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {\frac {x (4 A b-7 a B)}{\sqrt {a+b x^2}}-3 \left (\frac {1}{2} (2 A b-5 a B) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} B x \sqrt {a+b x^2}\right )}{3 b^3}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {\frac {x (4 A b-7 a B)}{\sqrt {a+b x^2}}-3 \left (\frac {1}{2} (2 A b-5 a B) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} B x \sqrt {a+b x^2}\right )}{3 b^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac {\frac {x (4 A b-7 a B)}{\sqrt {a+b x^2}}-3 \left (\frac {(2 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} B x \sqrt {a+b x^2}\right )}{3 b^3}\)

Input:

Int[(x^4*(A + B*x^2))/(a + b*x^2)^(5/2),x]
 

Output:

(a*(A*b - a*B)*x)/(3*b^3*(a + b*x^2)^(3/2)) - (((4*A*b - 7*a*B)*x)/Sqrt[a 
+ b*x^2] - 3*((B*x*Sqrt[a + b*x^2])/2 + ((2*A*b - 5*a*B)*ArcTanh[(Sqrt[b]* 
x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(3*b^3)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 224

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]
 

rule 299

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]
 

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 
2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-\frac {\left (-10 x^{2} B +3 A \right ) a x \,b^{\frac {3}{2}}}{3}-\frac {x^{3} \left (-3 x^{2} B +8 A \right ) b^{\frac {5}{2}}}{6}+\frac {5 B \sqrt {b}\, a^{2} x}{2}+\left (A b -\frac {5 B a}{2}\right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{\frac {7}{2}}}\)	\(95\)
default	\(A \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+B \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )\)	\(146\)
risch	\(\frac {B x \sqrt {b \,x^{2}+a}}{2 b^{3}}+\frac {-\frac {5 B a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+2 A \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )-\frac {a \left (A b -B a \right ) \left (-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x -\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b}-\frac {a \left (A b -B a \right ) \left (\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x +\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b}-\frac {\left (3 A b -5 B a \right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{2 b \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {\left (3 A b -5 B a \right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 b \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 b^{3}}\)	\(480\)

Input:

int(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(-1/3*(-10*B*x^2+3*A)*a*x*b^(3/2)-1/6*x^3*(-3*B*x^2+8*A)*b^(5/2)+5/2*B*b^( 
1/2)*a^2*x+(A*b-5/2*B*a)*(b*x^2+a)^(3/2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2) 
))/(b*x^2+a)^(3/2)/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.92 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{4} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, B b^{3} x^{5} + 4 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {3 \, {\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{4} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, B b^{3} x^{5} + 4 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \] Input:

integrate(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*((5*B*a*b^2 - 2*A*b^3)*x^4 + 5*B*a^3 - 2*A*a^2*b + 2*(5*B*a^2*b 
- 2*A*a*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) 
- 2*(3*B*b^3*x^5 + 4*(5*B*a*b^2 - 2*A*b^3)*x^3 + 3*(5*B*a^2*b - 2*A*a*b^2) 
*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4), 1/6*(3*((5*B*a*b^2 
 - 2*A*b^3)*x^4 + 5*B*a^3 - 2*A*a^2*b + 2*(5*B*a^2*b - 2*A*a*b^2)*x^2)*sqr 
t(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*B*b^3*x^5 + 4*(5*B*a*b^2 - 2 
*A*b^3)*x^3 + 3*(5*B*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a 
*b^5*x^2 + a^2*b^4)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 675 vs. \(2 (105) = 210\).

Time = 7.66 (sec) , antiderivative size = 675, normalized size of antiderivative = 5.92 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x**4*(B*x**2+A)/(b*x**2+a)**(5/2),x)
 

Output:

A*(3*a**(39/2)*b**11*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39 
/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x 
**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a 
))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2* 
sqrt(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39/2)*b**(27/2)*sqrt(1 + 
b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(2 
5/2)*x**3/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2 
)*x**2*sqrt(1 + b*x**2/a))) + B*(-15*a**(81/2)*b**22*sqrt(1 + b*x**2/a)*as 
inh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(7 
7/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) - 15*a**(79/2)*b**23*x**2*sqrt(1 + 
 b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2 
/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 15*a**40*b**(45/2)* 
x/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*s 
qrt(1 + b*x**2/a)) + 20*a**39*b**(47/2)*x**3/(6*a**(79/2)*b**(51/2)*sqrt(1 
 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 3*a**38*b* 
*(49/2)*x**5/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(5 
3/2)*x**2*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.40 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {B x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {1}{3} \, A x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {5 \, B a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {5 \, B a x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {A x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {5 \, B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} \] Input:

integrate(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*B*x^5/((b*x^2 + a)^(3/2)*b) - 1/3*A*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2 
*a/((b*x^2 + a)^(3/2)*b^2)) + 5/6*B*a*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a 
/((b*x^2 + a)^(3/2)*b^2))/b + 5/6*B*a*x/(sqrt(b*x^2 + a)*b^3) - 1/3*A*x/(s 
qrt(b*x^2 + a)*b^2) - 5/2*B*a*arcsinh(b*x/sqrt(a*b))/b^(7/2) + A*arcsinh(b 
*x/sqrt(a*b))/b^(5/2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (\frac {3 \, B x^{2}}{b} + \frac {4 \, {\left (5 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )}}{a b^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )}}{a b^{5}}\right )} x}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, B a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

integrate(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

1/6*((3*B*x^2/b + 4*(5*B*a^2*b^3 - 2*A*a*b^4)/(a*b^5))*x^2 + 3*(5*B*a^3*b^ 
2 - 2*A*a^2*b^3)/(a*b^5))*x/(b*x^2 + a)^(3/2) + 1/2*(5*B*a - 2*A*b)*log(ab 
s(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (B\,x^2+A\right )}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((x^4*(A + B*x^2))/(a + b*x^2)^(5/2),x)
 

Output:

int((x^4*(A + B*x^2))/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {12 \sqrt {b \,x^{2}+a}\, a b x +4 \sqrt {b \,x^{2}+a}\, b^{2} x^{3}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}+9 \sqrt {b}\, a^{2}+9 \sqrt {b}\, a b \,x^{2}}{8 b^{3} \left (b \,x^{2}+a \right )} \] Input:

int(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x)
 

Output:

(12*sqrt(a + b*x**2)*a*b*x + 4*sqrt(a + b*x**2)*b**2*x**3 - 12*sqrt(b)*log 
((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2 - 12*sqrt(b)*log((sqrt(a + b 
*x**2) + sqrt(b)*x)/sqrt(a))*a*b*x**2 + 9*sqrt(b)*a**2 + 9*sqrt(b)*a*b*x** 
2)/(8*b**3*(a + b*x**2))