\(\int \frac {x^5 (A+B x)}{(a+b x^2)^{5/2}} \, dx\) [1257]

Optimal result

Integrand size = 20, antiderivative size = 120 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac {a (6 A+7 B x)}{3 b^3 \sqrt {a+b x^2}}+\frac {A \sqrt {a+b x^2}}{b^3}+\frac {B x \sqrt {a+b x^2}}{2 b^3}-\frac {5 a B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3 b^2 x^4 (2 A+B x)+4 a b x^2 (6 A+5 B x)+a^2 (16 A+15 B x)}{6 b^3 \left (a+b x^2\right )^{3/2}}+\frac {5 a B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{7/2}} \] Input:

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

Output:

(3*b^2*x^4*(2*A + B*x) + 4*a*b*x^2*(6*A + 5*B*x) + a^2*(16*A + 15*B*x))/(6 
*b^3*(a + b*x^2)^(3/2)) + (5*a*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b 
^(7/2))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {530, 25, 2345, 27, 2346, 27, 455, 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.

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

\(\Big \downarrow \) 530

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2345

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {\int \frac {3 \left (\frac {2 B a^3}{b^3}-\frac {B x^2 a^2}{b^2}-\frac {A x a^2}{b^2}\right )}{\sqrt {b x^2+a}}dx}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {3 \int \frac {\frac {2 B a^3}{b^3}-\frac {B x^2 a^2}{b^2}-\frac {A x a^2}{b^2}}{\sqrt {b x^2+a}}dx}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {3 \left (\frac {\int \frac {a^2 (5 a B-2 A b x)}{b^2 \sqrt {b x^2+a}}dx}{2 b}-\frac {a^2 B x \sqrt {a+b x^2}}{2 b^3}\right )}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {3 \left (\frac {a^2 \int \frac {5 a B-2 A b x}{\sqrt {b x^2+a}}dx}{2 b^3}-\frac {a^2 B x \sqrt {a+b x^2}}{2 b^3}\right )}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {3 \left (\frac {a^2 \left (5 a B \int \frac {1}{\sqrt {b x^2+a}}dx-2 A \sqrt {a+b x^2}\right )}{2 b^3}-\frac {a^2 B x \sqrt {a+b x^2}}{2 b^3}\right )}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {3 \left (\frac {a^2 \left (5 a B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-2 A \sqrt {a+b x^2}\right )}{2 b^3}-\frac {a^2 B x \sqrt {a+b x^2}}{2 b^3}\right )}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {a^2 (6 A+7 B x)}{b^3 \sqrt {a+b x^2}}-\frac {3 \left (\frac {a^2 \left (\frac {5 a B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-2 A \sqrt {a+b x^2}\right )}{2 b^3}-\frac {a^2 B x \sqrt {a+b x^2}}{2 b^3}\right )}{a}}{3 a}-\frac {a^2 (A+B x)}{3 b^3 \left (a+b x^2\right )^{3/2}}\)

Input:

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

Output:

-1/3*(a^2*(A + B*x))/(b^3*(a + b*x^2)^(3/2)) + ((a^2*(6*A + 7*B*x))/(b^3*S 
qrt[a + b*x^2]) - (3*(-1/2*(a^2*B*x*Sqrt[a + b*x^2])/b^3 + (a^2*(-2*A*Sqrt 
[a + b*x^2] + (5*a*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/(2*b^ 
3)))/a)/(3*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 530

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co 
eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po 
lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x 
)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(a 
 + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; 
FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 
1] && IntegerQ[2*p]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.20

method	result	size
default	\(A \left (\frac {x^{4}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )}{b}\right )+B \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )\)	\(144\)
risch	\(\frac {\left (B x +2 A \right ) \sqrt {b \,x^{2}+a}}{2 b^{3}}-\frac {a \left (\frac {5 B \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {\left (A \sqrt {-a b}-B a \right ) \left (-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x -\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b}-\frac {\left (A \sqrt {-a b}+B a \right ) \left (\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x +\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b}+\frac {\left (4 A \sqrt {-a b}-5 B a \right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{2 a b \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {\left (4 A \sqrt {-a b}+5 B a \right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 a b \left (x +\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b^{3}}\)	\(486\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.44 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (B a b^{2} x^{4} + 2 \, B a^{2} b x^{2} + B a^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (3 \, B b^{3} x^{5} + 6 \, A b^{3} x^{4} + 20 \, B a b^{2} x^{3} + 24 \, A a b^{2} x^{2} + 15 \, B a^{2} b x + 16 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{12 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {15 \, {\left (B a b^{2} x^{4} + 2 \, B a^{2} b x^{2} + B a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, B b^{3} x^{5} + 6 \, A b^{3} x^{4} + 20 \, B a b^{2} x^{3} + 24 \, A a b^{2} x^{2} + 15 \, B a^{2} b x + 16 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \] Input:

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

Output:

[1/12*(15*(B*a*b^2*x^4 + 2*B*a^2*b*x^2 + B*a^3)*sqrt(b)*log(-2*b*x^2 + 2*s 
qrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(3*B*b^3*x^5 + 6*A*b^3*x^4 + 20*B*a*b^2* 
x^3 + 24*A*a*b^2*x^2 + 15*B*a^2*b*x + 16*A*a^2*b)*sqrt(b*x^2 + a))/(b^6*x^ 
4 + 2*a*b^5*x^2 + a^2*b^4), 1/6*(15*(B*a*b^2*x^4 + 2*B*a^2*b*x^2 + B*a^3)* 
sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*B*b^3*x^5 + 6*A*b^3*x^4 + 
 20*B*a*b^2*x^3 + 24*A*a*b^2*x^2 + 15*B*a^2*b*x + 16*A*a^2*b)*sqrt(b*x^2 + 
 a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4)]
 

Sympy [A] (verification not implemented)

Time = 7.72 (sec) , antiderivative size = 510, normalized size of antiderivative = 4.25 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=A \left (\begin {cases} \frac {8 a^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {12 a b x^{2}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} + \frac {3 b^{2} x^{4}}{3 a b^{3} \sqrt {a + b x^{2}} + 3 b^{4} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (- \frac {15 a^{\frac {81}{2}} b^{22} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {15 a^{\frac {79}{2}} b^{23} x^{2} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{40} b^{\frac {45}{2}} x}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {20 a^{39} b^{\frac {47}{2}} x^{3}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{38} b^{\frac {49}{2}} x^{5}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:

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

Output:

A*Piecewise((8*a**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x* 
*2)) + 12*a*b*x**2/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x** 
2)) + 3*b**2*x**4/(3*a*b**3*sqrt(a + b*x**2) + 3*b**4*x**2*sqrt(a + b*x**2 
)), Ne(b, 0)), (x**6/(6*a**(5/2)), True)) + B*(-15*a**(81/2)*b**22*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**(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)*s 
qrt(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*sqrt(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**(53/2)*x**2*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.22 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {B x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {5 \, B a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {A x^{4}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {4 \, A a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {5 \, B a x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {5 \, B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {8 \, A a^{2}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} \] Input:

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

Output:

1/2*B*x^5/((b*x^2 + a)^(3/2)*b) + 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 + A*x^4/((b*x^2 + a)^(3/2)*b) + 4*A*a*x^2/ 
((b*x^2 + a)^(3/2)*b^2) + 5/6*B*a*x/(sqrt(b*x^2 + a)*b^3) - 5/2*B*a*arcsin 
h(b*x/sqrt(a*b))/b^(7/2) + 8/3*A*a^2/((b*x^2 + a)^(3/2)*b^3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.79 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left ({\left (3 \, {\left (\frac {B x}{b} + \frac {2 \, A}{b}\right )} x + \frac {20 \, B a}{b^{2}}\right )} x + \frac {24 \, A a}{b^{2}}\right )} x + \frac {15 \, B a^{2}}{b^{3}}\right )} x + \frac {16 \, A a^{2}}{b^{3}}}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {5 \, B a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

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

Output:

1/6*((((3*(B*x/b + 2*A/b)*x + 20*B*a/b^2)*x + 24*A*a/b^2)*x + 15*B*a^2/b^3 
)*x + 16*A*a^2/b^3)/(b*x^2 + a)^(3/2) + 5/2*B*a*log(abs(-sqrt(b)*x + sqrt( 
b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.97 \[ \int \frac {x^5 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {32 \sqrt {b \,x^{2}+a}\, a^{3}+48 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}+30 \sqrt {b \,x^{2}+a}\, a^{2} b x +12 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+40 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+6 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}-60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,x^{2}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} x^{4}-5 \sqrt {b}\, a^{3}-10 \sqrt {b}\, a^{2} b \,x^{2}-5 \sqrt {b}\, a \,b^{2} x^{4}}{12 b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

(32*sqrt(a + b*x**2)*a**3 + 48*sqrt(a + b*x**2)*a**2*b*x**2 + 30*sqrt(a + 
b*x**2)*a**2*b*x + 12*sqrt(a + b*x**2)*a*b**2*x**4 + 40*sqrt(a + b*x**2)*a 
*b**2*x**3 + 6*sqrt(a + b*x**2)*b**3*x**5 - 30*sqrt(b)*log((sqrt(a + b*x** 
2) + sqrt(b)*x)/sqrt(a))*a**3 - 60*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b) 
*x)/sqrt(a))*a**2*b*x**2 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/s 
qrt(a))*a*b**2*x**4 - 5*sqrt(b)*a**3 - 10*sqrt(b)*a**2*b*x**2 - 5*sqrt(b)* 
a*b**2*x**4)/(12*b**3*(a**2 + 2*a*b*x**2 + b**2*x**4))