\(\int \frac {(A+B x) (a+b x^2)^{5/2}}{x^2} \, dx\) [1119]

Optimal result

Integrand size = 20, antiderivative size = 136 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\frac {1}{8} a (8 a B+15 A b x) \sqrt {a+b x^2}+\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}+\frac {15}{8} a^2 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-a^{5/2} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:

1/8*a*(15*A*b*x+8*B*a)*(b*x^2+a)^(1/2)+1/12*(15*A*b*x+4*B*a)*(b*x^2+a)^(3/ 
2)-1/5*(-B*x+5*A)*(b*x^2+a)^(5/2)/x+15/8*a^2*A*b^(1/2)*arctanh(b^(1/2)*x/( 
b*x^2+a)^(1/2))-a^(5/2)*B*arctanh((b*x^2+a)^(1/2)/a^(1/2))
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-8 a^2 (15 A-23 B x)+6 b^2 x^4 (5 A+4 B x)+a b x^2 (135 A+88 B x)\right )}{120 x}+2 a^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {15}{8} a^2 A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-8*a^2*(15*A - 23*B*x) + 6*b^2*x^4*(5*A + 4*B*x) + a*b*x 
^2*(135*A + 88*B*x)))/(120*x) + 2*a^(5/2)*B*ArcTanh[(Sqrt[b]*x - Sqrt[a + 
b*x^2])/Sqrt[a]] - (15*a^2*A*Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/ 
8
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {536, 535, 535, 538, 224, 219, 243, 73, 221}

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.

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

\(\Big \downarrow \) 536

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

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{4} a \int \frac {(4 a B+15 A b x) \sqrt {b x^2+a}}{x}dx-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \int \frac {8 a B+15 A b x}{x \sqrt {b x^2+a}}dx+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (15 A b \int \frac {1}{\sqrt {b x^2+a}}dx+8 a B \int \frac {1}{x \sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (15 A b \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+8 a B \int \frac {1}{x \sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (8 a B \int \frac {1}{x \sqrt {b x^2+a}}dx+15 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (4 a B \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+15 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (\frac {8 a B \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+15 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (15 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-8 \sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {1}{2} \sqrt {a+b x^2} (8 a B+15 A b x)\right )-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x)\)

Input:

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

Output:

((4*a*B + 15*A*b*x)*(a + b*x^2)^(3/2))/12 - ((5*A - B*x)*(a + b*x^2)^(5/2) 
)/(5*x) + (a*(((8*a*B + 15*A*b*x)*Sqrt[a + b*x^2])/2 + (a*(15*A*Sqrt[b]*Ar 
cTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]] - 8*Sqrt[a]*B*ArcTanh[Sqrt[a + b*x^2]/S 
qrt[a]]))/2))/4
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 535

Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim 
p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p 
 + 1)   Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free 
Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
 

rule 536

Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S 
imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( 
a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer 
Q[2*p]
 

rule 538

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15

method	result	size
risch	\(-\frac {a^{2} A \sqrt {b \,x^{2}+a}}{x}+\frac {A \,b^{2} x^{3} \sqrt {b \,x^{2}+a}}{4}+\frac {9 A b a x \sqrt {b \,x^{2}+a}}{8}+\frac {15 A \,a^{2} \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8}-B \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )+\frac {B \,b^{2} x^{4} \sqrt {b \,x^{2}+a}}{5}+\frac {11 B b a \,x^{2} \sqrt {b \,x^{2}+a}}{15}+\frac {23 B \,a^{2} \sqrt {b \,x^{2}+a}}{15}\)	\(157\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )+B \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\)	\(163\)

Input:

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

Output:

-a^2*A*(b*x^2+a)^(1/2)/x+1/4*A*b^2*x^3*(b*x^2+a)^(1/2)+9/8*A*b*a*x*(b*x^2+ 
a)^(1/2)+15/8*A*a^2*b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))-B*a^(5/2)*ln((2* 
a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+1/5*B*b^2*x^4*(b*x^2+a)^(1/2)+11/15*B*b*a* 
x^2*(b*x^2+a)^(1/2)+23/15*B*a^2*(b*x^2+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 525, normalized size of antiderivative = 3.86 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\left [\frac {225 \, A a^{2} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 120 \, B a^{\frac {5}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, x}, -\frac {225 \, A a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 60 \, B a^{\frac {5}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, x}, \frac {240 \, B \sqrt {-a} a^{2} x \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + 225 \, A a^{2} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, x}, -\frac {225 \, A a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 120 \, B \sqrt {-a} a^{2} x \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, x}\right ] \] Input:

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

Output:

[1/240*(225*A*a^2*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a 
) + 120*B*a^(5/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 
2*(24*B*b^2*x^5 + 30*A*b^2*x^4 + 88*B*a*b*x^3 + 135*A*a*b*x^2 + 184*B*a^2* 
x - 120*A*a^2)*sqrt(b*x^2 + a))/x, -1/120*(225*A*a^2*sqrt(-b)*x*arctan(sqr 
t(-b)*x/sqrt(b*x^2 + a)) - 60*B*a^(5/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)* 
sqrt(a) + 2*a)/x^2) - (24*B*b^2*x^5 + 30*A*b^2*x^4 + 88*B*a*b*x^3 + 135*A* 
a*b*x^2 + 184*B*a^2*x - 120*A*a^2)*sqrt(b*x^2 + a))/x, 1/240*(240*B*sqrt(- 
a)*a^2*x*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + 225*A*a^2*sqrt(b)*x*log(-2*b 
*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(24*B*b^2*x^5 + 30*A*b^2*x^4 + 
 88*B*a*b*x^3 + 135*A*a*b*x^2 + 184*B*a^2*x - 120*A*a^2)*sqrt(b*x^2 + a))/ 
x, -1/120*(225*A*a^2*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 120*B 
*sqrt(-a)*a^2*x*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) - (24*B*b^2*x^5 + 30*A* 
b^2*x^4 + 88*B*a*b*x^3 + 135*A*a*b*x^2 + 184*B*a^2*x - 120*A*a^2)*sqrt(b*x 
^2 + a))/x]
 

Sympy [A] (verification not implemented)

Time = 2.83 (sec) , antiderivative size = 420, normalized size of antiderivative = 3.09 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=- \frac {A a^{\frac {5}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{\frac {3}{2}} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + A a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + 2 A a b \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + A b^{2} \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a x \sqrt {a + b x^{2}}}{8 b} + \frac {x^{3} \sqrt {a + b x^{2}}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) - B a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{3}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a^{2} \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + 2 B a b \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-A*a**(5/2)/(x*sqrt(1 + b*x**2/a)) - A*a**(3/2)*b*x/sqrt(1 + b*x**2/a) + A 
*a**2*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) + 2*A*a*b*Piecewise((a*Piecewise((l 
og(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 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) + 
A*b**2*Piecewise((-a**2*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) + a*x*sqrt(a + b 
*x**2)/(8*b) + x**3*sqrt(a + b*x**2)/4, Ne(b, 0)), (sqrt(a)*x**3/3, True)) 
 - B*a**(5/2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**3/(sqrt(b)*x*sqrt(a/(b*x** 
2) + 1)) + B*a**2*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + 2*B*a*b*Piecewise((a*sq 
rt(a + b*x**2)/(3*b) + x**2*sqrt(a + b*x**2)/3, Ne(b, 0)), (sqrt(a)*x**2/2 
, True)) + B*b**2*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*s 
qrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4 
/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\frac {5}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b x + \frac {15}{8} \, \sqrt {b x^{2} + a} A a b x + \frac {15}{8} \, A a^{2} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a + \sqrt {b x^{2} + a} B a^{2} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{x} \] Input:

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

Output:

5/4*(b*x^2 + a)^(3/2)*A*b*x + 15/8*sqrt(b*x^2 + a)*A*a*b*x + 15/8*A*a^2*sq 
rt(b)*arcsinh(b*x/sqrt(a*b)) - B*a^(5/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1 
/5*(b*x^2 + a)^(5/2)*B + 1/3*(b*x^2 + a)^(3/2)*B*a + sqrt(b*x^2 + a)*B*a^2 
 - (b*x^2 + a)^(5/2)*A/x
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\frac {2 \, B a^{3} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {15}{8} \, A a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {2 \, A a^{3} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{120} \, {\left (184 \, B a^{2} + {\left (135 \, A a b + 2 \, {\left (44 \, B a b + 3 \, {\left (4 \, B b^{2} x + 5 \, A b^{2}\right )} x\right )} x\right )} x\right )} \sqrt {b x^{2} + a} \] Input:

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

Output:

2*B*a^3*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 15/8*A* 
a^2*sqrt(b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a))) + 2*A*a^3*sqrt(b)/((sqr 
t(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/120*(184*B*a^2 + (135*A*a*b + 2*(44*B 
*a*b + 3*(4*B*b^2*x + 5*A*b^2)*x)*x)*x)*sqrt(b*x^2 + a)
 

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5}+B\,a^2\,\sqrt {b\,x^2+a}+\frac {B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}}+B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \] Input:

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

Output:

(B*(a + b*x^2)^(5/2))/5 + B*a^2*(a + b*x^2)^(1/2) + B*a^(5/2)*atan(((a + b 
*x^2)^(1/2)*1i)/a^(1/2))*1i + (B*a*(a + b*x^2)^(3/2))/3 - (A*(a + b*x^2)^( 
5/2)*hypergeom([-5/2, -1/2], 1/2, -(b*x^2)/a))/(x*((b*x^2)/a + 1)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.46 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^2} \, dx=\frac {-120 \sqrt {b \,x^{2}+a}\, a^{3}+135 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}+184 \sqrt {b \,x^{2}+a}\, a^{2} b x +30 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+88 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+24 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+120 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b x -120 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b x +225 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} x -150 \sqrt {b}\, a^{3} x}{120 x} \] Input:

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

Output:

( - 120*sqrt(a + b*x**2)*a**3 + 135*sqrt(a + b*x**2)*a**2*b*x**2 + 184*sqr 
t(a + b*x**2)*a**2*b*x + 30*sqrt(a + b*x**2)*a*b**2*x**4 + 88*sqrt(a + b*x 
**2)*a*b**2*x**3 + 24*sqrt(a + b*x**2)*b**3*x**5 + 120*sqrt(a)*log((sqrt(a 
 + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*b*x - 120*sqrt(a)*log((sqr 
t(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*b*x + 225*sqrt(b)*log(( 
sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*x - 150*sqrt(b)*a**3*x)/(120*x 
)