\(\int \frac {(A+B x) (a+b x^2)^{5/2}}{x} \, dx\) [1118]

Optimal result

Integrand size = 20, antiderivative size = 132 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+b x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {5 a^3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}-a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:

1/16*a^2*(5*B*x+16*A)*(b*x^2+a)^(1/2)+1/24*a*(5*B*x+8*A)*(b*x^2+a)^(3/2)+1 
/30*(5*B*x+6*A)*(b*x^2+a)^(5/2)+5/16*a^3*B*arctanh(b^(1/2)*x/(b*x^2+a)^(1/ 
2))/b^(1/2)-a^(5/2)*A*arctanh((b*x^2+a)^(1/2)/a^(1/2))
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {1}{240} \left (\sqrt {a+b x^2} \left (8 b^2 x^4 (6 A+5 B x)+2 a b x^2 (88 A+65 B x)+a^2 (368 A+165 B x)\right )-\frac {75 a^3 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b}}\right ) \] Input:

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

Output:

2*a^(5/2)*A*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] + (Sqrt[a + b*x 
^2]*(8*b^2*x^4*(6*A + 5*B*x) + 2*a*b*x^2*(88*A + 65*B*x) + a^2*(368*A + 16 
5*B*x)) - (75*a^3*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/Sqrt[b])/240
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {535, 535, 27, 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.

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

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{6} a \int \frac {(6 A+5 B x) \left (b x^2+a\right )^{3/2}}{x}dx+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{6} a \left (\frac {1}{4} a \int \frac {3 (8 A+5 B x) \sqrt {b x^2+a}}{x}dx+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \int \frac {(8 A+5 B x) \sqrt {b x^2+a}}{x}dx+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {16 A+5 B x}{x \sqrt {b x^2+a}}dx+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \left (16 A \int \frac {1}{x \sqrt {b x^2+a}}dx+5 B \int \frac {1}{\sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \left (16 A \int \frac {1}{x \sqrt {b x^2+a}}dx+5 B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \left (16 A \int \frac {1}{x \sqrt {b x^2+a}}dx+\frac {5 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \left (8 A \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+\frac {5 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \left (\frac {16 A \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {5 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \left (\frac {5 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-\frac {16 A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (16 A+5 B x)\right )+\frac {1}{4} \left (a+b x^2\right )^{3/2} (8 A+5 B x)\right )+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x)\)

Input:

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

Output:

((6*A + 5*B*x)*(a + b*x^2)^(5/2))/30 + (a*(((8*A + 5*B*x)*(a + b*x^2)^(3/2 
))/4 + (3*a*(((16*A + 5*B*x)*Sqrt[a + b*x^2])/2 + (a*((5*B*ArcTanh[(Sqrt[b 
]*x)/Sqrt[a + b*x^2]])/Sqrt[b] - (16*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/S 
qrt[a]))/2))/4))/6
 

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 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 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 = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05

method	result	size
default	\(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )+A \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\)	\(139\)

Input:

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

Output:

B*(1/6*x*(b*x^2+a)^(5/2)+5/6*a*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+ 
a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))))+A*(1/5*(b*x^2+a)^( 
5/2)+a*(1/3*(b*x^2+a)^(3/2)+a*((b*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*( 
b*x^2+a)^(1/2))/x))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 545, normalized size of antiderivative = 4.13 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=\left [\frac {75 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 240 \, A a^{\frac {5}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b}, -\frac {75 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 120 \, A a^{\frac {5}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b}, \frac {480 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + 75 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b}, -\frac {75 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 240 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b}\right ] \] Input:

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

Output:

[1/480*(75*B*a^3*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 
 240*A*a^(5/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*( 
40*B*b^3*x^5 + 48*A*b^3*x^4 + 130*B*a*b^2*x^3 + 176*A*a*b^2*x^2 + 165*B*a^ 
2*b*x + 368*A*a^2*b)*sqrt(b*x^2 + a))/b, -1/240*(75*B*a^3*sqrt(-b)*arctan( 
sqrt(-b)*x/sqrt(b*x^2 + a)) - 120*A*a^(5/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + 
 a)*sqrt(a) + 2*a)/x^2) - (40*B*b^3*x^5 + 48*A*b^3*x^4 + 130*B*a*b^2*x^3 + 
 176*A*a*b^2*x^2 + 165*B*a^2*b*x + 368*A*a^2*b)*sqrt(b*x^2 + a))/b, 1/480* 
(480*A*sqrt(-a)*a^2*b*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + 75*B*a^3*sqrt(b 
)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*B*b^3*x^5 + 48*A 
*b^3*x^4 + 130*B*a*b^2*x^3 + 176*A*a*b^2*x^2 + 165*B*a^2*b*x + 368*A*a^2*b 
)*sqrt(b*x^2 + a))/b, -1/240*(75*B*a^3*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x 
^2 + a)) - 240*A*sqrt(-a)*a^2*b*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) - (40*B 
*b^3*x^5 + 48*A*b^3*x^4 + 130*B*a*b^2*x^3 + 176*A*a*b^2*x^2 + 165*B*a^2*b* 
x + 368*A*a^2*b)*sqrt(b*x^2 + a))/b]
 

Sympy [A] (verification not implemented)

Time = 6.36 (sec) , antiderivative size = 474, normalized size of antiderivative = 3.59 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=- A a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {A a^{3}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A a^{2} \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + 2 A 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 ) + A 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 ) + B a^{2} \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 ) + 2 B a b \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 b^{2} \left (\begin {cases} \frac {a^{3} \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 )}{16 b^{2}} - \frac {a^{2} x \sqrt {a + b x^{2}}}{16 b^{2}} + \frac {a x^{3} \sqrt {a + b x^{2}}}{24 b} + \frac {x^{5} \sqrt {a + b x^{2}}}{6} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{5}}{5} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-A*a**(5/2)*asinh(sqrt(a)/(sqrt(b)*x)) + A*a**3/(sqrt(b)*x*sqrt(a/(b*x**2) 
 + 1)) + A*a**2*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + 2*A*a*b*Piecewise((a*sqrt 
(a + b*x**2)/(3*b) + x**2*sqrt(a + b*x**2)/3, Ne(b, 0)), (sqrt(a)*x**2/2, 
True)) + A*b**2*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqr 
t(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4 
, True)) + B*a**2*Piecewise((a*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 + x*sqrt(a + 
b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) + 2*B*a*b*Piecewise((-a**2*Piecew 
ise((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*b**2*Piecewise((a**3*Piec 
ewise((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))/(16*b**2) - a**2*x*sqrt(a + b*x**2)/(16*b**2) + a* 
x**3*sqrt(a + b*x**2)/(24*b) + x**5*sqrt(a + b*x**2)/6, Ne(b, 0)), (sqrt(a 
)*x**5/5, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x + \frac {5}{16} \, \sqrt {b x^{2} + a} B a^{2} x + \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} - A a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a + \sqrt {b x^{2} + a} A a^{2} \] Input:

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

Output:

1/6*(b*x^2 + a)^(5/2)*B*x + 5/24*(b*x^2 + a)^(3/2)*B*a*x + 5/16*sqrt(b*x^2 
 + a)*B*a^2*x + 5/16*B*a^3*arcsinh(b*x/sqrt(a*b))/sqrt(b) - A*a^(5/2)*arcs 
inh(a/(sqrt(a*b)*abs(x))) + 1/5*(b*x^2 + a)^(5/2)*A + 1/3*(b*x^2 + a)^(3/2 
)*A*a + sqrt(b*x^2 + a)*A*a^2
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{5}+A\,a^2\,\sqrt {b\,x^2+a}+\frac {A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {B\,x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}}+A\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x} \, dx=\frac {23 \sqrt {b \,x^{2}+a}\, a^{3}}{15}+\frac {11 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}}{15}+\frac {11 \sqrt {b \,x^{2}+a}\, a^{2} b x}{16}+\frac {\sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}}{5}+\frac {13 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}}{24}+\frac {\sqrt {b \,x^{2}+a}\, b^{3} x^{5}}{6}+\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}-\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}+\frac {5 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}}{16} \] Input:

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

Output:

(368*sqrt(a + b*x**2)*a**3 + 176*sqrt(a + b*x**2)*a**2*b*x**2 + 165*sqrt(a 
 + b*x**2)*a**2*b*x + 48*sqrt(a + b*x**2)*a*b**2*x**4 + 130*sqrt(a + b*x** 
2)*a*b**2*x**3 + 40*sqrt(a + b*x**2)*b**3*x**5 + 240*sqrt(a)*log((sqrt(a + 
 b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**3 - 240*sqrt(a)*log((sqrt(a + 
b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a**3 + 75*sqrt(b)*log((sqrt(a + b* 
x**2) + sqrt(b)*x)/sqrt(a))*a**3)/240