\(\int \frac {(A+B x) (a+b x^2)^{5/2}}{x^6} \, dx\) [1123]

Optimal result

Integrand size = 20, antiderivative size = 140 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=-\frac {b^2 (8 A-15 B x) \sqrt {a+b x^2}}{8 x}-\frac {b (8 A+15 B x) \left (a+b x^2\right )^{3/2}}{24 x^3}-\frac {(4 A+5 B x) \left (a+b x^2\right )^{5/2}}{20 x^5}+A b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {15}{8} \sqrt {a} b^2 B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \] Output:

-1/8*b^2*(-15*B*x+8*A)*(b*x^2+a)^(1/2)/x-1/24*b*(15*B*x+8*A)*(b*x^2+a)^(3/ 
2)/x^3-1/20*(5*B*x+4*A)*(b*x^2+a)^(5/2)/x^5+A*b^(5/2)*arctanh(b^(1/2)*x/(b 
*x^2+a)^(1/2))-15/8*a^(1/2)*b^2*B*arctanh((b*x^2+a)^(1/2)/a^(1/2))
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=-\frac {\sqrt {a+b x^2} \left (8 b^2 x^4 (23 A-15 B x)+6 a^2 (4 A+5 B x)+a b x^2 (88 A+135 B x)\right )}{120 x^5}+2 A b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )-\frac {15}{8} \sqrt {a} b^2 B \log (x)+\frac {15}{8} \sqrt {a} b^2 B \log \left (-\sqrt {a}+\sqrt {a+b x^2}\right ) \] Input:

Integrate[((A + B*x)*(a + b*x^2)^(5/2))/x^6,x]
 

Output:

-1/120*(Sqrt[a + b*x^2]*(8*b^2*x^4*(23*A - 15*B*x) + 6*a^2*(4*A + 5*B*x) + 
 a*b*x^2*(88*A + 135*B*x)))/x^5 + 2*A*b^(5/2)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a 
] + Sqrt[a + b*x^2])] - (15*Sqrt[a]*b^2*B*Log[x])/8 + (15*Sqrt[a]*b^2*B*Lo 
g[-Sqrt[a] + Sqrt[a + b*x^2]])/8
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {537, 25, 537, 25, 536, 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.

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

\(\Big \downarrow \) 537

\(\displaystyle -\frac {1}{4} b \int -\frac {(4 A+5 B x) \left (b x^2+a\right )^{3/2}}{x^4}dx-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} b \int \frac {(4 A+5 B x) \left (b x^2+a\right )^{3/2}}{x^4}dx-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 537

\(\displaystyle \frac {1}{4} b \left (-\frac {1}{2} b \int -\frac {(8 A+15 B x) \sqrt {b x^2+a}}{x^2}dx-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \int \frac {(8 A+15 B x) \sqrt {b x^2+a}}{x^2}dx-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 536

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (\int \frac {15 a B+8 A b x}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (8 A b \int \frac {1}{\sqrt {b x^2+a}}dx+15 a B \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (8 A b \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+15 a B \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (15 a B \int \frac {1}{x \sqrt {b x^2+a}}dx+8 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (\frac {15}{2} a B \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+8 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (\frac {15 a B \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+8 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} b \left (\frac {1}{2} b \left (8 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2} (8 A-15 B x)}{x}-15 \sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )-\frac {\left (a+b x^2\right )^{3/2} (8 A+15 B x)}{6 x^3}\right )-\frac {\left (a+b x^2\right )^{5/2} (4 A+5 B x)}{20 x^5}\)

Input:

Int[((A + B*x)*(a + b*x^2)^(5/2))/x^6,x]
 

Output:

-1/20*((4*A + 5*B*x)*(a + b*x^2)^(5/2))/x^5 + (b*(-1/6*((8*A + 15*B*x)*(a 
+ b*x^2)^(3/2))/x^3 + (b*(-(((8*A - 15*B*x)*Sqrt[a + b*x^2])/x) + 8*A*Sqrt 
[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]] - 15*Sqrt[a]*B*ArcTanh[Sqrt[a + b 
*x^2]/Sqrt[a]]))/2))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 537

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), 
 x] - Simp[2*b*(p/((m + 1)*(m + 2)))   Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) 
*x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && 
 GtQ[p, 0] &&  !ILtQ[m + 2*p + 3, 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 = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (184 b^{2} A \,x^{4}+135 a b B \,x^{3}+88 a b A \,x^{2}+30 a^{2} B x +24 a^{2} A \right )}{120 x^{5}}+A \,b^{\frac {5}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )-\frac {15 B \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) \sqrt {a}\, b^{2}}{8}+B \,b^{2} \sqrt {b \,x^{2}+a}\)	\(122\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{5 a \,x^{5}}+\frac {2 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )}{5 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\right )\)	\(259\)

Input:

int((B*x+A)*(b*x^2+a)^(5/2)/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/120*(b*x^2+a)^(1/2)*(184*A*b^2*x^4+135*B*a*b*x^3+88*A*a*b*x^2+30*B*a^2* 
x+24*A*a^2)/x^5+A*b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))-15/8*B*ln((2*a+2*a 
^(1/2)*(b*x^2+a)^(1/2))/x)*a^(1/2)*b^2+B*b^2*(b*x^2+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 540, normalized size of antiderivative = 3.86 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=\left [\frac {120 \, A b^{\frac {5}{2}} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 225 \, B \sqrt {a} b^{2} x^{5} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (120 \, B b^{2} x^{5} - 184 \, A b^{2} x^{4} - 135 \, B a b x^{3} - 88 \, A a b x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, x^{5}}, -\frac {240 \, A \sqrt {-b} b^{2} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 225 \, B \sqrt {a} b^{2} x^{5} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (120 \, B b^{2} x^{5} - 184 \, A b^{2} x^{4} - 135 \, B a b x^{3} - 88 \, A a b x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, x^{5}}, \frac {225 \, B \sqrt {-a} b^{2} x^{5} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + 60 \, A b^{\frac {5}{2}} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + {\left (120 \, B b^{2} x^{5} - 184 \, A b^{2} x^{4} - 135 \, B a b x^{3} - 88 \, A a b x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, x^{5}}, -\frac {120 \, A \sqrt {-b} b^{2} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 225 \, B \sqrt {-a} b^{2} x^{5} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (120 \, B b^{2} x^{5} - 184 \, A b^{2} x^{4} - 135 \, B a b x^{3} - 88 \, A a b x^{2} - 30 \, B a^{2} x - 24 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, x^{5}}\right ] \] Input:

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^6,x, algorithm="fricas")
 

Output:

[1/240*(120*A*b^(5/2)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) 
+ 225*B*sqrt(a)*b^2*x^5*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2 
) + 2*(120*B*b^2*x^5 - 184*A*b^2*x^4 - 135*B*a*b*x^3 - 88*A*a*b*x^2 - 30*B 
*a^2*x - 24*A*a^2)*sqrt(b*x^2 + a))/x^5, -1/240*(240*A*sqrt(-b)*b^2*x^5*ar 
ctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 225*B*sqrt(a)*b^2*x^5*log(-(b*x^2 - 2*s 
qrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(120*B*b^2*x^5 - 184*A*b^2*x^4 - 13 
5*B*a*b*x^3 - 88*A*a*b*x^2 - 30*B*a^2*x - 24*A*a^2)*sqrt(b*x^2 + a))/x^5, 
1/120*(225*B*sqrt(-a)*b^2*x^5*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + 60*A*b^ 
(5/2)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + (120*B*b^2*x^5 
 - 184*A*b^2*x^4 - 135*B*a*b*x^3 - 88*A*a*b*x^2 - 30*B*a^2*x - 24*A*a^2)*s 
qrt(b*x^2 + a))/x^5, -1/120*(120*A*sqrt(-b)*b^2*x^5*arctan(sqrt(-b)*x/sqrt 
(b*x^2 + a)) - 225*B*sqrt(-a)*b^2*x^5*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) - 
 (120*B*b^2*x^5 - 184*A*b^2*x^4 - 135*B*a*b*x^3 - 88*A*a*b*x^2 - 30*B*a^2* 
x - 24*A*a^2)*sqrt(b*x^2 + a))/x^5]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (128) = 256\).

Time = 5.61 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.10 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=- \frac {A \sqrt {a} b^{2}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {8 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} + A b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b^{3} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {15 B \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8} - \frac {B a^{3}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 B a^{2} \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{x} + \frac {7 B a b^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B b^{\frac {5}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \] Input:

integrate((B*x+A)*(b*x**2+a)**(5/2)/x**6,x)
 

Output:

-A*sqrt(a)*b**2/(x*sqrt(1 + b*x**2/a)) - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 
1)/(5*x**4) - 11*A*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/(15*x**2) - 8*A*b**(5/2 
)*sqrt(a/(b*x**2) + 1)/15 + A*b**(5/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**3*x 
/(sqrt(a)*sqrt(1 + b*x**2/a)) - 15*B*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*x 
))/8 - B*a**3/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 3*B*a**2*sqrt(b)/(8* 
x**3*sqrt(a/(b*x**2) + 1)) - B*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/x + 7*B*a*b 
**(3/2)/(8*x*sqrt(a/(b*x**2) + 1)) + B*b**(5/2)*x/sqrt(a/(b*x**2) + 1)
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.56 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} A b^{3} x}{a} + A b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {15}{8} \, B \sqrt {a} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {15}{8} \, \sqrt {b x^{2} + a} B b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{8 \, a^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{8 \, a} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{15 \, a^{2} x} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B b}{8 \, a^{2} x^{2}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{4 \, a x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{5 \, a x^{5}} \] Input:

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^6,x, algorithm="maxima")
 

Output:

2/3*(b*x^2 + a)^(3/2)*A*b^3*x/a^2 + sqrt(b*x^2 + a)*A*b^3*x/a + A*b^(5/2)* 
arcsinh(b*x/sqrt(a*b)) - 15/8*B*sqrt(a)*b^2*arcsinh(a/(sqrt(a*b)*abs(x))) 
+ 15/8*sqrt(b*x^2 + a)*B*b^2 + 3/8*(b*x^2 + a)^(5/2)*B*b^2/a^2 + 5/8*(b*x^ 
2 + a)^(3/2)*B*b^2/a - 8/15*(b*x^2 + a)^(5/2)*A*b^2/(a^2*x) - 3/8*(b*x^2 + 
 a)^(7/2)*B*b/(a^2*x^2) - 2/15*(b*x^2 + a)^(7/2)*A*b/(a^2*x^3) - 1/4*(b*x^ 
2 + a)^(7/2)*B/(a*x^4) - 1/5*(b*x^2 + a)^(7/2)*A/(a*x^5)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (114) = 228\).

Time = 0.15 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.36 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=\frac {15 \, B a b^{2} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} - A b^{\frac {5}{2}} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \sqrt {b x^{2} + a} B b^{2} + \frac {135 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{9} B a b^{2} + 360 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a b^{\frac {5}{2}} - 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{7} B a^{2} b^{2} - 720 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A a^{2} b^{\frac {5}{2}} + 1120 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{3} b^{\frac {5}{2}} + 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} B a^{4} b^{2} - 560 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{4} b^{\frac {5}{2}} - 135 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} B a^{5} b^{2} + 184 \, A a^{5} b^{\frac {5}{2}}}{60 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \] Input:

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^6,x, algorithm="giac")
 

Output:

15/4*B*a*b^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - A* 
b^(5/2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a))) + sqrt(b*x^2 + a)*B*b^2 + 1 
/60*(135*(sqrt(b)*x - sqrt(b*x^2 + a))^9*B*a*b^2 + 360*(sqrt(b)*x - sqrt(b 
*x^2 + a))^8*A*a*b^(5/2) - 150*(sqrt(b)*x - sqrt(b*x^2 + a))^7*B*a^2*b^2 - 
 720*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a^2*b^(5/2) + 1120*(sqrt(b)*x - sqr 
t(b*x^2 + a))^4*A*a^3*b^(5/2) + 150*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*a^4* 
b^2 - 560*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^4*b^(5/2) - 135*(sqrt(b)*x - 
 sqrt(b*x^2 + a))*B*a^5*b^2 + 184*A*a^5*b^(5/2))/((sqrt(b)*x - sqrt(b*x^2 
+ a))^2 - a)^5
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right )}{x^6} \,d x \] Input:

int(((a + b*x^2)^(5/2)*(A + B*x))/x^6,x)
 

Output:

int(((a + b*x^2)^(5/2)*(A + B*x))/x^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{5/2}}{x^6} \, dx=\frac {-24 \sqrt {b \,x^{2}+a}\, a^{3}-88 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}-30 \sqrt {b \,x^{2}+a}\, a^{2} b x -184 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}-135 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+120 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+225 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{5}-225 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{5}+120 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} x^{5}+40 \sqrt {b}\, a \,b^{2} x^{5}}{120 x^{5}} \] Input:

int((B*x+A)*(b*x^2+a)^(5/2)/x^6,x)
 

Output:

( - 24*sqrt(a + b*x**2)*a**3 - 88*sqrt(a + b*x**2)*a**2*b*x**2 - 30*sqrt(a 
 + b*x**2)*a**2*b*x - 184*sqrt(a + b*x**2)*a*b**2*x**4 - 135*sqrt(a + b*x* 
*2)*a*b**2*x**3 + 120*sqrt(a + b*x**2)*b**3*x**5 + 225*sqrt(a)*log((sqrt(a 
 + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**5 - 225*sqrt(a)*log((sq 
rt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**5 + 120*sqrt(b)*log 
((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*x**5 + 40*sqrt(b)*a*b**2*x 
**5)/(120*x**5)