\(\int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx\) [982]

Optimal result

Integrand size = 20, antiderivative size = 144 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=-\frac {A \sqrt {a+b x^2}}{6 x^6}-\frac {A b \sqrt {a+b x^2}}{24 a x^4}+\frac {A b^2 \sqrt {a+b x^2}}{16 a^2 x^2}-\frac {B \left (a+b x^2\right )^{3/2}}{5 a x^5}+\frac {2 b B \left (a+b x^2\right )^{3/2}}{15 a^2 x^3}-\frac {A b^3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{5/2}} \] Output:

-1/6*A*(b*x^2+a)^(1/2)/x^6-1/24*A*b*(b*x^2+a)^(1/2)/a/x^4+1/16*A*b^2*(b*x^ 
2+a)^(1/2)/a^2/x^2-1/5*B*(b*x^2+a)^(3/2)/a/x^5+2/15*b*B*(b*x^2+a)^(3/2)/a^ 
2/x^3-1/16*A*b^3*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(5/2)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=\frac {\sqrt {a+b x^2} \left (-8 a^2 (5 A+6 B x)-2 a b x^2 (5 A+8 B x)+b^2 x^4 (15 A+32 B x)\right )}{240 a^2 x^6}+\frac {A b^3 \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \] Input:

Integrate[((A + B*x)*Sqrt[a + b*x^2])/x^7,x]
 

Output:

(Sqrt[a + b*x^2]*(-8*a^2*(5*A + 6*B*x) - 2*a*b*x^2*(5*A + 8*B*x) + b^2*x^4 
*(15*A + 32*B*x)))/(240*a^2*x^6) + (A*b^3*ArcTanh[(Sqrt[b]*x - Sqrt[a + b* 
x^2])/Sqrt[a]])/(8*a^(5/2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {539, 27, 539, 27, 539, 25, 534, 243, 51, 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.

\(\displaystyle \int \frac {\sqrt {a+b x^2} (A+B x)}{x^7} \, dx\)

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[((A + B*x)*Sqrt[a + b*x^2])/x^7,x]
 

Output:

-1/6*(A*(a + b*x^2)^(3/2))/(a*x^6) + ((-2*B*(a + b*x^2)^(3/2))/(5*x^5) - ( 
b*((-5*A*(a + b*x^2)^(3/2))/(4*a*x^4) + ((-16*B*(a + b*x^2)^(3/2))/(3*x^3) 
 - (5*A*b*(-(Sqrt[a + b*x^2]/x^2) - (b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/S 
qrt[a]))/2)/(4*a)))/5)/(2*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

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 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 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 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 539

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) 
   Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] 
/; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.69

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-32 B \,b^{2} x^{5}-15 b^{2} A \,x^{4}+16 a b B \,x^{3}+10 a b A \,x^{2}+48 a^{2} B x +40 a^{2} A \right )}{240 x^{6} a^{2}}-\frac {A \,b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{16 a^{\frac {5}{2}}}\)	\(99\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )}{2 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 a \,x^{5}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 a^{2} x^{3}}\right )\)	\(152\)

Input:

int((B*x+A)*(b*x^2+a)^(1/2)/x^7,x,method=_RETURNVERBOSE)
 

Output:

-1/240*(b*x^2+a)^(1/2)*(-32*B*b^2*x^5-15*A*b^2*x^4+16*B*a*b*x^3+10*A*a*b*x 
^2+48*B*a^2*x+40*A*a^2)/x^6/a^2-1/16*A*b^3/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^ 
2+a)^(1/2))/x)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.54 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=\left [\frac {15 \, A \sqrt {a} b^{3} x^{6} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (32 \, B a b^{2} x^{5} + 15 \, A a b^{2} x^{4} - 16 \, B a^{2} b x^{3} - 10 \, A a^{2} b x^{2} - 48 \, B a^{3} x - 40 \, A a^{3}\right )} \sqrt {b x^{2} + a}}{480 \, a^{3} x^{6}}, \frac {15 \, A \sqrt {-a} b^{3} x^{6} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (32 \, B a b^{2} x^{5} + 15 \, A a b^{2} x^{4} - 16 \, B a^{2} b x^{3} - 10 \, A a^{2} b x^{2} - 48 \, B a^{3} x - 40 \, A a^{3}\right )} \sqrt {b x^{2} + a}}{240 \, a^{3} x^{6}}\right ] \] Input:

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^7,x, algorithm="fricas")
 

Output:

[1/480*(15*A*sqrt(a)*b^3*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a 
)/x^2) + 2*(32*B*a*b^2*x^5 + 15*A*a*b^2*x^4 - 16*B*a^2*b*x^3 - 10*A*a^2*b* 
x^2 - 48*B*a^3*x - 40*A*a^3)*sqrt(b*x^2 + a))/(a^3*x^6), 1/240*(15*A*sqrt( 
-a)*b^3*x^6*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (32*B*a*b^2*x^5 + 15*A*a* 
b^2*x^4 - 16*B*a^2*b*x^3 - 10*A*a^2*b*x^2 - 48*B*a^3*x - 40*A*a^3)*sqrt(b* 
x^2 + a))/(a^3*x^6)]
 

Sympy [A] (verification not implemented)

Time = 5.72 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.40 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=- \frac {A a}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A \sqrt {b}}{24 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {3}{2}}}{48 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {5}{2}}}{16 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {5}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {B b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a x^{2}} + \frac {2 B b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{2}} \] Input:

integrate((B*x+A)*(b*x**2+a)**(1/2)/x**7,x)
 

Output:

-A*a/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 5*A*sqrt(b)/(24*x**5*sqrt(a/( 
b*x**2) + 1)) + A*b**(3/2)/(48*a*x**3*sqrt(a/(b*x**2) + 1)) + A*b**(5/2)/( 
16*a**2*x*sqrt(a/(b*x**2) + 1)) - A*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*a* 
*(5/2)) - B*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - B*b**(3/2)*sqrt(a/(b*x 
**2) + 1)/(15*a*x**2) + 2*B*b**(5/2)*sqrt(a/(b*x**2) + 1)/(15*a**2)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=-\frac {A b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {5}{2}}} + \frac {\sqrt {b x^{2} + a} A b^{3}}{16 \, a^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{16 \, a^{3} x^{2}} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{15 \, a^{2} x^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{8 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{5 \, a x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{6 \, a x^{6}} \] Input:

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^7,x, algorithm="maxima")
 

Output:

-1/16*A*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 1/16*sqrt(b*x^2 + a)*A 
*b^3/a^3 - 1/16*(b*x^2 + a)^(3/2)*A*b^2/(a^3*x^2) + 2/15*(b*x^2 + a)^(3/2) 
*B*b/(a^2*x^3) + 1/8*(b*x^2 + a)^(3/2)*A*b/(a^2*x^4) - 1/5*(b*x^2 + a)^(3/ 
2)*B/(a*x^5) - 1/6*(b*x^2 + a)^(3/2)*A/(a*x^6)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (116) = 232\).

Time = 0.13 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.26 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=\frac {A b^{3} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{2}} - \frac {15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{11} A b^{3} - 85 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{9} A a b^{3} - 480 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{2} b^{\frac {5}{2}} - 570 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{7} A a^{2} b^{3} + 320 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{3} b^{\frac {5}{2}} - 570 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{5} A a^{3} b^{3} - 85 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A a^{4} b^{3} + 192 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{5} b^{\frac {5}{2}} + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a^{5} b^{3} - 32 \, B a^{6} b^{\frac {5}{2}}}{120 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{6} a^{2}} \] Input:

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^7,x, algorithm="giac")
 

Output:

1/8*A*b^3*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 
 1/120*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^11*A*b^3 - 85*(sqrt(b)*x - sqrt(b 
*x^2 + a))^9*A*a*b^3 - 480*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^2*b^(5/2) - 
 570*(sqrt(b)*x - sqrt(b*x^2 + a))^7*A*a^2*b^3 + 320*(sqrt(b)*x - sqrt(b*x 
^2 + a))^6*B*a^3*b^(5/2) - 570*(sqrt(b)*x - sqrt(b*x^2 + a))^5*A*a^3*b^3 - 
 85*(sqrt(b)*x - sqrt(b*x^2 + a))^3*A*a^4*b^3 + 192*(sqrt(b)*x - sqrt(b*x^ 
2 + a))^2*B*a^5*b^(5/2) + 15*(sqrt(b)*x - sqrt(b*x^2 + a))*A*a^5*b^3 - 32* 
B*a^6*b^(5/2))/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^6*a^2)
 

Mupad [B] (verification not implemented)

Time = 10.75 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^2\,x^6}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a\,x^6}-\frac {A\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {B\,\sqrt {b\,x^2+a}\,\left (3\,a^2+a\,b\,x^2-2\,b^2\,x^4\right )}{15\,a^2\,x^5}+\frac {A\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{5/2}} \] Input:

int(((a + b*x^2)^(1/2)*(A + B*x))/x^7,x)
 

Output:

(A*b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(5/2)) - (A*(a + b*x 
^2)^(1/2))/(16*x^6) - (A*(a + b*x^2)^(3/2))/(6*a*x^6) + (A*(a + b*x^2)^(5/ 
2))/(16*a^2*x^6) - (B*(a + b*x^2)^(1/2)*(3*a^2 - 2*b^2*x^4 + a*b*x^2))/(15 
*a^2*x^5)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \sqrt {a+b x^2}}{x^7} \, dx=\frac {-40 \sqrt {b \,x^{2}+a}\, a^{3}-10 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}-48 \sqrt {b \,x^{2}+a}\, a^{2} b x +15 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}-16 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+32 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{6}-15 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{6}-32 \sqrt {b}\, b^{3} x^{6}}{240 a^{2} x^{6}} \] Input:

int((B*x+A)*(b*x^2+a)^(1/2)/x^7,x)
 

Output:

( - 40*sqrt(a + b*x**2)*a**3 - 10*sqrt(a + b*x**2)*a**2*b*x**2 - 48*sqrt(a 
 + b*x**2)*a**2*b*x + 15*sqrt(a + b*x**2)*a*b**2*x**4 - 16*sqrt(a + b*x**2 
)*a*b**2*x**3 + 32*sqrt(a + b*x**2)*b**3*x**5 + 15*sqrt(a)*log((sqrt(a + b 
*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**6 - 15*sqrt(a)*log((sqrt(a 
+ b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**6 - 32*sqrt(b)*b**3*x**6 
)/(240*a**2*x**6)