\(\int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx\) [407]

Optimal result

Integrand size = 18, antiderivative size = 146 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=\frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {b^3 (3 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}} \]

[Out]

1/24*(3*A*b-8*B*a)*(b*x+a)^(3/2)/a/x^3-1/4*A*(b*x+a)^(5/2)/a/x^4-1/64*b^3*(3*A*b-8*B*a)*arctanh((b*x+a)^(1/2)/
a^(1/2))/a^(5/2)+1/32*b*(3*A*b-8*B*a)*(b*x+a)^(1/2)/a/x^2+1/64*b^2*(3*A*b-8*B*a)*(b*x+a)^(1/2)/a^2/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {79, 43, 44, 65, 214} \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=-\frac {b^3 (3 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}}+\frac {b^2 \sqrt {a+b x} (3 A b-8 a B)}{64 a^2 x}+\frac {(a+b x)^{3/2} (3 A b-8 a B)}{24 a x^3}+\frac {b \sqrt {a+b x} (3 A b-8 a B)}{32 a x^2}-\frac {A (a+b x)^{5/2}}{4 a x^4} \]

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^5,x]

[Out]

(b*(3*A*b - 8*a*B)*Sqrt[a + b*x])/(32*a*x^2) + (b^2*(3*A*b - 8*a*B)*Sqrt[a + b*x])/(64*a^2*x) + ((3*A*b - 8*a*
B)*(a + b*x)^(3/2))/(24*a*x^3) - (A*(a + b*x)^(5/2))/(4*a*x^4) - (b^3*(3*A*b - 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sq
rt[a]])/(64*a^(5/2))

Rule 43

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {A (a+b x)^{5/2}}{4 a x^4}+\frac {\left (-\frac {3 A b}{2}+4 a B\right ) \int \frac {(a+b x)^{3/2}}{x^4} \, dx}{4 a} \\ & = \frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {(b (3 A b-8 a B)) \int \frac {\sqrt {a+b x}}{x^3} \, dx}{16 a} \\ & = \frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {\left (b^2 (3 A b-8 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{64 a} \\ & = \frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}+\frac {\left (b^3 (3 A b-8 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a^2} \\ & = \frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}+\frac {\left (b^2 (3 A b-8 a B)\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a^2} \\ & = \frac {b (3 A b-8 a B) \sqrt {a+b x}}{32 a x^2}+\frac {b^2 (3 A b-8 a B) \sqrt {a+b x}}{64 a^2 x}+\frac {(3 A b-8 a B) (a+b x)^{3/2}}{24 a x^3}-\frac {A (a+b x)^{5/2}}{4 a x^4}-\frac {b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=-\frac {\sqrt {a+b x} \left (-9 A b^3 x^3+6 a b^2 x^2 (A+4 B x)+16 a^3 (3 A+4 B x)+8 a^2 b x (9 A+14 B x)\right )}{192 a^2 x^4}+\frac {b^3 (-3 A b+8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{5/2}} \]

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^5,x]

[Out]

-1/192*(Sqrt[a + b*x]*(-9*A*b^3*x^3 + 6*a*b^2*x^2*(A + 4*B*x) + 16*a^3*(3*A + 4*B*x) + 8*a^2*b*x*(9*A + 14*B*x
)))/(a^2*x^4) + (b^3*(-3*A*b + 8*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(5/2))

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {3 \left (\frac {x^{4} \left (A b -\frac {8 B a}{3}\right ) b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8}+\sqrt {b x +a}\, \left (\frac {b^{2} x^{2} \left (4 B x +A \right ) a^{\frac {3}{2}}}{12}+b x \left (\frac {14 B x}{9}+A \right ) a^{\frac {5}{2}}+\left (\frac {8 B x}{9}+\frac {2 A}{3}\right ) a^{\frac {7}{2}}-\frac {A \sqrt {a}\, b^{3} x^{3}}{8}\right )\right )}{8 a^{\frac {5}{2}} x^{4}}\)	\(100\)
risch	\(-\frac {\sqrt {b x +a}\, \left (-9 A \,b^{3} x^{3}+24 B a \,b^{2} x^{3}+6 a A \,b^{2} x^{2}+112 B \,a^{2} b \,x^{2}+72 a^{2} A b x +64 a^{3} B x +48 a^{3} A \right )}{192 x^{4} a^{2}}-\frac {b^{3} \left (3 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64 a^{\frac {5}{2}}}\)	\(107\)
derivativedivides	\(2 b^{3} \left (-\frac {-\frac {\left (3 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{2}}+\frac {\left (33 A b +40 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a}+\left (\frac {11 A b}{128}-\frac {11 B a}{48}\right ) \left (b x +a \right )^{\frac {3}{2}}-\frac {a \left (3 A b -8 B a \right ) \sqrt {b x +a}}{128}}{b^{4} x^{4}}-\frac {\left (3 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {5}{2}}}\right )\)	\(120\)
default	\(2 b^{3} \left (-\frac {-\frac {\left (3 A b -8 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{128 a^{2}}+\frac {\left (33 A b +40 B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{384 a}+\left (\frac {11 A b}{128}-\frac {11 B a}{48}\right ) \left (b x +a \right )^{\frac {3}{2}}-\frac {a \left (3 A b -8 B a \right ) \sqrt {b x +a}}{128}}{b^{4} x^{4}}-\frac {\left (3 A b -8 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {5}{2}}}\right )\)	\(120\)

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^5,x,method=_RETURNVERBOSE)

[Out]

-3/8*(1/8*x^4*(A*b-8/3*B*a)*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+a)^(1/2)*(1/12*b^2*x^2*(4*B*x+A)*a^(3/2)+b
*x*(14/9*B*x+A)*a^(5/2)+(8/9*B*x+2/3*A)*a^(7/2)-1/8*A*a^(1/2)*b^3*x^3))/a^(5/2)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.77 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=\left [-\frac {3 \, {\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt {a} x^{4} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (48 \, A a^{4} + 3 \, {\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 2 \, {\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{384 \, a^{3} x^{4}}, -\frac {3 \, {\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (48 \, A a^{4} + 3 \, {\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{3} + 2 \, {\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{192 \, a^{3} x^{4}}\right ] \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^5,x, algorithm="fricas")

[Out]

[-1/384*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(a)*x^4*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(48*A*a^4 + 3*(8
*B*a^2*b^2 - 3*A*a*b^3)*x^3 + 2*(56*B*a^3*b + 3*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 9*A*a^3*b)*x)*sqrt(b*x + a))/(a^
3*x^4), -1/192*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(-a)*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*A*a^4 + 3*(8*B*a^2
*b^2 - 3*A*a*b^3)*x^3 + 2*(56*B*a^3*b + 3*A*a^2*b^2)*x^2 + 8*(8*B*a^4 + 9*A*a^3*b)*x)*sqrt(b*x + a))/(a^3*x^4)
]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (131) = 262\).

Time = 108.92 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.04 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=- \frac {A a^{2}}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 A a \sqrt {b}}{8 x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {13 A b^{\frac {3}{2}}}{32 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A b^{\frac {5}{2}}}{64 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {3 A b^{\frac {7}{2}}}{64 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {3 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {5}{2}}} - \frac {B a^{2}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {11 B a \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {17 B b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {B b^{\frac {5}{2}}}{8 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {3}{2}}} \]

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**5,x)

[Out]

-A*a**2/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) - 5*A*a*sqrt(b)/(8*x**(7/2)*sqrt(a/(b*x) + 1)) - 13*A*b**(3/2)/
(32*x**(5/2)*sqrt(a/(b*x) + 1)) + A*b**(5/2)/(64*a*x**(3/2)*sqrt(a/(b*x) + 1)) + 3*A*b**(7/2)/(64*a**2*sqrt(x)
*sqrt(a/(b*x) + 1)) - 3*A*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(5/2)) - B*a**2/(3*sqrt(b)*x**(7/2)*sqr
t(a/(b*x) + 1)) - 11*B*a*sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1)) - 17*B*b**(3/2)/(24*x**(3/2)*sqrt(a/(b*x) + 1
)) - B*b**(5/2)/(8*a*sqrt(x)*sqrt(a/(b*x) + 1)) + B*b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*a**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=-\frac {1}{384} \, b^{4} {\left (\frac {2 \, {\left (3 \, {\left (8 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + {\left (40 \, B a^{2} + 33 \, A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}} - 11 \, {\left (8 \, B a^{3} - 3 \, A a^{2} b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 3 \, {\left (8 \, B a^{4} - 3 \, A a^{3} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{4} a^{2} b - 4 \, {\left (b x + a\right )}^{3} a^{3} b + 6 \, {\left (b x + a\right )}^{2} a^{4} b - 4 \, {\left (b x + a\right )} a^{5} b + a^{6} b} + \frac {3 \, {\left (8 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^5,x, algorithm="maxima")

[Out]

-1/384*b^4*(2*(3*(8*B*a - 3*A*b)*(b*x + a)^(7/2) + (40*B*a^2 + 33*A*a*b)*(b*x + a)^(5/2) - 11*(8*B*a^3 - 3*A*a
^2*b)*(b*x + a)^(3/2) + 3*(8*B*a^4 - 3*A*a^3*b)*sqrt(b*x + a))/((b*x + a)^4*a^2*b - 4*(b*x + a)^3*a^3*b + 6*(b
*x + a)^2*a^4*b - 4*(b*x + a)*a^5*b + a^6*b) + 3*(8*B*a - 3*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a)
+ sqrt(a)))/(a^(5/2)*b))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=-\frac {\frac {3 \, {\left (8 \, B a b^{4} - 3 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {24 \, {\left (b x + a\right )}^{\frac {7}{2}} B a b^{4} + 40 \, {\left (b x + a\right )}^{\frac {5}{2}} B a^{2} b^{4} - 88 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{3} b^{4} + 24 \, \sqrt {b x + a} B a^{4} b^{4} - 9 \, {\left (b x + a\right )}^{\frac {7}{2}} A b^{5} + 33 \, {\left (b x + a\right )}^{\frac {5}{2}} A a b^{5} + 33 \, {\left (b x + a\right )}^{\frac {3}{2}} A a^{2} b^{5} - 9 \, \sqrt {b x + a} A a^{3} b^{5}}{a^{2} b^{4} x^{4}}}{192 \, b} \]

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^5,x, algorithm="giac")

[Out]

-1/192*(3*(8*B*a*b^4 - 3*A*b^5)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (24*(b*x + a)^(7/2)*B*a*b^4 +
40*(b*x + a)^(5/2)*B*a^2*b^4 - 88*(b*x + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x + a)*B*a^4*b^4 - 9*(b*x + a)^(7/2)*A
*b^5 + 33*(b*x + a)^(5/2)*A*a*b^5 + 33*(b*x + a)^(3/2)*A*a^2*b^5 - 9*sqrt(b*x + a)*A*a^3*b^5)/(a^2*b^4*x^4))/b

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^5} \, dx=-\frac {\left (\frac {11\,A\,b^4}{64}-\frac {11\,B\,a\,b^3}{24}\right )\,{\left (a+b\,x\right )}^{3/2}+\left (\frac {B\,a^2\,b^3}{8}-\frac {3\,A\,a\,b^4}{64}\right )\,\sqrt {a+b\,x}-\frac {\left (3\,A\,b^4-8\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{7/2}}{64\,a^2}+\frac {\left (33\,A\,b^4+40\,B\,a\,b^3\right )\,{\left (a+b\,x\right )}^{5/2}}{192\,a}}{{\left (a+b\,x\right )}^4-4\,a^3\,\left (a+b\,x\right )-4\,a\,{\left (a+b\,x\right )}^3+6\,a^2\,{\left (a+b\,x\right )}^2+a^4}-\frac {b^3\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-8\,B\,a\right )}{64\,a^{5/2}} \]

[In]

int(((A + B*x)*(a + b*x)^(3/2))/x^5,x)

[Out]

- (((11*A*b^4)/64 - (11*B*a*b^3)/24)*(a + b*x)^(3/2) + ((B*a^2*b^3)/8 - (3*A*a*b^4)/64)*(a + b*x)^(1/2) - ((3*
A*b^4 - 8*B*a*b^3)*(a + b*x)^(7/2))/(64*a^2) + ((33*A*b^4 + 40*B*a*b^3)*(a + b*x)^(5/2))/(192*a))/((a + b*x)^4
- 4*a^3*(a + b*x) - 4*a*(a + b*x)^3 + 6*a^2*(a + b*x)^2 + a^4) - (b^3*atanh((a + b*x)^(1/2)/a^(1/2))*(3*A*b -
8*B*a))/(64*a^(5/2))