\(\int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx\) [442]

Optimal result

Integrand size = 18, antiderivative size = 143 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=-\frac {5 b^2 (7 A b-6 a B)}{8 a^4 \sqrt {a+b x}}-\frac {A}{3 a x^3 \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}-\frac {5 b (7 A b-6 a B)}{24 a^3 x \sqrt {a+b x}}+\frac {5 b^2 (7 A b-6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{9/2}} \]

[Out]

5/8*b^2*(7*A*b-6*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(9/2)-5/8*b^2*(7*A*b-6*B*a)/a^4/(b*x+a)^(1/2)-1/3*A/a/x
^3/(b*x+a)^(1/2)+1/12*(7*A*b-6*B*a)/a^2/x^2/(b*x+a)^(1/2)-5/24*b*(7*A*b-6*B*a)/a^3/x/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 143, 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, 44, 53, 65, 214} \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=\frac {5 b^2 (7 A b-6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{9/2}}-\frac {5 b^2 (7 A b-6 a B)}{8 a^4 \sqrt {a+b x}}-\frac {5 b (7 A b-6 a B)}{24 a^3 x \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}-\frac {A}{3 a x^3 \sqrt {a+b x}} \]

[In]

Int[(A + B*x)/(x^4*(a + b*x)^(3/2)),x]

[Out]

(-5*b^2*(7*A*b - 6*a*B))/(8*a^4*Sqrt[a + b*x]) - A/(3*a*x^3*Sqrt[a + b*x]) + (7*A*b - 6*a*B)/(12*a^2*x^2*Sqrt[
a + b*x]) - (5*b*(7*A*b - 6*a*B))/(24*a^3*x*Sqrt[a + b*x]) + (5*b^2*(7*A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt
[a]])/(8*a^(9/2))

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 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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}{3 a x^3 \sqrt {a+b x}}+\frac {\left (-\frac {7 A b}{2}+3 a B\right ) \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx}{3 a} \\ & = -\frac {A}{3 a x^3 \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}+\frac {(5 b (7 A b-6 a B)) \int \frac {1}{x^2 (a+b x)^{3/2}} \, dx}{24 a^2} \\ & = -\frac {A}{3 a x^3 \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}-\frac {5 b (7 A b-6 a B)}{24 a^3 x \sqrt {a+b x}}-\frac {\left (5 b^2 (7 A b-6 a B)\right ) \int \frac {1}{x (a+b x)^{3/2}} \, dx}{16 a^3} \\ & = -\frac {5 b^2 (7 A b-6 a B)}{8 a^4 \sqrt {a+b x}}-\frac {A}{3 a x^3 \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}-\frac {5 b (7 A b-6 a B)}{24 a^3 x \sqrt {a+b x}}-\frac {\left (5 b^2 (7 A b-6 a B)\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{16 a^4} \\ & = -\frac {5 b^2 (7 A b-6 a B)}{8 a^4 \sqrt {a+b x}}-\frac {A}{3 a x^3 \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}-\frac {5 b (7 A b-6 a B)}{24 a^3 x \sqrt {a+b x}}-\frac {(5 b (7 A b-6 a B)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{8 a^4} \\ & = -\frac {5 b^2 (7 A b-6 a B)}{8 a^4 \sqrt {a+b x}}-\frac {A}{3 a x^3 \sqrt {a+b x}}+\frac {7 A b-6 a B}{12 a^2 x^2 \sqrt {a+b x}}-\frac {5 b (7 A b-6 a B)}{24 a^3 x \sqrt {a+b x}}+\frac {5 b^2 (7 A b-6 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=\frac {-105 A b^3 x^3-4 a^3 (2 A+3 B x)+2 a^2 b x (7 A+15 B x)+5 a b^2 x^2 (-7 A+18 B x)}{24 a^4 x^3 \sqrt {a+b x}}+\frac {5 b^2 (7 A b-6 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{9/2}} \]

[In]

Integrate[(A + B*x)/(x^4*(a + b*x)^(3/2)),x]

[Out]

(-105*A*b^3*x^3 - 4*a^3*(2*A + 3*B*x) + 2*a^2*b*x*(7*A + 15*B*x) + 5*a*b^2*x^2*(-7*A + 18*B*x))/(24*a^4*x^3*Sq
rt[a + b*x]) + (5*b^2*(7*A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(9/2))

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {\frac {35 x^{3} \left (A b -\frac {6 B a}{7}\right ) \sqrt {b x +a}\, b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8}-\frac {35 x^{2} b^{2} \left (-\frac {18 B x}{7}+A \right ) a^{\frac {3}{2}}}{24}+\frac {7 b x \left (\frac {15 B x}{7}+A \right ) a^{\frac {5}{2}}}{12}+\frac {\left (-3 B x -2 A \right ) a^{\frac {7}{2}}}{6}-\frac {35 A \sqrt {a}\, b^{3} x^{3}}{8}}{a^{\frac {9}{2}} \sqrt {b x +a}\, x^{3}}\)	\(106\)
risch	\(-\frac {\sqrt {b x +a}\, \left (57 A \,b^{2} x^{2}-42 B a b \,x^{2}-22 a A b x +12 a^{2} B x +8 a^{2} A \right )}{24 a^{4} x^{3}}-\frac {b^{2} \left (-\frac {2 \left (-16 A b +16 B a \right )}{\sqrt {b x +a}}-\frac {2 \left (35 A b -30 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{16 a^{4}}\)	\(107\)
derivativedivides	\(2 b^{2} \left (\frac {-\frac {\left (\frac {19 A b}{16}-\frac {7 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {17}{6} a b A +2 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {29}{16} a^{2} b A -\frac {9}{8} a^{3} B \right ) \sqrt {b x +a}}{b^{3} x^{3}}+\frac {5 \left (7 A b -6 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{4}}-\frac {A b -B a}{a^{4} \sqrt {b x +a}}\right )\)	\(126\)
default	\(2 b^{2} \left (\frac {-\frac {\left (\frac {19 A b}{16}-\frac {7 B a}{8}\right ) \left (b x +a \right )^{\frac {5}{2}}+\left (-\frac {17}{6} a b A +2 a^{2} B \right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {29}{16} a^{2} b A -\frac {9}{8} a^{3} B \right ) \sqrt {b x +a}}{b^{3} x^{3}}+\frac {5 \left (7 A b -6 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{4}}-\frac {A b -B a}{a^{4} \sqrt {b x +a}}\right )\)	\(126\)

[In]

int((B*x+A)/x^4/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

7/12*(15/2*x^3*(A*b-6/7*B*a)*(b*x+a)^(1/2)*b^2*arctanh((b*x+a)^(1/2)/a^(1/2))-5/2*x^2*b^2*(-18/7*B*x+A)*a^(3/2
)+b*x*(15/7*B*x+A)*a^(5/2)+2/7*(-3*B*x-2*A)*a^(7/2)-15/2*A*a^(1/2)*b^3*x^3)/(b*x+a)^(1/2)/a^(9/2)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.31 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=\left [-\frac {15 \, {\left ({\left (6 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} + {\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3}\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, A a^{4} - 15 \, {\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 5 \, {\left (6 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{48 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}, \frac {15 \, {\left ({\left (6 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} + {\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (8 \, A a^{4} - 15 \, {\left (6 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} - 5 \, {\left (6 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b x^{4} + a^{6} x^{3}\right )}}\right ] \]

[In]

integrate((B*x+A)/x^4/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(15*((6*B*a*b^3 - 7*A*b^4)*x^4 + (6*B*a^2*b^2 - 7*A*a*b^3)*x^3)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt
(a) + 2*a)/x) + 2*(8*A*a^4 - 15*(6*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 5*(6*B*a^3*b - 7*A*a^2*b^2)*x^2 + 2*(6*B*a^4 -
7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3), 1/24*(15*((6*B*a*b^3 - 7*A*b^4)*x^4 + (6*B*a^2*b^2 - 7*A*
a*b^3)*x^3)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8*A*a^4 - 15*(6*B*a^2*b^2 - 7*A*a*b^3)*x^3 - 5*(6*B*a
^3*b - 7*A*a^2*b^2)*x^2 + 2*(6*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b*x^4 + a^6*x^3)]

Sympy [A] (verification not implemented)

Time = 77.13 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.72 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=A \left (- \frac {1}{3 a \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {7 \sqrt {b}}{12 a^{2} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {35 b^{\frac {3}{2}}}{24 a^{3} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {35 b^{\frac {5}{2}}}{8 a^{4} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {35 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {9}{2}}}\right ) + B \left (- \frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}}\right ) \]

[In]

integrate((B*x+A)/x**4/(b*x+a)**(3/2),x)

[Out]

A*(-1/(3*a*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) + 7*sqrt(b)/(12*a**2*x**(5/2)*sqrt(a/(b*x) + 1)) - 35*b**(3/2)/
(24*a**3*x**(3/2)*sqrt(a/(b*x) + 1)) - 35*b**(5/2)/(8*a**4*sqrt(x)*sqrt(a/(b*x) + 1)) + 35*b**3*asinh(sqrt(a)/
(sqrt(b)*sqrt(x)))/(8*a**(9/2))) + B*(-1/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2)
*sqrt(a/(b*x) + 1)) + 15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) - 15*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x))
)/(4*a**(7/2)))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=-\frac {1}{48} \, b^{3} {\left (\frac {2 \, {\left (48 \, B a^{4} - 48 \, A a^{3} b - 15 \, {\left (6 \, B a - 7 \, A b\right )} {\left (b x + a\right )}^{3} + 40 \, {\left (6 \, B a^{2} - 7 \, A a b\right )} {\left (b x + a\right )}^{2} - 33 \, {\left (6 \, B a^{3} - 7 \, A a^{2} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {7}{2}} a^{4} b - 3 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} b + 3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} b - \sqrt {b x + a} a^{7} b} - \frac {15 \, {\left (6 \, B a - 7 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {9}{2}} b}\right )} \]

[In]

integrate((B*x+A)/x^4/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

-1/48*b^3*(2*(48*B*a^4 - 48*A*a^3*b - 15*(6*B*a - 7*A*b)*(b*x + a)^3 + 40*(6*B*a^2 - 7*A*a*b)*(b*x + a)^2 - 33
*(6*B*a^3 - 7*A*a^2*b)*(b*x + a))/((b*x + a)^(7/2)*a^4*b - 3*(b*x + a)^(5/2)*a^5*b + 3*(b*x + a)^(3/2)*a^6*b -
sqrt(b*x + a)*a^7*b) - 15*(6*B*a - 7*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(9/2)*b
))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=\frac {5 \, {\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{4}} + \frac {2 \, {\left (B a b^{2} - A b^{3}\right )}}{\sqrt {b x + a} a^{4}} + \frac {42 \, {\left (b x + a\right )}^{\frac {5}{2}} B a b^{2} - 96 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{2} b^{2} + 54 \, \sqrt {b x + a} B a^{3} b^{2} - 57 \, {\left (b x + a\right )}^{\frac {5}{2}} A b^{3} + 136 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{3} - 87 \, \sqrt {b x + a} A a^{2} b^{3}}{24 \, a^{4} b^{3} x^{3}} \]

[In]

integrate((B*x+A)/x^4/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

5/8*(6*B*a*b^2 - 7*A*b^3)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) + 2*(B*a*b^2 - A*b^3)/(sqrt(b*x + a)*a
^4) + 1/24*(42*(b*x + a)^(5/2)*B*a*b^2 - 96*(b*x + a)^(3/2)*B*a^2*b^2 + 54*sqrt(b*x + a)*B*a^3*b^2 - 57*(b*x +
a)^(5/2)*A*b^3 + 136*(b*x + a)^(3/2)*A*a*b^3 - 87*sqrt(b*x + a)*A*a^2*b^3)/(a^4*b^3*x^3)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x}{x^4 (a+b x)^{3/2}} \, dx=\frac {5\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (7\,A\,b-6\,B\,a\right )}{8\,a^{9/2}}-\frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{a}-\frac {11\,\left (7\,A\,b^3-6\,B\,a\,b^2\right )\,\left (a+b\,x\right )}{8\,a^2}+\frac {5\,\left (7\,A\,b^3-6\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^2}{3\,a^3}-\frac {5\,\left (7\,A\,b^3-6\,B\,a\,b^2\right )\,{\left (a+b\,x\right )}^3}{8\,a^4}}{3\,a\,{\left (a+b\,x\right )}^{5/2}-{\left (a+b\,x\right )}^{7/2}+a^3\,\sqrt {a+b\,x}-3\,a^2\,{\left (a+b\,x\right )}^{3/2}} \]

[In]

int((A + B*x)/(x^4*(a + b*x)^(3/2)),x)

[Out]

(5*b^2*atanh((a + b*x)^(1/2)/a^(1/2))*(7*A*b - 6*B*a))/(8*a^(9/2)) - ((2*(A*b^3 - B*a*b^2))/a - (11*(7*A*b^3 -
6*B*a*b^2)*(a + b*x))/(8*a^2) + (5*(7*A*b^3 - 6*B*a*b^2)*(a + b*x)^2)/(3*a^3) - (5*(7*A*b^3 - 6*B*a*b^2)*(a +
b*x)^3)/(8*a^4))/(3*a*(a + b*x)^(5/2) - (a + b*x)^(7/2) + a^3*(a + b*x)^(1/2) - 3*a^2*(a + b*x)^(3/2))