\(\int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx\) [451]

Optimal result

Integrand size = 18, antiderivative size = 136 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=\frac {5 b (7 A b-4 a B)}{12 a^3 (a+b x)^{3/2}}-\frac {A}{2 a x^2 (a+b x)^{3/2}}+\frac {7 A b-4 a B}{4 a^2 x (a+b x)^{3/2}}+\frac {5 b (7 A b-4 a B)}{4 a^4 \sqrt {a+b x}}-\frac {5 b (7 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 136, 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} \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=-\frac {5 b (7 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {5 b (7 A b-4 a B)}{4 a^4 \sqrt {a+b x}}+\frac {5 b (7 A b-4 a B)}{12 a^3 (a+b x)^{3/2}}+\frac {7 A b-4 a B}{4 a^2 x (a+b x)^{3/2}}-\frac {A}{2 a x^2 (a+b x)^{3/2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=\frac {105 A b^3 x^3+a^2 b x (21 A-80 B x)+20 a b^2 x^2 (7 A-3 B x)-6 a^3 (A+2 B x)}{12 a^4 x^2 (a+b x)^{3/2}}+\frac {5 b (-7 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{9/2}} \]

[In]

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

[Out]

(105*A*b^3*x^3 + a^2*b*x*(21*A - 80*B*x) + 20*a*b^2*x^2*(7*A - 3*B*x) - 6*a^3*(A + 2*B*x))/(12*a^4*x^2*(a + b*
x)^(3/2)) + (5*b*(-7*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(9/2))

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-11 A b x +4 B a x +2 A a \right )}{4 a^{4} x^{2}}+\frac {b \left (-\frac {2 \left (-24 A b +16 B a \right )}{\sqrt {b x +a}}+\frac {16 a \left (A b -B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}-\frac {2 \left (35 A b -20 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{8 a^{4}}\)	\(101\)
pseudoelliptic	\(\frac {-\frac {35 x^{2} \left (b x +a \right )^{\frac {3}{2}} b \left (A b -\frac {4 B a}{7}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4}+\frac {35 x^{2} \left (-\frac {3 B x}{7}+A \right ) b^{2} a^{\frac {3}{2}}}{3}+\frac {7 b x \left (-\frac {80 B x}{21}+A \right ) a^{\frac {5}{2}}}{4}+\frac {\left (-2 B x -A \right ) a^{\frac {7}{2}}}{2}+\frac {35 A \sqrt {a}\, b^{3} x^{3}}{4}}{a^{\frac {9}{2}} \left (b x +a \right )^{\frac {3}{2}} x^{2}}\)	\(104\)
derivativedivides	\(2 b \left (-\frac {\frac {\left (-\frac {11 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {13}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {5 \left (7 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{4}}-\frac {-3 A b +2 B a}{a^{4} \sqrt {b x +a}}-\frac {-A b +B a}{3 a^{3} \left (b x +a \right )^{\frac {3}{2}}}\right )\)	\(123\)
default	\(2 b \left (-\frac {\frac {\left (-\frac {11 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {13}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {5 \left (7 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{4}}-\frac {-3 A b +2 B a}{a^{4} \sqrt {b x +a}}-\frac {-A b +B a}{3 a^{3} \left (b x +a \right )^{\frac {3}{2}}}\right )\)	\(123\)

[In]

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

[Out]

-1/4*(b*x+a)^(1/2)*(-11*A*b*x+4*B*a*x+2*A*a)/a^4/x^2+1/8/a^4*b*(-2*(-24*A*b+16*B*a)/(b*x+a)^(1/2)+16/3*a*(A*b-
B*a)/(b*x+a)^(3/2)-2*(35*A*b-20*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.90 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=\left [-\frac {15 \, {\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 2 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (6 \, A a^{4} + 15 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 20 \, {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{24 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}, -\frac {15 \, {\left ({\left (4 \, B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 2 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (6 \, A a^{4} + 15 \, {\left (4 \, B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 20 \, {\left (4 \, B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left (4 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x + a}}{12 \, {\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/24*(15*((4*B*a*b^3 - 7*A*b^4)*x^4 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + (4*B*a^3*b - 7*A*a^2*b^2)*x^2)*sqrt(
a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(6*A*a^4 + 15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + 20*(4*B*a^3*
b - 7*A*a^2*b^2)*x^2 + 3*(4*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2), -1/12*
(15*((4*B*a*b^3 - 7*A*b^4)*x^4 + 2*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + (4*B*a^3*b - 7*A*a^2*b^2)*x^2)*sqrt(-a)*arc
tan(sqrt(b*x + a)*sqrt(-a)/a) + (6*A*a^4 + 15*(4*B*a^2*b^2 - 7*A*a*b^3)*x^3 + 20*(4*B*a^3*b - 7*A*a^2*b^2)*x^2
+ 3*(4*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x + a))/(a^5*b^2*x^4 + 2*a^6*b*x^3 + a^7*x^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1287 vs. \(2 (129) = 258\).

Time = 59.68 (sec) , antiderivative size = 1287, normalized size of antiderivative = 9.46 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

A*(-6*a**(89/2)*b**75*x**75/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x*
*(157/2)*sqrt(a/(b*x) + 1)) + 21*a**(87/2)*b**76*x**76/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) +
12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) + 140*a**(85/2)*b**77*x**77/(12*a**(93/2)*b**(151/2)*x*
*(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) + 105*a**(83/2)*b**78*x**78
/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1))
- 105*a**42*b**(155/2)*x**(155/2)*sqrt(a/(b*x) + 1)*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)
*x**(155/2)*sqrt(a/(b*x) + 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1)) - 105*a**41*b**(157/2)*x
**(157/2)*sqrt(a/(b*x) + 1)*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(12*a**(93/2)*b**(151/2)*x**(155/2)*sqrt(a/(b*x)
+ 1) + 12*a**(91/2)*b**(153/2)*x**(157/2)*sqrt(a/(b*x) + 1))) + B*(-6*a**17*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 1
8*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 46*a**16*b*x*sqrt(1 + b*x/a)/(6*a**(39/
2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 15*a**16*b*x*log(b*x/a)/(6*a**(
39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 30*a**16*b*x*log(sqrt(1 + b*
x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 70*a**15*b*
*2*x**2*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)
- 45*a**15*b**2*x**2*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b
**3*x**4) + 90*a**15*b**2*x**2*log(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b*
*2*x**3 + 6*a**(33/2)*b**3*x**4) - 30*a**14*b**3*x**3*sqrt(1 + b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 1
8*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 45*a**14*b**3*x**3*log(b*x/a)/(6*a**(39/2)*x + 18*a**(37/2)*b
*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 90*a**14*b**3*x**3*log(sqrt(1 + b*x/a) + 1)/(6*a**(3
9/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) - 15*a**13*b**4*x**4*log(b*x/a)
/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4) + 30*a**13*b**4*x**4*l
og(sqrt(1 + b*x/a) + 1)/(6*a**(39/2)*x + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x**3 + 6*a**(33/2)*b**3*x**4)
)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=-\frac {1}{24} \, b^{2} {\left (\frac {2 \, {\left (8 \, B a^{4} - 8 \, A a^{3} b + 15 \, {\left (4 \, B a - 7 \, A b\right )} {\left (b x + a\right )}^{3} - 25 \, {\left (4 \, B a^{2} - 7 \, A a b\right )} {\left (b x + a\right )}^{2} + 8 \, {\left (4 \, B a^{3} - 7 \, A a^{2} b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {7}{2}} a^{4} b - 2 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} b + {\left (b x + a\right )}^{\frac {3}{2}} a^{6} b} + \frac {15 \, {\left (4 \, B a - 7 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {9}{2}} b}\right )} \]

[In]

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

[Out]

-1/24*b^2*(2*(8*B*a^4 - 8*A*a^3*b + 15*(4*B*a - 7*A*b)*(b*x + a)^3 - 25*(4*B*a^2 - 7*A*a*b)*(b*x + a)^2 + 8*(4
*B*a^3 - 7*A*a^2*b)*(b*x + a))/((b*x + a)^(7/2)*a^4*b - 2*(b*x + a)^(5/2)*a^5*b + (b*x + a)^(3/2)*a^6*b) + 15*
(4*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 = 0.29 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=-\frac {5 \, {\left (4 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4}} - \frac {2 \, {\left (6 \, {\left (b x + a\right )} B a b + B a^{2} b - 9 \, {\left (b x + a\right )} A b^{2} - A a b^{2}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4}} - \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x + a} B a^{2} b - 11 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{2} + 13 \, \sqrt {b x + a} A a b^{2}}{4 \, a^{4} b^{2} x^{2}} \]

[In]

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

[Out]

-5/4*(4*B*a*b - 7*A*b^2)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4) - 2/3*(6*(b*x + a)*B*a*b + B*a^2*b - 9*
(b*x + a)*A*b^2 - A*a*b^2)/((b*x + a)^(3/2)*a^4) - 1/4*(4*(b*x + a)^(3/2)*B*a*b - 4*sqrt(b*x + a)*B*a^2*b - 11
*(b*x + a)^(3/2)*A*b^2 + 13*sqrt(b*x + a)*A*a*b^2)/(a^4*b^2*x^2)

Mupad [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x}{x^3 (a+b x)^{5/2}} \, dx=\frac {\frac {2\,\left (A\,b^2-B\,a\,b\right )}{3\,a}+\frac {2\,\left (7\,A\,b^2-4\,B\,a\,b\right )\,\left (a+b\,x\right )}{3\,a^2}-\frac {25\,\left (7\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^2}{12\,a^3}+\frac {5\,\left (7\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^3}{4\,a^4}}{{\left (a+b\,x\right )}^{7/2}-2\,a\,{\left (a+b\,x\right )}^{5/2}+a^2\,{\left (a+b\,x\right )}^{3/2}}-\frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (7\,A\,b-4\,B\,a\right )}{4\,a^{9/2}} \]

[In]

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

[Out]

((2*(A*b^2 - B*a*b))/(3*a) + (2*(7*A*b^2 - 4*B*a*b)*(a + b*x))/(3*a^2) - (25*(7*A*b^2 - 4*B*a*b)*(a + b*x)^2)/
(12*a^3) + (5*(7*A*b^2 - 4*B*a*b)*(a + b*x)^3)/(4*a^4))/((a + b*x)^(7/2) - 2*a*(a + b*x)^(5/2) + a^2*(a + b*x)
^(3/2)) - (5*b*atanh((a + b*x)^(1/2)/a^(1/2))*(7*A*b - 4*B*a))/(4*a^(9/2))