\(\int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx\) [441]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {3 b (5 A b-4 a B)}{4 a^3 \sqrt {a+b x}}-\frac {A}{2 a x^2 \sqrt {a+b x}}+\frac {5 A b-4 a B}{4 a^2 x \sqrt {a+b x}}-\frac {3 b (5 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {79, 44, 53, 65, 214} \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=-\frac {3 b (5 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {3 b (5 A b-4 a B)}{4 a^3 \sqrt {a+b x}}+\frac {5 A b-4 a B}{4 a^2 x \sqrt {a+b x}}-\frac {A}{2 a x^2 \sqrt {a+b x}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 A b^2 x^2+a b x (5 A-12 B x)-2 a^2 (A+2 B x)}{4 a^3 x^2 \sqrt {a+b x}}+\frac {3 b (-5 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[In]

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

[Out]

(15*A*b^2*x^2 + a*b*x*(5*A - 12*B*x) - 2*a^2*(A + 2*B*x))/(4*a^3*x^2*Sqrt[a + b*x]) + (3*b*(-5*A*b + 4*a*B)*Ar
cTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(7/2))

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(-\frac {15 \left (-\frac {x \left (-\frac {12 B x}{5}+A \right ) b \,a^{\frac {3}{2}}}{3}+\frac {2 \left (2 B x +A \right ) a^{\frac {5}{2}}}{15}+x^{2} b \left (-A b \sqrt {a}+\sqrt {b x +a}\, \left (A b -\frac {4 B a}{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )\right )}{4 \sqrt {b x +a}\, a^{\frac {7}{2}} x^{2}}\)	\(82\)
risch	\(-\frac {\sqrt {b x +a}\, \left (-7 A b x +4 B a x +2 A a \right )}{4 a^{3} x^{2}}+\frac {b \left (-\frac {2 \left (-8 A b +8 B a \right )}{\sqrt {b x +a}}-\frac {2 \left (15 A b -12 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{8 a^{3}}\)	\(83\)
derivativedivides	\(2 b \left (-\frac {\frac {\left (-\frac {7 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {9}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {3 \left (5 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}-\frac {-A b +B a}{a^{3} \sqrt {b x +a}}\right )\)	\(102\)
default	\(2 b \left (-\frac {\frac {\left (-\frac {7 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (\frac {9}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {3 \left (5 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}-\frac {-A b +B a}{a^{3} \sqrt {b x +a}}\right )\)	\(102\)

[In]

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

[Out]

-15/4/(b*x+a)^(1/2)/a^(7/2)*(-1/3*x*(-12/5*B*x+A)*b*a^(3/2)+2/15*(2*B*x+A)*a^(5/2)+x^2*b*(-A*b*a^(1/2)+(b*x+a)
^(1/2)*(A*b-4/5*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.50 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, A a^{3} + 3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, -\frac {3 \, {\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{3} + {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, A a^{3} + 3 \, {\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 45.74 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.68 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=A \left (- \frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}}\right ) + B \left (- \frac {1}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}}\right ) \]

[In]

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

[Out]

A*(-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))) + B*(-1/(a*sqrt(b)*
x**(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x))
)/a**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=-\frac {1}{8} \, b^{2} {\left (\frac {2 \, {\left (8 \, B a^{3} - 8 \, A a^{2} b + 3 \, {\left (4 \, B a - 5 \, A b\right )} {\left (b x + a\right )}^{2} - 5 \, {\left (4 \, B a^{2} - 5 \, A a b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {5}{2}} a^{3} b - 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} b + \sqrt {b x + a} a^{5} b} + \frac {3 \, {\left (4 \, B a - 5 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}} b}\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} - \frac {2 \, {\left (B a b - A b^{2}\right )}}{\sqrt {b x + a} a^{3}} - \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b - 4 \, \sqrt {b x + a} B a^{2} b - 7 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{2} + 9 \, \sqrt {b x + a} A a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x}{x^3 (a+b x)^{3/2}} \, dx=\frac {\frac {2\,\left (A\,b^2-B\,a\,b\right )}{a}-\frac {5\,\left (5\,A\,b^2-4\,B\,a\,b\right )\,\left (a+b\,x\right )}{4\,a^2}+\frac {3\,\left (5\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^2}{4\,a^3}}{{\left (a+b\,x\right )}^{5/2}-2\,a\,{\left (a+b\,x\right )}^{3/2}+a^2\,\sqrt {a+b\,x}}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (5\,A\,b-4\,B\,a\right )}{4\,a^{7/2}} \]

[In]

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

[Out]

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