\(\int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx\) [430]

Optimal result

Integrand size = 18, antiderivative size = 84 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=-\frac {A \sqrt {a+b x}}{2 a x^2}+\frac {(3 A b-4 a B) \sqrt {a+b x}}{4 a^2 x}-\frac {b (3 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 44, 65, 214} \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=-\frac {b (3 A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {\sqrt {a+b x} (3 A b-4 a B)}{4 a^2 x}-\frac {A \sqrt {a+b x}}{2 a x^2} \]

[In]

Int[(A + B*x)/(x^3*Sqrt[a + b*x]),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=\frac {\frac {\sqrt {a} \sqrt {a+b x} (3 A b x-2 a (A+2 B x))}{x^2}+b (-3 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-3 A b x +4 B a x +2 A a \right )}{4 a^{2} x^{2}}-\frac {b \left (3 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 a^{\frac {5}{2}}}\)	\(59\)
pseudoelliptic	\(\frac {-\frac {3 x^{2} \left (A b -\frac {4 B a}{3}\right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4}+\frac {3 \sqrt {b x +a}\, \left (\frac {2 \left (-2 B x -A \right ) a^{\frac {3}{2}}}{3}+A \sqrt {a}\, b x \right )}{4}}{a^{\frac {5}{2}} x^{2}}\)	\(65\)
derivativedivides	\(2 b \left (-\frac {-\frac {\left (3 A b -4 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a^{2}}+\frac {\left (5 A b -4 B a \right ) \sqrt {b x +a}}{8 a}}{b^{2} x^{2}}-\frac {\left (3 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {5}{2}}}\right )\)	\(82\)
default	\(2 b \left (-\frac {-\frac {\left (3 A b -4 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a^{2}}+\frac {\left (5 A b -4 B a \right ) \sqrt {b x +a}}{8 a}}{b^{2} x^{2}}-\frac {\left (3 A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {5}{2}}}\right )\)	\(82\)

[In]

int((B*x+A)/x^3/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.89 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=\left [-\frac {{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt {a} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {b x + a}}{8 \, a^{3} x^{2}}, -\frac {{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {b x + a}}{4 \, a^{3} x^{2}}\right ] \]

[In]

integrate((B*x+A)/x^3/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (73) = 146\).

Time = 14.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.86 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=- \frac {A}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {A \sqrt {b}}{4 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {3 A b^{\frac {3}{2}}}{4 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {3 A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {5}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{a \sqrt {x}} + \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {3}{2}}} \]

[In]

integrate((B*x+A)/x**3/(b*x+a)**(1/2),x)

[Out]

-A/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + A*sqrt(b)/(4*a*x**(3/2)*sqrt(a/(b*x) + 1)) + 3*A*b**(3/2)/(4*a**2*
sqrt(x)*sqrt(a/(b*x) + 1)) - 3*A*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(5/2)) - B*sqrt(b)*sqrt(a/(b*x) +
1)/(a*sqrt(x)) + B*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(3/2)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.45 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=-\frac {1}{8} \, b^{2} {\left (\frac {2 \, {\left ({\left (4 \, B a - 3 \, A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - {\left (4 \, B a^{2} - 5 \, A a b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{2} a^{2} b - 2 \, {\left (b x + a\right )} a^{3} b + a^{4} b} + \frac {{\left (4 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]

[In]

integrate((B*x+A)/x^3/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

-1/8*b^2*(2*((4*B*a - 3*A*b)*(b*x + a)^(3/2) - (4*B*a^2 - 5*A*a*b)*sqrt(b*x + a))/((b*x + a)^2*a^2*b - 2*(b*x
+ a)*a^3*b + a^4*b) + (4*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 = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.32 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=-\frac {\frac {{\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x + a} B a^{2} b^{2} - 3 \, {\left (b x + a\right )}^{\frac {3}{2}} A b^{3} + 5 \, \sqrt {b x + a} A a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \]

[In]

integrate((B*x+A)/x^3/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{x^3 \sqrt {a+b x}} \, dx=-\frac {\frac {\left (5\,A\,b^2-4\,B\,a\,b\right )\,\sqrt {a+b\,x}}{4\,a}-\frac {\left (3\,A\,b^2-4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^{3/2}}{4\,a^2}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}-\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-4\,B\,a\right )}{4\,a^{5/2}} \]

[In]

int((A + B*x)/(x^3*(a + b*x)^(1/2)),x)

[Out]

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