\(\int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx\) [394]

Optimal result

Integrand size = 18, antiderivative size = 82 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=\frac {(A b-4 a B) \sqrt {a+b x}}{4 a x}-\frac {A (a+b x)^{3/2}}{2 a x^2}+\frac {b (A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, 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, 43, 65, 214} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=\frac {b (A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}+\frac {\sqrt {a+b x} (A b-4 a B)}{4 a x}-\frac {A (a+b x)^{3/2}}{2 a x^2} \]

[In]

Int[(Sqrt[a + b*x]*(A + B*x))/x^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=-\frac {\sqrt {a+b x} (A b x+2 a (A+2 B x))}{4 a x^2}+\frac {b (A b-4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (A b x +4 B a x +2 A a \right )}{4 x^{2} a}+\frac {b \left (A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 a^{\frac {3}{2}}}\)	\(57\)
pseudoelliptic	\(-\frac {-b \,x^{2} \left (A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\left (\left (4 B x +2 A \right ) a^{\frac {3}{2}}+A \sqrt {a}\, b x \right ) \sqrt {b x +a}}{4 a^{\frac {3}{2}} x^{2}}\)	\(64\)
derivativedivides	\(2 b \left (-\frac {\frac {\left (A b +4 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a}+\left (-\frac {B a}{2}+\frac {A b}{8}\right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {\left (A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(76\)
default	\(2 b \left (-\frac {\frac {\left (A b +4 B a \right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a}+\left (-\frac {B a}{2}+\frac {A b}{8}\right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {\left (A b -4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(76\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.93 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=\left [-\frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {a} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x\right )} \sqrt {b x + a}}{8 \, a^{2} x^{2}}, \frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x\right )} \sqrt {b x + a}}{4 \, a^{2} x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (68) = 136\).

Time = 31.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=- \frac {A a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 A \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {A b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} - \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=-\frac {1}{8} \, b^{2} {\left (\frac {2 \, {\left ({\left (4 \, B a + A b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - {\left (4 \, B a^{2} - A a b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{2} a b - 2 \, {\left (b x + a\right )} a^{2} b + a^{3} b} - \frac {{\left (4 \, B a - A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}} b}\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=\frac {\frac {{\left (4 \, B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x + a} B a^{2} b^{2} + {\left (b x + a\right )}^{\frac {3}{2}} A b^{3} + \sqrt {b x + a} A a b^{3}}{a b^{2} x^{2}}}{4 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx=\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (A\,b-4\,B\,a\right )}{4\,a^{3/2}}-\frac {\left (\frac {A\,b^2}{4}-B\,a\,b\right )\,\sqrt {a+b\,x}+\frac {\left (A\,b^2+4\,B\,a\,b\right )\,{\left (a+b\,x\right )}^{3/2}}{4\,a}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2} \]

[In]

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

[Out]

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