\(\int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx\) [519]

Optimal result

Integrand size = 20, antiderivative size = 50 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=-\frac {2 A \sqrt {a+b x}}{a \sqrt {x}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {79, 65, 223, 212} \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}-\frac {2 A \sqrt {a+b x}}{a \sqrt {x}} \]

[In]

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

[Out]

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

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A \sqrt {a+b x}}{a \sqrt {x}}+B \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx \\ & = -\frac {2 A \sqrt {a+b x}}{a \sqrt {x}}+(2 B) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 A \sqrt {a+b x}}{a \sqrt {x}}+(2 B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right ) \\ & = -\frac {2 A \sqrt {a+b x}}{a \sqrt {x}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=-\frac {2 A \sqrt {a+b x}}{a \sqrt {x}}-\frac {2 B \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {b}} \]

[In]

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

[Out]

(-2*A*Sqrt[a + b*x])/(a*Sqrt[x]) - (2*B*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/Sqrt[b]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.32

method	result	size
risch	\(-\frac {2 A \sqrt {b x +a}}{a \sqrt {x}}+\frac {B \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {b}\, \sqrt {x}\, \sqrt {b x +a}}\)	\(66\)
default	\(\frac {\left (B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a x -2 A \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right ) \sqrt {b x +a}}{a \sqrt {x}\, \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(73\)

[In]

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

[Out]

-2*A*(b*x+a)^(1/2)/a/x^(1/2)+B*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))/b^(1/2)*(x*(b*x+a))^(1/2)/x^(1/2)/(b*
x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.18 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=\left [\frac {B a \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, \sqrt {b x + a} A b \sqrt {x}}{a b x}, -\frac {2 \, {\left (B a \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + \sqrt {b x + a} A b \sqrt {x}\right )}}{a b x}\right ] \]

[In]

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

[Out]

[(B*a*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*sqrt(b*x + a)*A*b*sqrt(x))/(a*b*x), -2*(B
*a*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + sqrt(b*x + a)*A*b*sqrt(x))/(a*b*x)]

Sympy [A] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=- \frac {2 A \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{a} + \frac {2 B \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=\frac {B \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{a x} \]

[In]

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

[Out]

B*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) - 2*sqrt(b*x^2 + a*x)*A/(a*x)

Giac [A] (verification not implemented)

none

Time = 77.78 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.40 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=-\frac {2 \, b^{2} {\left (\frac {B \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {b x + a} A}{\sqrt {{\left (b x + a\right )} b - a b} a}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx=-\frac {4\,B\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {-b}\,\sqrt {x}}\right )}{\sqrt {-b}}-\frac {2\,A\,\sqrt {a+b\,x}}{a\,\sqrt {x}} \]

[In]

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

[Out]

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