\(\int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx\) [530]

Optimal result

Integrand size = 20, antiderivative size = 60 \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\frac {2 (A b-a B) \sqrt {x}}{a b \sqrt {a+b x}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, 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} \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\frac {2 \sqrt {x} (A b-a B)}{a b \sqrt {a+b x}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\frac {2 (A b-a B) \sqrt {x}}{a b \sqrt {a+b x}}-\frac {2 B \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{b^{3/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(120\) vs. \(2(48)=96\).

Time = 1.46 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.02

method	result	size
default	\(\frac {\left (B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b x +2 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}+B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2}-2 B a \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right ) \sqrt {x}}{a \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}} \sqrt {b x +a}}\)	\(121\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.62 \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\left [\frac {{\left (B a b x + B a^{2}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (B a b - A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{a b^{3} x + a^{2} b^{2}}, -\frac {2 \, {\left ({\left (B a b x + B a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (B a b - A b^{2}\right )} \sqrt {b x + a} \sqrt {x}\right )}}{a b^{3} x + a^{2} b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\frac {2 A}{a \sqrt {b} \sqrt {\frac {a}{b x} + 1}} + B \left (\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {2 \sqrt {x}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.35 \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x^{2} + a x} A}{a b x + a^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} B}{b^{2} x + a b} + \frac {B \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (48) = 96\).

Time = 15.57 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.62 \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=-\frac {B \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt {b} {\left | b \right |}} - \frac {4 \, {\left (B a \sqrt {b} - A b^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx=\int \frac {A+B\,x}{\sqrt {x}\,{\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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