\(\int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx\) [518]

Optimal result

Integrand size = 20, antiderivative size = 56 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\frac {B \sqrt {x} \sqrt {a+b x}}{b}+\frac {(2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, 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 = {81, 65, 223, 212} \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\frac {(2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}+\frac {B \sqrt {x} \sqrt {a+b x}}{b} \]

[In]

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

[Out]

(B*Sqrt[x]*Sqrt[a + b*x])/b + ((2*A*b - a*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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\frac {B \sqrt {x} \sqrt {a+b x}}{b}+\frac {2 (2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {B \sqrt {x}\, \sqrt {b x +a}}{b}+\frac {\left (2 A b -B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(74\)
default	\(\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (2 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b +2 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}-B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a \right )}{2 \sqrt {x \left (b x +a \right )}\, b^{\frac {3}{2}}}\)	\(101\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.93 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\left [\frac {2 \, \sqrt {b x + a} B b \sqrt {x} - {\left (B a - 2 \, A b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right )}{2 \, b^{2}}, \frac {\sqrt {b x + a} B b \sqrt {x} + {\left (B a - 2 \, A b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right )}{b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.96 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\frac {2 A \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} + 2 B \left (\begin {cases} - \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{2 b} + \frac {\sqrt {x} \sqrt {a + b x}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{\frac {3}{2}}}{3 \sqrt {a}} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

2*A*asinh(sqrt(b)*sqrt(x)/sqrt(a))/sqrt(b) + 2*B*Piecewise((-a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sq
rt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(2*b) + sqrt(x)*sqrt(a + b*x)/(2*b), Ne(b,
0)), (x**(3/2)/(3*sqrt(a)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=-\frac {B a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} + \frac {\sqrt {b x^{2} + a x} B}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 75.49 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.32 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\frac {b {\left (\frac {{\left (B a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} B}{b^{2}}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.38 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a+b x}} \, dx=\frac {B\,\sqrt {x}\,\sqrt {a+b\,x}}{b}-\frac {2\,B\,a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )}{b^{3/2}}-\frac {4\,A\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {-b}\,\sqrt {x}}\right )}{\sqrt {-b}} \]

[In]

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

[Out]

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