\(\int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx\) [482]

Optimal result

Integrand size = 20, antiderivative size = 93 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=\frac {(4 A b-a B) \sqrt {x} \sqrt {a+b x}}{4 b}+\frac {B \sqrt {x} (a+b x)^{3/2}}{2 b}+\frac {a (4 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {81, 52, 65, 223, 212} \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=\frac {a (4 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{3/2}}+\frac {\sqrt {x} \sqrt {a+b x} (4 A b-a B)}{4 b}+\frac {B \sqrt {x} (a+b x)^{3/2}}{2 b} \]

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} (4 A b+a B+2 b B x)}{4 b}+\frac {a (-4 A b+a B) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{4 b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.42 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {\left (2 b B x +4 A b +B a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b}+\frac {a \left (4 A b -B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{8 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(88\)
default	\(\frac {\sqrt {b x +a}\, \sqrt {x}\, \left (4 B \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+4 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b +8 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}-B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2}+2 B a \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right )}{8 b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}}\)	\(136\)

[In]

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

[Out]

1/4*(2*B*b*x+4*A*b+B*a)*x^(1/2)*(b*x+a)^(1/2)/b+1/8*a*(4*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 = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.57 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=\left [-\frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (2 \, B b^{2} x + B a b + 4 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b^{2}}, \frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (2 \, B b^{2} x + B a b + 4 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=A \sqrt {a} \sqrt {x} \sqrt {1 + \frac {b x}{a}} + \frac {A a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} + 2 B \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a \sqrt {x} \sqrt {a + b x}}{8 b} + \frac {x^{\frac {3}{2}} \sqrt {a + b x}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

A*sqrt(a)*sqrt(x)*sqrt(1 + b*x/a) + A*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/sqrt(b) + 2*B*Piecewise((-a**2*Piecewis
e((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (sqrt(x)*log(sqrt(x))/sqrt(b*x), True))/(8*b
) + a*sqrt(x)*sqrt(a + b*x)/(8*b) + x**(3/2)*sqrt(a + b*x)/4, Ne(b, 0)), (sqrt(a)*x**(3/2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a x} B x - \frac {B a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, \sqrt {b}} + \sqrt {b x^{2} + a x} A + \frac {\sqrt {b x^{2} + a x} B a}{4 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 75.72 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=\frac {{\left (\sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B}{b^{2}} - \frac {B a b^{2} - 4 \, A b^{3}}{b^{4}}\right )} + \frac {{\left (B a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}}\right )} b}{4 \, {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {x}} \, dx=A\,\sqrt {x}\,\sqrt {a+b\,x}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )}{\sqrt {b}}+B\,\sqrt {x}\,\left (\frac {x}{2}+\frac {a}{4\,b}\right )\,\sqrt {a+b\,x}-\frac {B\,a^2\,\ln \left (a+2\,b\,x+2\,\sqrt {b}\,\sqrt {x}\,\sqrt {a+b\,x}\right )}{8\,b^{3/2}} \]

[In]

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

[Out]

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