\(\int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx\) [517]

Optimal result

Integrand size = 20, antiderivative size = 93 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {(4 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b}-\frac {a (4 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{5/2}} \]

[Out]

-1/4*a*(4*A*b-3*B*a)*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)+1/2*B*x^(3/2)*(b*x+a)^(1/2)/b+1/4*(4*A*b-3
*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^2

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} \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=-\frac {a (4 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{5/2}}+\frac {\sqrt {x} \sqrt {a+b x} (4 A b-3 a B)}{4 b^2}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b} \]

[In]

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

[Out]

((4*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(4*b^2) + (B*x^(3/2)*Sqrt[a + b*x])/(2*b) - (a*(4*A*b - 3*a*B)*ArcTanh
[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(5/2))

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 x^{3/2} \sqrt {a+b x}}{2 b}+\frac {\left (2 A b-\frac {3 a B}{2}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{2 b} \\ & = \frac {(4 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b}-\frac {(a (4 A b-3 a B)) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{8 b^2} \\ & = \frac {(4 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b}-\frac {(a (4 A b-3 a B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^2} \\ & = \frac {(4 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b}-\frac {(a (4 A b-3 a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^2} \\ & = \frac {(4 A b-3 a B) \sqrt {x} \sqrt {a+b x}}{4 b^2}+\frac {B x^{3/2} \sqrt {a+b x}}{2 b}-\frac {a (4 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {x} \sqrt {a+b x} (4 A b-3 a B+2 b B x)}{4 b^2}+\frac {a (-4 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 b^{5/2}} \]

[In]

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

[Out]

(Sqrt[x]*Sqrt[a + b*x]*(4*A*b - 3*a*B + 2*b*B*x))/(4*b^2) + (a*(-4*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sq
rt[a] + Sqrt[a + b*x])])/(2*b^(5/2))

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {\left (2 b B x +4 A b -3 B a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{2}}-\frac {a \left (4 A b -3 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +a}}\)	\(89\)
default	\(-\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (-4 B \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+4 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a b -8 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}-3 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2}+6 B a \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right )}{8 b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}}\)	\(136\)

[In]

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

[Out]

1/4*(2*B*b*x+4*A*b-3*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^2-1/8*a*(4*A*b-3*B*a)/b^(5/2)*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 = 151, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\left [-\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (2 \, B b^{2} x - 3 \, B a b + 4 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b^{3}}, -\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, B b^{2} x - 3 \, B a b + 4 \, A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b^{3}}\right ] \]

[In]

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

[Out]

[-1/8*((3*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 - 3*B*a*b +
4*A*b^2)*sqrt(b*x + a)*sqrt(x))/b^3, -1/4*((3*B*a^2 - 4*A*a*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) - (2*B*b^2*x - 3*B*a*b + 4*A*b^2)*sqrt(b*x + a)*sqrt(x))/b^3]

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\begin {cases} - \frac {a \left (2 A - \frac {3 B a}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x} + 2 b \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {\sqrt {x} \log {\left (\sqrt {x} \right )}}{\sqrt {b x}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x} \left (\frac {B x^{\frac {3}{2}}}{2 b} + \frac {\sqrt {x} \left (2 A - \frac {3 B a}{2 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a*(2*A - 3*B*a/(2*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x) + 2*b*sqrt(x))/sqrt(b), Ne(a, 0)), (s
qrt(x)*log(sqrt(x))/sqrt(b*x), True))/(2*b) + sqrt(a + b*x)*(B*x**(3/2)/(2*b) + sqrt(x)*(2*A - 3*B*a/(2*b))/(2
*b)), Ne(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\sqrt {b x^{2} + a x} B x}{2 \, b} + \frac {3 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} B a}{4 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A}{b} \]

[In]

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

[Out]

1/2*sqrt(b*x^2 + a*x)*B*x/b + 3/8*B*a^2*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(5/2) - 1/2*A*a*log(2*x
+ a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(3/2) - 3/4*sqrt(b*x^2 + a*x)*B*a/b^2 + sqrt(b*x^2 + a*x)*A/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (71) = 142\).

Time = 156.25 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\frac {4 \, {\left (a \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) + \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a}\right )} A {\left | b \right |}}{b^{2}} - \frac {{\left (3 \, a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) - \sqrt {{\left (b x + a\right )} b - a b} {\left (2 \, b x - 3 \, a\right )} \sqrt {b x + a}\right )} B {\left | b \right |}}{b^{3}}}{4 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.13 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.87 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a+b x}} \, dx=\frac {\frac {x^{7/2}\,\left (2\,A\,a\,b^2-\frac {3\,B\,a^2\,b}{2}\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}+\frac {x^{5/2}\,\left (\frac {11\,B\,a^2}{2}-2\,A\,a\,b\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}-\frac {\sqrt {x}\,\left (3\,B\,a^2-4\,A\,a\,b\right )}{2\,b^2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}+\frac {x^{3/2}\,\left (11\,B\,a^2-4\,A\,a\,b\right )}{2\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{\frac {6\,b^2\,x^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}-\frac {4\,b^3\,x^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}+\frac {b^4\,x^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}-\frac {4\,b\,x}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+1}-\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )\,\left (4\,A\,b-3\,B\,a\right )}{2\,b^{5/2}} \]

[In]

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

[Out]

((x^(7/2)*(2*A*a*b^2 - (3*B*a^2*b)/2))/((a + b*x)^(1/2) - a^(1/2))^7 + (x^(5/2)*((11*B*a^2)/2 - 2*A*a*b))/((a
+ b*x)^(1/2) - a^(1/2))^5 - (x^(1/2)*(3*B*a^2 - 4*A*a*b))/(2*b^2*((a + b*x)^(1/2) - a^(1/2))) + (x^(3/2)*(11*B
*a^2 - 4*A*a*b))/(2*b*((a + b*x)^(1/2) - a^(1/2))^3))/((6*b^2*x^2)/((a + b*x)^(1/2) - a^(1/2))^4 - (4*b^3*x^3)
/((a + b*x)^(1/2) - a^(1/2))^6 + (b^4*x^4)/((a + b*x)^(1/2) - a^(1/2))^8 - (4*b*x)/((a + b*x)^(1/2) - a^(1/2))
^2 + 1) - (a*atanh((b^(1/2)*x^(1/2))/((a + b*x)^(1/2) - a^(1/2)))*(4*A*b - 3*B*a))/(2*b^(5/2))