\(\int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx\) [483]

Optimal result

Integrand size = 20, antiderivative size = 82 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=\frac {(2 A b+a B) \sqrt {x} \sqrt {a+b x}}{a}-\frac {2 A (a+b x)^{3/2}}{a \sqrt {x}}+\frac {(2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, 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 = {79, 52, 65, 223, 212} \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=\frac {(a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}+\frac {\sqrt {x} \sqrt {a+b x} (a B+2 A b)}{a}-\frac {2 A (a+b x)^{3/2}}{a \sqrt {x}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=\frac {\sqrt {a+b x} (-2 A+B x)}{\sqrt {x}}+\frac {2 (2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-B x +2 A \right )}{\sqrt {x}}+\frac {\left (A b +\frac {B a}{2}\right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {b}\, \sqrt {x}\, \sqrt {b x +a}}\)	\(77\)
default	\(\frac {\sqrt {b x +a}\, \left (2 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) b x +B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a x +2 B x \sqrt {x \left (b x +a \right )}\, \sqrt {b}-4 A \sqrt {b}\, \sqrt {x \left (b x +a \right )}\right )}{2 \sqrt {x}\, \sqrt {x \left (b x +a \right )}\, \sqrt {b}}\)	\(118\)

[In]

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

[Out]

-(b*x+a)^(1/2)*(-B*x+2*A)/x^(1/2)+(A*b+1/2*B*a)*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 = 131, normalized size of antiderivative = 1.60 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=\left [\frac {{\left (B a + 2 \, A b\right )} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (B b x - 2 \, A b\right )} \sqrt {b x + a} \sqrt {x}}{2 \, b x}, -\frac {{\left (B a + 2 \, A b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (B b x - 2 \, A b\right )} \sqrt {b x + a} \sqrt {x}}{b x}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=- \frac {2 A \sqrt {a}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} + 2 A \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} - \frac {2 A b \sqrt {x}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}} + B \sqrt {a} \sqrt {x} \sqrt {1 + \frac {b x}{a}} + \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=\frac {B a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, \sqrt {b}} + A \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \sqrt {b x^{2} + a x} B - \frac {2 \, \sqrt {b x^{2} + a x} A}{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 76.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {a+b x} (A+B x)}{x^{3/2}} \, dx=\frac {{\left (\frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} B}{b} - \frac {B a b + 2 \, A b^{2}}{b^{2}}\right )}}{\sqrt {{\left (b x + a\right )} b - a b}} - \frac {{\left (B a + 2 \, A b\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right )}{b^{\frac {3}{2}}}\right )} b^{2}}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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