\(\int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx\) [357]

Optimal result

Integrand size = 18, antiderivative size = 85 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=-\frac {(A b-3 a B) \sqrt {x}}{a b^2}+\frac {(A b-a B) x^{3/2}}{a b (a+b x)}+\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 52, 65, 211} \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}}-\frac {\sqrt {x} (A b-3 a B)}{a b^2}+\frac {x^{3/2} (A b-a B)}{a b (a+b x)} \]

[In]

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

[Out]

-(((A*b - 3*a*B)*Sqrt[x])/(a*b^2)) + ((A*b - a*B)*x^(3/2))/(a*b*(a + b*x)) + ((A*b - 3*a*B)*ArcTan[(Sqrt[b]*Sq
rt[x])/Sqrt[a]])/(Sqrt[a]*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 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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\frac {\sqrt {x} (-A b+3 a B+2 b B x)}{b^2 (a+b x)}+\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {2 B \sqrt {x}}{b^{2}}+\frac {\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}}{b^{2}}\)	\(62\)
derivativedivides	\(\frac {2 B \sqrt {x}}{b^{2}}+\frac {\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}}{b^{2}}\)	\(63\)
default	\(\frac {2 B \sqrt {x}}{b^{2}}+\frac {\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}}{b^{2}}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.33 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\left [\frac {{\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (2 \, B a b^{2} x + 3 \, B a^{2} b - A a b^{2}\right )} \sqrt {x}}{2 \, {\left (a b^{4} x + a^{2} b^{3}\right )}}, \frac {{\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (2 \, B a b^{2} x + 3 \, B a^{2} b - A a b^{2}\right )} \sqrt {x}}{a b^{4} x + a^{2} b^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 634 vs. \(2 (73) = 146\).

Time = 2.00 (sec) , antiderivative size = 634, normalized size of antiderivative = 7.46 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{a^{2}} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\\frac {A a b \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {A a b \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {2 A b^{2} \sqrt {x} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {A b^{2} x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {A b^{2} x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {3 B a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {3 B a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {6 B a b \sqrt {x} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {3 B a b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {3 B a b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {4 B b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x**(5/2)/5)/a**2, Eq
(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x))/b**2, Eq(a, 0)), (A*a*b*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b)
+ 2*b**4*x*sqrt(-a/b)) - A*a*b*log(sqrt(x) + sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) - 2*A*b**
2*sqrt(x)*sqrt(-a/b)/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + A*b**2*x*log(sqrt(x) - sqrt(-a/b))/(2*a*b**
3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) - A*b**2*x*log(sqrt(x) + sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(
-a/b)) - 3*B*a**2*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 3*B*a**2*log(sqrt(x)
+ sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 6*B*a*b*sqrt(x)*sqrt(-a/b)/(2*a*b**3*sqrt(-a/b) +
2*b**4*x*sqrt(-a/b)) - 3*B*a*b*x*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 3*B*
a*b*x*log(sqrt(x) + sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 4*B*b**2*x**(3/2)*sqrt(-a/b)/(2*
a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\frac {{\left (B a - A b\right )} \sqrt {x}}{b^{3} x + a b^{2}} + \frac {2 \, B \sqrt {x}}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\frac {2 \, B \sqrt {x}}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {B a \sqrt {x} - A b \sqrt {x}}{{\left (b x + a\right )} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2} \, dx=\frac {2\,B\,\sqrt {x}}{b^2}-\frac {\sqrt {x}\,\left (A\,b-B\,a\right )}{x\,b^3+a\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-3\,B\,a\right )}{\sqrt {a}\,b^{5/2}} \]

[In]

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

[Out]

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