\(\int \frac {\sqrt {x} (A+B x)}{(b x+c x^2)^2} \, dx\) [181]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^2} \, dx=\frac {b B-3 A c}{b^2 c \sqrt {x}}-\frac {b B-A c}{b c \sqrt {x} (b+c x)}+\frac {(b B-3 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2} \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {795, 79, 53, 65, 211} \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^2} \, dx=\frac {(b B-3 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{5/2} \sqrt {c}}+\frac {b B-3 A c}{b^2 c \sqrt {x}}-\frac {b B-A c}{b c \sqrt {x} (b+c x)} \]

[In]

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

[Out]

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

Rule 53

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

Rule 795

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

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

Mathematica [A] (verified)

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

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) \sqrt {x}}{c x +b}+\frac {\left (3 A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{2}}-\frac {2 A}{b^{2} \sqrt {x}}\)	\(64\)
default	\(-\frac {2 \left (\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) \sqrt {x}}{c x +b}+\frac {\left (3 A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{2}}-\frac {2 A}{b^{2} \sqrt {x}}\)	\(64\)
risch	\(-\frac {2 A}{b^{2} \sqrt {x}}-\frac {\frac {2 \left (\frac {A c}{2}-\frac {B b}{2}\right ) \sqrt {x}}{c x +b}+\frac {\left (3 A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}}}{b^{2}}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.53 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^2} \, dx=\left [\frac {{\left ({\left (B b c - 3 \, A c^{2}\right )} x^{2} + {\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b + 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) - 2 \, {\left (2 \, A b^{2} c - {\left (B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt {x}}{2 \, {\left (b^{3} c^{2} x^{2} + b^{4} c x\right )}}, -\frac {{\left ({\left (B b c - 3 \, A c^{2}\right )} x^{2} + {\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (2 \, A b^{2} c - {\left (B b^{2} c - 3 \, A b c^{2}\right )} x\right )} \sqrt {x}}{b^{3} c^{2} x^{2} + b^{4} c x}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 794 vs. \(2 (73) = 146\).

Time = 12.68 (sec) , antiderivative size = 794, normalized size of antiderivative = 9.34 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^2} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b^{2}} & \text {for}\: c = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{c^{2}} & \text {for}\: b = 0 \\- \frac {3 A b c \sqrt {x} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} + \frac {3 A b c \sqrt {x} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} - \frac {4 A b c \sqrt {- \frac {b}{c}}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} - \frac {3 A c^{2} x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} + \frac {3 A c^{2} x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} - \frac {6 A c^{2} x \sqrt {- \frac {b}{c}}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} + \frac {B b^{2} \sqrt {x} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} - \frac {B b^{2} \sqrt {x} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} + \frac {B b c x^{\frac {3}{2}} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} - \frac {B b c x^{\frac {3}{2}} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} + \frac {2 B b c x \sqrt {- \frac {b}{c}}}{2 b^{3} c \sqrt {x} \sqrt {- \frac {b}{c}} + 2 b^{2} c^{2} x^{\frac {3}{2}} \sqrt {- \frac {b}{c}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^2} \, dx=\frac {{\left (B b - 3 \, A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b^{2}} + \frac {B b x - 3 \, A c x - 2 \, A b}{{\left (c x^{\frac {3}{2}} + b \sqrt {x}\right )} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.93 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {x} (A+B x)}{\left (b x+c x^2\right )^2} \, dx=-\frac {\frac {2\,A}{b}+\frac {x\,\left (3\,A\,c-B\,b\right )}{b^2}}{b\,\sqrt {x}+c\,x^{3/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (3\,A\,c-B\,b\right )}{b^{5/2}\,\sqrt {c}} \]

[In]

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

[Out]

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