\(\int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx\) [170]

Optimal result

Integrand size = 22, antiderivative size = 69 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=-\frac {2 (b B-A c) \sqrt {x}}{c^2}+\frac {2 B x^{3/2}}{3 c}+\frac {2 \sqrt {b} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, 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, 81, 52, 65, 211} \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \sqrt {b} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}}-\frac {2 \sqrt {x} (b B-A c)}{c^2}+\frac {2 B x^{3/2}}{3 c} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \sqrt {x} (-3 b B+3 A c+B c x)}{3 c^2}+\frac {2 \sqrt {b} (b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {2 \left (B c x +3 A c -3 B b \right ) \sqrt {x}}{3 c^{2}}-\frac {2 b \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{2} \sqrt {b c}}\)	\(53\)
derivativedivides	\(\frac {\frac {2 B c \,x^{\frac {3}{2}}}{3}+2 A c \sqrt {x}-2 \sqrt {x}\, B b}{c^{2}}-\frac {2 b \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{2} \sqrt {b c}}\)	\(58\)
default	\(\frac {\frac {2 B c \,x^{\frac {3}{2}}}{3}+2 A c \sqrt {x}-2 \sqrt {x}\, B b}{c^{2}}-\frac {2 b \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c^{2} \sqrt {b c}}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.87 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\left [-\frac {3 \, {\left (B b - A c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x - 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt {x}}{3 \, c^{2}}, \frac {2 \, {\left (3 \, {\left (B b - A c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (B c x - 3 \, B b + 3 \, A c\right )} \sqrt {x}\right )}}{3 \, c^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (66) = 132\).

Time = 2.33 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.20 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: c = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{c} & \text {for}\: b = 0 \\- \frac {A b \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{2} \sqrt {- \frac {b}{c}}} + \frac {A b \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{2} \sqrt {- \frac {b}{c}}} + \frac {2 A \sqrt {x}}{c} + \frac {B b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{3} \sqrt {- \frac {b}{c}}} - \frac {B b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{3} \sqrt {- \frac {b}{c}}} - \frac {2 B b \sqrt {x}}{c^{2}} + \frac {2 B x^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \, {\left (B b^{2} - A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {2 \, {\left (B c x^{\frac {3}{2}} - 3 \, {\left (B b - A c\right )} \sqrt {x}\right )}}{3 \, c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\frac {2 \, {\left (B b^{2} - A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{2}} + \frac {2 \, {\left (B c^{2} x^{\frac {3}{2}} - 3 \, B b c \sqrt {x} + 3 \, A c^{2} \sqrt {x}\right )}}{3 \, c^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {x^{3/2} (A+B x)}{b x+c x^2} \, dx=\sqrt {x}\,\left (\frac {2\,A}{c}-\frac {2\,B\,b}{c^2}\right )+\frac {2\,B\,x^{3/2}}{3\,c}+\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,\sqrt {x}\,\left (A\,c-B\,b\right )}{B\,b^2-A\,b\,c}\right )\,\left (A\,c-B\,b\right )}{c^{5/2}} \]

[In]

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

[Out]

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