3.2.71 \(\int \frac {\sqrt {x} (A+B x)}{b x+c x^2} \, dx\) [171]

Optimal. Leaf size=49 \[ \frac {2 B \sqrt {x}}{c}-\frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {795, 81, 65, 211} \begin {gather*} \frac {2 B \sqrt {x}}{c}-\frac {2 (b B-A c) \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 49, normalized size = 1.00 \begin {gather*} \frac {2 B \sqrt {x}}{c}-\frac {2 (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b} c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.53, size = 40, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {2 B \sqrt {x}}{c}+\frac {2 \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c \sqrt {b c}}\)	\(40\)
default	\(\frac {2 B \sqrt {x}}{c}+\frac {2 \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{c \sqrt {b c}}\)	\(40\)
risch	\(\frac {2 B \sqrt {x}}{c}+\frac {2 \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right ) A}{\sqrt {b c}}-\frac {2 \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right ) B b}{c \sqrt {b c}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, B \sqrt {x}}{c} - \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.81, size = 102, normalized size = 2.08 \begin {gather*} \left [\frac {2 \, B b c \sqrt {x} + {\left (B b - A c\right )} \sqrt {-b c} \log \left (\frac {c x - b - 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right )}{b c^{2}}, \frac {2 \, {\left (B b c \sqrt {x} + {\left (B b - A c\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right )\right )}}{b c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (46) = 92\).
time = 1.06, size = 180, normalized size = 3.67 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{b} & \text {for}\: c = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{c} & \text {for}\: b = 0 \\\frac {A \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c \sqrt {- \frac {b}{c}}} - \frac {A \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c \sqrt {- \frac {b}{c}}} - \frac {B b \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c^{2} \sqrt {- \frac {b}{c}}} + \frac {B b \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c^{2} \sqrt {- \frac {b}{c}}} + \frac {2 B \sqrt {x}}{c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.12, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, B \sqrt {x}}{c} - \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 37, normalized size = 0.76 \begin {gather*} \frac {2\,B\,\sqrt {x}}{c}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{\sqrt {b}\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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