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

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*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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Rubi [A]  time = 0.026395, antiderivative size = 49, normalized size of antiderivative = 1., 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 = {781, 80, 63, 205} \[ \frac{2 B \sqrt{x}}{c}-\frac{2 (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} c^{3/2}} \]

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 781

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]

Rule 80

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 63

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 205

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

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 ) \operatorname{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*}

Mathematica [A]  time = 0.0283581, size = 49, normalized size = 1. \[ \frac{2 B \sqrt{x}}{c}-\frac{2 (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} c^{3/2}} \]

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 = 0.008, size = 53, normalized size = 1.1 \begin{align*} 2\,{\frac{B\sqrt{x}}{c}}+2\,{\frac{A}{\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }-2\,{\frac{bB}{c\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.11272, size = 250, normalized size = 5.1 \begin{align*} \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{align*}

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 [A]  time = 3.91469, size = 218, normalized size = 4.45 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{2 A}{\sqrt{x}} + 2 B \sqrt{x}\right ) & \text{for}\: b = 0 \wedge c = 0 \\\frac{- \frac{2 A}{\sqrt{x}} + 2 B \sqrt{x}}{c} & \text{for}\: b = 0 \\\frac{2 A \sqrt{x} + \frac{2 B x^{\frac{3}{2}}}{3}}{b} & \text{for}\: c = 0 \\- \frac{i A \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{\sqrt{b} c \sqrt{\frac{1}{c}}} + \frac{i A \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{\sqrt{b} c \sqrt{\frac{1}{c}}} + \frac{i B \sqrt{b} \log{\left (- i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{c^{2} \sqrt{\frac{1}{c}}} - \frac{i B \sqrt{b} \log{\left (i \sqrt{b} \sqrt{\frac{1}{c}} + \sqrt{x} \right )}}{c^{2} \sqrt{\frac{1}{c}}} + \frac{2 B \sqrt{x}}{c} & \text{otherwise} \end{cases} \end{align*}

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*sqrt(x))/c, Eq(b, 0)),
((2*A*sqrt(x) + 2*B*x**(3/2)/3)/b, Eq(c, 0)), (-I*A*log(-I*sqrt(b)*sqrt(1/c) + sqrt(x))/(sqrt(b)*c*sqrt(1/c))
+ I*A*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(sqrt(b)*c*sqrt(1/c)) + I*B*sqrt(b)*log(-I*sqrt(b)*sqrt(1/c) + sqrt(
x))/(c**2*sqrt(1/c)) - I*B*sqrt(b)*log(I*sqrt(b)*sqrt(1/c) + sqrt(x))/(c**2*sqrt(1/c)) + 2*B*sqrt(x)/c, True))

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Giac [A]  time = 1.11409, size = 53, normalized size = 1.08 \begin{align*} \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{align*}

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)