3.2.72 \(\int \frac {A+B x}{\sqrt {x} (b x+c x^2)} \, dx\) [172]

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

[Out]

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

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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, 79, 65, 211} \begin {gather*} \frac {2 (b B-A c) \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2} \sqrt {c}}-\frac {2 A}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b} - \frac {2 \, A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.82, size = 112, normalized size = 2.29 \begin {gather*} \left [-\frac {2 \, A b c \sqrt {x} - {\left (B b - A c\right )} \sqrt {-b c} x \log \left (\frac {c x - b + 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right )}{b^{2} c x}, -\frac {2 \, {\left (A b c \sqrt {x} + {\left (B b - A c\right )} \sqrt {b c} x \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right )\right )}}{b^{2} c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (46) = 92\).
time = 1.16, size = 178, normalized size = 3.63 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b} & \text {for}\: c = 0 \\- \frac {A \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{b \sqrt {- \frac {b}{c}}} + \frac {A \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{b \sqrt {- \frac {b}{c}}} - \frac {2 A}{b \sqrt {x}} + \frac {B \log {\left (\sqrt {x} - \sqrt {- \frac {b}{c}} \right )}}{c \sqrt {- \frac {b}{c}}} - \frac {B \log {\left (\sqrt {x} + \sqrt {- \frac {b}{c}} \right )}}{c \sqrt {- \frac {b}{c}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.14, size = 39, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} b} - \frac {2 \, A}{b \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.07, size = 50, normalized size = 1.02 \begin {gather*} \frac {2\,B\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )}{\sqrt {b}\,\sqrt {c}}-\frac {2\,A\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2\,A}{b\,\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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