3.348 \(\int \frac {A+B x}{\sqrt {x} (a+b x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {80, 63, 205} \[ \frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}+\frac {2 B \sqrt {x}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(Sqrt[x]*(a + b*x)),x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 49, normalized size = 1.00 \[ \frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}+\frac {2 B \sqrt {x}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(Sqrt[x]*(a + b*x)),x]

[Out]

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

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fricas [A]  time = 0.65, size = 102, normalized size = 2.08 \[ \left [\frac {2 \, B a b \sqrt {x} + {\left (B a - A b\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{a b^{2}}, \frac {2 \, {\left (B a b \sqrt {x} + {\left (B a - A b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )\right )}}{a b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.26, size = 39, normalized size = 0.80 \[ \frac {2 \, B \sqrt {x}}{b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 53, normalized size = 1.08 \[ \frac {2 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}-\frac {2 B a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}+\frac {2 B \sqrt {x}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.97, size = 39, normalized size = 0.80 \[ \frac {2 \, B \sqrt {x}}{b} - \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.36, size = 37, normalized size = 0.76 \[ \frac {2\,B\,\sqrt {x}}{b}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{\sqrt {a}\,b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(1/2)*(a + b*x)),x)

[Out]

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

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sympy [A]  time = 2.11, size = 218, normalized size = 4.45 \[ \begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{a} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b} & \text {for}\: a = 0 \\- \frac {i A \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{\sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i A \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{\sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i B \sqrt {a} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{2} \sqrt {\frac {1}{b}}} - \frac {i B \sqrt {a} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{2} \sqrt {\frac {1}{b}}} + \frac {2 B \sqrt {x}}{b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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