3.519 \(\int \frac {A+B x}{x^{3/2} \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=50 \[ \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}-\frac {2 A \sqrt {a+b x}}{a \sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {78, 63, 217, 206} \[ \frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{\sqrt {b}}-\frac {2 A \sqrt {a+b x}}{a \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[a + b*x])/(a*Sqrt[x]) + (2*B*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/Sqrt[b]

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 78

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] || LtQ
[p, n]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 69, normalized size = 1.38 \[ \frac {2 \left (\frac {a^{3/2} B \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {A (a+b x)}{\sqrt {x}}\right )}{a \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-((A*(a + b*x))/Sqrt[x]) + (a^(3/2)*B*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[b]))/(a*S
qrt[a + b*x])

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fricas [A]  time = 0.69, size = 109, normalized size = 2.18 \[ \left [\frac {B a \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, \sqrt {b x + a} A b \sqrt {x}}{a b x}, -\frac {2 \, {\left (B a \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + \sqrt {b x + a} A b \sqrt {x}\right )}}{a b x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 73, normalized size = 1.46 \[ \frac {\left (B a x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-2 \sqrt {\left (b x +a \right ) x}\, A \sqrt {b}\right ) \sqrt {b x +a}}{\sqrt {\left (b x +a \right ) x}\, a \sqrt {b}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(3/2)/(b*x+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.86, size = 49, normalized size = 0.98 \[ \frac {B \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

B*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) - 2*sqrt(b*x^2 + a*x)*A/(a*x)

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mupad [B]  time = 1.07, size = 48, normalized size = 0.96 \[ -\frac {4\,B\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {-b}\,\sqrt {x}}\right )}{\sqrt {-b}}-\frac {2\,A\,\sqrt {a+b\,x}}{a\,\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(3/2)*(a + b*x)^(1/2)),x)

[Out]

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

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sympy [A]  time = 15.06, size = 44, normalized size = 0.88 \[ - \frac {2 A \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{a} + \frac {2 B \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(3/2)/(b*x+a)**(1/2),x)

[Out]

-2*A*sqrt(b)*sqrt(a/(b*x) + 1)/a + 2*B*asinh(sqrt(b)*sqrt(x)/sqrt(a))/sqrt(b)

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