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

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Rubi [A]  time = 0.0186293, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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 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 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]

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])

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*}

Mathematica [A]  time = 0.141566, 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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Maple [A]  time = 0.012, size = 73, normalized size = 1.5 \begin{align*}{\frac{1}{a} \left ( B\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a \right ){\frac{1}{\sqrt{b}}}} \right ) xa-2\,A\sqrt{b}\sqrt{x \left ( bx+a \right ) } \right ) \sqrt{bx+a}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{\frac{1}{\sqrt{b}}}} \end{align*}

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.75624, size = 281, normalized size = 5.62 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 12.4442, size = 44, normalized size = 0.88 \begin{align*} - \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}} \end{align*}

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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