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

Optimal. Leaf size=56 \[ \frac{(2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{3/2}}+\frac{B \sqrt{x} \sqrt{a+b x}}{b} \]

[Out]

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

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Rubi [A]  time = 0.0245248, antiderivative size = 56, 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 = {80, 63, 217, 206} \[ \frac{(2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{3/2}}+\frac{B \sqrt{x} \sqrt{a+b x}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0548013, size = 77, normalized size = 1.38 \[ \frac{\sqrt{b} B \sqrt{x} (a+b x)-\sqrt{a} \sqrt{\frac{b x}{a}+1} (a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{3/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 101, normalized size = 1.8 \begin{align*}{\frac{1}{2}\sqrt{x}\sqrt{bx+a} \left ( 2\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) b+2\,B\sqrt{b}\sqrt{x \left ( bx+a \right ) }-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 ) a \right ){\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.62627, size = 293, normalized size = 5.23 \begin{align*} \left [\frac{2 \, \sqrt{b x + a} B b \sqrt{x} -{\left (B a - 2 \, A b\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right )}{2 \, b^{2}}, \frac{\sqrt{b x + a} B b \sqrt{x} +{\left (B a - 2 \, A b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right )}{b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*sqrt(b*x + a)*B*b*sqrt(x) - (B*a - 2*A*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a))/b^
2, (sqrt(b*x + a)*B*b*sqrt(x) + (B*a - 2*A*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))))/b^2]

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Sympy [A]  time = 8.3857, size = 73, normalized size = 1.3 \begin{align*} \frac{2 A \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{\sqrt{b}} + \frac{B \sqrt{a} \sqrt{x} \sqrt{1 + \frac{b x}{a}}}{b} - \frac{B a \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*asinh(sqrt(b)*sqrt(x)/sqrt(a))/sqrt(b) + B*sqrt(a)*sqrt(x)*sqrt(1 + b*x/a)/b - B*a*asinh(sqrt(b)*sqrt(x)/s
qrt(a))/b**(3/2)

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

[Out]

Timed out