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

Optimal. Leaf size=82 \[ \frac{\sqrt{x} \sqrt{a+b x} (a B+2 A b)}{a}+\frac{(a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{\sqrt{b}}-\frac{2 A (a+b x)^{3/2}}{a \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0402522, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 50, 63, 217, 206} \[ \frac{\sqrt{x} \sqrt{a+b x} (a B+2 A b)}{a}+\frac{(a B+2 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{\sqrt{b}}-\frac{2 A (a+b x)^{3/2}}{a \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b + a*B)*Sqrt[x]*Sqrt[a + b*x])/a - (2*A*(a + b*x)^(3/2))/(a*Sqrt[x]) + ((2*A*b + a*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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A]  time = 0.149647, size = 72, normalized size = 0.88 \[ \sqrt{a+b x} \left (\frac{(a B+2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \sqrt{\frac{b x}{a}+1}}+\frac{B x-2 A}{\sqrt{x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 118, normalized size = 1.4 \begin{align*}{\frac{1}{2}\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 ) xb+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\,Bx\sqrt{x \left ( bx+a \right ) }\sqrt{b}-4\,A\sqrt{b}\sqrt{x \left ( bx+a \right ) } \right ){\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)*(b*x+a)^(1/2)/x^(3/2),x)

[Out]

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))*x*b+B*ln(1/2*(2*(x*(b*x+a))^(1/2)
*b^(1/2)+2*b*x+a)/b^(1/2))*x*a+2*B*x*(x*(b*x+a))^(1/2)*b^(1/2)-4*A*b^(1/2)*(x*(b*x+a))^(1/2))/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)*(b*x+a)^(1/2)/x^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.65591, size = 338, normalized size = 4.12 \begin{align*} \left [\frac{{\left (B a + 2 \, A b\right )} \sqrt{b} x \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (B b x - 2 \, A b\right )} \sqrt{b x + a} \sqrt{x}}{2 \, b x}, -\frac{{\left (B a + 2 \, A b\right )} \sqrt{-b} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (B b x - 2 \, A b\right )} \sqrt{b x + a} \sqrt{x}}{b x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.86491, size = 116, normalized size = 1.41 \begin{align*} A \left (- \frac{2 \sqrt{a}}{\sqrt{x} \sqrt{1 + \frac{b x}{a}}} + 2 \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )} - \frac{2 b \sqrt{x}}{\sqrt{a} \sqrt{1 + \frac{b x}{a}}}\right ) + B \left (\sqrt{a} \sqrt{x} \sqrt{1 + \frac{b x}{a}} + \frac{a \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{\sqrt{b}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out