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

Optimal. Leaf size=93 \[ \frac{\sqrt{x} \sqrt{a+b x} (4 A b-3 a B)}{4 b^2}-\frac{a (4 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{5/2}}+\frac{B x^{3/2} \sqrt{a+b x}}{2 b} \]

[Out]

((4*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(4*b^2) + (B*x^(3/2)*Sqrt[a + b*x])/(2*b) - (a*(4*A*b - 3*a*B)*ArcTanh
[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0388393, antiderivative size = 93, 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 = {80, 50, 63, 217, 206} \[ \frac{\sqrt{x} \sqrt{a+b x} (4 A b-3 a B)}{4 b^2}-\frac{a (4 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{5/2}}+\frac{B x^{3/2} \sqrt{a+b x}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(4*b^2) + (B*x^(3/2)*Sqrt[a + b*x])/(2*b) - (a*(4*A*b - 3*a*B)*ArcTanh
[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(5/2))

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 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{x} (A+B x)}{\sqrt{a+b x}} \, dx &=\frac{B x^{3/2} \sqrt{a+b x}}{2 b}+\frac{\left (2 A b-\frac{3 a B}{2}\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{2 b}\\ &=\frac{(4 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{4 b^2}+\frac{B x^{3/2} \sqrt{a+b x}}{2 b}-\frac{(a (4 A b-3 a B)) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{8 b^2}\\ &=\frac{(4 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{4 b^2}+\frac{B x^{3/2} \sqrt{a+b x}}{2 b}-\frac{(a (4 A b-3 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{4 b^2}\\ &=\frac{(4 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{4 b^2}+\frac{B x^{3/2} \sqrt{a+b x}}{2 b}-\frac{(a (4 A b-3 a B)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^2}\\ &=\frac{(4 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{4 b^2}+\frac{B x^{3/2} \sqrt{a+b x}}{2 b}-\frac{a (4 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0658834, size = 93, normalized size = 1. \[ \frac{a^{3/2} \sqrt{\frac{b x}{a}+1} (3 a B-4 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )+\sqrt{b} \sqrt{x} (a+b x) (-3 a B+4 A b+2 b B x)}{4 b^{5/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*(a + b*x)*(4*A*b - 3*a*B + 2*b*B*x) + a^(3/2)*(-4*A*b + 3*a*B)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqr
t[b]*Sqrt[x])/Sqrt[a]])/(4*b^(5/2)*Sqrt[a + b*x])

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 136, normalized size = 1.5 \begin{align*} -{\frac{1}{8}\sqrt{x}\sqrt{bx+a} \left ( -4\,Bx{b}^{3/2}\sqrt{x \left ( bx+a \right ) }+4\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) ab-8\,A{b}^{3/2}\sqrt{x \left ( bx+a \right ) }-3\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{2}+6\,Ba\sqrt{b}\sqrt{x \left ( bx+a \right ) } \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \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]

-1/8*x^(1/2)*(b*x+a)^(1/2)/b^(5/2)*(-4*B*x*b^(3/2)*(x*(b*x+a))^(1/2)+4*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2
*b*x+a)/b^(1/2))*a*b-8*A*b^(3/2)*(x*(b*x+a))^(1/2)-3*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a
^2+6*B*a*b^(1/2)*(x*(b*x+a))^(1/2))/(x*(b*x+a))^(1/2)

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [A]  time = 2.72875, size = 393, normalized size = 4.23 \begin{align*} \left [-\frac{{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (2 \, B b^{2} x - 3 \, B a b + 4 \, A b^{2}\right )} \sqrt{b x + a} \sqrt{x}}{8 \, b^{3}}, -\frac{{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, B b^{2} x - 3 \, B a b + 4 \, A b^{2}\right )} \sqrt{b x + a} \sqrt{x}}{4 \, b^{3}}\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]

[-1/8*((3*B*a^2 - 4*A*a*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(2*B*b^2*x - 3*B*a*b +
4*A*b^2)*sqrt(b*x + a)*sqrt(x))/b^3, -1/4*((3*B*a^2 - 4*A*a*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) - (2*B*b^2*x - 3*B*a*b + 4*A*b^2)*sqrt(b*x + a)*sqrt(x))/b^3]

________________________________________________________________________________________

Sympy [A]  time = 10.1651, size = 156, normalized size = 1.68 \begin{align*} \frac{A \sqrt{a} \sqrt{x} \sqrt{1 + \frac{b x}{a}}}{b} - \frac{A a \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{b^{\frac{3}{2}}} - \frac{3 B a^{\frac{3}{2}} \sqrt{x}}{4 b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{B \sqrt{a} x^{\frac{3}{2}}}{4 b \sqrt{1 + \frac{b x}{a}}} + \frac{3 B a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4 b^{\frac{5}{2}}} + \frac{B x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \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]

A*sqrt(a)*sqrt(x)*sqrt(1 + b*x/a)/b - A*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(3/2) - 3*B*a**(3/2)*sqrt(x)/(4*b*
*2*sqrt(1 + b*x/a)) - B*sqrt(a)*x**(3/2)/(4*b*sqrt(1 + b*x/a)) + 3*B*a**2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(4*b*
*(5/2)) + B*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a))

________________________________________________________________________________________

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