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

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

[Out]

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

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Rubi [A]  time = 0.039612, 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} \[ \frac{a (4 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{3/2}}+\frac{\sqrt{x} \sqrt{a+b x} (4 A b-a B)}{4 b}+\frac{B \sqrt{x} (a+b x)^{3/2}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.159979, size = 87, normalized size = 0.94 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} (B (a+2 b x)+4 A b)-\frac{\sqrt{a} (a B-4 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 136, normalized size = 1.5 \begin{align*}{\frac{1}{8}\sqrt{bx+a}\sqrt{x} \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 ) }-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}^{2}+2\,Ba\sqrt{b}\sqrt{x \left ( bx+a \right ) } \right ){b}^{-{\frac{3}{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)*(b*x+a)^(1/2)/x^(1/2),x)

[Out]

1/8*(b*x+a)^(1/2)*x^(1/2)/b^(3/2)*(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)-B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^2+2
*B*a*b^(1/2)*(x*(b*x+a))^(1/2))/(x*(b*x+a))^(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^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.70597, size = 381, normalized size = 4.1 \begin{align*} \left [-\frac{{\left (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 + B a b + 4 \, A b^{2}\right )} \sqrt{b x + a} \sqrt{x}}{8 \, b^{2}}, \frac{{\left (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 + B a b + 4 \, A b^{2}\right )} \sqrt{b x + a} \sqrt{x}}{4 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 10.7039, size = 573, normalized size = 6.16 \begin{align*} \frac{2 A \left (\begin{cases} \frac{\sqrt{a} \sqrt{b} \sqrt{\frac{b x}{a}} \sqrt{a + b x}}{2} + \frac{a \sqrt{b} \operatorname{acosh}{\left (\frac{\sqrt{a + b x}}{\sqrt{a}} \right )}}{2} & \text{for}\: \frac{\left |{a + b x}\right |}{\left |{a}\right |} > 1 \\\frac{i \sqrt{a} \sqrt{b} \sqrt{a + b x}}{2 \sqrt{- \frac{b x}{a}}} - \frac{i a \sqrt{b} \operatorname{asin}{\left (\frac{\sqrt{a + b x}}{\sqrt{a}} \right )}}{2} - \frac{i \sqrt{b} \left (a + b x\right )^{\frac{3}{2}}}{2 \sqrt{a} \sqrt{- \frac{b x}{a}}} & \text{otherwise} \end{cases}\right )}{b} - \frac{2 B a \left (\begin{cases} \frac{\sqrt{a} \sqrt{b} \sqrt{\frac{b x}{a}} \sqrt{a + b x}}{2} + \frac{a \sqrt{b} \operatorname{acosh}{\left (\frac{\sqrt{a + b x}}{\sqrt{a}} \right )}}{2} & \text{for}\: \frac{\left |{a + b x}\right |}{\left |{a}\right |} > 1 \\\frac{i \sqrt{a} \sqrt{b} \sqrt{a + b x}}{2 \sqrt{- \frac{b x}{a}}} - \frac{i a \sqrt{b} \operatorname{asin}{\left (\frac{\sqrt{a + b x}}{\sqrt{a}} \right )}}{2} - \frac{i \sqrt{b} \left (a + b x\right )^{\frac{3}{2}}}{2 \sqrt{a} \sqrt{- \frac{b x}{a}}} & \text{otherwise} \end{cases}\right )}{b^{2}} + \frac{2 B \left (\begin{cases} - \frac{3 a^{\frac{3}{2}} \sqrt{b} \sqrt{a + b x}}{8 \sqrt{\frac{b x}{a}}} + \frac{\sqrt{a} \sqrt{b} \left (a + b x\right )^{\frac{3}{2}}}{8 \sqrt{\frac{b x}{a}}} + \frac{3 a^{2} \sqrt{b} \operatorname{acosh}{\left (\frac{\sqrt{a + b x}}{\sqrt{a}} \right )}}{8} + \frac{\sqrt{b} \left (a + b x\right )^{\frac{5}{2}}}{4 \sqrt{a} \sqrt{\frac{b x}{a}}} & \text{for}\: \frac{\left |{a + b x}\right |}{\left |{a}\right |} > 1 \\\frac{3 i a^{\frac{3}{2}} \sqrt{b} \sqrt{a + b x}}{8 \sqrt{- \frac{b x}{a}}} - \frac{i \sqrt{a} \sqrt{b} \left (a + b x\right )^{\frac{3}{2}}}{8 \sqrt{- \frac{b x}{a}}} - \frac{3 i a^{2} \sqrt{b} \operatorname{asin}{\left (\frac{\sqrt{a + b x}}{\sqrt{a}} \right )}}{8} - \frac{i \sqrt{b} \left (a + b x\right )^{\frac{5}{2}}}{4 \sqrt{a} \sqrt{- \frac{b x}{a}}} & \text{otherwise} \end{cases}\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*Piecewise((sqrt(a)*sqrt(b)*sqrt(b*x/a)*sqrt(a + b*x)/2 + a*sqrt(b)*acosh(sqrt(a + b*x)/sqrt(a))/2, Abs(a +
b*x)/Abs(a) > 1), (I*sqrt(a)*sqrt(b)*sqrt(a + b*x)/(2*sqrt(-b*x/a)) - I*a*sqrt(b)*asin(sqrt(a + b*x)/sqrt(a))
/2 - I*sqrt(b)*(a + b*x)**(3/2)/(2*sqrt(a)*sqrt(-b*x/a)), True))/b - 2*B*a*Piecewise((sqrt(a)*sqrt(b)*sqrt(b*x
/a)*sqrt(a + b*x)/2 + a*sqrt(b)*acosh(sqrt(a + b*x)/sqrt(a))/2, Abs(a + b*x)/Abs(a) > 1), (I*sqrt(a)*sqrt(b)*s
qrt(a + b*x)/(2*sqrt(-b*x/a)) - I*a*sqrt(b)*asin(sqrt(a + b*x)/sqrt(a))/2 - I*sqrt(b)*(a + b*x)**(3/2)/(2*sqrt
(a)*sqrt(-b*x/a)), True))/b**2 + 2*B*Piecewise((-3*a**(3/2)*sqrt(b)*sqrt(a + b*x)/(8*sqrt(b*x/a)) + sqrt(a)*sq
rt(b)*(a + b*x)**(3/2)/(8*sqrt(b*x/a)) + 3*a**2*sqrt(b)*acosh(sqrt(a + b*x)/sqrt(a))/8 + sqrt(b)*(a + b*x)**(5
/2)/(4*sqrt(a)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (3*I*a**(3/2)*sqrt(b)*sqrt(a + b*x)/(8*sqrt(-b*x/a)) -
I*sqrt(a)*sqrt(b)*(a + b*x)**(3/2)/(8*sqrt(-b*x/a)) - 3*I*a**2*sqrt(b)*asin(sqrt(a + b*x)/sqrt(a))/8 - I*sqrt(
b)*(a + b*x)**(5/2)/(4*sqrt(a)*sqrt(-b*x/a)), True))/b**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)*(b*x+a)^(1/2)/x^(1/2),x, algorithm="giac")

[Out]

Timed out