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

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

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Rubi [A]  time = 0.04, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {80, 50, 63, 217, 206} \begin {gather*} \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} \end {gather*}

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 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 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 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])

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]

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*}

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Mathematica [A]  time = 0.22, size = 87, normalized size = 0.94 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.12, size = 87, normalized size = 0.94 \begin {gather*} \frac {\left (a^2 B-4 a A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{4 b^{3/2}}+\frac {\sqrt {a+b x} \left (a B \sqrt {x}+4 A b \sqrt {x}+2 b B x^{3/2}\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.81, size = 146, normalized size = 1.57 \begin {gather*} \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 {gather*}

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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maple [A]  time = 0.01, size = 136, normalized size = 1.46 \begin {gather*} \frac {\sqrt {b x +a}\, \left (4 A a b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-B \,a^{2} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+4 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {3}{2}} x +8 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {3}{2}}+2 \sqrt {\left (b x +a \right ) x}\, B a \sqrt {b}\right ) \sqrt {x}}{8 \sqrt {\left (b x +a \right ) x}\, b^{\frac {3}{2}}} \end {gather*}

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

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maxima [A]  time = 0.86, size = 109, normalized size = 1.17 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a x} B x - \frac {B a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} + \frac {A a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, \sqrt {b}} + \sqrt {b x^{2} + a x} A + \frac {\sqrt {b x^{2} + a x} B a}{4 \, b} \end {gather*}

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]

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

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mupad [B]  time = 1.32, size = 96, normalized size = 1.03 \begin {gather*} A\,\sqrt {x}\,\sqrt {a+b\,x}+\frac {2\,A\,a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )}{\sqrt {b}}+B\,\sqrt {x}\,\left (\frac {x}{2}+\frac {a}{4\,b}\right )\,\sqrt {a+b\,x}-\frac {B\,a^2\,\ln \left (a+2\,b\,x+2\,\sqrt {b}\,\sqrt {x}\,\sqrt {a+b\,x}\right )}{8\,b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*x^(1/2)*(a + b*x)^(1/2) + (2*A*a*atanh((b^(1/2)*x^(1/2))/((a + b*x)^(1/2) - a^(1/2))))/b^(1/2) + B*x^(1/2)*(
x/2 + a/(4*b))*(a + b*x)^(1/2) - (B*a^2*log(a + 2*b*x + 2*b^(1/2)*x^(1/2)*(a + b*x)^(1/2)))/(8*b^(3/2))

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sympy [C]  time = 9.45, size = 568, normalized size = 6.11 \begin {gather*} \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}\: \left |{1 + \frac {b x}{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}\: \left |{1 + \frac {b x}{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}\: \left |{1 + \frac {b x}{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 {gather*}

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(1 +
b*x/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)*s
qrt(a + b*x)/2 + a*sqrt(b)*acosh(sqrt(a + b*x)/sqrt(a))/2, Abs(1 + b*x/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 + 2*B*Piecewise((-3*a**(3/2)*sqrt(b)*sqrt(a + b*x)/(8*sqrt(b*x/a)) + sqrt(a)*sqrt(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*sqr
t(a)*sqrt(b*x/a)), Abs(1 + b*x/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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