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

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {80, 63, 217, 206} \begin {gather*} \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} \end {gather*}

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 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 {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 = 0.08, size = 77, normalized size = 1.38 \begin {gather*} \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}} \end {gather*}

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])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.08, size = 57, normalized size = 1.02 \begin {gather*} \frac {(a B-2 A b) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{3/2}}+\frac {B \sqrt {x} \sqrt {a+b x}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.55, size = 108, normalized size = 1.93 \begin {gather*} \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 {gather*}

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]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.01, size = 101, normalized size = 1.80 \begin {gather*} \frac {\sqrt {b x +a}\, \left (2 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 \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+2 \sqrt {\left (b x +a \right ) x}\, B \sqrt {b}\right ) \sqrt {x}}{2 \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)/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*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*b+2*B*b^(1/2)*((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)/((b*x+a)*x)^(1/2)/b^(3/2)

________________________________________________________________________________________

maxima [A]  time = 0.91, size = 75, normalized size = 1.34 \begin {gather*} -\frac {B a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} + \frac {\sqrt {b x^{2} + a x} B}{b} \end {gather*}

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]

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

________________________________________________________________________________________

mupad [B]  time = 1.35, size = 77, normalized size = 1.38 \begin {gather*} \frac {B\,\sqrt {x}\,\sqrt {a+b\,x}}{b}-\frac {2\,B\,a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )}{b^{3/2}}-\frac {4\,A\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {-b}\,\sqrt {x}}\right )}{\sqrt {-b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 8.29, size = 73, normalized size = 1.30 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________