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

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

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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 {\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} \end {gather*}

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 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 {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*}

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Mathematica [A]  time = 0.12, size = 93, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

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])

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IntegrateAlgebraic [A]  time = 0.15, size = 89, normalized size = 0.96 \begin {gather*} \frac {\left (4 a A b-3 a^2 B\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{4 b^{5/2}}+\frac {\sqrt {a+b x} \left (-3 a B \sqrt {x}+4 A b \sqrt {x}+2 b B x^{3/2}\right )}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 151, normalized size = 1.62 \begin {gather*} \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 {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]

[-1/8*((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]

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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)*x^(1/2)/(b*x+a)^(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 )-3 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}}+6 \sqrt {\left (b x +a \right ) x}\, B a \sqrt {b}\right ) \sqrt {x}}{8 \sqrt {\left (b x +a \right ) x}\, b^{\frac {5}{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]

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

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maxima [A]  time = 0.92, size = 121, normalized size = 1.30 \begin {gather*} \frac {\sqrt {b x^{2} + a x} B x}{2 \, b} + \frac {3 \, B a^{2} \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {A a \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{2 \, b^{\frac {3}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} B a}{4 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} A}{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*sqrt(b*x^2 + a*x)*B*x/b + 3/8*B*a^2*log(2*x + a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(5/2) - 1/2*A*a*log(2*x
+ a/b + 2*sqrt(b*x^2 + a*x)/sqrt(b))/b^(3/2) - 3/4*sqrt(b*x^2 + a*x)*B*a/b^2 + sqrt(b*x^2 + a*x)*A/b

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mupad [B]  time = 5.06, size = 267, normalized size = 2.87 \begin {gather*} \frac {\frac {x^{7/2}\,\left (2\,A\,a\,b^2-\frac {3\,B\,a^2\,b}{2}\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}+\frac {x^{5/2}\,\left (\frac {11\,B\,a^2}{2}-2\,A\,a\,b\right )}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}-\frac {\sqrt {x}\,\left (3\,B\,a^2-4\,A\,a\,b\right )}{2\,b^2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}+\frac {x^{3/2}\,\left (11\,B\,a^2-4\,A\,a\,b\right )}{2\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{\frac {6\,b^2\,x^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}-\frac {4\,b^3\,x^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}+\frac {b^4\,x^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}-\frac {4\,b\,x}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+1}-\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a+b\,x}-\sqrt {a}}\right )\,\left (4\,A\,b-3\,B\,a\right )}{2\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^(7/2)*(2*A*a*b^2 - (3*B*a^2*b)/2))/((a + b*x)^(1/2) - a^(1/2))^7 + (x^(5/2)*((11*B*a^2)/2 - 2*A*a*b))/((a
+ b*x)^(1/2) - a^(1/2))^5 - (x^(1/2)*(3*B*a^2 - 4*A*a*b))/(2*b^2*((a + b*x)^(1/2) - a^(1/2))) + (x^(3/2)*(11*B
*a^2 - 4*A*a*b))/(2*b*((a + b*x)^(1/2) - a^(1/2))^3))/((6*b^2*x^2)/((a + b*x)^(1/2) - a^(1/2))^4 - (4*b^3*x^3)
/((a + b*x)^(1/2) - a^(1/2))^6 + (b^4*x^4)/((a + b*x)^(1/2) - a^(1/2))^8 - (4*b*x)/((a + b*x)^(1/2) - a^(1/2))
^2 + 1) - (a*atanh((b^(1/2)*x^(1/2))/((a + b*x)^(1/2) - a^(1/2)))*(4*A*b - 3*B*a))/(2*b^(5/2))

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sympy [A]  time = 7.82, size = 156, normalized size = 1.68 \begin {gather*} \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 {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]

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))

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