3.6.5 \(\int \frac {A+B x}{\sqrt {x} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 \sqrt {x} (A b-a B)}{a b \sqrt {a+b x}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 60, 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 = {78, 63, 217, 206} \begin {gather*} \frac {2 \sqrt {x} (A b-a B)}{a b \sqrt {a+b x}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*Sqrt[x])/(a*b*Sqrt[a + b*x]) + (2*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [A]  time = 0.08, size = 76, normalized size = 1.27 \begin {gather*} \frac {2 a^{3/2} B \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+2 \sqrt {b} \sqrt {x} (A b-a B)}{a b^{3/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 62, normalized size = 1.03 \begin {gather*} -\frac {2 \sqrt {x} (a B-A b)}{a b \sqrt {a+b x}}-\frac {2 B \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 157, normalized size = 2.62 \begin {gather*} \left [\frac {{\left (B a b x + B a^{2}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (B a b - A b^{2}\right )} \sqrt {b x + a} \sqrt {x}}{a b^{3} x + a^{2} b^{2}}, -\frac {2 \, {\left ({\left (B a b x + B a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (B a b - A b^{2}\right )} \sqrt {b x + a} \sqrt {x}\right )}}{a b^{3} x + a^{2} b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(3/2)/x^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 86.16, size = 97, normalized size = 1.62 \begin {gather*} -\frac {B \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt {b} {\left | b \right |}} - \frac {4 \, {\left (B a \sqrt {b} - A b^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(3/2)/x^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 121, normalized size = 2.02 \begin {gather*} \frac {\left (B a b x \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 )+2 \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}}{\sqrt {\left (b x +a \right ) x}\, \sqrt {b x +a}\, a \,b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)^(3/2)/x^(1/2),x)

[Out]

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

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maxima [A]  time = 0.90, size = 81, normalized size = 1.35 \begin {gather*} \frac {2 \, \sqrt {b x^{2} + a x} A}{a b x + a^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} B}{b^{2} x + a b} + \frac {B \log \left (2 \, x + \frac {a}{b} + \frac {2 \, \sqrt {b x^{2} + a x}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)^(3/2)/x^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {A+B\,x}{\sqrt {x}\,{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(1/2)*(a + b*x)^(3/2)),x)

[Out]

int((A + B*x)/(x^(1/2)*(a + b*x)^(3/2)), x)

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sympy [A]  time = 19.50, size = 68, normalized size = 1.13 \begin {gather*} \frac {2 A}{a \sqrt {b} \sqrt {\frac {a}{b x} + 1}} + B \left (\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {2 \sqrt {x}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)**(3/2)/x**(1/2),x)

[Out]

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

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