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

Optimal. Leaf size=99 \[ -\frac {2 A (a+b x)}{a \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {770, 78, 63, 205} \begin {gather*} -\frac {2 (a+b x) (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A (a+b x)}{a \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*(a + b*x))/(a*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*(A*b - a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])
/Sqrt[a]])/(a^(3/2)*Sqrt[b]*Sqrt[a^2 + 2*a*b*x + b^2*x^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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {A+B x}{x^{3/2} \left (a b+b^2 x\right )} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 A (a+b x)}{a \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (-\frac {A b^2}{2}+\frac {a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )} \, dx}{a b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 A (a+b x)}{a \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 \left (-\frac {A b^2}{2}+\frac {a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b+b^2 x^2} \, dx,x,\sqrt {x}\right )}{a b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 A (a+b x)}{a \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 79, normalized size = 0.80 \begin {gather*} \frac {2 (a+b x) \left (-\sqrt {x} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-\sqrt {a} A \sqrt {b}\right )}{a^{3/2} \sqrt {b} \sqrt {x} \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 5.92, size = 66, normalized size = 0.67 \begin {gather*} \frac {(a+b x) \left (\frac {2 (a B-A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}-\frac {2 A}{a \sqrt {x}}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 112, normalized size = 1.13 \begin {gather*} \left [-\frac {2 \, A a b \sqrt {x} - {\left (B a - A b\right )} \sqrt {-a b} x \log \left (\frac {b x - a + 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{a^{2} b x}, -\frac {2 \, {\left (A a b \sqrt {x} + {\left (B a - A b\right )} \sqrt {a b} x \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )\right )}}{a^{2} b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 57, normalized size = 0.58 \begin {gather*} \frac {2 \, {\left (B a \mathrm {sgn}\left (b x + a\right ) - A b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {2 \, A \mathrm {sgn}\left (b x + a\right )}{a \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*a*sgn(b*x + a) - A*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a) - 2*A*sgn(b*x + a)/(a*sqrt(x
))

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maple [A]  time = 0.06, size = 71, normalized size = 0.72 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (A b \sqrt {x}\, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-B a \sqrt {x}\, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+\sqrt {a b}\, A \right )}{\sqrt {\left (b x +a \right )^{2}}\, \sqrt {a b}\, a \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.61, size = 180, normalized size = 1.82 \begin {gather*} \frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} - \frac {{\left ({\left (B a b^{2} + A b^{3}\right )} x^{2} - 3 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {x} - \frac {2 \, {\left ({\left (B a^{2} b + A a b^{2}\right )} x^{2} + 3 \, {\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )}}{\sqrt {x}} - \frac {6 \, {\left (A a^{2} b x^{2} - A a^{3} x\right )}}{x^{\frac {3}{2}}}}{3 \, {\left (a^{3} b x + a^{4}\right )}} + \frac {{\left (B a b + A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{2} - A a b\right )} \sqrt {x}}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a) - 1/3*(((B*a*b^2 + A*b^3)*x^2 - 3*(B*a^2*b - 3*A*a*b^2
)*x)*sqrt(x) - 2*((B*a^2*b + A*a*b^2)*x^2 + 3*(B*a^3 - 3*A*a^2*b)*x)/sqrt(x) - 6*(A*a^2*b*x^2 - A*a^3*x)/x^(3/
2))/(a^3*b*x + a^4) + 1/3*((B*a*b + A*b^2)*x^(3/2) - 6*(B*a^2 - A*a*b)*sqrt(x))/a^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x^{3/2}\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{\frac {3}{2}} \sqrt {\left (a + b x\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/(x**(3/2)*sqrt((a + b*x)**2)), x)

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