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

Optimal. Leaf size=63 \[ \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} b^{3/2}}+\frac {\sqrt {x} (A b-a B)}{a b (a+b x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {27, 78, 63, 205} \begin {gather*} \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} b^{3/2}}+\frac {\sqrt {x} (A b-a B)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 63, normalized size = 1.00 \begin {gather*} \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} b^{3/2}}+\frac {\sqrt {x} (A b-a B)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.10, size = 64, normalized size = 1.02 \begin {gather*} \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} b^{3/2}}-\frac {\sqrt {x} (a B-A b)}{a b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 177, normalized size = 2.81 \begin {gather*} \left [-\frac {{\left (B a^{2} + A a b + {\left (B a b + A b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (B a^{2} b - A a b^{2}\right )} \sqrt {x}}{2 \, {\left (a^{2} b^{3} x + a^{3} b^{2}\right )}}, -\frac {{\left (B a^{2} + A a b + {\left (B a b + A b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (B a^{2} b - A a b^{2}\right )} \sqrt {x}}{a^{2} b^{3} x + a^{3} b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 60, normalized size = 0.95 \begin {gather*} \frac {{\left (B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a b} - \frac {B a \sqrt {x} - A b \sqrt {x}}{{\left (b x + a\right )} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 69, normalized size = 1.10 \begin {gather*} \frac {A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a}+\frac {B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}+\frac {\left (A b -B a \right ) \sqrt {x}}{\left (b x +a \right ) a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.43, size = 58, normalized size = 0.92 \begin {gather*} -\frac {{\left (B a - A b\right )} \sqrt {x}}{a b^{2} x + a^{2} b} + \frac {{\left (B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.18, size = 51, normalized size = 0.81 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b+B\,a\right )}{a^{3/2}\,b^{3/2}}+\frac {\sqrt {x}\,\left (A\,b-B\,a\right )}{a\,b\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 8.36, size = 716, normalized size = 11.37 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{a^{2}} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{b^{2}} & \text {for}\: a = 0 \\\frac {2 i A \sqrt {a} b^{2} \sqrt {x} \sqrt {\frac {1}{b}}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} + \frac {A a b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} - \frac {A a b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} + \frac {A b^{2} x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} - \frac {A b^{2} x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} - \frac {2 i B a^{\frac {3}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} + \frac {B a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} - \frac {B a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} + \frac {B a b x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} - \frac {B a b x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {5}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i a^{\frac {3}{2}} b^{3} x \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-2*A/(3*x**(3/2)) - 2*B/sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/a**2,
Eq(b, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/b**2, Eq(a, 0)), (2*I*A*sqrt(a)*b**2*sqrt(x)*sqrt(1/b)/(2*I*a**(
5/2)*b**2*sqrt(1/b) + 2*I*a**(3/2)*b**3*x*sqrt(1/b)) + A*a*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)
*b**2*sqrt(1/b) + 2*I*a**(3/2)*b**3*x*sqrt(1/b)) - A*a*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2
*sqrt(1/b) + 2*I*a**(3/2)*b**3*x*sqrt(1/b)) + A*b**2*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2*
sqrt(1/b) + 2*I*a**(3/2)*b**3*x*sqrt(1/b)) - A*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2*sq
rt(1/b) + 2*I*a**(3/2)*b**3*x*sqrt(1/b)) - 2*I*B*a**(3/2)*b*sqrt(x)*sqrt(1/b)/(2*I*a**(5/2)*b**2*sqrt(1/b) + 2
*I*a**(3/2)*b**3*x*sqrt(1/b)) + B*a**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2*sqrt(1/b) + 2*I*
a**(3/2)*b**3*x*sqrt(1/b)) - B*a**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2*sqrt(1/b) + 2*I*a**(
3/2)*b**3*x*sqrt(1/b)) + B*a*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2*sqrt(1/b) + 2*I*a**(3/
2)*b**3*x*sqrt(1/b)) - B*a*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(5/2)*b**2*sqrt(1/b) + 2*I*a**(3/2)*
b**3*x*sqrt(1/b)), True))

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