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

Optimal. Leaf size=49 \[ -\frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}-\frac {2 A}{a \sqrt {x}} \]

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Rubi [A]  time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {78, 63, 205} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*(a + b*x)),x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 49, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 49, normalized size = 1.00 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 112, normalized size = 2.29 \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),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 = 1.26, size = 39, normalized size = 0.80 \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 {2 \, A}{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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 53, normalized size = 1.08 \begin {gather*} -\frac {2 A b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a}+\frac {2 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}-\frac {2 A}{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),x)

[Out]

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

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maxima [A]  time = 2.00, size = 39, normalized size = 0.80 \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 {2 \, A}{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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.08, size = 50, normalized size = 1.02 \begin {gather*} \frac {2\,B\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}}-\frac {2\,A\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2\,A}{a\,\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(3/2)*(a + b*x)),x)

[Out]

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

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sympy [A]  time = 3.29, size = 216, normalized size = 4.41 \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 {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{a \sqrt {x}} + \frac {i A \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i A \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i B \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{\sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i B \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{\sqrt {a} b \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)/x**(3/2)/(b*x+a),x)

[Out]

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

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