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

Optimal. Leaf size=73 \[ \frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {3 A b-2 a B}{a^2 \sqrt {a+b x}}-\frac {A}{a x \sqrt {a+b x}} \]

[Out]

(3*A*b-2*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(5/2)+(-3*A*b+2*B*a)/a^2/(b*x+a)^(1/2)-A/a/x/(b*x+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 51, 63, 208} \[ -\frac {3 A b-2 a B}{a^2 \sqrt {a+b x}}+\frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {A}{a x \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A*b - 2*a*B)/(a^2*Sqrt[a + b*x])) - A/(a*x*Sqrt[a + b*x]) + ((3*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a
]])/a^(5/2)

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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 208

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

Rubi steps

\begin {align*} \int \frac {A+B x}{x^2 (a+b x)^{3/2}} \, dx &=-\frac {A}{a x \sqrt {a+b x}}+\frac {\left (-\frac {3 A b}{2}+a B\right ) \int \frac {1}{x (a+b x)^{3/2}} \, dx}{a}\\ &=-\frac {3 A b-2 a B}{a^2 \sqrt {a+b x}}-\frac {A}{a x \sqrt {a+b x}}-\frac {(3 A b-2 a B) \int \frac {1}{x \sqrt {a+b x}} \, dx}{2 a^2}\\ &=-\frac {3 A b-2 a B}{a^2 \sqrt {a+b x}}-\frac {A}{a x \sqrt {a+b x}}-\frac {(3 A b-2 a B) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{a^2 b}\\ &=-\frac {3 A b-2 a B}{a^2 \sqrt {a+b x}}-\frac {A}{a x \sqrt {a+b x}}+\frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.67 \[ \frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b x}{a}+1\right ) (2 a B x-3 A b x)-a A}{a^2 x \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*A) + (-3*A*b*x + 2*a*B*x)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x)/a])/(a^2*x*Sqrt[a + b*x])

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fricas [A]  time = 0.67, size = 211, normalized size = 2.89 \[ \left [-\frac {{\left ({\left (2 \, B a b - 3 \, A b^{2}\right )} x^{2} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (A a^{2} - {\left (2 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {b x + a}}{2 \, {\left (a^{3} b x^{2} + a^{4} x\right )}}, \frac {{\left ({\left (2 \, B a b - 3 \, A b^{2}\right )} x^{2} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (A a^{2} - {\left (2 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {b x + a}}{a^{3} b x^{2} + a^{4} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((2*B*a*b - 3*A*b^2)*x^2 + (2*B*a^2 - 3*A*a*b)*x)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x)
+ 2*(A*a^2 - (2*B*a^2 - 3*A*a*b)*x)*sqrt(b*x + a))/(a^3*b*x^2 + a^4*x), (((2*B*a*b - 3*A*b^2)*x^2 + (2*B*a^2 -
3*A*a*b)*x)*sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (A*a^2 - (2*B*a^2 - 3*A*a*b)*x)*sqrt(b*x + a))/(a^3*b
*x^2 + a^4*x)]

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giac [A]  time = 1.30, size = 87, normalized size = 1.19 \[ \frac {{\left (2 \, B a - 3 \, A b\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {2 \, {\left (b x + a\right )} B a - 2 \, B a^{2} - 3 \, {\left (b x + a\right )} A b + 2 \, A a b}{{\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 67, normalized size = 0.92 \[ -\frac {2 \left (A b -B a \right )}{\sqrt {b x +a}\, a^{2}}-\frac {2 \left (-\frac {\left (3 A b -2 B a \right ) \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}+\frac {\sqrt {b x +a}\, A}{2 x}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.98, size = 107, normalized size = 1.47 \[ -\frac {1}{2} \, b {\left (\frac {2 \, {\left (2 \, B a^{2} - 2 \, A a b - {\left (2 \, B a - 3 \, A b\right )} {\left (b x + a\right )}\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} a^{2} b - \sqrt {b x + a} a^{3} b} - \frac {{\left (2 \, B a - 3 \, A b\right )} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*(2*(2*B*a^2 - 2*A*a*b - (2*B*a - 3*A*b)*(b*x + a))/((b*x + a)^(3/2)*a^2*b - sqrt(b*x + a)*a^3*b) - (2*B
*a - 3*A*b)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/(a^(5/2)*b))

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mupad [B]  time = 0.42, size = 82, normalized size = 1.12 \[ \frac {\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )\,\left (3\,A\,b-2\,B\,a\right )}{a^{5/2}}-\frac {\frac {2\,\left (A\,b-B\,a\right )}{a}-\frac {\left (3\,A\,b-2\,B\,a\right )\,\left (a+b\,x\right )}{a^2}}{a\,\sqrt {a+b\,x}-{\left (a+b\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((a + b*x)^(1/2)/a^(1/2))*(3*A*b - 2*B*a))/a^(5/2) - ((2*(A*b - B*a))/a - ((3*A*b - 2*B*a)*(a + b*x))/a^
2)/(a*(a + b*x)^(1/2) - (a + b*x)^(3/2))

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sympy [B]  time = 48.22, size = 224, normalized size = 3.07 \[ A \left (- \frac {1}{a \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {5}{2}}}\right ) + B \left (\frac {2 a^{3} \sqrt {1 + \frac {b x}{a}}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} + \frac {a^{3} \log {\left (\frac {b x}{a} \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} + \frac {a^{2} b x \log {\left (\frac {b x}{a} \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x} - \frac {2 a^{2} b x \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )}}{a^{\frac {9}{2}} + a^{\frac {7}{2}} b x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-1/(a*sqrt(b)*x**(3/2)*sqrt(a/(b*x) + 1)) - 3*sqrt(b)/(a**2*sqrt(x)*sqrt(a/(b*x) + 1)) + 3*b*asinh(sqrt(a)/
(sqrt(b)*sqrt(x)))/a**(5/2)) + B*(2*a**3*sqrt(1 + b*x/a)/(a**(9/2) + a**(7/2)*b*x) + a**3*log(b*x/a)/(a**(9/2)
+ a**(7/2)*b*x) - 2*a**3*log(sqrt(1 + b*x/a) + 1)/(a**(9/2) + a**(7/2)*b*x) + a**2*b*x*log(b*x/a)/(a**(9/2) +
a**(7/2)*b*x) - 2*a**2*b*x*log(sqrt(1 + b*x/a) + 1)/(a**(9/2) + a**(7/2)*b*x))

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