\(\int \frac {A+B x}{x^{5/2} (a+b x)} \, dx\) [350]

Optimal result

Integrand size = 18, antiderivative size = 69 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=-\frac {2 A}{3 a x^{3/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {79, 53, 65, 211} \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=\frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {2 A}{3 a x^{3/2}} \]

[In]

Int[(A + B*x)/(x^(5/2)*(a + b*x)),x]

[Out]

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

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 211

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=-\frac {2 (a A-3 A b x+3 a B x)}{3 a^2 x^{3/2}}-\frac {2 \sqrt {b} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}} \]

[In]

Integrate[(A + B*x)/(x^(5/2)*(a + b*x)),x]

[Out]

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

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {2 \left (-3 A b x +3 B a x +A a \right )}{3 a^{2} x^{\frac {3}{2}}}+\frac {2 b \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\)	\(54\)
derivativedivides	\(\frac {2 b \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}-\frac {2 A}{3 a \,x^{\frac {3}{2}}}-\frac {2 \left (-A b +B a \right )}{a^{2} \sqrt {x}}\)	\(57\)
default	\(\frac {2 b \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}-\frac {2 A}{3 a \,x^{\frac {3}{2}}}-\frac {2 \left (-A b +B a \right )}{a^{2} \sqrt {x}}\)	\(57\)

[In]

int((B*x+A)/x^(5/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.12 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}}{3 \, a^{2} x^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}\right )}}{3 \, a^{2} x^{2}}\right ] \]

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (66) = 132\).

Time = 1.84 (sec) , antiderivative size = 218, normalized size of antiderivative = 3.16 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{3 a x^{\frac {3}{2}}} + \frac {A b \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{2} \sqrt {- \frac {a}{b}}} - \frac {A b \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{2} \sqrt {- \frac {a}{b}}} + \frac {2 A b}{a^{2} \sqrt {x}} - \frac {B \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a \sqrt {- \frac {a}{b}}} + \frac {B \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a \sqrt {- \frac {a}{b}}} - \frac {2 B}{a \sqrt {x}} & \text {otherwise} \end {cases} \]

[In]

integrate((B*x+A)/x**(5/2)/(b*x+a),x)

[Out]

Piecewise((zoo*(-2*A/(5*x**(5/2)) - 2*B/(3*x**(3/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(5*x**(5/2)) - 2*B/(3*x**(
3/2)))/b, Eq(a, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/a, Eq(b, 0)), (-2*A/(3*a*x**(3/2)) + A*b*log(sqrt(x) -
sqrt(-a/b))/(a**2*sqrt(-a/b)) - A*b*log(sqrt(x) + sqrt(-a/b))/(a**2*sqrt(-a/b)) + 2*A*b/(a**2*sqrt(x)) - B*lo
g(sqrt(x) - sqrt(-a/b))/(a*sqrt(-a/b)) + B*log(sqrt(x) + sqrt(-a/b))/(a*sqrt(-a/b)) - 2*B/(a*sqrt(x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=-\frac {2 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} - \frac {2 \, {\left (A a + 3 \, {\left (B a - A b\right )} x\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=-\frac {2 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} - \frac {2 \, {\left (3 \, B a x - 3 \, A b x + A a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x}{x^{5/2} (a+b x)} \, dx=\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{5/2}}-\frac {\frac {2\,A}{3\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{a^2}}{x^{3/2}} \]

[In]

int((A + B*x)/(x^(5/2)*(a + b*x)),x)

[Out]

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