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

Optimal. Leaf size=69 \[ \frac {2 \sqrt {b} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 (A b-a B)}{a^2 \sqrt {x}}-\frac {2 A}{3 a x^{3/2}} \]

[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)

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Rubi [A]  time = 0.04, antiderivative size = 69, 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, 205} \[ \frac {2 (A b-a B)}{a^2 \sqrt {x}}+\frac {2 \sqrt {b} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {2 A}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[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 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 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^{5/2} (a+b x)} \, dx &=-\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)) \operatorname {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*}

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Mathematica [C]  time = 0.01, size = 43, normalized size = 0.62 \[ \frac {6 x (A b-a B) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {b x}{a}\right )-2 a A}{3 a^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*A + 6*(A*b - a*B)*x*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x)/a)])/(3*a^2*x^(3/2))

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fricas [A]  time = 0.72, size = 146, normalized size = 2.12 \[ \left [-\frac {3 \, {\left (B a - A b\right )} x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}}{3 \, a^{2} x^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}\right )}}{3 \, a^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 1.23, size = 55, normalized size = 0.80 \[ -\frac {2 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} - \frac {2 \, {\left (3 \, B a x - 3 \, A b x + A a\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.01, size = 78, normalized size = 1.13 \[ \frac {2 A \,b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}-\frac {2 B b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a}+\frac {2 A b}{a^{2} \sqrt {x}}-\frac {2 B}{a \sqrt {x}}-\frac {2 A}{3 a \,x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.97, size = 56, normalized size = 0.81 \[ -\frac {2 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} - \frac {2 \, {\left (A a + 3 \, {\left (B a - A b\right )} x\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.39, size = 54, normalized size = 0.78 \[ \frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{5/2}}-\frac {\frac {2\,A}{3\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{a^2}}{x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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
)

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sympy [A]  time = 9.45, size = 248, normalized size = 3.59 \[ \begin {cases} \tilde {\infty } \left (- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{3 a x^{\frac {3}{2}}} + \frac {2 A b}{a^{2} \sqrt {x}} - \frac {i A b \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {i A b \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {2 B}{a \sqrt {x}} + \frac {i B \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i B \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)) + 2*A*b/(a**2*sqrt(
x)) - I*A*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(5/2)*sqrt(1/b)) + I*A*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x)
)/(a**(5/2)*sqrt(1/b)) - 2*B/(a*sqrt(x)) + I*B*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(3/2)*sqrt(1/b)) - I*B*
log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(3/2)*sqrt(1/b)), True))

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