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

Optimal. Leaf size=107 \[ \frac {\sqrt {b} (5 A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 A b-3 a B}{a^3 \sqrt {x}}-\frac {5 A b-3 a B}{3 a^2 b x^{3/2}}+\frac {A b-a B}{a b x^{3/2} (a+b x)} \]

[Out]

1/3*(-5*A*b+3*B*a)/a^2/b/x^(3/2)+(A*b-B*a)/a/b/x^(3/2)/(b*x+a)+(5*A*b-3*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))*b
^(1/2)/a^(7/2)+(5*A*b-3*B*a)/a^3/x^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {78, 51, 63, 205} \[ -\frac {5 A b-3 a B}{3 a^2 b x^{3/2}}+\frac {5 A b-3 a B}{a^3 \sqrt {x}}+\frac {\sqrt {b} (5 A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {A b-a B}{a b x^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(5*A*b - 3*a*B)/(3*a^2*b*x^(3/2)) + (5*A*b - 3*a*B)/(a^3*Sqrt[x]) + (A*b - a*B)/(a*b*x^(3/2)*(a + b*x)) + (Sq
rt[b]*(5*A*b - 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(7/2)

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)^2} \, dx &=\frac {A b-a B}{a b x^{3/2} (a+b x)}-\frac {\left (-\frac {5 A b}{2}+\frac {3 a B}{2}\right ) \int \frac {1}{x^{5/2} (a+b x)} \, dx}{a b}\\ &=-\frac {5 A b-3 a B}{3 a^2 b x^{3/2}}+\frac {A b-a B}{a b x^{3/2} (a+b x)}-\frac {(5 A b-3 a B) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{2 a^2}\\ &=-\frac {5 A b-3 a B}{3 a^2 b x^{3/2}}+\frac {5 A b-3 a B}{a^3 \sqrt {x}}+\frac {A b-a B}{a b x^{3/2} (a+b x)}+\frac {(b (5 A b-3 a B)) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{2 a^3}\\ &=-\frac {5 A b-3 a B}{3 a^2 b x^{3/2}}+\frac {5 A b-3 a B}{a^3 \sqrt {x}}+\frac {A b-a B}{a b x^{3/2} (a+b x)}+\frac {(b (5 A b-3 a B)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{a^3}\\ &=-\frac {5 A b-3 a B}{3 a^2 b x^{3/2}}+\frac {5 A b-3 a B}{a^3 \sqrt {x}}+\frac {A b-a B}{a b x^{3/2} (a+b x)}+\frac {\sqrt {b} (5 A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 64, normalized size = 0.60 \[ \frac {(a+b x) (3 a B-5 A b) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {b x}{a}\right )+3 a (A b-a B)}{3 a^2 b x^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(A*b - a*B) + (-5*A*b + 3*a*B)*(a + b*x)*Hypergeometric2F1[-3/2, 1, -1/2, -((b*x)/a)])/(3*a^2*b*x^(3/2)*(
a + b*x))

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 262, normalized size = 2.45 \[ \left [-\frac {3 \, {\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (2 \, A a^{2} + 3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, \frac {3 \, {\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (2 \, A a^{2} + 3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*((3*B*a*b - 5*A*b^2)*x^3 + (3*B*a^2 - 5*A*a*b)*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)
/(b*x + a)) + 2*(2*A*a^2 + 3*(3*B*a*b - 5*A*b^2)*x^2 + 2*(3*B*a^2 - 5*A*a*b)*x)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)
, 1/3*(3*((3*B*a*b - 5*A*b^2)*x^3 + (3*B*a^2 - 5*A*a*b)*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (2*A*
a^2 + 3*(3*B*a*b - 5*A*b^2)*x^2 + 2*(3*B*a^2 - 5*A*a*b)*x)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)]

________________________________________________________________________________________

giac [A]  time = 1.32, size = 85, normalized size = 0.79 \[ -\frac {{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {B a b \sqrt {x} - A b^{2} \sqrt {x}}{{\left (b x + a\right )} a^{3}} - \frac {2 \, {\left (3 \, B a x - 6 \, A b x + A a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*B*a*b - 5*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - (B*a*b*sqrt(x) - A*b^2*sqrt(x))/((b*x + a)*
a^3) - 2/3*(3*B*a*x - 6*A*b*x + A*a)/(a^3*x^(3/2))

________________________________________________________________________________________

maple [A]  time = 0.02, size = 113, normalized size = 1.06 \[ \frac {5 A \,b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{3}}-\frac {3 B b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}+\frac {A \,b^{2} \sqrt {x}}{\left (b x +a \right ) a^{3}}-\frac {B b \sqrt {x}}{\left (b x +a \right ) a^{2}}+\frac {4 A b}{a^{3} \sqrt {x}}-\frac {2 B}{a^{2} \sqrt {x}}-\frac {2 A}{3 a^{2} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*b^2*x^(1/2)/(b*x+a)*A-1/a^2*b*x^(1/2)/(b*x+a)*B+5/a^3*b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*A-
3/a^2*b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))*B-2/3*A/a^2/x^(3/2)+4/a^3/x^(1/2)*A*b-2/a^2/x^(1/2)*B

________________________________________________________________________________________

maxima [A]  time = 1.94, size = 93, normalized size = 0.87 \[ -\frac {2 \, A a^{2} + 3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x}{3 \, {\left (a^{3} b x^{\frac {5}{2}} + a^{4} x^{\frac {3}{2}}\right )}} - \frac {{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*A*a^2 + 3*(3*B*a*b - 5*A*b^2)*x^2 + 2*(3*B*a^2 - 5*A*a*b)*x)/(a^3*b*x^(5/2) + a^4*x^(3/2)) - (3*B*a*b
- 5*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3)

________________________________________________________________________________________

mupad [B]  time = 0.43, size = 81, normalized size = 0.76 \[ \frac {\frac {2\,x\,\left (5\,A\,b-3\,B\,a\right )}{3\,a^2}-\frac {2\,A}{3\,a}+\frac {b\,x^2\,\left (5\,A\,b-3\,B\,a\right )}{a^3}}{a\,x^{3/2}+b\,x^{5/2}}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (5\,A\,b-3\,B\,a\right )}{a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*x*(5*A*b - 3*B*a))/(3*a^2) - (2*A)/(3*a) + (b*x^2*(5*A*b - 3*B*a))/a^3)/(a*x^(3/2) + b*x^(5/2)) + (b^(1/2)
*atan((b^(1/2)*x^(1/2))/a^(1/2))*(5*A*b - 3*B*a))/a^(7/2)

________________________________________________________________________________________

sympy [A]  time = 54.14, size = 983, normalized size = 9.19 \[ \begin {cases} \tilde {\infty } \left (- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}}{b^{2}} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} - \frac {2 B}{\sqrt {x}}}{a^{2}} & \text {for}\: b = 0 \\- \frac {4 i A a^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {20 i A a^{\frac {3}{2}} b x \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {30 i A \sqrt {a} b^{2} x^{2} \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 A a b x^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 A a b x^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {15 A b^{2} x^{\frac {5}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {15 A b^{2} x^{\frac {5}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {12 i B a^{\frac {5}{2}} x \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {18 i B a^{\frac {3}{2}} b x^{2} \sqrt {\frac {1}{b}}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {9 B a^{2} x^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {9 B a^{2} x^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} - \frac {9 B a b x^{\frac {5}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}}} + \frac {9 B a b x^{\frac {5}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{6 i a^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 6 i a^{\frac {7}{2}} b x^{\frac {5}{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)**2,x)

[Out]

Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(
5/2)))/b**2, Eq(a, 0)), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/a**2, Eq(b, 0)), (-4*I*A*a**(5/2)*sqrt(1/b)/(6*I*a*
*(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 20*I*A*a**(3/2)*b*x*sqrt(1/b)/(6*I*a**(9/2)*x
**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 30*I*A*sqrt(a)*b**2*x**2*sqrt(1/b)/(6*I*a**(9/2)*x**(
3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 15*A*a*b*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6
*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 15*A*a*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b
) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 15*A*b**2*x**(5/2)*log(-I
*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 15*A*b**
2*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(
1/b)) - 12*I*B*a**(5/2)*x*sqrt(1/b)/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 18
*I*B*a**(3/2)*b*x**2*sqrt(1/b)/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) - 9*B*a**
2*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt
(1/b)) + 9*B*a**2*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b) + 6*I*a**(7/2)*
b*x**(5/2)*sqrt(1/b)) - 9*B*a*b*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3/2)*sqrt(1/b)
+ 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)) + 9*B*a*b*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(9/2)*x**(3
/2)*sqrt(1/b) + 6*I*a**(7/2)*b*x**(5/2)*sqrt(1/b)), True))

________________________________________________________________________________________