3.532 \(\int \frac {A+B x}{x^{5/2} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=83 \[ \frac {4 \sqrt {a+b x} (4 A b-3 a B)}{3 a^3 \sqrt {x}}-\frac {2 (4 A b-3 a B)}{3 a^2 \sqrt {x} \sqrt {a+b x}}-\frac {2 A}{3 a x^{3/2} \sqrt {a+b x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {78, 45, 37} \[ \frac {4 \sqrt {a+b x} (4 A b-3 a B)}{3 a^3 \sqrt {x}}-\frac {2 (4 A b-3 a B)}{3 a^2 \sqrt {x} \sqrt {a+b x}}-\frac {2 A}{3 a x^{3/2} \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(3*a*x^(3/2)*Sqrt[a + b*x]) - (2*(4*A*b - 3*a*B))/(3*a^2*Sqrt[x]*Sqrt[a + b*x]) + (4*(4*A*b - 3*a*B)*Sq
rt[a + b*x])/(3*a^3*Sqrt[x])

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

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*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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]))))

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{5/2} (a+b x)^{3/2}} \, dx &=-\frac {2 A}{3 a x^{3/2} \sqrt {a+b x}}+\frac {\left (2 \left (-2 A b+\frac {3 a B}{2}\right )\right ) \int \frac {1}{x^{3/2} (a+b x)^{3/2}} \, dx}{3 a}\\ &=-\frac {2 A}{3 a x^{3/2} \sqrt {a+b x}}-\frac {2 (4 A b-3 a B)}{3 a^2 \sqrt {x} \sqrt {a+b x}}-\frac {(2 (4 A b-3 a B)) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{3 a^2}\\ &=-\frac {2 A}{3 a x^{3/2} \sqrt {a+b x}}-\frac {2 (4 A b-3 a B)}{3 a^2 \sqrt {x} \sqrt {a+b x}}+\frac {4 (4 A b-3 a B) \sqrt {a+b x}}{3 a^3 \sqrt {x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 54, normalized size = 0.65 \[ -\frac {2 \left (a^2 (A+3 B x)+2 a b x (3 B x-2 A)-8 A b^2 x^2\right )}{3 a^3 x^{3/2} \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-8*A*b^2*x^2 + 2*a*b*x*(-2*A + 3*B*x) + a^2*(A + 3*B*x)))/(3*a^3*x^(3/2)*Sqrt[a + b*x])

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 67, normalized size = 0.81 \[ -\frac {2 \, {\left (A a^{2} + 2 \, {\left (3 \, B a b - 4 \, A b^{2}\right )} x^{2} + {\left (3 \, B a^{2} - 4 \, A a b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.23, size = 215, normalized size = 2.59 \[ -\frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (3 \, B a^{3} b^{3} {\left | b \right |} - 5 \, A a^{2} b^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{a^{5} b^{2}} - \frac {3 \, {\left (B a^{4} b^{3} {\left | b \right |} - 2 \, A a^{3} b^{4} {\left | b \right |}\right )}}{a^{5} b^{2}}\right )}}{3 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {3}{2}}} - \frac {4 \, {\left (B^{2} a^{2} b^{5} - 2 \, A B a b^{6} + A^{2} b^{7}\right )}}{{\left (B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {5}{2}} + B a^{2} b^{\frac {7}{2}} - A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {7}{2}} - A a b^{\frac {9}{2}}\right )} a^{2} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

-2/3*sqrt(b*x + a)*((3*B*a^3*b^3*abs(b) - 5*A*a^2*b^4*abs(b))*(b*x + a)/(a^5*b^2) - 3*(B*a^4*b^3*abs(b) - 2*A*
a^3*b^4*abs(b))/(a^5*b^2))/((b*x + a)*b - a*b)^(3/2) - 4*(B^2*a^2*b^5 - 2*A*B*a*b^6 + A^2*b^7)/((B*a*(sqrt(b*x
+ a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(5/2) + B*a^2*b^(7/2) - A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)
*b - a*b))^2*b^(7/2) - A*a*b^(9/2))*a^2*abs(b))

________________________________________________________________________________________

maple [A]  time = 0.00, size = 52, normalized size = 0.63 \[ -\frac {2 \left (-8 A \,b^{2} x^{2}+6 B a b \,x^{2}-4 A a b x +3 B \,a^{2} x +A \,a^{2}\right )}{3 \sqrt {b x +a}\, a^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(5/2)/(b*x+a)^(3/2),x)

[Out]

-2/3*(-8*A*b^2*x^2+6*B*a*b*x^2-4*A*a*b*x+3*B*a^2*x+A*a^2)/(b*x+a)^(1/2)/x^(3/2)/a^3

________________________________________________________________________________________

maxima [A]  time = 0.83, size = 96, normalized size = 1.16 \[ -\frac {4 \, B b x}{\sqrt {b x^{2} + a x} a^{2}} + \frac {16 \, A b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {2 \, B}{\sqrt {b x^{2} + a x} a} + \frac {8 \, A b}{3 \, \sqrt {b x^{2} + a x} a^{2}} - \frac {2 \, A}{3 \, \sqrt {b x^{2} + a x} a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(5/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

-4*B*b*x/(sqrt(b*x^2 + a*x)*a^2) + 16/3*A*b^2*x/(sqrt(b*x^2 + a*x)*a^3) - 2*B/(sqrt(b*x^2 + a*x)*a) + 8/3*A*b/
(sqrt(b*x^2 + a*x)*a^2) - 2/3*A/(sqrt(b*x^2 + a*x)*a*x)

________________________________________________________________________________________

mupad [B]  time = 0.93, size = 77, normalized size = 0.93 \[ -\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{3\,a\,b}+\frac {x\,\left (6\,B\,a^2-8\,A\,a\,b\right )}{3\,a^3\,b}-\frac {x^2\,\left (16\,A\,b^2-12\,B\,a\,b\right )}{3\,a^3\,b}\right )}{x^{5/2}+\frac {a\,x^{3/2}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(5/2)*(a + b*x)^(3/2)),x)

[Out]

-((a + b*x)^(1/2)*((2*A)/(3*a*b) + (x*(6*B*a^2 - 8*A*a*b))/(3*a^3*b) - (x^2*(16*A*b^2 - 12*B*a*b))/(3*a^3*b)))
/(x^(5/2) + (a*x^(3/2))/b)

________________________________________________________________________________________

sympy [B]  time = 68.06, size = 265, normalized size = 3.19 \[ A \left (- \frac {2 a^{3} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {6 a^{2} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {24 a b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {16 b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}}\right ) + B \left (- \frac {2}{a \sqrt {b} x \sqrt {\frac {a}{b x} + 1}} - \frac {4 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x} + 1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**(5/2)/(b*x+a)**(3/2),x)

[Out]

A*(-2*a**3*b**(9/2)*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 6*a**2*b**(11/2)
*x*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 24*a*b**(13/2)*x**2*sqrt(a/(b*x)
+ 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 16*b**(15/2)*x**3*sqrt(a/(b*x) + 1)/(3*a**5*b**4*
x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3)) + B*(-2/(a*sqrt(b)*x*sqrt(a/(b*x) + 1)) - 4*sqrt(b)/(a**2*sqrt(a/(b*
x) + 1)))

________________________________________________________________________________________