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

Optimal. Leaf size=113 \[ \frac{16 \sqrt{a+b x} (2 A b-a B)}{3 a^4 \sqrt{x}}-\frac{8 (2 A b-a B)}{3 a^3 \sqrt{x} \sqrt{a+b x}}-\frac{2 (2 A b-a B)}{3 a^2 \sqrt{x} (a+b x)^{3/2}}-\frac{2 A}{3 a x^{3/2} (a+b x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0360582, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ \frac{16 \sqrt{a+b x} (2 A b-a B)}{3 a^4 \sqrt{x}}-\frac{8 (2 A b-a B)}{3 a^3 \sqrt{x} \sqrt{a+b x}}-\frac{2 (2 A b-a B)}{3 a^2 \sqrt{x} (a+b x)^{3/2}}-\frac{2 A}{3 a x^{3/2} (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \frac{A+B x}{x^{5/2} (a+b x)^{5/2}} \, dx &=-\frac{2 A}{3 a x^{3/2} (a+b x)^{3/2}}+\frac{\left (2 \left (-3 A b+\frac{3 a B}{2}\right )\right ) \int \frac{1}{x^{3/2} (a+b x)^{5/2}} \, dx}{3 a}\\ &=-\frac{2 A}{3 a x^{3/2} (a+b x)^{3/2}}-\frac{2 (2 A b-a B)}{3 a^2 \sqrt{x} (a+b x)^{3/2}}-\frac{(4 (2 A b-a B)) \int \frac{1}{x^{3/2} (a+b x)^{3/2}} \, dx}{3 a^2}\\ &=-\frac{2 A}{3 a x^{3/2} (a+b x)^{3/2}}-\frac{2 (2 A b-a B)}{3 a^2 \sqrt{x} (a+b x)^{3/2}}-\frac{8 (2 A b-a B)}{3 a^3 \sqrt{x} \sqrt{a+b x}}-\frac{(8 (2 A b-a B)) \int \frac{1}{x^{3/2} \sqrt{a+b x}} \, dx}{3 a^3}\\ &=-\frac{2 A}{3 a x^{3/2} (a+b x)^{3/2}}-\frac{2 (2 A b-a B)}{3 a^2 \sqrt{x} (a+b x)^{3/2}}-\frac{8 (2 A b-a B)}{3 a^3 \sqrt{x} \sqrt{a+b x}}+\frac{16 (2 A b-a B) \sqrt{a+b x}}{3 a^4 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0268231, size = 70, normalized size = 0.62 \[ -\frac{2 \left (-6 a^2 b x (A-2 B x)+a^3 (A+3 B x)+8 a b^2 x^2 (B x-3 A)-16 A b^3 x^3\right )}{3 a^4 x^{3/2} (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-16*A*b^3*x^3 - 6*a^2*b*x*(A - 2*B*x) + 8*a*b^2*x^2*(-3*A + B*x) + a^3*(A + 3*B*x)))/(3*a^4*x^(3/2)*(a +
b*x)^(3/2))

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Maple [A]  time = 0.005, size = 76, normalized size = 0.7 \begin{align*} -{\frac{-32\,A{b}^{3}{x}^{3}+16\,B{x}^{3}a{b}^{2}-48\,aA{b}^{2}{x}^{2}+24\,B{x}^{2}{a}^{2}b-12\,{a}^{2}Abx+6\,{a}^{3}Bx+2\,A{a}^{3}}{3\,{a}^{4}}{x}^{-{\frac{3}{2}}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-16*A*b^3*x^3+8*B*a*b^2*x^3-24*A*a*b^2*x^2+12*B*a^2*b*x^2-6*A*a^2*b*x+3*B*a^3*x+A*a^3)/x^(3/2)/(b*x+a)^(
3/2)/a^4

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Maxima [A]  time = 1.34909, size = 176, normalized size = 1.56 \begin{align*} \frac{2 \, B x}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} a} - \frac{16 \, B b x}{3 \, \sqrt{b x^{2} + a x} a^{3}} - \frac{4 \, A b x}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} a^{2}} + \frac{32 \, A b^{2} x}{3 \, \sqrt{b x^{2} + a x} a^{4}} - \frac{8 \, B}{3 \, \sqrt{b x^{2} + a x} a^{2}} - \frac{2 \, A}{3 \,{\left (b x^{2} + a x\right )}^{\frac{3}{2}} a} + \frac{16 \, A b}{3 \, \sqrt{b x^{2} + a x} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.84618, size = 215, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (A a^{3} + 8 \,{\left (B a b^{2} - 2 \, A b^{3}\right )} x^{3} + 12 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} x^{2} + 3 \,{\left (B a^{3} - 2 \, A a^{2} b\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(A*a^3 + 8*(B*a*b^2 - 2*A*b^3)*x^3 + 12*(B*a^2*b - 2*A*a*b^2)*x^2 + 3*(B*a^3 - 2*A*a^2*b)*x)*sqrt(b*x + a
)*sqrt(x)/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.45137, size = 409, normalized size = 3.62 \begin{align*} \frac{\sqrt{b x + a}{\left (\frac{{\left (3 \, B a^{4} b^{3}{\left | b \right |} - 8 \, A a^{3} b^{4}{\left | b \right |}\right )}{\left (b x + a\right )}}{a^{2} b^{6}} - \frac{3 \,{\left (B a^{5} b^{3}{\left | b \right |} - 3 \, A a^{4} b^{4}{\left | b \right |}\right )}}{a^{2} b^{6}}\right )}}{48 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{3}{2}}} - \frac{4 \,{\left (3 \, B a{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac{5}{2}} + 12 \, B a^{2}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{7}{2}} - 6 \, A{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac{7}{2}} + 5 \, B a^{3} b^{\frac{9}{2}} - 18 \, A a{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{9}{2}} - 8 \, A a^{2} b^{\frac{11}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*sqrt(b*x + a)*((3*B*a^4*b^3*abs(b) - 8*A*a^3*b^4*abs(b))*(b*x + a)/(a^2*b^6) - 3*(B*a^5*b^3*abs(b) - 3*A*
a^4*b^4*abs(b))/(a^2*b^6))/((b*x + a)*b - a*b)^(3/2) - 4/3*(3*B*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b -
a*b))^4*b^(5/2) + 12*B*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(7/2) - 6*A*(sqrt(b*x + a)*sq
rt(b) - sqrt((b*x + a)*b - a*b))^4*b^(7/2) + 5*B*a^3*b^(9/2) - 18*A*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*
b - a*b))^2*b^(9/2) - 8*A*a^2*b^(11/2))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*a^3*abs
(b))