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

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

[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)

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Rubi [A]  time = 0.04892, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {27, 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^2 + 2*a*b*x + b^2*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 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} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\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 = 0.0194671, size = 64, normalized size = 0.6 \[ \frac{(a+b x) (3 a B-5 A b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{b x}{a}\right )+3 a (A b-a B)}{3 a^2 b x^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.017, size = 113, normalized size = 1.1 \begin{align*} -{\frac{2\,A}{3\,{a}^{2}}{x}^{-{\frac{3}{2}}}}+4\,{\frac{Ab}{{a}^{3}\sqrt{x}}}-2\,{\frac{B}{{a}^{2}\sqrt{x}}}+{\frac{A{b}^{2}}{{a}^{3} \left ( bx+a \right ) }\sqrt{x}}-{\frac{bB}{{a}^{2} \left ( bx+a \right ) }\sqrt{x}}+5\,{\frac{A{b}^{2}}{{a}^{3}\sqrt{ab}}\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) }-3\,{\frac{bB}{{a}^{2}\sqrt{ab}}\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62882, size = 572, normalized size = 5.35 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 90.6224, size = 983, normalized size = 9.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]  time = 1.14622, size = 115, normalized size = 1.07 \begin{align*} -\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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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