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

Optimal. Leaf size=67 \[ -\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 A}{a^2 \sqrt {a+b x}}+\frac {2 (A b-a B)}{3 a b (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 67, 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, 208} \begin {gather*} \frac {2 A}{a^2 \sqrt {a+b x}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 (A b-a B)}{3 a b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x*(a + b*x)^(5/2)),x]

[Out]

(2*(A*b - a*B))/(3*a*b*(a + b*x)^(3/2)) + (2*A)/(a^2*Sqrt[a + b*x]) - (2*A*ArcTanh[Sqrt[a + b*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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.84 \begin {gather*} \frac {2 a (A b-a B)+6 A b (a+b x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b x}{a}+1\right )}{3 a^2 b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 64, normalized size = 0.96 \begin {gather*} \frac {2 \left (a^2 (-B)+3 A b (a+b x)+a A b\right )}{3 a^2 b (a+b x)^{3/2}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*A*b - a^2*B + 3*A*b*(a + b*x)))/(3*a^2*b*(a + b*x)^(3/2)) - (2*A*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(5/2)

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fricas [A]  time = 1.31, size = 221, normalized size = 3.30 \begin {gather*} \left [\frac {3 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, A a b^{2} x - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x + a}}{3 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}}, \frac {2 \, {\left (3 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, A a b^{2} x - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x + a}\right )}}{3 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.32, size = 61, normalized size = 0.91 \begin {gather*} \frac {2 \, A \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {2 \, {\left (B a^{2} - 3 \, {\left (b x + a\right )} A b - A a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 59, normalized size = 0.88 \begin {gather*} \frac {-\frac {2 A b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {2 A b}{\sqrt {b x +a}\, a^{2}}-\frac {2 \left (-A b +B a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} a}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.98, size = 69, normalized size = 1.03 \begin {gather*} \frac {A \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {2 \, {\left (B a^{2} - 3 \, {\left (b x + a\right )} A b - A a b\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.41, size = 56, normalized size = 0.84 \begin {gather*} \frac {\frac {2\,\left (A\,b-B\,a\right )}{3\,a}+\frac {2\,A\,b\,\left (a+b\,x\right )}{a^2}}{b\,{\left (a+b\,x\right )}^{3/2}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x*(a + b*x)^(5/2)),x)

[Out]

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

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sympy [A]  time = 23.77, size = 68, normalized size = 1.01 \begin {gather*} \frac {2 A}{a^{2} \sqrt {a + b x}} + \frac {2 A \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{a^{2} \sqrt {- a}} - \frac {2 \left (- A b + B a\right )}{3 a b \left (a + b x\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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