3.469 \(\int \frac {x^{5/2}}{-a+b x} \, dx\)

Optimal. Leaf size=68 \[ -\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {2 a^2 \sqrt {x}}{b^3}+\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b} \]

[Out]

2/3*a*x^(3/2)/b^2+2/5*x^(5/2)/b-2*a^(5/2)*arctanh(b^(1/2)*x^(1/2)/a^(1/2))/b^(7/2)+2*a^2*x^(1/2)/b^3

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Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {50, 63, 208} \[ \frac {2 a^2 \sqrt {x}}{b^3}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(-a + b*x),x]

[Out]

(2*a^2*Sqrt[x])/b^3 + (2*a*x^(3/2))/(3*b^2) + (2*x^(5/2))/(5*b) - (2*a^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]
])/b^(7/2)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 {x^{5/2}}{-a+b x} \, dx &=\frac {2 x^{5/2}}{5 b}+\frac {a \int \frac {x^{3/2}}{-a+b x} \, dx}{b}\\ &=\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b}+\frac {a^2 \int \frac {\sqrt {x}}{-a+b x} \, dx}{b^2}\\ &=\frac {2 a^2 \sqrt {x}}{b^3}+\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b}+\frac {a^3 \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{b^3}\\ &=\frac {2 a^2 \sqrt {x}}{b^3}+\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b}+\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {2 a^2 \sqrt {x}}{b^3}+\frac {2 a x^{3/2}}{3 b^2}+\frac {2 x^{5/2}}{5 b}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 61, normalized size = 0.90 \[ \frac {2 \sqrt {x} \left (15 a^2+5 a b x+3 b^2 x^2\right )}{15 b^3}-\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(-a + b*x),x]

[Out]

(2*Sqrt[x]*(15*a^2 + 5*a*b*x + 3*b^2*x^2))/(15*b^3) - (2*a^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(7/2)

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fricas [A]  time = 0.43, size = 131, normalized size = 1.93 \[ \left [\frac {15 \, a^{2} \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (3 \, b^{2} x^{2} + 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{15 \, b^{3}}, \frac {2 \, {\left (15 \, a^{2} \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (3 \, b^{2} x^{2} + 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*a^2*sqrt(a/b)*log((b*x - 2*b*sqrt(x)*sqrt(a/b) + a)/(b*x - a)) + 2*(3*b^2*x^2 + 5*a*b*x + 15*a^2)*sq
rt(x))/b^3, 2/15*(15*a^2*sqrt(-a/b)*arctan(b*sqrt(x)*sqrt(-a/b)/a) + (3*b^2*x^2 + 5*a*b*x + 15*a^2)*sqrt(x))/b
^3]

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giac [A]  time = 1.03, size = 61, normalized size = 0.90 \[ \frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} b^{3}} + \frac {2 \, {\left (3 \, b^{4} x^{\frac {5}{2}} + 5 \, a b^{3} x^{\frac {3}{2}} + 15 \, a^{2} b^{2} \sqrt {x}\right )}}{15 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*b^3) + 2/15*(3*b^4*x^(5/2) + 5*a*b^3*x^(3/2) + 15*a^2*b^2*sqrt(
x))/b^5

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maple [A]  time = 0.01, size = 54, normalized size = 0.79 \[ -\frac {2 a^{3} \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b^{3}}+\frac {\frac {2 b^{2} x^{\frac {5}{2}}}{5}+\frac {2 a b \,x^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {x}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x-a),x)

[Out]

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

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maxima [A]  time = 2.94, size = 70, normalized size = 1.03 \[ \frac {a^{3} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{2} x^{\frac {5}{2}} + 5 \, a b x^{\frac {3}{2}} + 15 \, a^{2} \sqrt {x}\right )}}{15 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*b^3) + 2/15*(3*b^2*x^(5/2) + 5*a*b*x^(3/2)
+ 15*a^2*sqrt(x))/b^3

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mupad [B]  time = 0.15, size = 51, normalized size = 0.75 \[ \frac {2\,x^{5/2}}{5\,b}+\frac {2\,a\,x^{3/2}}{3\,b^2}+\frac {2\,a^2\,\sqrt {x}}{b^3}+\frac {a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i}}{b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^(5/2)/(a - b*x),x)

[Out]

(2*x^(5/2))/(5*b) + (2*a*x^(3/2))/(3*b^2) + (2*a^2*x^(1/2))/b^3 + (a^(5/2)*atan((b^(1/2)*x^(1/2)*1i)/a^(1/2))*
2i)/b^(7/2)

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sympy [A]  time = 7.10, size = 116, normalized size = 1.71 \[ \begin {cases} \frac {a^{\frac {5}{2}} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{4} \sqrt {\frac {1}{b}}} - \frac {a^{\frac {5}{2}} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{b^{4} \sqrt {\frac {1}{b}}} + \frac {2 a^{2} \sqrt {x}}{b^{3}} + \frac {2 a x^{\frac {3}{2}}}{3 b^{2}} + \frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\- \frac {2 x^{\frac {7}{2}}}{7 a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(b*x-a),x)

[Out]

Piecewise((a**(5/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(b**4*sqrt(1/b)) - a**(5/2)*log(sqrt(a)*sqrt(1/b) + sqrt
(x))/(b**4*sqrt(1/b)) + 2*a**2*sqrt(x)/b**3 + 2*a*x**(3/2)/(3*b**2) + 2*x**(5/2)/(5*b), Ne(b, 0)), (-2*x**(7/2
)/(7*a), True))

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