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

Optimal. Leaf size=70 \[ -\frac {5 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {5 a \sqrt {x}}{b^3}+\frac {x^{5/2}}{b (a-b x)}+\frac {5 x^{3/2}}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {47, 50, 63, 208} \[ -\frac {5 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}+\frac {5 a \sqrt {x}}{b^3}+\frac {x^{5/2}}{b (a-b x)}+\frac {5 x^{3/2}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(-a + b*x)^2,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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)^2} \, dx &=\frac {x^{5/2}}{b (a-b x)}+\frac {5 \int \frac {x^{3/2}}{-a+b x} \, dx}{2 b}\\ &=\frac {5 x^{3/2}}{3 b^2}+\frac {x^{5/2}}{b (a-b x)}+\frac {(5 a) \int \frac {\sqrt {x}}{-a+b x} \, dx}{2 b^2}\\ &=\frac {5 a \sqrt {x}}{b^3}+\frac {5 x^{3/2}}{3 b^2}+\frac {x^{5/2}}{b (a-b x)}+\frac {\left (5 a^2\right ) \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{2 b^3}\\ &=\frac {5 a \sqrt {x}}{b^3}+\frac {5 x^{3/2}}{3 b^2}+\frac {x^{5/2}}{b (a-b x)}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {5 a \sqrt {x}}{b^3}+\frac {5 x^{3/2}}{3 b^2}+\frac {x^{5/2}}{b (a-b x)}-\frac {5 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.37 \[ \frac {2 x^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {b x}{a}\right )}{7 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(-a + b*x)^2,x]

[Out]

(2*x^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, (b*x)/a])/(7*a^2)

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fricas [A]  time = 0.48, size = 167, normalized size = 2.39 \[ \left [\frac {15 \, {\left (a b x - a^{2}\right )} \sqrt {\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {\frac {a}{b}} + a}{b x - a}\right ) + 2 \, {\left (2 \, b^{2} x^{2} + 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (b^{4} x - a b^{3}\right )}}, \frac {15 \, {\left (a b x - a^{2}\right )} \sqrt {-\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {-\frac {a}{b}}}{a}\right ) + {\left (2 \, b^{2} x^{2} + 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (b^{4} x - a b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b*x - a^2)*sqrt(a/b)*log((b*x - 2*b*sqrt(x)*sqrt(a/b) + a)/(b*x - a)) + 2*(2*b^2*x^2 + 10*a*b*x -
15*a^2)*sqrt(x))/(b^4*x - a*b^3), 1/3*(15*(a*b*x - a^2)*sqrt(-a/b)*arctan(b*sqrt(x)*sqrt(-a/b)/a) + (2*b^2*x^2
+ 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x - a*b^3)]

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giac [A]  time = 0.98, size = 69, normalized size = 0.99 \[ \frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} b^{3}} - \frac {a^{2} \sqrt {x}}{{\left (b x - a\right )} b^{3}} + \frac {2 \, {\left (b^{4} x^{\frac {3}{2}} + 6 \, a b^{3} \sqrt {x}\right )}}{3 \, b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a^2*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*b^3) - a^2*sqrt(x)/((b*x - a)*b^3) + 2/3*(b^4*x^(3/2) + 6*a*b^3
*sqrt(x))/b^6

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maple [A]  time = 0.01, size = 61, normalized size = 0.87 \[ \frac {2 \left (-\frac {5 \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}-\frac {\sqrt {x}}{2 \left (b x -a \right )}\right ) a^{2}}{b^{3}}+\frac {\frac {2 b \,x^{\frac {3}{2}}}{3}+4 a \sqrt {x}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.06, size = 81, normalized size = 1.16 \[ -\frac {a^{2} \sqrt {x}}{b^{4} x - a b^{3}} + \frac {5 \, a^{2} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + 6 \, a \sqrt {x}\right )}}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*sqrt(x)/(b^4*x - a*b^3) + 5/2*a^2*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*b^3) +
2/3*(b*x^(3/2) + 6*a*sqrt(x))/b^3

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mupad [B]  time = 0.07, size = 61, normalized size = 0.87 \[ \frac {2\,x^{3/2}}{3\,b^2}+\frac {4\,a\,\sqrt {x}}{b^3}+\frac {a^2\,\sqrt {x}}{a\,b^3-b^4\,x}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a - b*x)^2,x)

[Out]

(2*x^(3/2))/(3*b^2) + (4*a*x^(1/2))/b^3 + (a^2*x^(1/2))/(a*b^3 - b^4*x) + (a^(3/2)*atan((b^(1/2)*x^(1/2)*1i)/a
^(1/2))*5i)/b^(7/2)

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sympy [A]  time = 24.75, size = 444, normalized size = 6.34 \[ \begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{2}} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b^{2}} & \text {for}\: a = 0 \\- \frac {30 a^{\frac {5}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {20 a^{\frac {3}{2}} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {4 \sqrt {a} b^{3} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {15 a^{3} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {15 a^{3} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} + \frac {15 a^{2} b x \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} - \frac {15 a^{2} b x \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{- 6 a^{\frac {3}{2}} b^{4} \sqrt {\frac {1}{b}} + 6 \sqrt {a} b^{5} x \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**2), Eq(b, 0)), (2*x**(3/2)/(3*b**2), Eq(a, 0)
), (-30*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(-6*a**(3/2)*b**4*sqrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)) + 20*a**(3/2)*b
**2*x**(3/2)*sqrt(1/b)/(-6*a**(3/2)*b**4*sqrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)) + 4*sqrt(a)*b**3*x**(5/2)*sqr
t(1/b)/(-6*a**(3/2)*b**4*sqrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)) - 15*a**3*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(
-6*a**(3/2)*b**4*sqrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)) + 15*a**3*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(-6*a**(3/
2)*b**4*sqrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)) + 15*a**2*b*x*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(-6*a**(3/2)*b
**4*sqrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)) - 15*a**2*b*x*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(-6*a**(3/2)*b**4*s
qrt(1/b) + 6*sqrt(a)*b**5*x*sqrt(1/b)), True))

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