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

Optimal. Leaf size=95 \[ \frac {5 a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{7/2}}-\frac {5 \sqrt {x} \sqrt {a-b x}}{b^3}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}+\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {47, 50, 63, 217, 203} \begin {gather*} -\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}-\frac {5 \sqrt {x} \sqrt {a-b x}}{b^3}+\frac {5 a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{7/2}}+\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a - b*x)^(5/2),x]

[Out]

(2*x^(5/2))/(3*b*(a - b*x)^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[a - b*x]) - (5*Sqrt[x]*Sqrt[a - b*x])/b^3 + (5*a*
ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{(a-b x)^{5/2}} \, dx &=\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}}-\frac {5 \int \frac {x^{3/2}}{(a-b x)^{3/2}} \, dx}{3 b}\\ &=\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx}{b^2}\\ &=\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}-\frac {5 \sqrt {x} \sqrt {a-b x}}{b^3}+\frac {(5 a) \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{2 b^3}\\ &=\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}-\frac {5 \sqrt {x} \sqrt {a-b x}}{b^3}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}-\frac {5 \sqrt {x} \sqrt {a-b x}}{b^3}+\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )}{b^3}\\ &=\frac {2 x^{5/2}}{3 b (a-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a-b x}}-\frac {5 \sqrt {x} \sqrt {a-b x}}{b^3}+\frac {5 a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.54 \begin {gather*} \frac {2 x^{7/2} \sqrt {1-\frac {b x}{a}} \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};\frac {b x}{a}\right )}{7 a^2 \sqrt {a-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a - b*x)^(5/2),x]

[Out]

(2*x^(7/2)*Sqrt[1 - (b*x)/a]*Hypergeometric2F1[5/2, 7/2, 9/2, (b*x)/a])/(7*a^2*Sqrt[a - b*x])

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IntegrateAlgebraic [A]  time = 0.20, size = 96, normalized size = 1.01 \begin {gather*} \frac {\sqrt {a-b x} \left (-15 a^2 \sqrt {x}+20 a b x^{3/2}-3 b^2 x^{5/2}\right )}{3 b^3 (b x-a)^2}+\frac {5 a \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/(a - b*x)^(5/2),x]

[Out]

(Sqrt[a - b*x]*(-15*a^2*Sqrt[x] + 20*a*b*x^(3/2) - 3*b^2*x^(5/2)))/(3*b^3*(-a + b*x)^2) + (5*a*Sqrt[-b]*Log[-(
Sqrt[-b]*Sqrt[x]) + Sqrt[a - b*x]])/b^4

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fricas [A]  time = 1.35, size = 215, normalized size = 2.26 \begin {gather*} \left [-\frac {15 \, {\left (a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) + 2 \, {\left (3 \, b^{3} x^{2} - 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{6 \, {\left (b^{6} x^{2} - 2 \, a b^{5} x + a^{2} b^{4}\right )}}, -\frac {15 \, {\left (a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (3 \, b^{3} x^{2} - 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{3 \, {\left (b^{6} x^{2} - 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(15*(a*b^2*x^2 - 2*a^2*b*x + a^3)*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) + 2*(3*b^
3*x^2 - 20*a*b^2*x + 15*a^2*b)*sqrt(-b*x + a)*sqrt(x))/(b^6*x^2 - 2*a*b^5*x + a^2*b^4), -1/3*(15*(a*b^2*x^2 -
2*a^2*b*x + a^3)*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + (3*b^3*x^2 - 20*a*b^2*x + 15*a^2*b)*sqrt(-
b*x + a)*sqrt(x))/(b^6*x^2 - 2*a*b^5*x + a^2*b^4)]

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giac [B]  time = 112.52, size = 221, normalized size = 2.33 \begin {gather*} \frac {{\left (\frac {15 \, a \log \left ({\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2}\right )}{\sqrt {-b} b^{2}} - \frac {6 \, \sqrt {{\left (b x - a\right )} b + a b} \sqrt {-b x + a}}{b^{3}} - \frac {8 \, {\left (9 \, a^{2} {\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{4} - 12 \, a^{3} {\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2} b + 7 \, a^{4} b^{2}\right )}}{{\left ({\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2} - a b\right )}^{3} \sqrt {-b} b}\right )} {\left | b \right |}}{6 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a*log((sqrt(-b*x + a)*sqrt(-b) - sqrt((b*x - a)*b + a*b))^2)/(sqrt(-b)*b^2) - 6*sqrt((b*x - a)*b + a*b
)*sqrt(-b*x + a)/b^3 - 8*(9*a^2*(sqrt(-b*x + a)*sqrt(-b) - sqrt((b*x - a)*b + a*b))^4 - 12*a^3*(sqrt(-b*x + a)
*sqrt(-b) - sqrt((b*x - a)*b + a*b))^2*b + 7*a^4*b^2)/(((sqrt(-b*x + a)*sqrt(-b) - sqrt((b*x - a)*b + a*b))^2
- a*b)^3*sqrt(-b)*b))*abs(b)/b^2

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maple [B]  time = 0.04, size = 160, normalized size = 1.68 \begin {gather*} \frac {\left (\frac {5 a \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{2 b^{\frac {7}{2}}}+\frac {2 \sqrt {-\left (x -\frac {a}{b}\right ) a -\left (x -\frac {a}{b}\right )^{2} b}\, a^{2}}{3 \left (x -\frac {a}{b}\right )^{2} b^{5}}+\frac {14 \sqrt {-\left (x -\frac {a}{b}\right ) a -\left (x -\frac {a}{b}\right )^{2} b}\, a}{3 \left (x -\frac {a}{b}\right ) b^{4}}\right ) \sqrt {\left (-b x +a \right ) x}}{\sqrt {-b x +a}\, \sqrt {x}}-\frac {\sqrt {-b x +a}\, \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^(1/2)*(-b*x+a)^(1/2)/b^3+(5/2/b^(7/2)*a*arctan((x-1/2*a/b)/(-b*x^2+a*x)^(1/2)*b^(1/2))+2/3/b^5*a^2/(x-a/b)^
2*(-(x-a/b)*a-(x-a/b)^2*b)^(1/2)+14/3/b^4*a/(x-a/b)*(-(x-a/b)*a-(x-a/b)^2*b)^(1/2))*((-b*x+a)*x)^(1/2)/(-b*x+a
)^(1/2)/x^(1/2)

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maxima [A]  time = 3.02, size = 94, normalized size = 0.99 \begin {gather*} \frac {2 \, a b^{2} + \frac {10 \, {\left (b x - a\right )} a b}{x} - \frac {15 \, {\left (b x - a\right )}^{2} a}{x^{2}}}{3 \, {\left (\frac {{\left (-b x + a\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (-b x + a\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} - \frac {5 \, a \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*a*b^2 + 10*(b*x - a)*a*b/x - 15*(b*x - a)^2*a/x^2)/((-b*x + a)^(3/2)*b^4/x^(3/2) + (-b*x + a)^(5/2)*b^3
/x^(5/2)) - 5*a*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}}{{\left (a-b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a - b*x)^(5/2),x)

[Out]

int(x^(5/2)/(a - b*x)^(5/2), x)

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sympy [B]  time = 8.48, size = 971, normalized size = 10.22

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-30*I*a**(81/2)*b**22*x**(51/2)*sqrt(-1 + b*x/a)*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**(79/2)*b**(51
/2)*x**(51/2)*sqrt(-1 + b*x/a) - 6*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(-1 + b*x/a)) + 15*pi*a**(81/2)*b**22*x**
(51/2)*sqrt(-1 + b*x/a)/(6*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(-1 + b*x/a) - 6*a**(77/2)*b**(53/2)*x**(53/2)*sq
rt(-1 + b*x/a)) + 30*I*a**(79/2)*b**23*x**(53/2)*sqrt(-1 + b*x/a)*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**(79/2)*
b**(51/2)*x**(51/2)*sqrt(-1 + b*x/a) - 6*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(-1 + b*x/a)) - 15*pi*a**(79/2)*b**
23*x**(53/2)*sqrt(-1 + b*x/a)/(6*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(-1 + b*x/a) - 6*a**(77/2)*b**(53/2)*x**(53
/2)*sqrt(-1 + b*x/a)) + 30*I*a**40*b**(45/2)*x**26/(6*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(-1 + b*x/a) - 6*a**(7
7/2)*b**(53/2)*x**(53/2)*sqrt(-1 + b*x/a)) - 40*I*a**39*b**(47/2)*x**27/(6*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(
-1 + b*x/a) - 6*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(-1 + b*x/a)) + 6*I*a**38*b**(49/2)*x**28/(6*a**(79/2)*b**(5
1/2)*x**(51/2)*sqrt(-1 + b*x/a) - 6*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (15*a**(
81/2)*b**22*x**(51/2)*sqrt(1 - b*x/a)*asin(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 -
b*x/a) - 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 - b*x/a)) - 15*a**(79/2)*b**23*x**(53/2)*sqrt(1 - b*x/a)*asin(
sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 - b*x/a) - 3*a**(77/2)*b**(53/2)*x**(53/2)*sq
rt(1 - b*x/a)) - 15*a**40*b**(45/2)*x**26/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 - b*x/a) - 3*a**(77/2)*b**(5
3/2)*x**(53/2)*sqrt(1 - b*x/a)) + 20*a**39*b**(47/2)*x**27/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 - b*x/a) -
3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 - b*x/a)) - 3*a**38*b**(49/2)*x**28/(3*a**(79/2)*b**(51/2)*x**(51/2)*sq
rt(1 - b*x/a) - 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 - b*x/a)), True))

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