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

Optimal. Leaf size=91 \[ -\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{7/2}}+\frac {5 \sqrt {x} \sqrt {a+b x}}{b^3}-\frac {10 x^{3/2}}{3 b^2 \sqrt {a+b x}}-\frac {2 x^{5/2}}{3 b (a+b x)^{3/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {47, 50, 63, 217, 206} \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 \tanh ^{-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
*ArcTanh[(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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 \tanh ^{-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 = 50, normalized size = 0.55 \begin {gather*} \frac {2 x^{7/2} \sqrt {\frac {b x}{a}+1} \, _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 = 0.15, size = 78, normalized size = 0.86 \begin {gather*} \frac {15 a^2 \sqrt {x}+20 a b x^{3/2}+3 b^2 x^{5/2}}{3 b^3 (a+b x)^{3/2}}+\frac {5 a \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^2*Sqrt[x] + 20*a*b*x^(3/2) + 3*b^2*x^(5/2))/(3*b^3*(a + b*x)^(3/2)) + (5*a*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt
[a + b*x]])/b^(7/2)

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fricas [A]  time = 1.42, size = 214, normalized size = 2.35 \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}}{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)/(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 = 92.46, size = 197, normalized size = 2.16 \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 )}{b^{\frac {5}{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} \sqrt {b} + 12 \, a^{3} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {3}{2}} + 7 \, a^{4} b^{\frac {5}{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} b^{2}}\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)/b^(5/2) + 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*sqrt(b) + 12*a^3*(sqrt(b*x + a)*sqrt
(b) - sqrt((b*x + a)*b - a*b))^2*b^(3/2) + 7*a^4*b^(5/2))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^
2 + a*b)^3*b^2))*abs(b)/b^2

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maple [B]  time = 0.05, size = 147, normalized size = 1.62 \begin {gather*} \frac {\left (-\frac {5 a \ln \left (\frac {b x +\frac {a}{2}}{\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*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))+14/3/b^4*a/(x+a/b)*(-(x+a/
b)*a+(x+a/b)^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))*((b*x+a)*x)^(1/2)/(b*x+a)^(1/2)/
x^(1/2)

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maxima [A]  time = 2.94, size = 109, normalized size = 1.20 \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 \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{2 \, 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)) + 5/2*a*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/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 = 7.61, size = 396, normalized size = 4.35 \begin {gather*} - \frac {15 a^{\frac {81}{2}} b^{22} x^{\frac {51}{2}} \sqrt {1 + \frac {b x}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{3 a^{\frac {79}{2}} b^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {b x}{a}}} - \frac {15 a^{\frac {79}{2}} b^{23} x^{\frac {53}{2}} \sqrt {1 + \frac {b x}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{3 a^{\frac {79}{2}} b^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{40} b^{\frac {45}{2}} x^{26}}{3 a^{\frac {79}{2}} b^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {b x}{a}}} + \frac {20 a^{39} b^{\frac {47}{2}} x^{27}}{3 a^{\frac {79}{2}} b^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {b x}{a}}} + \frac {3 a^{38} b^{\frac {49}{2}} x^{28}}{3 a^{\frac {79}{2}} b^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*a**(81/2)*b**22*x**(51/2)*sqrt(1 + b*x/a)*asinh(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)*asinh(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**40*b**(45/2)*x**26/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 + b*x/a) + 3*a**(7
7/2)*b**(53/2)*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)*sqrt(1 + b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a))

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