3.4.4 \(\int \frac {(a+b x)^{5/2}}{x} \, dx\) [304]

Optimal. Leaf size=65 \[ 2 a^2 \sqrt {a+b x}+\frac {2}{3} a (a+b x)^{3/2}+\frac {2}{5} (a+b x)^{5/2}-2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {52, 65, 214} \begin {gather*} -2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 a^2 \sqrt {a+b x}+\frac {2}{3} a (a+b x)^{3/2}+\frac {2}{5} (a+b x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x,x]

[Out]

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

Rule 52

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 65

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 214

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 56, normalized size = 0.86 \begin {gather*} \frac {2}{15} \sqrt {a+b x} \left (23 a^2+11 a b x+3 b^2 x^2\right )-2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x,x]

[Out]

(2*Sqrt[a + b*x]*(23*a^2 + 11*a*b*x + 3*b^2*x^2))/15 - 2*a^(5/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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Maple [A]
time = 0.09, size = 50, normalized size = 0.77

method	result	size
derivativedivides	\(\frac {2 a \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}-2 a^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a^{2} \sqrt {b x +a}\)	\(50\)
default	\(\frac {2 a \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}-2 a^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a^{2} \sqrt {b x +a}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 64, normalized size = 0.98 \begin {gather*} a^{\frac {5}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} a + 2 \, \sqrt {b x + a} a^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.49, size = 114, normalized size = 1.75 \begin {gather*} \left [a^{\frac {5}{2}} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x + a}, 2 \, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \frac {2}{15} \, {\left (3 \, b^{2} x^{2} + 11 \, a b x + 23 \, a^{2}\right )} \sqrt {b x + a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[a^(5/2)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2/15*(3*b^2*x^2 + 11*a*b*x + 23*a^2)*sqrt(b*x + a), 2*
sqrt(-a)*a^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + 2/15*(3*b^2*x^2 + 11*a*b*x + 23*a^2)*sqrt(b*x + a)]

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Sympy [A]
time = 2.18, size = 97, normalized size = 1.49 \begin {gather*} \frac {46 a^{\frac {5}{2}} \sqrt {1 + \frac {b x}{a}}}{15} + a^{\frac {5}{2}} \log {\left (\frac {b x}{a} \right )} - 2 a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )} + \frac {22 a^{\frac {3}{2}} b x \sqrt {1 + \frac {b x}{a}}}{15} + \frac {2 \sqrt {a} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

46*a**(5/2)*sqrt(1 + b*x/a)/15 + a**(5/2)*log(b*x/a) - 2*a**(5/2)*log(sqrt(1 + b*x/a) + 1) + 22*a**(3/2)*b*x*s
qrt(1 + b*x/a)/15 + 2*sqrt(a)*b**2*x**2*sqrt(1 + b*x/a)/5

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Giac [A]
time = 0.97, size = 56, normalized size = 0.86 \begin {gather*} \frac {2 \, a^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} a + 2 \, \sqrt {b x + a} a^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 52, normalized size = 0.80 \begin {gather*} \frac {2\,a\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {2\,{\left (a+b\,x\right )}^{5/2}}{5}+2\,a^2\,\sqrt {a+b\,x}+a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/x,x)

[Out]

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

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