3.6.62 \(\int \frac {(2+b x)^{5/2}}{x^{5/2}} \, dx\) [562]

Optimal. Leaf size=81 \[ 5 b^2 \sqrt {x} \sqrt {2+b x}-\frac {10 b (2+b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2+b x)^{5/2}}{3 x^{3/2}}+10 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

-2/3*(b*x+2)^(5/2)/x^(3/2)+10*b^(3/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))-10/3*b*(b*x+2)^(3/2)/x^(1/2)+5*b^2*
x^(1/2)*(b*x+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 81, 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 = {49, 52, 56, 221} \begin {gather*} 10 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )+5 b^2 \sqrt {x} \sqrt {b x+2}-\frac {2 (b x+2)^{5/2}}{3 x^{3/2}}-\frac {10 b (b x+2)^{3/2}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + b*x)^(5/2)/x^(5/2),x]

[Out]

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

Rule 49

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 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 56

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

Rule 221

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 63, normalized size = 0.78 \begin {gather*} \frac {\sqrt {2+b x} \left (-8-28 b x+3 b^2 x^2\right )}{3 x^{3/2}}-10 b^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + b*x)^(5/2)/x^(5/2),x]

[Out]

(Sqrt[2 + b*x]*(-8 - 28*b*x + 3*b^2*x^2))/(3*x^(3/2)) - 10*b^(3/2)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]]

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Maple [A]
time = 0.14, size = 63, normalized size = 0.78

method	result	size
meijerg	\(-\frac {15 b^{\frac {3}{2}} \left (\frac {32 \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {3}{8} x^{2} b^{2}+\frac {7}{2} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{45 x^{\frac {3}{2}} b^{\frac {3}{2}}}-\frac {8 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{3}\right )}{4 \sqrt {\pi }}\)	\(63\)
risch	\(\frac {3 b^{3} x^{3}-22 x^{2} b^{2}-64 b x -16}{3 x^{\frac {3}{2}} \sqrt {b x +2}}+\frac {5 b^{\frac {3}{2}} \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*b^(3/2)/Pi^(1/2)*(32/45*Pi^(1/2)/x^(3/2)*2^(1/2)/b^(3/2)*(-3/8*x^2*b^2+7/2*b*x+1)*(1/2*b*x+1)^(1/2)-8/3*
Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [A]
time = 0.50, size = 96, normalized size = 1.19 \begin {gather*} -5 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {8 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {2 \, \sqrt {b x + 2} b^{2}}{{\left (b - \frac {b x + 2}{x}\right )} \sqrt {x}} - \frac {4 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*b^(3/2)*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x))) - 8*sqrt(b*x + 2)*b/sqrt(
x) - 2*sqrt(b*x + 2)*b^2/((b - (b*x + 2)/x)*sqrt(x)) - 4/3*(b*x + 2)^(3/2)/x^(3/2)

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Fricas [A]
time = 0.79, size = 123, normalized size = 1.52 \begin {gather*} \left [\frac {15 \, b^{\frac {3}{2}} x^{2} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (3 \, b^{2} x^{2} - 28 \, b x - 8\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}, -\frac {30 \, \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} - 28 \, b x - 8\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(15*b^(3/2)*x^2*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (3*b^2*x^2 - 28*b*x - 8)*sqrt(b*x + 2)*sqr
t(x))/x^2, -1/3*(30*sqrt(-b)*b*x^2*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) - (3*b^2*x^2 - 28*b*x - 8)*sqrt(
b*x + 2)*sqrt(x))/x^2]

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Sympy [A]
time = 3.42, size = 88, normalized size = 1.09 \begin {gather*} b^{\frac {5}{2}} x \sqrt {1 + \frac {2}{b x}} - \frac {28 b^{\frac {3}{2}} \sqrt {1 + \frac {2}{b x}}}{3} - 5 b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} + 10 b^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {2}{b x}} + 1 \right )} - \frac {8 \sqrt {b} \sqrt {1 + \frac {2}{b x}}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**(5/2)*x*sqrt(1 + 2/(b*x)) - 28*b**(3/2)*sqrt(1 + 2/(b*x))/3 - 5*b**(3/2)*log(1/(b*x)) + 10*b**(3/2)*log(sqr
t(1 + 2/(b*x)) + 1) - 8*sqrt(b)*sqrt(1 + 2/(b*x))/(3*x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,1]%%%}+%%%{-4,[1,0]%%%}+%%%{-4,[0,1]%%%}+%%%{-8,[0,0]%%%},0,%%%{6,[2,2]%%%}+%%%{4,[2,1]%%%}+%%%{6,[2,0]%%%}+%
%%{4,[1,2]%%%}+%%%{28

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + 2)^(5/2)/x^(5/2),x)

[Out]

int((b*x + 2)^(5/2)/x^(5/2), x)

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