3.6.57 \(\int x^{5/2} (2+b x)^{5/2} \, dx\) [557]

Optimal. Leaf size=144 \[ \frac {5 \sqrt {x} \sqrt {2+b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{48 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}-\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}} \]

[Out]

1/6*x^(7/2)*(b*x+2)^(3/2)+1/6*x^(7/2)*(b*x+2)^(5/2)-5/8*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-5/48*x^(3
/2)*(b*x+2)^(1/2)/b^2+1/24*x^(5/2)*(b*x+2)^(1/2)/b+1/8*x^(7/2)*(b*x+2)^(1/2)+5/16*x^(1/2)*(b*x+2)^(1/2)/b^3

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Rubi [A]
time = 0.03, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \begin {gather*} -\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}+\frac {5 \sqrt {x} \sqrt {b x+2}}{16 b^3}-\frac {5 x^{3/2} \sqrt {b x+2}}{48 b^2}+\frac {1}{6} x^{7/2} (b x+2)^{5/2}+\frac {1}{6} x^{7/2} (b x+2)^{3/2}+\frac {1}{8} x^{7/2} \sqrt {b x+2}+\frac {x^{5/2} \sqrt {b x+2}}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(2 + b*x)^(5/2),x]

[Out]

(5*Sqrt[x]*Sqrt[2 + b*x])/(16*b^3) - (5*x^(3/2)*Sqrt[2 + b*x])/(48*b^2) + (x^(5/2)*Sqrt[2 + b*x])/(24*b) + (x^
(7/2)*Sqrt[2 + b*x])/8 + (x^(7/2)*(2 + b*x)^(3/2))/6 + (x^(7/2)*(2 + b*x)^(5/2))/6 - (5*ArcSinh[(Sqrt[b]*Sqrt[
x])/Sqrt[2]])/(8*b^(7/2))

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 x^{5/2} (2+b x)^{5/2} \, dx &=\frac {1}{6} x^{7/2} (2+b x)^{5/2}+\frac {5}{6} \int x^{5/2} (2+b x)^{3/2} \, dx\\ &=\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}+\frac {1}{2} \int x^{5/2} \sqrt {2+b x} \, dx\\ &=\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}+\frac {1}{8} \int \frac {x^{5/2}}{\sqrt {2+b x}} \, dx\\ &=\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}-\frac {5 \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{24 b}\\ &=-\frac {5 x^{3/2} \sqrt {2+b x}}{48 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{16 b^2}\\ &=\frac {5 \sqrt {x} \sqrt {2+b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{48 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}-\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{16 b^3}\\ &=\frac {5 \sqrt {x} \sqrt {2+b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{48 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}-\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^3}\\ &=\frac {5 \sqrt {x} \sqrt {2+b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{48 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2+b x}+\frac {1}{6} x^{7/2} (2+b x)^{3/2}+\frac {1}{6} x^{7/2} (2+b x)^{5/2}-\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 92, normalized size = 0.64 \begin {gather*} \frac {\sqrt {x} \sqrt {2+b x} \left (15-5 b x+2 b^2 x^2+54 b^3 x^3+40 b^4 x^4+8 b^5 x^5\right )}{48 b^3}+\frac {5 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{8 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(2 + b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(15 - 5*b*x + 2*b^2*x^2 + 54*b^3*x^3 + 40*b^4*x^4 + 8*b^5*x^5))/(48*b^3) + (5*Log[-(Sqr
t[b]*Sqrt[x]) + Sqrt[2 + b*x]])/(8*b^(7/2))

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Mathics [A]
time = 171.19, size = 119, normalized size = 0.83 \begin {gather*} \frac {-30 b^6 \text {ArcSinh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (2+b x\right )^2+30 b^{\frac {13}{2}} \sqrt {x} \left (2+b x\right )^{\frac {3}{2}}+5 b^{\frac {15}{2}} x^{\frac {3}{2}} \left (2+b x\right )^{\frac {3}{2}}-b^{\frac {17}{2}} x^{\frac {5}{2}} \left (2+b x\right )^{\frac {3}{2}}+2 b^{\frac {19}{2}} x^{\frac {7}{2}} \left (55+67 b x+28 b^2 x^2+4 b^3 x^3\right ) \left (2+b x\right )^{\frac {3}{2}}}{48 b^{\frac {19}{2}} \left (2+b x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^(5/2)*(2 + b*x)^(5/2),x]')

[Out]

(-30 b ^ 6 ArcSinh[Sqrt[2] Sqrt[b] Sqrt[x] / 2] (2 + b x) ^ 2 + 30 b ^ (13 / 2) Sqrt[x] (2 + b x) ^ (3 / 2) +
5 b ^ (15 / 2) x ^ (3 / 2) (2 + b x) ^ (3 / 2) - b ^ (17 / 2) x ^ (5 / 2) (2 + b x) ^ (3 / 2) + 2 b ^ (19 / 2)
x ^ (7 / 2) (55 + 67 b x + 28 b ^ 2 x ^ 2 + 4 b ^ 3 x ^ 3) (2 + b x) ^ (3 / 2)) / (48 b ^ (19 / 2) (2 + b x)
^ 2)

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Maple [A]
time = 0.11, size = 147, normalized size = 1.02

method	result	size
meijerg	\(-\frac {120 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (56 b^{5} x^{5}+280 b^{4} x^{4}+378 b^{3} x^{3}+14 x^{2} b^{2}-35 b x +105\right ) \sqrt {\frac {b x}{2}+1}}{40320}+\frac {\sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{192}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\)	\(87\)
risch	\(\frac {\left (8 b^{5} x^{5}+40 b^{4} x^{4}+54 b^{3} x^{3}+2 x^{2} b^{2}-5 b x +15\right ) \sqrt {x}\, \sqrt {b x +2}}{48 b^{3}}-\frac {5 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{16 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(101\)
default	\(\frac {x^{\frac {5}{2}} \left (b x +2\right )^{\frac {7}{2}}}{6 b}-\frac {5 \left (\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {7}{2}}}{5 b}-\frac {3 \left (\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {7}{2}}}{4 b}-\frac {\frac {\left (b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {5 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{4 b}\right )}{5 b}\right )}{6 b}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/b*x^(5/2)*(b*x+2)^(7/2)-5/6/b*(1/5/b*x^(3/2)*(b*x+2)^(7/2)-3/5/b*(1/4/b*x^(1/2)*(b*x+2)^(7/2)-1/4/b*(1/3*(
b*x+2)^(5/2)*x^(1/2)+5/6*(b*x+2)^(3/2)*x^(1/2)+5/2*x^(1/2)*(b*x+2)^(1/2)+5/2*(x*(b*x+2))^(1/2)/(b*x+2)^(1/2)/x
^(1/2)*ln((b*x+1)/b^(1/2)+(b*x^2+2*x)^(1/2))/b^(1/2))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (99) = 198\).
time = 0.35, size = 223, normalized size = 1.55 \begin {gather*} \frac {\frac {15 \, \sqrt {b x + 2} b^{5}}{\sqrt {x}} - \frac {85 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {198 \, {\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}} + \frac {198 \, {\left (b x + 2\right )}^{\frac {7}{2}} b^{2}}{x^{\frac {7}{2}}} - \frac {85 \, {\left (b x + 2\right )}^{\frac {9}{2}} b}{x^{\frac {9}{2}}} + \frac {15 \, {\left (b x + 2\right )}^{\frac {11}{2}}}{x^{\frac {11}{2}}}}{24 \, {\left (b^{9} - \frac {6 \, {\left (b x + 2\right )} b^{8}}{x} + \frac {15 \, {\left (b x + 2\right )}^{2} b^{7}}{x^{2}} - \frac {20 \, {\left (b x + 2\right )}^{3} b^{6}}{x^{3}} + \frac {15 \, {\left (b x + 2\right )}^{4} b^{5}}{x^{4}} - \frac {6 \, {\left (b x + 2\right )}^{5} b^{4}}{x^{5}} + \frac {{\left (b x + 2\right )}^{6} b^{3}}{x^{6}}\right )}} + \frac {5 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{16 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*sqrt(b*x + 2)*b^5/sqrt(x) - 85*(b*x + 2)^(3/2)*b^4/x^(3/2) + 198*(b*x + 2)^(5/2)*b^3/x^(5/2) + 198*(b
*x + 2)^(7/2)*b^2/x^(7/2) - 85*(b*x + 2)^(9/2)*b/x^(9/2) + 15*(b*x + 2)^(11/2)/x^(11/2))/(b^9 - 6*(b*x + 2)*b^
8/x + 15*(b*x + 2)^2*b^7/x^2 - 20*(b*x + 2)^3*b^6/x^3 + 15*(b*x + 2)^4*b^5/x^4 - 6*(b*x + 2)^5*b^4/x^5 + (b*x
+ 2)^6*b^3/x^6) + 5/16*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(7/2)

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Fricas [A]
time = 0.31, size = 172, normalized size = 1.19 \begin {gather*} \left [\frac {{\left (8 \, b^{6} x^{5} + 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 15 \, \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{48 \, b^{4}}, \frac {{\left (8 \, b^{6} x^{5} + 40 \, b^{5} x^{4} + 54 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 30 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{48 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*((8*b^6*x^5 + 40*b^5*x^4 + 54*b^4*x^3 + 2*b^3*x^2 - 5*b^2*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 15*sqrt(b)*l
og(b*x - sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b^4, 1/48*((8*b^6*x^5 + 40*b^5*x^4 + 54*b^4*x^3 + 2*b^3*x^2 - 5*b
^2*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 30*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))))/b^4]

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Sympy [A]
time = 176.88, size = 158, normalized size = 1.10 \begin {gather*} \frac {b^{3} x^{\frac {13}{2}}}{6 \sqrt {b x + 2}} + \frac {7 b^{2} x^{\frac {11}{2}}}{6 \sqrt {b x + 2}} + \frac {67 b x^{\frac {9}{2}}}{24 \sqrt {b x + 2}} + \frac {55 x^{\frac {7}{2}}}{24 \sqrt {b x + 2}} - \frac {x^{\frac {5}{2}}}{48 b \sqrt {b x + 2}} + \frac {5 x^{\frac {3}{2}}}{48 b^{2} \sqrt {b x + 2}} + \frac {5 \sqrt {x}}{8 b^{3} \sqrt {b x + 2}} - \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{8 b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**3*x**(13/2)/(6*sqrt(b*x + 2)) + 7*b**2*x**(11/2)/(6*sqrt(b*x + 2)) + 67*b*x**(9/2)/(24*sqrt(b*x + 2)) + 55*
x**(7/2)/(24*sqrt(b*x + 2)) - x**(5/2)/(48*b*sqrt(b*x + 2)) + 5*x**(3/2)/(48*b**2*sqrt(b*x + 2)) + 5*sqrt(x)/(
8*b**3*sqrt(b*x + 2)) - 5*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(8*b**(7/2))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (99) = 198\).
time = 0.02, size = 500, normalized size = 3.47 \begin {gather*} 2 b^{2} \left (2 \left (\left (\left (\left (\left (\frac {\frac {1}{5443200}\cdot 226800 b^{10} \sqrt {x} \sqrt {x}}{b^{10}}+\frac {\frac {1}{5443200}\cdot 45360 b^{9}}{b^{10}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{5443200}\cdot 102060 b^{8}}{b^{10}}\right ) \sqrt {x} \sqrt {x}+\frac {\frac {1}{5443200}\cdot 238140 b^{7}}{b^{10}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{5443200}\cdot 595350 b^{6}}{b^{10}}\right ) \sqrt {x} \sqrt {x}+\frac {\frac {1}{5443200}\cdot 1786050 b^{5}}{b^{10}}\right ) \sqrt {x} \sqrt {b x+2}+\frac {21 \ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{16 b^{5} \sqrt {b}}\right )+8 b \left (2 \left (\left (\left (\left (\frac {\frac {1}{100800}\cdot 5040 b^{8} \sqrt {x} \sqrt {x}}{b^{8}}+\frac {\frac {1}{100800}\cdot 1260 b^{7}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 2940 b^{6}}{b^{8}}\right ) \sqrt {x} \sqrt {x}+\frac {\frac {1}{100800}\cdot 7350 b^{5}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 22050 b^{4}}{b^{8}}\right ) \sqrt {x} \sqrt {b x+2}-\frac {7 \ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{8 b^{4} \sqrt {b}}\right )+8 \left (2 \left (\left (\left (\frac {\frac {1}{2880}\cdot 180 b^{6} \sqrt {x} \sqrt {x}}{b^{6}}+\frac {\frac {1}{2880}\cdot 60 b^{5}}{b^{6}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{2880}\cdot 150 b^{4}}{b^{6}}\right ) \sqrt {x} \sqrt {x}+\frac {\frac {1}{2880}\cdot 450 b^{3}}{b^{6}}\right ) \sqrt {x} \sqrt {b x+2}+\frac {5 \ln \left (\sqrt {b x+2}-\sqrt {b} \sqrt {x}\right )}{8 b^{3} \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(((2*((4*(5*x + 1/b)*x - 9/b^2)*x + 21/b^3)*x - 105/b^4)*x + 315/b^5)*sqrt(b*x + 2)*sqrt(x) + 630*log(-s
qrt(b)*sqrt(x) + sqrt(b*x + 2))/b^(11/2))*b^2 + 1/30*(((2*(3*(4*x + 1/b)*x - 7/b^2)*x + 35/b^3)*x - 105/b^4)*s
qrt(b*x + 2)*sqrt(x) - 210*log(-sqrt(b)*sqrt(x) + sqrt(b*x + 2))/b^(9/2))*b + 1/6*((2*(3*x + 1/b)*x - 5/b^2)*x
+ 15/b^3)*sqrt(b*x + 2)*sqrt(x) + 5*log(-sqrt(b)*sqrt(x) + sqrt(b*x + 2))/b^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x + 2)^(5/2),x)

[Out]

int(x^(5/2)*(b*x + 2)^(5/2), x)

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