3.6.58 \(\int x^{3/2} (2+b x)^{5/2} \, dx\) [558]

Optimal. Leaf size=123 \[ -\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}} \]

[Out]

1/4*x^(5/2)*(b*x+2)^(3/2)+1/5*x^(5/2)*(b*x+2)^(5/2)+3/4*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)+1/8*x^(3/
2)*(b*x+2)^(1/2)/b+1/4*x^(5/2)*(b*x+2)^(1/2)-3/8*x^(1/2)*(b*x+2)^(1/2)/b^2

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Rubi [A]
time = 0.02, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}}-\frac {3 \sqrt {x} \sqrt {b x+2}}{8 b^2}+\frac {1}{5} x^{5/2} (b x+2)^{5/2}+\frac {1}{4} x^{5/2} (b x+2)^{3/2}+\frac {1}{4} x^{5/2} \sqrt {b x+2}+\frac {x^{3/2} \sqrt {b x+2}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(2 + b*x)^(5/2),x]

[Out]

(-3*Sqrt[x]*Sqrt[2 + b*x])/(8*b^2) + (x^(3/2)*Sqrt[2 + b*x])/(8*b) + (x^(5/2)*Sqrt[2 + b*x])/4 + (x^(5/2)*(2 +
b*x)^(3/2))/4 + (x^(5/2)*(2 + b*x)^(5/2))/5 + (3*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(5/2))

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^{3/2} (2+b x)^{5/2} \, dx &=\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\int x^{3/2} (2+b x)^{3/2} \, dx\\ &=\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\frac {3}{4} \int x^{3/2} \sqrt {2+b x} \, dx\\ &=\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\frac {1}{4} \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx\\ &=\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}-\frac {3 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{8 b}\\ &=-\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{8 b^2}\\ &=-\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^2}\\ &=-\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{5} x^{5/2} (2+b x)^{5/2}+\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 84, normalized size = 0.68 \begin {gather*} \frac {\sqrt {x} \sqrt {2+b x} \left (-15+5 b x+62 b^2 x^2+42 b^3 x^3+8 b^4 x^4\right )}{40 b^2}-\frac {3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{4 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(2 + b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(-15 + 5*b*x + 62*b^2*x^2 + 42*b^3*x^3 + 8*b^4*x^4))/(40*b^2) - (3*Log[-(Sqrt[b]*Sqrt[x
]) + Sqrt[2 + b*x]])/(4*b^(5/2))

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Maple [A]
time = 0.11, size = 126, normalized size = 1.02

method	result	size
meijerg	\(-\frac {60 \left (\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-8 b^{4} x^{4}-42 b^{3} x^{3}-62 x^{2} b^{2}-5 b x +15\right ) \sqrt {\frac {b x}{2}+1}}{2400}-\frac {\sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{80}\right )}{b^{\frac {5}{2}} \sqrt {\pi }}\)	\(79\)
risch	\(\frac {\left (8 b^{4} x^{4}+42 b^{3} x^{3}+62 x^{2} b^{2}+5 b x -15\right ) \sqrt {x}\, \sqrt {b x +2}}{40 b^{2}}+\frac {3 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right ) \sqrt {x \left (b x +2\right )}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(93\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {7}{2}}}{5 b}-\frac {3 \left (\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {7}{2}}}{4 b}-\frac {\frac {\left (b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {5 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{4 b}\right )}{5 b}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x+2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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. 194 vs. \(2 (84) = 168\).
time = 0.51, size = 194, normalized size = 1.58 \begin {gather*} -\frac {\frac {15 \, \sqrt {b x + 2} b^{4}}{\sqrt {x}} - \frac {70 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{3}}{x^{\frac {3}{2}}} + \frac {128 \, {\left (b x + 2\right )}^{\frac {5}{2}} b^{2}}{x^{\frac {5}{2}}} + \frac {70 \, {\left (b x + 2\right )}^{\frac {7}{2}} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (b x + 2\right )}^{\frac {9}{2}}}{x^{\frac {9}{2}}}}{20 \, {\left (b^{7} - \frac {5 \, {\left (b x + 2\right )} b^{6}}{x} + \frac {10 \, {\left (b x + 2\right )}^{2} b^{5}}{x^{2}} - \frac {10 \, {\left (b x + 2\right )}^{3} b^{4}}{x^{3}} + \frac {5 \, {\left (b x + 2\right )}^{4} b^{3}}{x^{4}} - \frac {{\left (b x + 2\right )}^{5} b^{2}}{x^{5}}\right )}} - \frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{8 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="maxima")

[Out]

-1/20*(15*sqrt(b*x + 2)*b^4/sqrt(x) - 70*(b*x + 2)^(3/2)*b^3/x^(3/2) + 128*(b*x + 2)^(5/2)*b^2/x^(5/2) + 70*(b
*x + 2)^(7/2)*b/x^(7/2) - 15*(b*x + 2)^(9/2)/x^(9/2))/(b^7 - 5*(b*x + 2)*b^6/x + 10*(b*x + 2)^2*b^5/x^2 - 10*(
b*x + 2)^3*b^4/x^3 + 5*(b*x + 2)^4*b^3/x^4 - (b*x + 2)^5*b^2/x^5) - 3/8*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))
/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(5/2)

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Fricas [A]
time = 0.57, size = 155, normalized size = 1.26 \begin {gather*} \left [\frac {{\left (8 \, b^{5} x^{4} + 42 \, b^{4} x^{3} + 62 \, 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 )}{40 \, b^{3}}, \frac {{\left (8 \, b^{5} x^{4} + 42 \, b^{4} x^{3} + 62 \, 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 )}{40 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(b*x+2)^(5/2),x, algorithm="fricas")

[Out]

[1/40*((8*b^5*x^4 + 42*b^4*x^3 + 62*b^3*x^2 + 5*b^2*x - 15*b)*sqrt(b*x + 2)*sqrt(x) + 15*sqrt(b)*log(b*x + sqr
t(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b^3, 1/40*((8*b^5*x^4 + 42*b^4*x^3 + 62*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^3]

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Sympy [A]
time = 32.70, size = 138, normalized size = 1.12 \begin {gather*} \frac {b^{3} x^{\frac {11}{2}}}{5 \sqrt {b x + 2}} + \frac {29 b^{2} x^{\frac {9}{2}}}{20 \sqrt {b x + 2}} + \frac {73 b x^{\frac {7}{2}}}{20 \sqrt {b x + 2}} + \frac {129 x^{\frac {5}{2}}}{40 \sqrt {b x + 2}} - \frac {x^{\frac {3}{2}}}{8 b \sqrt {b x + 2}} - \frac {3 \sqrt {x}}{4 b^{2} \sqrt {b x + 2}} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(b*x+2)**(5/2),x)

[Out]

b**3*x**(11/2)/(5*sqrt(b*x + 2)) + 29*b**2*x**(9/2)/(20*sqrt(b*x + 2)) + 73*b*x**(7/2)/(20*sqrt(b*x + 2)) + 12
9*x**(5/2)/(40*sqrt(b*x + 2)) - x**(3/2)/(8*b*sqrt(b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(b*x + 2)) + 3*asinh(sqrt
(2)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2))

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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(x^(3/2)*(b*x+2)^(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 x^{3/2}\,{\left (b\,x+2\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x + 2)^(5/2),x)

[Out]

int(x^(3/2)*(b*x + 2)^(5/2), x)

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