3.7.42 \(\int \frac {x^{3/2}}{(2-b x)^{5/2}} \, dx\) [642]

Optimal. Leaf size=67 \[ \frac {2 x^{3/2}}{3 b (2-b x)^{3/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {2-b x}}+\frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {49, 56, 222} \begin {gather*} \frac {2 \text {ArcSin}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {2-b x}}+\frac {2 x^{3/2}}{3 b (2-b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(2 - b*x)^(5/2),x]

[Out]

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

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 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 222

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

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 64, normalized size = 0.96 \begin {gather*} \frac {4 \sqrt {x} (-3+2 b x)}{3 b^2 (2-b x)^{3/2}}-\frac {2 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{(-b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(2 - b*x)^(5/2),x]

[Out]

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

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Maple [A]
time = 0.13, size = 73, normalized size = 1.09

method result size
meijerg \(-\frac {4 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {5}{2}} \left (-10 b x +15\right )}{20 b^{2} \left (-\frac {b x}{2}+1\right )^{\frac {3}{2}}}+\frac {3 \sqrt {\pi }\, \left (-b \right )^{\frac {5}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{2 b^{\frac {5}{2}}}\right )}{3 \left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, b}\) \(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3/(-b)^(3/2)/Pi^(1/2)/b*(-1/20*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(5/2)*(-10*b*x+15)/b^2/(-1/2*b*x+1)^(3/2)+3/2*
Pi^(1/2)*(-b)^(5/2)/b^(5/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [A]
time = 0.50, size = 50, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (b + \frac {3 \, {\left (b x - 2\right )}}{x}\right )} x^{\frac {3}{2}}}{3 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{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]

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

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Fricas [A]
time = 1.47, size = 173, normalized size = 2.58 \begin {gather*} \left [-\frac {3 \, {\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) - 4 \, {\left (2 \, b^{2} x - 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, {\left (b^{5} x^{2} - 4 \, b^{4} x + 4 \, b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - 2 \, {\left (2 \, b^{2} x - 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x}\right )}}{3 \, {\left (b^{5} x^{2} - 4 \, b^{4} x + 4 \, b^{3}\right )}}\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]

[-1/3*(3*(b^2*x^2 - 4*b*x + 4)*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) - 4*(2*b^2*x - 3*b)*sq
rt(-b*x + 2)*sqrt(x))/(b^5*x^2 - 4*b^4*x + 4*b^3), -2/3*(3*(b^2*x^2 - 4*b*x + 4)*sqrt(b)*arctan(sqrt(-b*x + 2)
/(sqrt(b)*sqrt(x))) - 2*(2*b^2*x - 3*b)*sqrt(-b*x + 2)*sqrt(x))/(b^5*x^2 - 4*b^4*x + 4*b^3)]

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Sympy [C] Result contains complex when optimal does not.
time = 1.89, size = 648, normalized size = 9.67 \begin {gather*} \begin {cases} \frac {8 i b^{\frac {11}{2}} x^{8}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} - \frac {12 i b^{\frac {9}{2}} x^{7}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} - \frac {6 i b^{5} x^{\frac {15}{2}} \sqrt {b x - 2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} + \frac {3 \pi b^{5} x^{\frac {15}{2}} \sqrt {b x - 2}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} + \frac {12 i b^{4} x^{\frac {13}{2}} \sqrt {b x - 2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} - \frac {6 \pi b^{4} x^{\frac {13}{2}} \sqrt {b x - 2}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {b x - 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {b x - 2}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {8 b^{\frac {11}{2}} x^{8}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} + \frac {12 b^{\frac {9}{2}} x^{7}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} + \frac {6 b^{5} x^{\frac {15}{2}} \sqrt {- b x + 2} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} - \frac {12 b^{4} x^{\frac {13}{2}} \sqrt {- b x + 2} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {15}{2}} x^{\frac {15}{2}} \sqrt {- b x + 2} - 6 b^{\frac {13}{2}} x^{\frac {13}{2}} \sqrt {- b x + 2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*I*b**(11/2)*x**8/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)) - 12
*I*b**(9/2)*x**7/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)) - 6*I*b**5*x**(15
/2)*sqrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2
)*sqrt(b*x - 2)) + 3*pi*b**5*x**(15/2)*sqrt(b*x - 2)/(3*b**(15/2)*x**(15/2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13
/2)*sqrt(b*x - 2)) + 12*I*b**4*x**(13/2)*sqrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)
*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)) - 6*pi*b**4*x**(13/2)*sqrt(b*x - 2)/(3*b**(15/2)*x**(15/
2)*sqrt(b*x - 2) - 6*b**(13/2)*x**(13/2)*sqrt(b*x - 2)), Abs(b*x) > 2), (-8*b**(11/2)*x**8/(3*b**(15/2)*x**(15
/2)*sqrt(-b*x + 2) - 6*b**(13/2)*x**(13/2)*sqrt(-b*x + 2)) + 12*b**(9/2)*x**7/(3*b**(15/2)*x**(15/2)*sqrt(-b*x
 + 2) - 6*b**(13/2)*x**(13/2)*sqrt(-b*x + 2)) + 6*b**5*x**(15/2)*sqrt(-b*x + 2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2
)/(3*b**(15/2)*x**(15/2)*sqrt(-b*x + 2) - 6*b**(13/2)*x**(13/2)*sqrt(-b*x + 2)) - 12*b**4*x**(13/2)*sqrt(-b*x
+ 2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(15/2)*x**(15/2)*sqrt(-b*x + 2) - 6*b**(13/2)*x**(13/2)*sqrt(-b*x +
 2)), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (50) = 100\).
time = 4.90, size = 178, normalized size = 2.66 \begin {gather*} \frac {{\left (\frac {3 \, \log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b}} + \frac {16 \, {\left (3 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{4} \sqrt {-b} - 6 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} \sqrt {-b} b + 8 \, \sqrt {-b} b^{2}\right )}}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )}^{3}}\right )} {\left | b \right |}}{3 \, b^{3}} \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]

1/3*(3*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/sqrt(-b) + 16*(3*(sqrt(-b*x + 2)*sqrt(-b) -
sqrt((b*x - 2)*b + 2*b))^4*sqrt(-b) - 6*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2*sqrt(-b)*b + 8*s
qrt(-b)*b^2)/((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)^3)*abs(b)/b^3

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{3/2}}{{\left (2-b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(2 - b*x)^(5/2),x)

[Out]

int(x^(3/2)/(2 - b*x)^(5/2), x)

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