3.7.35 \(\int \frac {x^{5/2}}{(2-b x)^{3/2}} \, dx\)

Optimal. Leaf size=89 \[ -\frac {15 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}+\frac {2 x^{5/2}}{b \sqrt {2-b x}} \]

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Rubi [A]  time = 0.02, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {47, 50, 54, 216} \begin {gather*} \frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {15 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {2 x^{5/2}}{b \sqrt {2-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

(2*x^(5/2))/(b*Sqrt[2 - b*x]) + (15*Sqrt[x]*Sqrt[2 - b*x])/(2*b^3) + (5*x^(3/2)*Sqrt[2 - b*x])/(2*b^2) - (15*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

Rule 47

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 50

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 54

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 216

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^{5/2}}{(2-b x)^{3/2}} \, dx &=\frac {2 x^{5/2}}{b \sqrt {2-b x}}-\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{b}\\ &=\frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{2 b^2}\\ &=\frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{2 b^3}\\ &=\frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {2 x^{5/2}}{b \sqrt {2-b x}}+\frac {15 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 x^{3/2} \sqrt {2-b x}}{2 b^2}-\frac {15 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.34 \begin {gather*} \frac {x^{7/2} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};\frac {b x}{2}\right )}{7 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

(x^(7/2)*Hypergeometric2F1[3/2, 7/2, 9/2, (b*x)/2])/(7*Sqrt[2])

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IntegrateAlgebraic [A]  time = 0.17, size = 88, normalized size = 0.99 \begin {gather*} \frac {\sqrt {2-b x} \left (b^2 x^{5/2}+5 b x^{3/2}-30 \sqrt {x}\right )}{2 b^3 (b x-2)}-\frac {15 \sqrt {-b} \log \left (\sqrt {2-b x}-\sqrt {-b} \sqrt {x}\right )}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/(2 - b*x)^(3/2),x]

[Out]

(Sqrt[2 - b*x]*(-30*Sqrt[x] + 5*b*x^(3/2) + b^2*x^(5/2)))/(2*b^3*(-2 + b*x)) - (15*Sqrt[-b]*Log[-(Sqrt[-b]*Sqr
t[x]) + Sqrt[2 - b*x]])/b^4

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fricas [A]  time = 1.23, size = 155, normalized size = 1.74 \begin {gather*} \left [-\frac {15 \, {\left (b x - 2\right )} \sqrt {-b} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) - {\left (b^{3} x^{2} + 5 \, b^{2} x - 30 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x - 2 \, b^{4}\right )}}, \frac {30 \, {\left (b x - 2\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + {\left (b^{3} x^{2} + 5 \, b^{2} x - 30 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{2 \, {\left (b^{5} x - 2 \, b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(15*(b*x - 2)*sqrt(-b)*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) - (b^3*x^2 + 5*b^2*x - 30*b)*sqrt
(-b*x + 2)*sqrt(x))/(b^5*x - 2*b^4), 1/2*(30*(b*x - 2)*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (b^3
*x^2 + 5*b^2*x - 30*b)*sqrt(-b*x + 2)*sqrt(x))/(b^5*x - 2*b^4)]

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giac [B]  time = 10.85, size = 136, normalized size = 1.53 \begin {gather*} \frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left (\frac {b x - 2}{b^{3}} + \frac {9}{b^{3}}\right )} - \frac {15 \, \log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b} b^{2}} + \frac {64}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )} \sqrt {-b} b}\right )} {\left | b \right |}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)/b^3 + 9/b^3) - 15*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((
b*x - 2)*b + 2*b))^2)/(sqrt(-b)*b^2) + 64/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)*sqrt(
-b)*b))*abs(b)/b^2

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maple [B]  time = 0.03, size = 138, normalized size = 1.55 \begin {gather*} -\frac {\left (\frac {15 \arctan \left (\frac {\left (x -\frac {1}{b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 b^{\frac {7}{2}}}+\frac {8 \sqrt {-\left (x -\frac {2}{b}\right )^{2} b -2 x +\frac {4}{b}}}{\left (x -\frac {2}{b}\right ) b^{4}}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {-b x +2}\, \sqrt {x}}-\frac {\left (b x +7\right ) \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}\, \sqrt {x}}{2 \sqrt {-\left (b x -2\right ) x}\, \sqrt {-b x +2}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x+7)*(b*x-2)*x^(1/2)/b^3/(-(b*x-2)*x)^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)-(15/2/b^(7/2)*arctan((x-
1/b)/(-b*x^2+2*x)^(1/2)*b^(1/2))+8/b^4/(x-2/b)*(-b*(x-2/b)^2-2*x+4/b)^(1/2))*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)
/x^(1/2)

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maxima [A]  time = 2.99, size = 101, normalized size = 1.13 \begin {gather*} \frac {8 \, b^{2} - \frac {25 \, {\left (b x - 2\right )} b}{x} + \frac {15 \, {\left (b x - 2\right )}^{2}}{x^{2}}}{\frac {\sqrt {-b x + 2} b^{5}}{\sqrt {x}} + \frac {2 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (-b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}} + \frac {15 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*b^2 - 25*(b*x - 2)*b/x + 15*(b*x - 2)^2/x^2)/(sqrt(-b*x + 2)*b^5/sqrt(x) + 2*(-b*x + 2)^(3/2)*b^4/x^(3/2) +
(-b*x + 2)^(5/2)*b^3/x^(5/2)) + 15*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(2 - b*x)^(3/2),x)

[Out]

int(x^(5/2)/(2 - b*x)^(3/2), x)

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sympy [A]  time = 7.03, size = 173, normalized size = 1.94 \begin {gather*} \begin {cases} \frac {i x^{\frac {5}{2}}}{2 b \sqrt {b x - 2}} + \frac {5 i x^{\frac {3}{2}}}{2 b^{2} \sqrt {b x - 2}} - \frac {15 i \sqrt {x}}{b^{3} \sqrt {b x - 2}} + \frac {15 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {for}\: \frac {\left |{b x}\right |}{2} > 1 \\- \frac {x^{\frac {5}{2}}}{2 b \sqrt {- b x + 2}} - \frac {5 x^{\frac {3}{2}}}{2 b^{2} \sqrt {- b x + 2}} + \frac {15 \sqrt {x}}{b^{3} \sqrt {- b x + 2}} - \frac {15 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*x**(5/2)/(2*b*sqrt(b*x - 2)) + 5*I*x**(3/2)/(2*b**2*sqrt(b*x - 2)) - 15*I*sqrt(x)/(b**3*sqrt(b*x
- 2)) + 15*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), Abs(b*x)/2 > 1), (-x**(5/2)/(2*b*sqrt(-b*x + 2)) - 5*x
**(3/2)/(2*b**2*sqrt(-b*x + 2)) + 15*sqrt(x)/(b**3*sqrt(-b*x + 2)) - 15*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/
2), True))

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