3.1674 \(\int \frac {x^{3/2}}{(a+\frac {b}{x})^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}}+\frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (a x+b)} \]

[Out]

-7/3*b*x^(3/2)/a^3+7/5*x^(5/2)/a^2-x^(7/2)/a/(a*x+b)-7*b^(5/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2))/a^(9/2)+7*b^2*x
^(1/2)/a^4

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Rubi [A]  time = 0.03, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {263, 47, 50, 63, 205} \[ \frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (a x+b)} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b/x)^2,x]

[Out]

(7*b^2*Sqrt[x])/a^4 - (7*b*x^(3/2))/(3*a^3) + (7*x^(5/2))/(5*a^2) - x^(7/2)/(a*(b + a*x)) - (7*b^(5/2)*ArcTan[
(Sqrt[a]*Sqrt[x])/Sqrt[b]])/a^(9/2)

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rubi steps

\begin {align*} \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^2} \, dx &=\int \frac {x^{7/2}}{(b+a x)^2} \, dx\\ &=-\frac {x^{7/2}}{a (b+a x)}+\frac {7 \int \frac {x^{5/2}}{b+a x} \, dx}{2 a}\\ &=\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {(7 b) \int \frac {x^{3/2}}{b+a x} \, dx}{2 a^2}\\ &=-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}+\frac {\left (7 b^2\right ) \int \frac {\sqrt {x}}{b+a x} \, dx}{2 a^3}\\ &=\frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {\left (7 b^3\right ) \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{2 a^4}\\ &=\frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {\left (7 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{a^4}\\ &=\frac {7 b^2 \sqrt {x}}{a^4}-\frac {7 b x^{3/2}}{3 a^3}+\frac {7 x^{5/2}}{5 a^2}-\frac {x^{7/2}}{a (b+a x)}-\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.32 \[ \frac {2 x^{9/2} \, _2F_1\left (2,\frac {9}{2};\frac {11}{2};-\frac {a x}{b}\right )}{9 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b/x)^2,x]

[Out]

(2*x^(9/2)*Hypergeometric2F1[2, 9/2, 11/2, -((a*x)/b)])/(9*b^2)

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fricas [A]  time = 0.63, size = 188, normalized size = 2.21 \[ \left [\frac {105 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (6 \, a^{3} x^{3} - 14 \, a^{2} b x^{2} + 70 \, a b^{2} x + 105 \, b^{3}\right )} \sqrt {x}}{30 \, {\left (a^{5} x + a^{4} b\right )}}, -\frac {105 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - {\left (6 \, a^{3} x^{3} - 14 \, a^{2} b x^{2} + 70 \, a b^{2} x + 105 \, b^{3}\right )} \sqrt {x}}{15 \, {\left (a^{5} x + a^{4} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(105*(a*b^2*x + b^3)*sqrt(-b/a)*log((a*x - 2*a*sqrt(x)*sqrt(-b/a) - b)/(a*x + b)) + 2*(6*a^3*x^3 - 14*a^
2*b*x^2 + 70*a*b^2*x + 105*b^3)*sqrt(x))/(a^5*x + a^4*b), -1/15*(105*(a*b^2*x + b^3)*sqrt(b/a)*arctan(a*sqrt(x
)*sqrt(b/a)/b) - (6*a^3*x^3 - 14*a^2*b*x^2 + 70*a*b^2*x + 105*b^3)*sqrt(x))/(a^5*x + a^4*b)]

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giac [A]  time = 0.16, size = 76, normalized size = 0.89 \[ -\frac {7 \, b^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {b^{3} \sqrt {x}}{{\left (a x + b\right )} a^{4}} + \frac {2 \, {\left (3 \, a^{8} x^{\frac {5}{2}} - 10 \, a^{7} b x^{\frac {3}{2}} + 45 \, a^{6} b^{2} \sqrt {x}\right )}}{15 \, a^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(a+b/x)^2,x, algorithm="giac")

[Out]

-7*b^3*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + b^3*sqrt(x)/((a*x + b)*a^4) + 2/15*(3*a^8*x^(5/2) - 10*a^
7*b*x^(3/2) + 45*a^6*b^2*sqrt(x))/a^10

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maple [A]  time = 0.01, size = 71, normalized size = 0.84 \[ \frac {2 x^{\frac {5}{2}}}{5 a^{2}}-\frac {7 b^{3} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{4}}-\frac {4 b \,x^{\frac {3}{2}}}{3 a^{3}}+\frac {b^{3} \sqrt {x}}{\left (a x +b \right ) a^{4}}+\frac {6 b^{2} \sqrt {x}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a+b/x)^2,x)

[Out]

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

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maxima [A]  time = 2.56, size = 77, normalized size = 0.91 \[ \frac {6 \, a^{3} - \frac {14 \, a^{2} b}{x} + \frac {70 \, a b^{2}}{x^{2}} + \frac {105 \, b^{3}}{x^{3}}}{15 \, {\left (\frac {a^{5}}{x^{\frac {5}{2}}} + \frac {a^{4} b}{x^{\frac {7}{2}}}\right )}} + \frac {7 \, b^{3} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(6*a^3 - 14*a^2*b/x + 70*a*b^2/x^2 + 105*b^3/x^3)/(a^5/x^(5/2) + a^4*b/x^(7/2)) + 7*b^3*arctan(b/(sqrt(a*
b)*sqrt(x)))/(sqrt(a*b)*a^4)

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mupad [B]  time = 0.04, size = 68, normalized size = 0.80 \[ \frac {2\,x^{5/2}}{5\,a^2}-\frac {4\,b\,x^{3/2}}{3\,a^3}+\frac {6\,b^2\,\sqrt {x}}{a^4}+\frac {b^3\,\sqrt {x}}{x\,a^5+b\,a^4}-\frac {7\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a + b/x)^2,x)

[Out]

(2*x^(5/2))/(5*a^2) - (4*b*x^(3/2))/(3*a^3) + (6*b^2*x^(1/2))/a^4 + (b^3*x^(1/2))/(a^4*b + a^5*x) - (7*b^(5/2)
*atan((a^(1/2)*x^(1/2))/b^(1/2)))/a^(9/2)

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sympy [A]  time = 57.53, size = 542, normalized size = 6.38 \[ \begin {cases} \tilde {\infty } x^{\frac {9}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {9}{2}}}{9 b^{2}} & \text {for}\: a = 0 \\\frac {2 x^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: b = 0 \\\frac {12 i a^{4} \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {1}{a}}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {28 i a^{3} b^{\frac {3}{2}} x^{\frac {5}{2}} \sqrt {\frac {1}{a}}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {140 i a^{2} b^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{a}}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {210 i a b^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {1}{a}}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {105 a b^{3} x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {105 a b^{3} x \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} - \frac {105 b^{4} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} + \frac {105 b^{4} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{30 i a^{6} \sqrt {b} x \sqrt {\frac {1}{a}} + 30 i a^{5} b^{\frac {3}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(9/2), Eq(a, 0) & Eq(b, 0)), (2*x**(9/2)/(9*b**2), Eq(a, 0)), (2*x**(5/2)/(5*a**2), Eq(b, 0)
), (12*I*a**4*sqrt(b)*x**(7/2)*sqrt(1/a)/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) - 28*I
*a**3*b**(3/2)*x**(5/2)*sqrt(1/a)/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) + 140*I*a**2*
b**(5/2)*x**(3/2)*sqrt(1/a)/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) + 210*I*a*b**(7/2)*
sqrt(x)*sqrt(1/a)/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) - 105*a*b**3*x*log(-I*sqrt(b)
*sqrt(1/a) + sqrt(x))/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) + 105*a*b**3*x*log(I*sqrt
(b)*sqrt(1/a) + sqrt(x))/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) - 105*b**4*log(-I*sqrt
(b)*sqrt(1/a) + sqrt(x))/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)) + 105*b**4*log(I*sqrt(
b)*sqrt(1/a) + sqrt(x))/(30*I*a**6*sqrt(b)*x*sqrt(1/a) + 30*I*a**5*b**(3/2)*sqrt(1/a)), True))

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