3.6.78 \(\int \frac {x^{3/2}}{(a+b x)^{3/2}} \, dx\) [578]

Optimal. Leaf size=68 \[ -\frac {2 x^{3/2}}{b \sqrt {a+b x}}+\frac {3 \sqrt {x} \sqrt {a+b x}}{b^2}-\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {49, 52, 65, 223, 212} \begin {gather*} -\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}}+\frac {3 \sqrt {x} \sqrt {a+b x}}{b^2}-\frac {2 x^{3/2}}{b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b*x)^(3/2),x]

[Out]

(-2*x^(3/2))/(b*Sqrt[a + b*x]) + (3*Sqrt[x]*Sqrt[a + b*x])/b^2 - (3*a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]]
)/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 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 65

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 212

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

Rule 223

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

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 57, normalized size = 0.84 \begin {gather*} \frac {\sqrt {x} (3 a+b x)}{b^2 \sqrt {a+b x}}+\frac {3 a \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b*x)^(3/2),x]

[Out]

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

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Mathics [A]
time = 3.74, size = 77, normalized size = 1.13 \begin {gather*} \frac {\sqrt {a} \left (-3 \sqrt {a} b^3 \text {ArcSinh}\left [\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ] \left (a+b x\right )+3 a b^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {a+b x}{a}}+b^{\frac {9}{2}} x^{\frac {3}{2}} \sqrt {\frac {a+b x}{a}}\right )}{b^{\frac {11}{2}} \left (a+b x\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x^(3/2)/(a + b*x)^(3/2),x]')

[Out]

Sqrt[a] (-3 Sqrt[a] b ^ 3 ArcSinh[Sqrt[b] Sqrt[x] / Sqrt[a]] (a + b x) + 3 a b ^ (7 / 2) Sqrt[x] Sqrt[(a + b x
) / a] + b ^ (9 / 2) x ^ (3 / 2) Sqrt[(a + b x) / a]) / (b ^ (11 / 2) (a + b x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(105\) vs. \(2(52)=104\).
time = 0.13, size = 106, normalized size = 1.56

method	result	size
risch	\(\frac {\sqrt {x}\, \sqrt {b x +a}}{b^{2}}+\frac {\left (-\frac {3 a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {x^{2} b +a x}\right )}{2 b^{\frac {5}{2}}}+\frac {2 a \sqrt {\left (x +\frac {a}{b}\right )^{2} b -a \left (x +\frac {a}{b}\right )}}{b^{3} \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 92, normalized size = 1.35 \begin {gather*} \frac {2 \, a b - \frac {3 \, {\left (b x + a\right )} a}{x}}{\frac {\sqrt {b x + a} b^{3}}{\sqrt {x}} - \frac {{\left (b x + a\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}}} + \frac {3 \, a \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{2 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*b - 3*(b*x + a)*a/x)/(sqrt(b*x + a)*b^3/sqrt(x) - (b*x + a)^(3/2)*b^2/x^(3/2)) + 3/2*a*log(-(sqrt(b) - sq
rt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(5/2)

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Fricas [A]
time = 0.34, size = 145, normalized size = 2.13 \begin {gather*} \left [\frac {3 \, {\left (a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{2 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {3 \, {\left (a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{b^{4} x + a b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.99, size = 71, normalized size = 1.04 \begin {gather*} \frac {3 \sqrt {a} \sqrt {x}}{b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} + \frac {x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(a)*sqrt(x)/(b**2*sqrt(1 + b*x/a)) - 3*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(5/2) + x**(3/2)/(sqrt(a)*b*s
qrt(1 + b*x/a))

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Giac [A]
time = 0.01, size = 97, normalized size = 1.43 \begin {gather*} 2 \left (\frac {2 \left (\frac {\frac {1}{4} b^{2} \sqrt {x} \sqrt {x}}{b^{3}}+\frac {\frac {1}{4}\cdot 3 b a}{b^{3}}\right ) \sqrt {x} \sqrt {a+b x}}{a+b x}+\frac {6 a \ln \left |\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right |}{4 b^{2} \sqrt {b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x)*(x/b + 3*a/b^2)/sqrt(b*x + a) + 3*a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a + b*x)^(3/2),x)

[Out]

int(x^(3/2)/(a + b*x)^(3/2), x)

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