3.549 \(\int \frac{(a+b x)^{5/2}}{x^{3/2}} \, dx\)

Optimal. Leaf size=89 \[ \frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}+\frac{15}{4} a b \sqrt{x} \sqrt{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.0279171, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {47, 50, 63, 217, 206} \[ \frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}+\frac{15}{4} a b \sqrt{x} \sqrt{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x^(3/2),x]

[Out]

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

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 217

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]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^{5/2}}{x^{3/2}} \, dx &=-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+(5 b) \int \frac{(a+b x)^{3/2}}{\sqrt{x}} \, dx\\ &=\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{1}{4} (15 a b) \int \frac{\sqrt{a+b x}}{\sqrt{x}} \, dx\\ &=\frac{15}{4} a b \sqrt{x} \sqrt{a+b x}+\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{1}{8} \left (15 a^2 b\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx\\ &=\frac{15}{4} a b \sqrt{x} \sqrt{a+b x}+\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{15}{4} a b \sqrt{x} \sqrt{a+b x}+\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )\\ &=\frac{15}{4} a b \sqrt{x} \sqrt{a+b x}+\frac{5}{2} b \sqrt{x} (a+b x)^{3/2}-\frac{2 (a+b x)^{5/2}}{\sqrt{x}}+\frac{15}{4} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0119474, size = 48, normalized size = 0.54 \[ -\frac{2 a^2 \sqrt{a+b x} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x}{a}\right )}{\sqrt{x} \sqrt{\frac{b x}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^(3/2),x]

[Out]

(-2*a^2*Sqrt[a + b*x]*Hypergeometric2F1[-5/2, -1/2, 1/2, -((b*x)/a)])/(Sqrt[x]*Sqrt[1 + (b*x)/a])

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Maple [A]  time = 0.014, size = 84, normalized size = 0.9 \begin{align*} -{\frac{-2\,{b}^{2}{x}^{2}-9\,abx+8\,{a}^{2}}{4}\sqrt{bx+a}{\frac{1}{\sqrt{x}}}}+{\frac{15\,{a}^{2}}{8}\sqrt{b}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ) \sqrt{x \left ( bx+a \right ) }{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x+a)^(1/2)*(-2*b^2*x^2-9*a*b*x+8*a^2)/x^(1/2)+15/8*a^2*b^(1/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2
))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97287, size = 351, normalized size = 3.94 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{8 \, x}, -\frac{15 \, a^{2} \sqrt{-b} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, b^{2} x^{2} + 9 \, a b x - 8 \, a^{2}\right )} \sqrt{b x + a} \sqrt{x}}{4 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*a^2*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^2*x^2 + 9*a*b*x - 8*a^2)*sqrt
(b*x + a)*sqrt(x))/x, -1/4*(15*a^2*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (2*b^2*x^2 + 9*a*b*
x - 8*a^2)*sqrt(b*x + a)*sqrt(x))/x]

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Sympy [A]  time = 12.4347, size = 126, normalized size = 1.42 \begin{align*} - \frac{2 a^{\frac{5}{2}}}{\sqrt{x} \sqrt{1 + \frac{b x}{a}}} + \frac{a^{\frac{3}{2}} b \sqrt{x}}{4 \sqrt{1 + \frac{b x}{a}}} + \frac{11 \sqrt{a} b^{2} x^{\frac{3}{2}}}{4 \sqrt{1 + \frac{b x}{a}}} + \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4} + \frac{b^{3} x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**(5/2)/(sqrt(x)*sqrt(1 + b*x/a)) + a**(3/2)*b*sqrt(x)/(4*sqrt(1 + b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(4*s
qrt(1 + b*x/a)) + 15*a**2*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/4 + b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out