3.307 \(\int \frac{(a+b x)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=81 \[ -\frac{5 b^2 \sqrt{a+b x}}{8 x}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{5 b (a+b x)^{3/2}}{12 x^2}-\frac{(a+b x)^{5/2}}{3 x^3} \]

[Out]

(-5*b^2*Sqrt[a + b*x])/(8*x) - (5*b*(a + b*x)^(3/2))/(12*x^2) - (a + b*x)^(5/2)/(3*x^3) - (5*b^3*ArcTanh[Sqrt[
a + b*x]/Sqrt[a]])/(8*Sqrt[a])

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Rubi [A]  time = 0.0226134, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {47, 63, 208} \[ -\frac{5 b^2 \sqrt{a+b x}}{8 x}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{5 b (a+b x)^{3/2}}{12 x^2}-\frac{(a+b x)^{5/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x^4,x]

[Out]

(-5*b^2*Sqrt[a + b*x])/(8*x) - (5*b*(a + b*x)^(3/2))/(12*x^2) - (a + b*x)^(5/2)/(3*x^3) - (5*b^3*ArcTanh[Sqrt[
a + b*x]/Sqrt[a]])/(8*Sqrt[a])

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 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 208

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^{5/2}}{x^4} \, dx &=-\frac{(a+b x)^{5/2}}{3 x^3}+\frac{1}{6} (5 b) \int \frac{(a+b x)^{3/2}}{x^3} \, dx\\ &=-\frac{5 b (a+b x)^{3/2}}{12 x^2}-\frac{(a+b x)^{5/2}}{3 x^3}+\frac{1}{8} \left (5 b^2\right ) \int \frac{\sqrt{a+b x}}{x^2} \, dx\\ &=-\frac{5 b^2 \sqrt{a+b x}}{8 x}-\frac{5 b (a+b x)^{3/2}}{12 x^2}-\frac{(a+b x)^{5/2}}{3 x^3}+\frac{1}{16} \left (5 b^3\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=-\frac{5 b^2 \sqrt{a+b x}}{8 x}-\frac{5 b (a+b x)^{3/2}}{12 x^2}-\frac{(a+b x)^{5/2}}{3 x^3}+\frac{1}{8} \left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )\\ &=-\frac{5 b^2 \sqrt{a+b x}}{8 x}-\frac{5 b (a+b x)^{3/2}}{12 x^2}-\frac{(a+b x)^{5/2}}{3 x^3}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0569685, size = 79, normalized size = 0.98 \[ -\frac{34 a^2 b x+8 a^3+59 a b^2 x^2+15 b^3 x^3 \sqrt{\frac{b x}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )+33 b^3 x^3}{24 x^3 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^4,x]

[Out]

-(8*a^3 + 34*a^2*b*x + 59*a*b^2*x^2 + 33*b^3*x^3 + 15*b^3*x^3*Sqrt[1 + (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]])/(2
4*x^3*Sqrt[a + b*x])

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Maple [A]  time = 0.009, size = 63, normalized size = 0.8 \begin{align*} 2\,{b}^{3} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( -{\frac{11\, \left ( bx+a \right ) ^{5/2}}{16}}+5/6\,a \left ( bx+a \right ) ^{3/2}-{\frac{5\,{a}^{2}\sqrt{bx+a}}{16}} \right ) }-{\frac{5}{16\,\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^4,x)

[Out]

2*b^3*((-11/16*(b*x+a)^(5/2)+5/6*a*(b*x+a)^(3/2)-5/16*a^2*(b*x+a)^(1/2))/b^3/x^3-5/16*arctanh((b*x+a)^(1/2)/a^
(1/2))/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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59282, size = 350, normalized size = 4.32 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{3} x^{3} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{48 \, a x^{3}}, \frac{15 \, \sqrt{-a} b^{3} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt{b x + a}}{24 \, a x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(15*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(33*a*b^2*x^2 + 26*a^2*b*x + 8*a^3)
*sqrt(b*x + a))/(a*x^3), 1/24*(15*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (33*a*b^2*x^2 + 26*a^2*b
*x + 8*a^3)*sqrt(b*x + a))/(a*x^3)]

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Sympy [A]  time = 6.73303, size = 104, normalized size = 1.28 \begin{align*} - \frac{a^{2} \sqrt{b} \sqrt{\frac{a}{b x} + 1}}{3 x^{\frac{5}{2}}} - \frac{13 a b^{\frac{3}{2}} \sqrt{\frac{a}{b x} + 1}}{12 x^{\frac{3}{2}}} - \frac{11 b^{\frac{5}{2}} \sqrt{\frac{a}{b x} + 1}}{8 \sqrt{x}} - \frac{5 b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{8 \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)/x**4,x)

[Out]

-a**2*sqrt(b)*sqrt(a/(b*x) + 1)/(3*x**(5/2)) - 13*a*b**(3/2)*sqrt(a/(b*x) + 1)/(12*x**(3/2)) - 11*b**(5/2)*sqr
t(a/(b*x) + 1)/(8*sqrt(x)) - 5*b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*sqrt(a))

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Giac [A]  time = 1.28311, size = 107, normalized size = 1.32 \begin{align*} \frac{\frac{15 \, b^{4} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{33 \,{\left (b x + a\right )}^{\frac{5}{2}} b^{4} - 40 \,{\left (b x + a\right )}^{\frac{3}{2}} a b^{4} + 15 \, \sqrt{b x + a} a^{2} b^{4}}{b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^4,x, algorithm="giac")

[Out]

1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - (33*(b*x + a)^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b^4 + 15
*sqrt(b*x + a)*a^2*b^4)/(b^3*x^3))/b