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

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

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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*}

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.98 \begin {gather*} -\frac {8 a^3+34 a^2 b x+15 b^3 x^3 \sqrt {\frac {b x}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {b x}{a}+1}\right )+59 a b^2 x^2+33 b^3 x^3}{24 x^3 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^4,x]

[Out]

-1/24*(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]
])/(x^3*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.14, size = 68, normalized size = 0.84 \begin {gather*} -\frac {\sqrt {a+b x} \left (15 a^2-40 a (a+b x)+33 (a+b x)^2\right )}{24 x^3}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(5/2)/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*(15*a^2 - 40*a*(a + b*x) + 33*(a + b*x)^2))/x^3 - (5*b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/
(8*Sqrt[a])

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fricas [A]  time = 0.99, size = 146, normalized size = 1.80 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.13, size = 79, normalized size = 0.98 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.01, size = 63, normalized size = 0.78 \begin {gather*} 2 \left (-\frac {5 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}+\frac {-\frac {5 \sqrt {b x +a}\, a^{2}}{16}+\frac {5 \left (b x +a \right )^{\frac {3}{2}} a}{6}-\frac {11 \left (b x +a \right )^{\frac {5}{2}}}{16}}{b^{3} x^{3}}\right ) b^{3} \end {gather*}

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*(b*x+a)^(3/2)*a-5/16*(b*x+a)^(1/2)*a^2)/x^3/b^3-5/16*arctanh((b*x+a)^(1/2)/a^
(1/2))/a^(1/2))

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maxima [A]  time = 2.99, size = 115, normalized size = 1.42 \begin {gather*} \frac {5 \, b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, \sqrt {a}} - \frac {33 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 15 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/sqrt(a) - 1/24*(33*(b*x + a)^(5/2)*b^3 - 40*
(b*x + a)^(3/2)*a*b^3 + 15*sqrt(b*x + a)*a^2*b^3)/((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2 - a^3)

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mupad [B]  time = 0.05, size = 64, normalized size = 0.79 \begin {gather*} \frac {5\,a\,{\left (a+b\,x\right )}^{3/2}}{3\,x^3}-\frac {5\,a^2\,\sqrt {a+b\,x}}{8\,x^3}-\frac {11\,{\left (a+b\,x\right )}^{5/2}}{8\,x^3}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{8\,\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/x^4,x)

[Out]

(b^3*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*5i)/(8*a^(1/2)) - (5*a^2*(a + b*x)^(1/2))/(8*x^3) - (11*(a + b*x)^(5/2
))/(8*x^3) + (5*a*(a + b*x)^(3/2))/(3*x^3)

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sympy [A]  time = 5.16, size = 104, normalized size = 1.28 \begin {gather*} - \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 {gather*}

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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