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

Optimal. Leaf size=103 \[ \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {(a+b x)^{5/2}}{4 x^4}-\frac {5 b (a+b x)^{3/2}}{24 x^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {47, 51, 63, 208} \begin {gather*} \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/x^5,x]

[Out]

(-5*b^2*Sqrt[a + b*x])/(32*x^2) - (5*b^3*Sqrt[a + b*x])/(64*a*x) - (5*b*(a + b*x)^(3/2))/(24*x^3) - (a + b*x)^
(5/2)/(4*x^4) + (5*b^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(3/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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^5} \, dx &=-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {1}{8} (5 b) \int \frac {(a+b x)^{3/2}}{x^4} \, dx\\ &=-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {1}{16} \left (5 b^2\right ) \int \frac {\sqrt {a+b x}}{x^3} \, dx\\ &=-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {1}{64} \left (5 b^3\right ) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx\\ &=-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}-\frac {\left (5 b^4\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{128 a}\\ &=-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{64 a}\\ &=-\frac {5 b^2 \sqrt {a+b x}}{32 x^2}-\frac {5 b^3 \sqrt {a+b x}}{64 a x}-\frac {5 b (a+b x)^{3/2}}{24 x^3}-\frac {(a+b x)^{5/2}}{4 x^4}+\frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.34 \begin {gather*} -\frac {2 b^4 (a+b x)^{7/2} \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {b x}{a}+1\right )}{7 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/x^5,x]

[Out]

(-2*b^4*(a + b*x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, 1 + (b*x)/a])/(7*a^5)

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IntegrateAlgebraic [A]  time = 0.15, size = 83, normalized size = 0.81 \begin {gather*} \frac {5 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{64 a^{3/2}}-\frac {\sqrt {a+b x} \left (15 a^3-55 a^2 (a+b x)+73 a (a+b x)^2+15 (a+b x)^3\right )}{192 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(Sqrt[a + b*x]*(15*a^3 - 55*a^2*(a + b*x) + 73*a*(a + b*x)^2 + 15*(a + b*x)^3))/(a*x^4) + (5*b^4*ArcTan
h[Sqrt[a + b*x]/Sqrt[a]])/(64*a^(3/2))

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fricas [A]  time = 0.96, size = 167, normalized size = 1.62 \begin {gather*} \left [\frac {15 \, \sqrt {a} b^{4} x^{4} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{384 \, a^{2} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} x^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{3} x^{3} + 118 \, a^{2} b^{2} x^{2} + 136 \, a^{3} b x + 48 \, a^{4}\right )} \sqrt {b x + a}}{192 \, a^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*sqrt(a)*b^4*x^4*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(15*a*b^3*x^3 + 118*a^2*b^2*x^2 +
136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4), -1/192*(15*sqrt(-a)*b^4*x^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) +
(15*a*b^3*x^3 + 118*a^2*b^2*x^2 + 136*a^3*b*x + 48*a^4)*sqrt(b*x + a))/(a^2*x^4)]

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giac [A]  time = 1.09, size = 99, normalized size = 0.96 \begin {gather*} -\frac {\frac {15 \, b^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b x + a} a^{3} b^{5}}{a b^{4} x^{4}}}{192 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(15*b^5*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + (15*(b*x + a)^(7/2)*b^5 + 73*(b*x + a)^(5/2)*a*b^
5 - 55*(b*x + a)^(3/2)*a^2*b^5 + 15*sqrt(b*x + a)*a^3*b^5)/(a*b^4*x^4))/b

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maple [A]  time = 0.01, size = 75, normalized size = 0.73 \begin {gather*} 2 \left (\frac {5 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{128 a^{\frac {3}{2}}}+\frac {-\frac {5 \sqrt {b x +a}\, a^{2}}{128}+\frac {55 \left (b x +a \right )^{\frac {3}{2}} a}{384}-\frac {5 \left (b x +a \right )^{\frac {7}{2}}}{128 a}-\frac {73 \left (b x +a \right )^{\frac {5}{2}}}{384}}{b^{4} x^{4}}\right ) b^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^4*((-5/128/a*(b*x+a)^(7/2)-73/384*(b*x+a)^(5/2)+55/384*(b*x+a)^(3/2)*a-5/128*(b*x+a)^(1/2)*a^2)/x^4/b^4+5/
128*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2))

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maxima [A]  time = 2.93, size = 144, normalized size = 1.40 \begin {gather*} -\frac {5 \, b^{4} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{128 \, a^{\frac {3}{2}}} - \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {b x + a} a^{3} b^{4}}{192 \, {\left ({\left (b x + a\right )}^{4} a - 4 \, {\left (b x + a\right )}^{3} a^{2} + 6 \, {\left (b x + a\right )}^{2} a^{3} - 4 \, {\left (b x + a\right )} a^{4} + a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/128*b^4*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2) - 1/192*(15*(b*x + a)^(7/2)*b^4 +
73*(b*x + a)^(5/2)*a*b^4 - 55*(b*x + a)^(3/2)*a^2*b^4 + 15*sqrt(b*x + a)*a^3*b^4)/((b*x + a)^4*a - 4*(b*x + a)
^3*a^2 + 6*(b*x + a)^2*a^3 - 4*(b*x + a)*a^4 + a^5)

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mupad [B]  time = 0.11, size = 79, normalized size = 0.77 \begin {gather*} \frac {55\,a\,{\left (a+b\,x\right )}^{3/2}}{192\,x^4}-\frac {5\,a^2\,\sqrt {a+b\,x}}{64\,x^4}-\frac {5\,{\left (a+b\,x\right )}^{7/2}}{64\,a\,x^4}-\frac {73\,{\left (a+b\,x\right )}^{5/2}}{192\,x^4}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{64\,a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(5/2)/x^5,x)

[Out]

(55*a*(a + b*x)^(3/2))/(192*x^4) - (5*a^2*(a + b*x)^(1/2))/(64*x^4) - (5*(a + b*x)^(7/2))/(64*a*x^4) - (b^4*at
an(((a + b*x)^(1/2)*1i)/a^(1/2))*5i)/(64*a^(3/2)) - (73*(a + b*x)^(5/2))/(192*x^4)

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sympy [A]  time = 8.36, size = 155, normalized size = 1.50 \begin {gather*} - \frac {a^{3}}{4 \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {23 a^{2} \sqrt {b}}{24 x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {127 a b^{\frac {3}{2}}}{96 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {133 b^{\frac {5}{2}}}{192 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 b^{\frac {7}{2}}}{64 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{64 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**3/(4*sqrt(b)*x**(9/2)*sqrt(a/(b*x) + 1)) - 23*a**2*sqrt(b)/(24*x**(7/2)*sqrt(a/(b*x) + 1)) - 127*a*b**(3/2
)/(96*x**(5/2)*sqrt(a/(b*x) + 1)) - 133*b**(5/2)/(192*x**(3/2)*sqrt(a/(b*x) + 1)) - 5*b**(7/2)/(64*a*sqrt(x)*s
qrt(a/(b*x) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(64*a**(3/2))

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