3.299 \(\int \frac {(a+b x)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=84 \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}}-\frac {b^2 \sqrt {a+b x}}{8 a x}-\frac {(a+b x)^{3/2}}{3 x^3}-\frac {b \sqrt {a+b x}}{4 x^2} \]

[Out]

-1/3*(b*x+a)^(3/2)/x^3+1/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)-1/4*b*(b*x+a)^(1/2)/x^2-1/8*b^2*(b*x+a)^
(1/2)/a/x

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Rubi [A]  time = 0.02, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {47, 51, 63, 208} \[ \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}}-\frac {b^2 \sqrt {a+b x}}{8 a x}-\frac {b \sqrt {a+b x}}{4 x^2}-\frac {(a+b x)^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/x^4,x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.42 \[ \frac {2 b^3 (a+b x)^{5/2} \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {b x}{a}+1\right )}{5 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/x^4,x]

[Out]

(2*b^3*(a + b*x)^(5/2)*Hypergeometric2F1[5/2, 4, 7/2, 1 + (b*x)/a])/(5*a^4)

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fricas [A]  time = 0.54, size = 145, normalized size = 1.73 \[ \left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{2} x^{3}}, -\frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^4,x, algorithm="fricas")

[Out]

[1/48*(3*sqrt(a)*b^3*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(3*a*b^2*x^2 + 14*a^2*b*x + 8*a^3)*s
qrt(b*x + a))/(a^2*x^3), -1/24*(3*sqrt(-a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b^2*x^2 + 14*a^2*b*
x + 8*a^3)*sqrt(b*x + a))/(a^2*x^3)]

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giac [A]  time = 1.16, size = 84, normalized size = 1.00 \[ -\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x + a} a^{2} b^{4}}{a b^{3} x^{3}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^4,x, algorithm="giac")

[Out]

-1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) + (3*(b*x + a)^(5/2)*b^4 + 8*(b*x + a)^(3/2)*a*b^4 -
3*sqrt(b*x + a)*a^2*b^4)/(a*b^3*x^3))/b

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maple [A]  time = 0.01, size = 63, normalized size = 0.75 \[ 2 \left (\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {\frac {\sqrt {b x +a}\, a}{16}-\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a}-\frac {\left (b x +a \right )^{\frac {3}{2}}}{6}}{b^{3} x^{3}}\right ) b^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^4,x)

[Out]

2*b^3*((-1/16/a*(b*x+a)^(5/2)-1/6*(b*x+a)^(3/2)+1/16*(b*x+a)^(1/2)*a)/x^3/b^3+1/16*arctanh((b*x+a)^(1/2)/a^(1/
2))/a^(3/2))

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maxima [A]  time = 3.05, size = 119, normalized size = 1.42 \[ -\frac {b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} a - 3 \, {\left (b x + a\right )}^{2} a^{2} + 3 \, {\left (b x + a\right )} a^{3} - a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^4,x, algorithm="maxima")

[Out]

-1/16*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2) - 1/24*(3*(b*x + a)^(5/2)*b^3 + 8*(
b*x + a)^(3/2)*a*b^3 - 3*sqrt(b*x + a)*a^2*b^3)/((b*x + a)^3*a - 3*(b*x + a)^2*a^2 + 3*(b*x + a)*a^3 - a^4)

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mupad [B]  time = 0.10, size = 64, normalized size = 0.76 \[ \frac {a\,\sqrt {a+b\,x}}{8\,x^3}-\frac {{\left (a+b\,x\right )}^{5/2}}{8\,a\,x^3}-\frac {{\left (a+b\,x\right )}^{3/2}}{3\,x^3}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/x^4,x)

[Out]

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

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sympy [A]  time = 5.84, size = 124, normalized size = 1.48 \[ - \frac {a^{2}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {11 a \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {17 b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {b^{\frac {5}{2}}}{8 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/x**4,x)

[Out]

-a**2/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 11*a*sqrt(b)/(12*x**(5/2)*sqrt(a/(b*x) + 1)) - 17*b**(3/2)/(24*
x**(3/2)*sqrt(a/(b*x) + 1)) - b**(5/2)/(8*a*sqrt(x)*sqrt(a/(b*x) + 1)) + b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))
/(8*a**(3/2))

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