\(\int \frac {(a+b x)^{3/2}}{x^4} \, dx\) [299]

Optimal result

Integrand size = 13, antiderivative size = 84 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {b \sqrt {a+b x}}{4 x^2}-\frac {b^2 \sqrt {a+b x}}{8 a x}-\frac {(a+b x)^{3/2}}{3 x^3}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}} \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 44, 65, 214} \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}}-\frac {b^2 \sqrt {a+b x}}{8 a x}-\frac {(a+b x)^{3/2}}{3 x^3}-\frac {b \sqrt {a+b x}}{4 x^2} \]

[In]

Int[(a + b*x)^(3/2)/x^4,x]

[Out]

-1/4*(b*Sqrt[a + b*x])/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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rubi steps \begin{align*} \text {integral}& = -\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 \text {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*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {\sqrt {a+b x} \left (8 a^2+14 a b x+3 b^2 x^2\right )}{24 a x^3}+\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{3/2}} \]

[In]

Integrate[(a + b*x)^(3/2)/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*(8*a^2 + 14*a*b*x + 3*b^2*x^2))/(a*x^3) + (b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(3/2)
)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (3 b^{2} x^{2}+14 a b x +8 a^{2}\right )}{24 x^{3} a}+\frac {b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\)	\(56\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-\sqrt {b x +a}\, \left (\sqrt {a}\, b^{2} x^{2}+\frac {14 a^{\frac {3}{2}} b x}{3}+\frac {8 a^{\frac {5}{2}}}{3}\right )}{8 a^{\frac {3}{2}} x^{3}}\)	\(61\)
derivativedivides	\(2 b^{3} \left (-\frac {\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {a \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {3}{2}}}\right )\)	\(64\)
default	\(2 b^{3} \left (-\frac {\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6}-\frac {a \sqrt {b x +a}}{16}}{b^{3} x^{3}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {3}{2}}}\right )\)	\(64\)

[In]

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

[Out]

-1/24*(b*x+a)^(1/2)*(3*b^2*x^2+14*a*b*x+8*a^2)/x^3/a+1/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{2} x^{3}}, -\frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{2} + 14 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{2} x^{3}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 3.87 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.48 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=- \frac {a^{2}}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {11 a \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {17 b^{\frac {3}{2}}}{24 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {b^{\frac {5}{2}}}{8 a \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {3}{2}}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} a - 3 \, {\left (b x + a\right )}^{2} a^{2} + 3 \, {\left (b x + a\right )} a^{3} - a^{4}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=-\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} + 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x + a} a^{2} b^{4}}{a b^{3} x^{3}}}{24 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \frac {(a+b x)^{3/2}}{x^4} \, dx=\frac {a\,\sqrt {a+b\,x}}{8\,x^3}-\frac {{\left (a+b\,x\right )}^{5/2}}{8\,a\,x^3}-\frac {{\left (a+b\,x\right )}^{3/2}}{3\,x^3}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{3/2}} \]

[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)