\(\int \frac {(a+b x)^{5/2}}{x^4} \, dx\) [307]

Optimal result

Integrand size = 13, antiderivative size = 81 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=-\frac {5 b^2 \sqrt {a+b x}}{8 x}-\frac {5 b (a+b x)^{3/2}}{12 x^2}-\frac {(a+b x)^{5/2}}{3 x^3}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]

[Out]

-5/12*b*(b*x+a)^(3/2)/x^2-1/3*(b*x+a)^(5/2)/x^3-5/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2)-5/8*b^2*(b*x+a)
^(1/2)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {43, 65, 214} \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {5 b^2 \sqrt {a+b x}}{8 x}-\frac {(a+b x)^{5/2}}{3 x^3}-\frac {5 b (a+b x)^{3/2}}{12 x^2} \]

[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 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 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)^{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 ) \text {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] (verified)

Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=-\frac {\sqrt {a+b x} \left (8 a^2+26 a b x+33 b^2 x^2\right )}{24 x^3}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 \sqrt {a}} \]

[In]

Integrate[(a + b*x)^(5/2)/x^4,x]

[Out]

-1/24*(Sqrt[a + b*x]*(8*a^2 + 26*a*b*x + 33*b^2*x^2))/x^3 - (5*b^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*Sqrt[a])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (33 b^{2} x^{2}+26 a b x +8 a^{2}\right )}{24 x^{3}}-\frac {5 b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\)	\(53\)
derivativedivides	\(2 b^{3} \left (-\frac {\frac {11 \left (b x +a \right )^{\frac {5}{2}}}{16}-\frac {5 a \left (b x +a \right )^{\frac {3}{2}}}{6}+\frac {5 a^{2} \sqrt {b x +a}}{16}}{b^{3} x^{3}}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\)	\(64\)
default	\(2 b^{3} \left (-\frac {\frac {11 \left (b x +a \right )^{\frac {5}{2}}}{16}-\frac {5 a \left (b x +a \right )^{\frac {3}{2}}}{6}+\frac {5 a^{2} \sqrt {b x +a}}{16}}{b^{3} x^{3}}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 \sqrt {a}}\right )\)	\(64\)
pseudoelliptic	\(\frac {-15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-8 \sqrt {b x +a}\, a^{\frac {5}{2}}-26 a^{\frac {3}{2}} b x \sqrt {b x +a}-33 b^{2} x^{2} \sqrt {b x +a}\, \sqrt {a}}{24 x^{3} \sqrt {a}}\)	\(74\)

[In]

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

[Out]

-1/24*(b*x+a)^(1/2)*(33*b^2*x^2+26*a*b*x+8*a^2)/x^3-5/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.80 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=\left [\frac {15 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a x^{3}}, \frac {15 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - {\left (33 \, a b^{2} x^{2} + 26 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a x^{3}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 3.51 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.28 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=- \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 x^{\frac {5}{2}}} - \frac {13 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{12 x^{\frac {3}{2}}} - \frac {11 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{8 \sqrt {x}} - \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 \sqrt {a}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=\frac {5 \, b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, \sqrt {a}} - \frac {33 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} + 15 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2} - a^{3}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {33 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 15 \, \sqrt {b x + a} a^{2} b^{4}}{b^{3} x^{3}}}{24 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b x)^{5/2}}{x^4} \, dx=\frac {5\,a\,{\left (a+b\,x\right )}^{3/2}}{3\,x^3}-\frac {5\,a^2\,\sqrt {a+b\,x}}{8\,x^3}-\frac {11\,{\left (a+b\,x\right )}^{5/2}}{8\,x^3}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{8\,\sqrt {a}} \]

[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)