\(\int \frac {(a x+b x^2)^{5/2}}{x^5} \, dx\) [31]

Optimal result

Integrand size = 17, antiderivative size = 89 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) \]

[Out]

-10/3*b*(b*x^2+a*x)^(3/2)/x^2-2/3*(b*x^2+a*x)^(5/2)/x^4+5*a*b^(3/2)*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))+5*b^2
*(b*x^2+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {676, 678, 634, 212} \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=5 a b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )+5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4} \]

[In]

Int[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

5*b^2*Sqrt[a*x + b*x^2] - (10*b*(a*x + b*x^2)^(3/2))/(3*x^2) - (2*(a*x + b*x^2)^(5/2))/(3*x^4) + 5*a*b^(3/2)*A
rcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 676

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 678

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\frac {1}{3} (5 b) \int \frac {\left (a x+b x^2\right )^{3/2}}{x^3} \, dx \\ & = -\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\left (5 b^2\right ) \int \frac {\sqrt {a x+b x^2}}{x} \, dx \\ & = 5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\frac {1}{2} \left (5 a b^2\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx \\ & = 5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+\left (5 a b^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right ) \\ & = 5 b^2 \sqrt {a x+b x^2}-\frac {10 b \left (a x+b x^2\right )^{3/2}}{3 x^2}-\frac {2 \left (a x+b x^2\right )^{5/2}}{3 x^4}+5 a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {a+b x} \left (-2 a^2-14 a b x+3 b^2 x^2\right )+30 a b^{3/2} x^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{3 x^2 \sqrt {a+b x}} \]

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^5,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[a + b*x]*(-2*a^2 - 14*a*b*x + 3*b^2*x^2) + 30*a*b^(3/2)*x^(3/2)*ArcTanh[(Sqrt[b]*Sqrt
[x])/(-Sqrt[a] + Sqrt[a + b*x])]))/(3*x^2*Sqrt[a + b*x])

Maple [A] (verified)

Time = 2.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {\left (b x +a \right ) \left (-3 b^{2} x^{2}+14 a b x +2 a^{2}\right )}{3 x \sqrt {x \left (b x +a \right )}}+\frac {5 a \,b^{\frac {3}{2}} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2}\)	\(70\)
pseudoelliptic	\(\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a \,b^{2} x^{2}+3 \sqrt {x \left (b x +a \right )}\, b^{\frac {5}{2}} x^{2}-14 a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}-2 a^{2} \sqrt {x \left (b x +a \right )}\, \sqrt {b}}{3 x^{2} \sqrt {b}}\)	\(86\)
default	\(-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{5}}+\frac {4 b \left (-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{4}}+\frac {6 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{3}}-\frac {8 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}}-\frac {10 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\right )}{3 a}\right )}{a}\right )}{a}\right )}{3 a}\)	\(208\)

[In]

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

[Out]

-1/3*(b*x+a)*(-3*b^2*x^2+14*a*b*x+2*a^2)/x/(x*(b*x+a))^(1/2)+5/2*a*b^(3/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^
(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=\left [\frac {15 \, a b^{\frac {3}{2}} x^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{6 \, x^{2}}, -\frac {15 \, a \sqrt {-b} b x^{2} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (3 \, b^{2} x^{2} - 14 \, a b x - 2 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{3 \, x^{2}}\right ] \]

[In]

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

[Out]

[1/6*(15*a*b^(3/2)*x^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(3*b^2*x^2 - 14*a*b*x - 2*a^2)*sqrt(b*
x^2 + a*x))/x^2, -1/3*(15*a*sqrt(-b)*b*x^2*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (3*b^2*x^2 - 14*a*b*x -
2*a^2)*sqrt(b*x^2 + a*x))/x^2]

Sympy [F]

\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{5}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=\frac {5}{2} \, a b^{\frac {3}{2}} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - \frac {35 \, \sqrt {b x^{2} + a x} a b}{6 \, x} - \frac {5 \, \sqrt {b x^{2} + a x} a^{2}}{6 \, x^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{6 \, x^{3}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{x^{4}} \]

[In]

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

[Out]

5/2*a*b^(3/2)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 35/6*sqrt(b*x^2 + a*x)*a*b/x - 5/6*sqrt(b*x^2 + a
*x)*a^2/x^2 - 5/6*(b*x^2 + a*x)^(3/2)*a/x^3 + (b*x^2 + a*x)^(5/2)/x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=-\frac {5}{2} \, a b^{\frac {3}{2}} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right ) + \sqrt {b x^{2} + a x} b^{2} + \frac {2 \, {\left (9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{2} b + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{3} \sqrt {b} + a^{4}\right )}}{3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3}} \]

[In]

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

[Out]

-5/2*a*b^(3/2)*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + sqrt(b*x^2 + a*x)*b^2 + 2/3*(9*(sqrt
(b)*x - sqrt(b*x^2 + a*x))^2*a^2*b + 3*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^3*sqrt(b) + a^4)/(sqrt(b)*x - sqrt(b*
x^2 + a*x))^3

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^5} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^5} \,d x \]

[In]

int((a*x + b*x^2)^(5/2)/x^5,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^5, x)