\(\int \frac {(a x+b x^2)^{5/2}}{x^3} \, dx\) [29]

Optimal result

Integrand size = 17, antiderivative size = 94 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=-\frac {5}{8} a (a+2 b x) \sqrt {a x+b x^2}-\frac {5}{3} b \left (a x+b x^2\right )^{3/2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {676, 678, 626, 634, 212} \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 \sqrt {b}}-\frac {5}{8} a (a+2 b x) \sqrt {a x+b x^2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-\frac {5}{3} b \left (a x+b x^2\right )^{3/2} \]

[In]

Int[(a*x + b*x^2)^(5/2)/x^3,x]

[Out]

(-5*a*(a + 2*b*x)*Sqrt[a*x + b*x^2])/8 - (5*b*(a*x + b*x^2)^(3/2))/3 + (2*(a*x + b*x^2)^(5/2))/x^2 + (5*a^3*Ar
cTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(8*Sqrt[b])

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 626

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

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}}{x^2}-(5 b) \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx \\ & = -\frac {5}{3} b \left (a x+b x^2\right )^{3/2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}-\frac {1}{2} (5 a b) \int \sqrt {a x+b x^2} \, dx \\ & = -\frac {5}{8} a (a+2 b x) \sqrt {a x+b x^2}-\frac {5}{3} b \left (a x+b x^2\right )^{3/2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac {1}{16} \left (5 a^3\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx \\ & = -\frac {5}{8} a (a+2 b x) \sqrt {a x+b x^2}-\frac {5}{3} b \left (a x+b x^2\right )^{3/2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac {1}{8} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right ) \\ & = -\frac {5}{8} a (a+2 b x) \sqrt {a x+b x^2}-\frac {5}{3} b \left (a x+b x^2\right )^{3/2}+\frac {2 \left (a x+b x^2\right )^{5/2}}{x^2}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\frac {1}{24} \sqrt {x (a+b x)} \left (33 a^2+26 a b x+8 b^2 x^2-\frac {15 a^3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {b} \sqrt {x} \sqrt {a+b x}}\right ) \]

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^3,x]

[Out]

(Sqrt[x*(a + b*x)]*(33*a^2 + 26*a*b*x + 8*b^2*x^2 - (15*a^3*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(Sqrt[b]*
Sqrt[x]*Sqrt[a + b*x])))/24

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{3}}{8 \sqrt {b}}+\frac {11 \left (\sqrt {b}\, a^{2}+\frac {26 b^{\frac {3}{2}} a x}{33}+\frac {8 b^{\frac {5}{2}} x^{2}}{33}\right ) \sqrt {x \left (b x +a \right )}}{8 \sqrt {b}}\)	\(64\)
risch	\(\frac {\left (8 b^{2} x^{2}+26 a b x +33 a^{2}\right ) x \left (b x +a \right )}{24 \sqrt {x \left (b x +a \right )}}+\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 \sqrt {b}}\)	\(70\)
default	\(\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{a \,x^{3}}-\frac {8 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}}-\frac {10 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\right )}{3 a}\right )}{a}\)	\(156\)

[In]

int((b*x^2+a*x)^(5/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

5/8/b^(1/2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*a^3+11/8*(b^(1/2)*a^2+26/33*b^(3/2)*a*x+8/33*b^(5/2)*x^2)/b^(
1/2)*(x*(b*x+a))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{48 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (8 \, b^{3} x^{2} + 26 \, a b^{2} x + 33 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{24 \, b}\right ] \]

[In]

integrate((b*x^2+a*x)^(5/2)/x^3,x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(8*b^3*x^2 + 26*a*b^2*x + 33*a^2*b)*sqr
t(b*x^2 + a*x))/b, -1/24*(15*a^3*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (8*b^3*x^2 + 26*a*b^2*x +
33*a^2*b)*sqrt(b*x^2 + a*x))/b]

Sympy [F]

\[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{3}}\, dx \]

[In]

integrate((b*x**2+a*x)**(5/2)/x**3,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\frac {5 \, a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, \sqrt {b}} + \frac {5}{8} \, \sqrt {b x^{2} + a x} a^{2} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a}{12 \, x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{3 \, x^{2}} \]

[In]

integrate((b*x^2+a*x)^(5/2)/x^3,x, algorithm="maxima")

[Out]

5/16*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) + 5/8*sqrt(b*x^2 + a*x)*a^2 + 5/12*(b*x^2 + a*x)
^(3/2)*a/x + 1/3*(b*x^2 + a*x)^(5/2)/x^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=-\frac {5 \, a^{3} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (33 \, a^{2} + 2 \, {\left (4 \, b^{2} x + 13 \, a b\right )} x\right )} \]

[In]

integrate((b*x^2+a*x)^(5/2)/x^3,x, algorithm="giac")

[Out]

-5/16*a^3*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/sqrt(b) + 1/24*sqrt(b*x^2 + a*x)*(33*a^2 + 2
*(4*b^2*x + 13*a*b)*x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{x^3} \,d x \]

[In]

int((a*x + b*x^2)^(5/2)/x^3,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^3, x)