\(\int \frac {(a x+b x^2)^{5/2}}{x^2} \, dx\) [28]

Optimal result

Integrand size = 17, antiderivative size = 101 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {5 a^4 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{3/2}} \]

[Out]

5/24*a*(b*x^2+a*x)^(3/2)+1/4*(b*x^2+a*x)^(5/2)/x-5/64*a^4*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(3/2)+5/64*a^
2*(2*b*x+a)*(b*x^2+a*x)^(1/2)/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, 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 = {678, 626, 634, 212} \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=-\frac {5 a^4 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{3/2}}+\frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x} \]

[In]

Int[(a*x + b*x^2)^(5/2)/x^2,x]

[Out]

(5*a^2*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(64*b) + (5*a*(a*x + b*x^2)^(3/2))/24 + (a*x + b*x^2)^(5/2)/(4*x) - (5*a
^4*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(64*b^(3/2))

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 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 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 {\left (a x+b x^2\right )^{5/2}}{4 x}+\frac {1}{8} (5 a) \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx \\ & = \frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}+\frac {1}{16} \left (5 a^2\right ) \int \sqrt {a x+b x^2} \, dx \\ & = \frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{128 b} \\ & = \frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {\left (5 a^4\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{64 b} \\ & = \frac {5 a^2 (a+2 b x) \sqrt {a x+b x^2}}{64 b}+\frac {5}{24} a \left (a x+b x^2\right )^{3/2}+\frac {\left (a x+b x^2\right )^{5/2}}{4 x}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (15 a^3+118 a^2 b x+136 a b^2 x^2+48 b^3 x^3\right )+\frac {30 a^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{192 b^{3/2}} \]

[In]

Integrate[(a*x + b*x^2)^(5/2)/x^2,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(15*a^3 + 118*a^2*b*x + 136*a*b^2*x^2 + 48*b^3*x^3) + (30*a^4*ArcTanh[(Sqrt[b]*Sqr
t[x])/(Sqrt[a] - Sqrt[a + b*x])])/(Sqrt[x]*Sqrt[a + b*x])))/(192*b^(3/2))

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(-\frac {5 \left (\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{4}-\sqrt {x \left (b x +a \right )}\, \left (a^{3} \sqrt {b}+\frac {118 a^{2} b^{\frac {3}{2}} x}{15}+\frac {136 b^{\frac {5}{2}} a \,x^{2}}{15}+\frac {16 b^{\frac {7}{2}} x^{3}}{5}\right )\right )}{64 b^{\frac {3}{2}}}\)	\(73\)
risch	\(\frac {\left (48 b^{3} x^{3}+136 a \,b^{2} x^{2}+118 a^{2} b x +15 a^{3}\right ) x \left (b x +a \right )}{192 b \sqrt {x \left (b x +a \right )}}-\frac {5 a^{4} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 b^{\frac {3}{2}}}\)	\(84\)
default	\(\frac {2 \left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{3 a \,x^{2}}-\frac {10 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\right )}{3 a}\)	\(130\)

[In]

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

[Out]

-5/64*(arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*a^4-(x*(b*x+a))^(1/2)*(a^3*b^(1/2)+118/15*a^2*b^(3/2)*x+136/15*b^(
5/2)*a*x^2+16/5*b^(7/2)*x^3))/b^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.67 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=\left [\frac {15 \, a^{4} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{384 \, b^{2}}, \frac {15 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (48 \, b^{4} x^{3} + 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x + 15 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{192 \, b^{2}}\right ] \]

[In]

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

[Out]

[1/384*(15*a^4*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(48*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*
b^2*x + 15*a^3*b)*sqrt(b*x^2 + a*x))/b^2, 1/192*(15*a^4*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (4
8*b^4*x^3 + 136*a*b^3*x^2 + 118*a^2*b^2*x + 15*a^3*b)*sqrt(b*x^2 + a*x))/b^2]

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 369, normalized size of antiderivative = 3.65 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=a^{2} \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \left (\frac {a}{4 b} + \frac {x}{2}\right ) \sqrt {a x + b x^{2}} & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {3}{2}}}{3 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \frac {a^{3} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a x + b x^{2}} \left (- \frac {a^{2}}{8 b^{2}} + \frac {a x}{12 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} - \frac {5 a^{4} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{128 b^{3}} + \sqrt {a x + b x^{2}} \cdot \left (\frac {5 a^{3}}{64 b^{3}} - \frac {5 a^{2} x}{96 b^{2}} + \frac {a x^{2}}{24 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {7}{2}}}{7 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*Piecewise((-a**2*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), ((a/(2
*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(8*b) + (a/(4*b) + x/2)*sqrt(a*x + b*x**2), Ne(b, 0
)), (2*(a*x)**(3/2)/(3*a), Ne(a, 0)), (0, True)) + 2*a*b*Piecewise((a**3*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x
+ b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/
(16*b**2) + sqrt(a*x + b*x**2)*(-a**2/(8*b**2) + a*x/(12*b) + x**2/3), Ne(b, 0)), (2*(a*x)**(5/2)/(5*a**2), Ne
(a, 0)), (0, True)) + b**2*Piecewise((-5*a**4*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b)
, Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(128*b**3) + sqrt(a*x + b*x
**2)*(5*a**3/(64*b**3) - 5*a**2*x/(96*b**2) + a*x**2/(24*b) + x**3/4), Ne(b, 0)), (2*(a*x)**(7/2)/(7*a**3), Ne
(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=\frac {5}{32} \, \sqrt {b x^{2} + a x} a^{2} x - \frac {5 \, a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {3}{2}}} + \frac {5}{24} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a + \frac {5 \, \sqrt {b x^{2} + a x} a^{3}}{64 \, b} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{4 \, x} \]

[In]

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

[Out]

5/32*sqrt(b*x^2 + a*x)*a^2*x - 5/128*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) + 5/24*(b*x^2 +
a*x)^(3/2)*a + 5/64*sqrt(b*x^2 + a*x)*a^3/b + 1/4*(b*x^2 + a*x)^(5/2)/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x^2} \, dx=\frac {5 \, a^{4} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{128 \, b^{\frac {3}{2}}} + \frac {1}{192} \, \sqrt {b x^{2} + a x} {\left (\frac {15 \, a^{3}}{b} + 2 \, {\left (59 \, a^{2} + 4 \, {\left (6 \, b^{2} x + 17 \, a b\right )} x\right )} x\right )} \]

[In]

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

[Out]

5/128*a^4*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(3/2) + 1/192*sqrt(b*x^2 + a*x)*(15*a^3/b
+ 2*(59*a^2 + 4*(6*b^2*x + 17*a*b)*x)*x)

Mupad [F(-1)]

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

[In]

int((a*x + b*x^2)^(5/2)/x^2,x)

[Out]

int((a*x + b*x^2)^(5/2)/x^2, x)