\(\int \frac {(a x+b x^2)^{5/2}}{x} \, dx\) [27]

Optimal result

Integrand size = 17, antiderivative size = 107 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=-\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {3 a^5 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{5/2}} \]

[Out]

1/16*a*(2*b*x+a)*(b*x^2+a*x)^(3/2)/b+1/5*(b*x^2+a*x)^(5/2)+3/128*a^5*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(5
/2)-3/128*a^3*(2*b*x+a)*(b*x^2+a*x)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 107, 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} \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=\frac {3 a^5 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{5/2}}-\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2} \]

[In]

Int[(a*x + b*x^2)^(5/2)/x,x]

[Out]

(-3*a^3*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(128*b^2) + (a*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(16*b) + (a*x + b*x^2)^
(5/2)/5 + (3*a^5*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(128*b^(5/2))

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 {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {1}{2} a \int \left (a x+b x^2\right )^{3/2} \, dx \\ & = \frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}-\frac {\left (3 a^3\right ) \int \sqrt {a x+b x^2} \, dx}{32 b} \\ & = -\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {\left (3 a^5\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{256 b^2} \\ & = -\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {\left (3 a^5\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{128 b^2} \\ & = -\frac {3 a^3 (a+2 b x) \sqrt {a x+b x^2}}{128 b^2}+\frac {a (a+2 b x) \left (a x+b x^2\right )^{3/2}}{16 b}+\frac {1}{5} \left (a x+b x^2\right )^{5/2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (-15 a^4+10 a^3 b x+248 a^2 b^2 x^2+336 a b^3 x^3+128 b^4 x^4\right )+\frac {30 a^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{640 b^{5/2}} \]

[In]

Integrate[(a*x + b*x^2)^(5/2)/x,x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(-15*a^4 + 10*a^3*b*x + 248*a^2*b^2*x^2 + 336*a*b^3*x^3 + 128*b^4*x^4) + (30*a^5*A
rcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(Sqrt[x]*Sqrt[a + b*x])))/(640*b^(5/2))

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{5}}{128}-\frac {3 \sqrt {x \left (b x +a \right )}\, \left (a^{4} \sqrt {b}-\frac {2 a^{3} b^{\frac {3}{2}} x}{3}-\frac {248 b^{\frac {5}{2}} a^{2} x^{2}}{15}-\frac {112 b^{\frac {7}{2}} a \,x^{3}}{5}-\frac {128 x^{4} b^{\frac {9}{2}}}{15}\right )}{128}}{b^{\frac {5}{2}}}\)	\(84\)
risch	\(-\frac {\left (-128 b^{4} x^{4}-336 a \,b^{3} x^{3}-248 a^{2} b^{2} x^{2}-10 a^{3} b x +15 a^{4}\right ) x \left (b x +a \right )}{640 b^{2} \sqrt {x \left (b x +a \right )}}+\frac {3 a^{5} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{256 b^{\frac {5}{2}}}\)	\(95\)
default	\(\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5}+\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2}\)	\(104\)

[In]

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

[Out]

3/128*(arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*a^5-(x*(b*x+a))^(1/2)*(a^4*b^(1/2)-2/3*a^3*b^(3/2)*x-248/15*b^(5/2
)*a^2*x^2-112/5*b^(7/2)*a*x^3-128/15*x^4*b^(9/2)))/b^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=\left [\frac {15 \, a^{5} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt {b x^{2} + a x}}{1280 \, b^{3}}, -\frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (128 \, b^{5} x^{4} + 336 \, a b^{4} x^{3} + 248 \, a^{2} b^{3} x^{2} + 10 \, a^{3} b^{2} x - 15 \, a^{4} b\right )} \sqrt {b x^{2} + a x}}{640 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/1280*(15*a^5*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^
2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x^2 + a*x))/b^3, -1/640*(15*a^5*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*
sqrt(-b)/(b*x)) - (128*b^5*x^4 + 336*a*b^4*x^3 + 248*a^2*b^3*x^2 + 10*a^3*b^2*x - 15*a^4*b)*sqrt(b*x^2 + a*x))
/b^3]

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 415, normalized size of antiderivative = 3.88 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=a^{2} \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 ) + 2 a b \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 ) + b^{2} \left (\begin {cases} \frac {7 a^{5} \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 )}{256 b^{4}} + \sqrt {a x + b x^{2}} \left (- \frac {7 a^{4}}{128 b^{4}} + \frac {7 a^{3} x}{192 b^{3}} - \frac {7 a^{2} x^{2}}{240 b^{2}} + \frac {a x^{3}}{40 b} + \frac {x^{4}}{5}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {9}{2}}}{9 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*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)) + 2*a*b*Piecewise((-5*a**4*Piece
wise((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)/s
qrt(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)) + b**2*Piecewise((7*a**5*Piecew
ise((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)/sq
rt(b*(a/(2*b) + x)**2), True))/(256*b**4) + sqrt(a*x + b*x**2)*(-7*a**4/(128*b**4) + 7*a**3*x/(192*b**3) - 7*a
**2*x**2/(240*b**2) + a*x**3/(40*b) + x**4/5), Ne(b, 0)), (2*(a*x)**(9/2)/(9*a**4), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=\frac {1}{8} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x - \frac {3 \, \sqrt {b x^{2} + a x} a^{3} x}{64 \, b} + \frac {3 \, a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {5}{2}}} + \frac {1}{5} \, {\left (b x^{2} + a x\right )}^{\frac {5}{2}} - \frac {3 \, \sqrt {b x^{2} + a x} a^{4}}{128 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}}{16 \, b} \]

[In]

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

[Out]

1/8*(b*x^2 + a*x)^(3/2)*a*x - 3/64*sqrt(b*x^2 + a*x)*a^3*x/b + 3/256*a^5*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s
qrt(b))/b^(5/2) + 1/5*(b*x^2 + a*x)^(5/2) - 3/128*sqrt(b*x^2 + a*x)*a^4/b^2 + 1/16*(b*x^2 + a*x)^(3/2)*a^2/b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a x+b x^2\right )^{5/2}}{x} \, dx=-\frac {3 \, a^{5} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{256 \, b^{\frac {5}{2}}} - \frac {1}{640} \, \sqrt {b x^{2} + a x} {\left (\frac {15 \, a^{4}}{b^{2}} - 2 \, {\left (\frac {5 \, a^{3}}{b} + 4 \, {\left (31 \, a^{2} + 2 \, {\left (8 \, b^{2} x + 21 \, a b\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

-3/256*a^5*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(5/2) - 1/640*sqrt(b*x^2 + a*x)*(15*a^4/b
^2 - 2*(5*a^3/b + 4*(31*a^2 + 2*(8*b^2*x + 21*a*b)*x)*x)*x)

Mupad [F(-1)]

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

[In]

int((a*x + b*x^2)^(5/2)/x,x)

[Out]

int((a*x + b*x^2)^(5/2)/x, x)