\(\int x (a x+b x^2)^{5/2} \, dx\) [25]

Optimal result

Integrand size = 15, antiderivative size = 139 \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=-\frac {5 a^5 (a+2 b x) \sqrt {a x+b x^2}}{1024 b^4}+\frac {5 a^3 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{384 b^3}-\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b}+\frac {5 a^7 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{1024 b^{9/2}} \]

[Out]

5/384*a^3*(2*b*x+a)*(b*x^2+a*x)^(3/2)/b^3-1/24*a*(2*b*x+a)*(b*x^2+a*x)^(5/2)/b^2+1/7*(b*x^2+a*x)^(7/2)/b+5/102
4*a^7*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(9/2)-5/1024*a^5*(2*b*x+a)*(b*x^2+a*x)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {654, 626, 634, 212} \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=\frac {5 a^7 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{1024 b^{9/2}}-\frac {5 a^5 (a+2 b x) \sqrt {a x+b x^2}}{1024 b^4}+\frac {5 a^3 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{384 b^3}-\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b} \]

[In]

Int[x*(a*x + b*x^2)^(5/2),x]

[Out]

(-5*a^5*(a + 2*b*x)*Sqrt[a*x + b*x^2])/(1024*b^4) + (5*a^3*(a + 2*b*x)*(a*x + b*x^2)^(3/2))/(384*b^3) - (a*(a
+ 2*b*x)*(a*x + b*x^2)^(5/2))/(24*b^2) + (a*x + b*x^2)^(7/2)/(7*b) + (5*a^7*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x
^2]])/(1024*b^(9/2))

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 654

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a x+b x^2\right )^{7/2}}{7 b}-\frac {a \int \left (a x+b x^2\right )^{5/2} \, dx}{2 b} \\ & = -\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b}+\frac {\left (5 a^3\right ) \int \left (a x+b x^2\right )^{3/2} \, dx}{48 b^2} \\ & = \frac {5 a^3 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{384 b^3}-\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b}-\frac {\left (5 a^5\right ) \int \sqrt {a x+b x^2} \, dx}{256 b^3} \\ & = -\frac {5 a^5 (a+2 b x) \sqrt {a x+b x^2}}{1024 b^4}+\frac {5 a^3 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{384 b^3}-\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b}+\frac {\left (5 a^7\right ) \int \frac {1}{\sqrt {a x+b x^2}} \, dx}{2048 b^4} \\ & = -\frac {5 a^5 (a+2 b x) \sqrt {a x+b x^2}}{1024 b^4}+\frac {5 a^3 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{384 b^3}-\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b}+\frac {\left (5 a^7\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{1024 b^4} \\ & = -\frac {5 a^5 (a+2 b x) \sqrt {a x+b x^2}}{1024 b^4}+\frac {5 a^3 (a+2 b x) \left (a x+b x^2\right )^{3/2}}{384 b^3}-\frac {a (a+2 b x) \left (a x+b x^2\right )^{5/2}}{24 b^2}+\frac {\left (a x+b x^2\right )^{7/2}}{7 b}+\frac {5 a^7 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{1024 b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01 \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (-105 a^6+70 a^5 b x-56 a^4 b^2 x^2+48 a^3 b^3 x^3+4736 a^2 b^4 x^4+7424 a b^5 x^5+3072 b^6 x^6\right )+\frac {210 a^7 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{21504 b^{9/2}} \]

[In]

Integrate[x*(a*x + b*x^2)^(5/2),x]

[Out]

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(-105*a^6 + 70*a^5*b*x - 56*a^4*b^2*x^2 + 48*a^3*b^3*x^3 + 4736*a^2*b^4*x^4 + 7424
*a*b^5*x^5 + 3072*b^6*x^6) + (210*a^7*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(Sqrt[x]*Sqrt[a +
b*x])))/(21504*b^(9/2))

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {\left (-3072 b^{6} x^{6}-7424 a \,x^{5} b^{5}-4736 a^{2} x^{4} b^{4}-48 a^{3} x^{3} b^{3}+56 a^{4} x^{2} b^{2}-70 a^{5} x b +105 a^{6}\right ) x \left (b x +a \right )}{21504 b^{4} \sqrt {x \left (b x +a \right )}}+\frac {5 a^{7} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2048 b^{\frac {9}{2}}}\)	\(117\)
default	\(\frac {\left (b \,x^{2}+a x \right )^{\frac {7}{2}}}{7 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{12 b}-\frac {5 a^{2} \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{24 b}\right )}{2 b}\)	\(141\)

[In]

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

[Out]

-1/21504*(-3072*b^6*x^6-7424*a*b^5*x^5-4736*a^2*b^4*x^4-48*a^3*b^3*x^3+56*a^4*b^2*x^2-70*a^5*b*x+105*a^6)*x*(b
*x+a)/b^4/(x*(b*x+a))^(1/2)+5/2048*a^7/b^(9/2)*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 = 236, normalized size of antiderivative = 1.70 \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=\left [\frac {105 \, a^{7} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (3072 \, b^{7} x^{6} + 7424 \, a b^{6} x^{5} + 4736 \, a^{2} b^{5} x^{4} + 48 \, a^{3} b^{4} x^{3} - 56 \, a^{4} b^{3} x^{2} + 70 \, a^{5} b^{2} x - 105 \, a^{6} b\right )} \sqrt {b x^{2} + a x}}{43008 \, b^{5}}, -\frac {105 \, a^{7} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) - {\left (3072 \, b^{7} x^{6} + 7424 \, a b^{6} x^{5} + 4736 \, a^{2} b^{5} x^{4} + 48 \, a^{3} b^{4} x^{3} - 56 \, a^{4} b^{3} x^{2} + 70 \, a^{5} b^{2} x - 105 \, a^{6} b\right )} \sqrt {b x^{2} + a x}}{21504 \, b^{5}}\right ] \]

[In]

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

[Out]

[1/43008*(105*a^7*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(3072*b^7*x^6 + 7424*a*b^6*x^5 + 47
36*a^2*b^5*x^4 + 48*a^3*b^4*x^3 - 56*a^4*b^3*x^2 + 70*a^5*b^2*x - 105*a^6*b)*sqrt(b*x^2 + a*x))/b^5, -1/21504*
(105*a^7*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) - (3072*b^7*x^6 + 7424*a*b^6*x^5 + 4736*a^2*b^5*x^4
+ 48*a^3*b^4*x^3 - 56*a^4*b^3*x^2 + 70*a^5*b^2*x - 105*a^6*b)*sqrt(b*x^2 + a*x))/b^5]

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.24 \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=\begin {cases} \frac {5 a^{7} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2048 b^{4}} + \sqrt {a x + b x^{2}} \left (- \frac {5 a^{6}}{1024 b^{4}} + \frac {5 a^{5} x}{1536 b^{3}} - \frac {a^{4} x^{2}}{384 b^{2}} + \frac {a^{3} x^{3}}{448 b} + \frac {37 a^{2} x^{4}}{168} + \frac {29 a b x^{5}}{84} + \frac {b^{2} x^{6}}{7}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {9}{2}}}{9 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((5*a**7*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))/(2048*b**4) + sqrt(a*x + b*x**2)*(-5*a**6/(1024*b**4) +
5*a**5*x/(1536*b**3) - a**4*x**2/(384*b**2) + a**3*x**3/(448*b) + 37*a**2*x**4/168 + 29*a*b*x**5/84 + b**2*x*
*6/7), Ne(b, 0)), (2*(a*x)**(9/2)/(9*a**2), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.17 \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=-\frac {5 \, \sqrt {b x^{2} + a x} a^{5} x}{512 \, b^{3}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3} x}{192 \, b^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} a x}{12 \, b} + \frac {5 \, a^{7} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2048 \, b^{\frac {9}{2}}} - \frac {5 \, \sqrt {b x^{2} + a x} a^{6}}{1024 \, b^{4}} + \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{4}}{384 \, b^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}} a^{2}}{24 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {7}{2}}}{7 \, b} \]

[In]

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

[Out]

-5/512*sqrt(b*x^2 + a*x)*a^5*x/b^3 + 5/192*(b*x^2 + a*x)^(3/2)*a^3*x/b^2 - 1/12*(b*x^2 + a*x)^(5/2)*a*x/b + 5/
2048*a^7*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(9/2) - 5/1024*sqrt(b*x^2 + a*x)*a^6/b^4 + 5/384*(b*x^
2 + a*x)^(3/2)*a^4/b^3 - 1/24*(b*x^2 + a*x)^(5/2)*a^2/b^2 + 1/7*(b*x^2 + a*x)^(7/2)/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.85 \[ \int x \left (a x+b x^2\right )^{5/2} \, dx=-\frac {5 \, a^{7} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{2048 \, b^{\frac {9}{2}}} - \frac {1}{21504} \, \sqrt {b x^{2} + a x} {\left (\frac {105 \, a^{6}}{b^{4}} - 2 \, {\left (\frac {35 \, a^{5}}{b^{3}} - 4 \, {\left (\frac {7 \, a^{4}}{b^{2}} - 2 \, {\left (\frac {3 \, a^{3}}{b} + 8 \, {\left (37 \, a^{2} + 2 \, {\left (12 \, b^{2} x + 29 \, a b\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

-5/2048*a^7*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(9/2) - 1/21504*sqrt(b*x^2 + a*x)*(105*a
^6/b^4 - 2*(35*a^5/b^3 - 4*(7*a^4/b^2 - 2*(3*a^3/b + 8*(37*a^2 + 2*(12*b^2*x + 29*a*b)*x)*x)*x)*x)*x)

Mupad [F(-1)]

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

[In]

int(x*(a*x + b*x^2)^(5/2),x)

[Out]

int(x*(a*x + b*x^2)^(5/2), x)