\(\int \frac {x^6}{(a x+b x^2)^{5/2}} \, dx\) [56]

Optimal result

Integrand size = 17, antiderivative size = 128 \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 a^3 x^2}{3 b^4 \left (a x+b x^2\right )^{3/2}}-\frac {20 a^2 x}{3 b^4 \sqrt {a x+b x^2}}-\frac {13 a \sqrt {a x+b x^2}}{4 b^4}+\frac {\left (a x+b x^2\right )^{3/2}}{2 b^4 x}+\frac {35 a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{9/2}} \] Output:

2/3*a^3*x^2/b^4/(b*x^2+a*x)^(3/2)-20/3*a^2*x/b^4/(b*x^2+a*x)^(1/2)-13/4*a* 
(b*x^2+a*x)^(1/2)/b^4+1/2*(b*x^2+a*x)^(3/2)/b^4/x+35/4*a^2*arctanh(b^(1/2) 
*x/(b*x^2+a*x)^(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.85 \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {x \left (\sqrt {b} x \left (-105 a^3-140 a^2 b x-21 a b^2 x^2+6 b^3 x^3\right )+210 a^2 \sqrt {x} (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{12 b^{9/2} (x (a+b x))^{3/2}} \] Input:

Integrate[x^6/(a*x + b*x^2)^(5/2),x]
 

Output:

(x*(Sqrt[b]*x*(-105*a^3 - 140*a^2*b*x - 21*a*b^2*x^2 + 6*b^3*x^3) + 210*a^ 
2*Sqrt[x]*(a + b*x)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b 
*x])]))/(12*b^(9/2)*(x*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1133, 1124, 2192, 27, 1160, 1091, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1133

\(\displaystyle \frac {7 \int \frac {x^4}{\left (b x^2+a x\right )^{3/2}}dx}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1124

\(\displaystyle \frac {7 \left (\frac {\int \frac {a^2-b x a+b^2 x^2}{\sqrt {b x^2+a x}}dx}{b^3}-\frac {2 a^2 x}{b^3 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {7 \left (\frac {\frac {\int \frac {a b (4 a-7 b x)}{2 \sqrt {b x^2+a x}}dx}{2 b}+\frac {1}{2} b x \sqrt {a x+b x^2}}{b^3}-\frac {2 a^2 x}{b^3 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7 \left (\frac {\frac {1}{4} a \int \frac {4 a-7 b x}{\sqrt {b x^2+a x}}dx+\frac {1}{2} b x \sqrt {a x+b x^2}}{b^3}-\frac {2 a^2 x}{b^3 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {7 \left (\frac {\frac {1}{4} a \left (\frac {15}{2} a \int \frac {1}{\sqrt {b x^2+a x}}dx-7 \sqrt {a x+b x^2}\right )+\frac {1}{2} b x \sqrt {a x+b x^2}}{b^3}-\frac {2 a^2 x}{b^3 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {7 \left (\frac {\frac {1}{4} a \left (15 a \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}-7 \sqrt {a x+b x^2}\right )+\frac {1}{2} b x \sqrt {a x+b x^2}}{b^3}-\frac {2 a^2 x}{b^3 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {7 \left (\frac {\frac {1}{4} a \left (\frac {15 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b}}-7 \sqrt {a x+b x^2}\right )+\frac {1}{2} b x \sqrt {a x+b x^2}}{b^3}-\frac {2 a^2 x}{b^3 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^5}{3 b \left (a x+b x^2\right )^{3/2}}\)

Input:

Int[x^6/(a*x + b*x^2)^(5/2),x]
 

Output:

(-2*x^5)/(3*b*(a*x + b*x^2)^(3/2)) + (7*((-2*a^2*x)/(b^3*Sqrt[a*x + b*x^2] 
) + ((b*x*Sqrt[a*x + b*x^2])/2 + (a*(-7*Sqrt[a*x + b*x^2] + (15*a*ArcTanh[ 
(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/Sqrt[b]))/4)/b^3))/(3*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1091

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

rule 1124

Int[((d_.) + (e_.)*(x_))^(m_.)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[-2*e*(2*c*d - b*e)^(m - 2)*((d + e*x)/(c^(m - 1)*Sqrt[a + 
b*x + c*x^2])), x] + Simp[e^2/c^(m - 1)   Int[(1/Sqrt[a + b*x + c*x^2])*Exp 
andToSum[((2*c*d - b*e)^(m - 1) - c^(m - 1)*(d + e*x)^(m - 1))/(c*d - b*e - 
 c*e*x), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e 
^2, 0] && IGtQ[m, 0]
 

rule 1133

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), 
 x] - Simp[e^2*((m + p)/(c*(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] && EqQ[c*d^2 - b*d*e + a*e 
^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1160

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] + Simp[(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[p, -1]
 

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(\frac {-140 b^{\frac {3}{2}} a^{2} x^{2}-21 b^{\frac {5}{2}} a \,x^{3}+6 b^{\frac {7}{2}} x^{4}+105 a^{2} \left (-x a \sqrt {b}+\sqrt {x \left (b x +a \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) \left (b x +a \right )\right )}{b^{\frac {9}{2}} \sqrt {x \left (b x +a \right )}\, \left (12 b x +12 a \right )}\)	\(98\)
risch	\(-\frac {\left (-2 b x +11 a \right ) x \left (b x +a \right )}{4 b^{4} \sqrt {x \left (b x +a \right )}}+\frac {35 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {9}{2}}}-\frac {20 a^{2} \sqrt {\left (x +\frac {a}{b}\right )^{2} b -a \left (x +\frac {a}{b}\right )}}{3 b^{5} \left (x +\frac {a}{b}\right )}+\frac {2 a^{3} \sqrt {\left (x +\frac {a}{b}\right )^{2} b -a \left (x +\frac {a}{b}\right )}}{3 b^{6} \left (x +\frac {a}{b}\right )^{2}}\)	\(144\)
default	\(\frac {x^{5}}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {7 a \left (\frac {x^{4}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {a \left (-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\right )}{2 b}\right )}{2 b}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )}{4 b}\)	\(294\)

Input:

int(x^6/(b*x^2+a*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

105/b^(9/2)/(x*(b*x+a))^(1/2)*(-4/3*b^(3/2)*a^2*x^2-1/5*b^(5/2)*a*x^3+2/35 
*b^(7/2)*x^4+a^2*(-x*a*b^(1/2)+(x*(b*x+a))^(1/2)*arctanh((x*(b*x+a))^(1/2) 
/x/b^(1/2))*(b*x+a)))/(12*b*x+12*a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.95 \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\left [\frac {105 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (6 \, b^{4} x^{3} - 21 \, a b^{3} x^{2} - 140 \, a^{2} b^{2} x - 105 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{24 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac {105 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (6 \, b^{4} x^{3} - 21 \, a b^{3} x^{2} - 140 \, a^{2} b^{2} x - 105 \, a^{3} b\right )} \sqrt {b x^{2} + a x}}{12 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \] Input:

integrate(x^6/(b*x^2+a*x)^(5/2),x, algorithm="fricas")
 

Output:

[1/24*(105*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*sqrt(b)*log(2*b*x + a + 2*sqrt( 
b*x^2 + a*x)*sqrt(b)) + 2*(6*b^4*x^3 - 21*a*b^3*x^2 - 140*a^2*b^2*x - 105* 
a^3*b)*sqrt(b*x^2 + a*x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5), -1/12*(105*(a^2 
*b^2*x^2 + 2*a^3*b*x + a^4)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b* 
x + a)) - (6*b^4*x^3 - 21*a*b^3*x^2 - 140*a^2*b^2*x - 105*a^3*b)*sqrt(b*x^ 
2 + a*x))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)]
 

Sympy [F]

\[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {x^{6}}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**6/(b*x**2+a*x)**(5/2),x)
 

Output:

Integral(x**6/(x*(a + b*x))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.48 \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {x^{5}}{2 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} - \frac {35 \, a^{2} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {a x}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, x}{\sqrt {b x^{2} + a x} a b} - \frac {1}{\sqrt {b x^{2} + a x} b^{2}}\right )}}{24 \, b^{2}} - \frac {7 \, a x^{4}}{4 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {35 \, a^{2} x}{6 \, \sqrt {b x^{2} + a x} b^{4}} + \frac {35 \, a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {9}{2}}} - \frac {35 \, \sqrt {b x^{2} + a x} a}{12 \, b^{4}} \] Input:

integrate(x^6/(b*x^2+a*x)^(5/2),x, algorithm="maxima")
 

Output:

1/2*x^5/((b*x^2 + a*x)^(3/2)*b) - 35/24*a^2*x*(3*x^2/((b*x^2 + a*x)^(3/2)* 
b) + a*x/((b*x^2 + a*x)^(3/2)*b^2) - 2*x/(sqrt(b*x^2 + a*x)*a*b) - 1/(sqrt 
(b*x^2 + a*x)*b^2))/b^2 - 7/4*a*x^4/((b*x^2 + a*x)^(3/2)*b^2) - 35/6*a^2*x 
/(sqrt(b*x^2 + a*x)*b^4) + 35/8*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sq 
rt(b))/b^(9/2) - 35/12*sqrt(b*x^2 + a*x)*a/b^4
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.23 \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {1}{4} \, \sqrt {b x^{2} + a x} {\left (\frac {2 \, x}{b^{3}} - \frac {11 \, a}{b^{4}}\right )} - \frac {35 \, a^{2} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{8 \, b^{\frac {9}{2}}} - \frac {2 \, {\left (12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{3} b + 21 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{4} \sqrt {b} + 10 \, a^{5}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a\right )}^{3} b^{\frac {9}{2}}} \] Input:

integrate(x^6/(b*x^2+a*x)^(5/2),x, algorithm="giac")
 

Output:

1/4*sqrt(b*x^2 + a*x)*(2*x/b^3 - 11*a/b^4) - 35/8*a^2*log(abs(-2*(sqrt(b)* 
x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(9/2) - 2/3*(12*(sqrt(b)*x - sqrt(b 
*x^2 + a*x))^2*a^3*b + 21*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^4*sqrt(b) + 10 
*a^5)/(((sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a)^3*b^(9/2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {x^6}{{\left (b\,x^2+a\,x\right )}^{5/2}} \,d x \] Input:

int(x^6/(a*x + b*x^2)^(5/2),x)
 

Output:

int(x^6/(a*x + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.20 \[ \int \frac {x^6}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {840 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3}+840 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b x +175 \sqrt {b}\, \sqrt {b x +a}\, a^{3}+175 \sqrt {b}\, \sqrt {b x +a}\, a^{2} b x -840 \sqrt {x}\, a^{3} b -1120 \sqrt {x}\, a^{2} b^{2} x -168 \sqrt {x}\, a \,b^{3} x^{2}+48 \sqrt {x}\, b^{4} x^{3}}{96 \sqrt {b x +a}\, b^{5} \left (b x +a \right )} \] Input:

int(x^6/(b*x^2+a*x)^(5/2),x)
 

Output:

(840*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))* 
a**3 + 840*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqr 
t(a))*a**2*b*x + 175*sqrt(b)*sqrt(a + b*x)*a**3 + 175*sqrt(b)*sqrt(a + b*x 
)*a**2*b*x - 840*sqrt(x)*a**3*b - 1120*sqrt(x)*a**2*b**2*x - 168*sqrt(x)*a 
*b**3*x**2 + 48*sqrt(x)*b**4*x**3)/(96*sqrt(a + b*x)*b**5*(a + b*x))