\(\int \frac {x^5 (A+B x)}{(b x+c x^2)^{5/2}} \, dx\) [162]

Optimal result

Integrand size = 22, antiderivative size = 155 \[ \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 (b B-A c) x^4}{3 c^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 (8 b B-5 A c) x^2}{3 c^3 \sqrt {b x+c x^2}}-\frac {5 (7 b B-4 A c) \sqrt {b x+c x^2}}{4 c^4}+\frac {B x \sqrt {b x+c x^2}}{2 c^3}+\frac {5 b (7 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{9/2}} \] Output:

2/3*(-A*c+B*b)*x^4/c^2/(c*x^2+b*x)^(3/2)+2/3*(-5*A*c+8*B*b)*x^2/c^3/(c*x^2 
+b*x)^(1/2)-5/4*(-4*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/c^4+1/2*B*x*(c*x^2+b*x)^( 
1/2)/c^3+5/4*b*(-4*A*c+7*B*b)*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.16 \[ \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {x \left (\sqrt {c} x \left (-105 b^3 B+b c^2 x (80 A-21 B x)+20 b^2 c (3 A-7 B x)+6 c^3 x^2 (2 A+B x)\right )+120 A b c \sqrt {x} (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+210 b^2 B \sqrt {x} (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{12 c^{9/2} (x (b+c x))^{3/2}} \] Input:

Integrate[(x^5*(A + B*x))/(b*x + c*x^2)^(5/2),x]
 

Output:

(x*(Sqrt[c]*x*(-105*b^3*B + b*c^2*x*(80*A - 21*B*x) + 20*b^2*c*(3*A - 7*B* 
x) + 6*c^3*x^2*(2*A + B*x)) + 120*A*b*c*Sqrt[x]*(b + c*x)^(3/2)*ArcTanh[(S 
qrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])] + 210*b^2*B*Sqrt[x]*(b + c*x)^( 
3/2)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(12*c^(9/2)*( 
x*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1218, 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.

\(\displaystyle \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1218

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

\(\Big \downarrow \) 1124

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

\(\Big \downarrow \) 2192

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

Int[(x^5*(A + B*x))/(b*x + c*x^2)^(5/2),x]
 

Output:

(-2*(b*B - A*c)*x^5)/(3*b*c*(b*x + c*x^2)^(3/2)) - (((4*A)/b - (7*B)/c)*(( 
-2*b^2*x)/(c^3*Sqrt[b*x + c*x^2]) + ((c*x*Sqrt[b*x + c*x^2])/2 + (b*(-7*Sq 
rt[b*x + c*x^2] + (15*b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/Sqrt[c]))/ 
4)/c^3))/3
 

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 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 1218

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + ( 
c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(( 
a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(2*c*d - b*e))), x] - Simp[e*((m*(g*(c* 
d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g))/(c*(p + 1)*(2*c*d - b*e)))   I 
nt[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d 
, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
 

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 = 1.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(-\frac {5 \left (-x \left (-\frac {7 B x}{3}+A \right ) b^{2} c^{\frac {3}{2}}-\frac {4 x^{2} b \left (-\frac {21 B x}{80}+A \right ) c^{\frac {5}{2}}}{3}-\frac {\left (\frac {B x}{2}+A \right ) x^{3} c^{\frac {7}{2}}}{5}+\left (\frac {7 B \sqrt {c}\, b^{2} x}{4}+\left (c x +b \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) \sqrt {x \left (c x +b \right )}\, \left (A c -\frac {7 B b}{4}\right )\right ) b \right )}{\sqrt {x \left (c x +b \right )}\, c^{\frac {9}{2}} \left (c x +b \right )}\)	\(120\)
risch	\(\frac {\left (2 B c x +4 A c -11 B b \right ) x \left (c x +b \right )}{4 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {b \left (20 A \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-\frac {35 B b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-\frac {16 \left (3 A c -4 B b \right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{c \left (\frac {b}{c}+x \right )}+\frac {8 b^{2} \left (A c -B b \right ) \left (\frac {2 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{3 b \left (\frac {b}{c}+x \right )^{2}}+\frac {4 c \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{3 b^{2} \left (\frac {b}{c}+x \right )}\right )}{c^{2}}\right )}{8 c^{4}}\)	\(245\)
default	\(A \left (\frac {x^{4}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {5 b \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )}{2 c}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}}{c}\right )}{2 c}\right )+B \left (\frac {x^{5}}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {7 b \left (\frac {x^{4}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {5 b \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )}{2 c}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}}{c}\right )}{2 c}\right )}{4 c}\right )\)	\(566\)

Input:

int(x^5*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-5*(-x*(-7/3*B*x+A)*b^2*c^(3/2)-4/3*x^2*b*(-21/80*B*x+A)*c^(5/2)-1/5*(1/2* 
B*x+A)*x^3*c^(7/2)+(7/4*B*c^(1/2)*b^2*x+(c*x+b)*arctanh((x*(c*x+b))^(1/2)/ 
x/c^(1/2))*(x*(c*x+b))^(1/2)*(A*c-7/4*B*b))*b)/(x*(c*x+b))^(1/2)/c^(9/2)/( 
c*x+b)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.46 \[ \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{4} - 4 \, A b^{3} c + {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{2} + 2 \, {\left (7 \, B b^{3} c - 4 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (6 \, B c^{4} x^{3} - 105 \, B b^{3} c + 60 \, A b^{2} c^{2} - 3 \, {\left (7 \, B b c^{3} - 4 \, A c^{4}\right )} x^{2} - 20 \, {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}, -\frac {15 \, {\left (7 \, B b^{4} - 4 \, A b^{3} c + {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{2} + 2 \, {\left (7 \, B b^{3} c - 4 \, A b^{2} c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (6 \, B c^{4} x^{3} - 105 \, B b^{3} c + 60 \, A b^{2} c^{2} - 3 \, {\left (7 \, B b c^{3} - 4 \, A c^{4}\right )} x^{2} - 20 \, {\left (7 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{12 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}\right ] \] Input:

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

Output:

[-1/24*(15*(7*B*b^4 - 4*A*b^3*c + (7*B*b^2*c^2 - 4*A*b*c^3)*x^2 + 2*(7*B*b 
^3*c - 4*A*b^2*c^2)*x)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c) 
) - 2*(6*B*c^4*x^3 - 105*B*b^3*c + 60*A*b^2*c^2 - 3*(7*B*b*c^3 - 4*A*c^4)* 
x^2 - 20*(7*B*b^2*c^2 - 4*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^7*x^2 + 2*b*c^ 
6*x + b^2*c^5), -1/12*(15*(7*B*b^4 - 4*A*b^3*c + (7*B*b^2*c^2 - 4*A*b*c^3) 
*x^2 + 2*(7*B*b^3*c - 4*A*b^2*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sq 
rt(-c)/(c*x + b)) - (6*B*c^4*x^3 - 105*B*b^3*c + 60*A*b^2*c^2 - 3*(7*B*b*c 
^3 - 4*A*c^4)*x^2 - 20*(7*B*b^2*c^2 - 4*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^ 
7*x^2 + 2*b*c^6*x + b^2*c^5)]
 

Sympy [F]

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

integrate(x**5*(B*x+A)/(c*x**2+b*x)**(5/2),x)
 

Output:

Integral(x**5*(A + B*x)/(x*(b + c*x))**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (131) = 262\).

Time = 0.04 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.34 \[ \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {B x^{5}}{2 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {35 \, B b^{2} x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )}}{24 \, c^{2}} + \frac {5 \, A b x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {b x}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} - \frac {2 \, x}{\sqrt {c x^{2} + b x} b c} - \frac {1}{\sqrt {c x^{2} + b x} c^{2}}\right )}}{6 \, c} - \frac {7 \, B b x^{4}}{4 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {A x^{4}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {35 \, B b^{2} x}{6 \, \sqrt {c x^{2} + b x} c^{4}} + \frac {10 \, A b x}{3 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {35 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {9}{2}}} - \frac {5 \, A b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {7}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} B b}{12 \, c^{4}} + \frac {5 \, \sqrt {c x^{2} + b x} A}{3 \, c^{3}} \] Input:

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

Output:

1/2*B*x^5/((c*x^2 + b*x)^(3/2)*c) - 35/24*B*b^2*x*(3*x^2/((c*x^2 + b*x)^(3 
/2)*c) + b*x/((c*x^2 + b*x)^(3/2)*c^2) - 2*x/(sqrt(c*x^2 + b*x)*b*c) - 1/( 
sqrt(c*x^2 + b*x)*c^2))/c^2 + 5/6*A*b*x*(3*x^2/((c*x^2 + b*x)^(3/2)*c) + b 
*x/((c*x^2 + b*x)^(3/2)*c^2) - 2*x/(sqrt(c*x^2 + b*x)*b*c) - 1/(sqrt(c*x^2 
 + b*x)*c^2))/c - 7/4*B*b*x^4/((c*x^2 + b*x)^(3/2)*c^2) + A*x^4/((c*x^2 + 
b*x)^(3/2)*c) - 35/6*B*b^2*x/(sqrt(c*x^2 + b*x)*c^4) + 10/3*A*b*x/(sqrt(c* 
x^2 + b*x)*c^3) + 35/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/ 
c^(9/2) - 5/2*A*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 3 
5/12*sqrt(c*x^2 + b*x)*B*b/c^4 + 5/3*sqrt(c*x^2 + b*x)*A/c^3
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.59 \[ \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, B x}{c^{3}} - \frac {11 \, B b c^{7} - 4 \, A c^{8}}{c^{11}}\right )} - \frac {5 \, {\left (7 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {9}{2}}} - \frac {2 \, {\left (12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} c - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c^{2} + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} B b^{4} \sqrt {c} - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} c^{\frac {3}{2}} + 10 \, B b^{5} - 7 \, A b^{4} c\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )}^{3} c^{\frac {9}{2}}} \] Input:

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

Output:

1/4*sqrt(c*x^2 + b*x)*(2*B*x/c^3 - (11*B*b*c^7 - 4*A*c^8)/c^11) - 5/8*(7*B 
*b^2 - 4*A*b*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^ 
(9/2) - 2/3*(12*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3*c - 9*(sqrt(c)*x - 
 sqrt(c*x^2 + b*x))^2*A*b^2*c^2 + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))*B*b^4 
*sqrt(c) - 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*c^(3/2) + 10*B*b^5 - 7 
*A*b^4*c)/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)^3*c^(9/2))
 

Mupad [F(-1)]

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

int((x^5*(A + B*x))/(b*x + c*x^2)^(5/2),x)
 

Output:

int((x^5*(A + B*x))/(b*x + c*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.85 \[ \int \frac {x^5 (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {-480 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a \,b^{2} c -480 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a b \,c^{2} x +840 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{4}+840 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{3} c x -80 \sqrt {c}\, \sqrt {c x +b}\, a \,b^{2} c -80 \sqrt {c}\, \sqrt {c x +b}\, a b \,c^{2} x +175 \sqrt {c}\, \sqrt {c x +b}\, b^{4}+175 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c x +480 \sqrt {x}\, a \,b^{2} c^{2}+640 \sqrt {x}\, a b \,c^{3} x +96 \sqrt {x}\, a \,c^{4} x^{2}-840 \sqrt {x}\, b^{4} c -1120 \sqrt {x}\, b^{3} c^{2} x -168 \sqrt {x}\, b^{2} c^{3} x^{2}+48 \sqrt {x}\, b \,c^{4} x^{3}}{96 \sqrt {c x +b}\, c^{5} \left (c x +b \right )} \] Input:

int(x^5*(B*x+A)/(c*x^2+b*x)^(5/2),x)
 

Output:

( - 480*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b 
))*a*b**2*c - 480*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt( 
c))/sqrt(b))*a*b*c**2*x + 840*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + s 
qrt(x)*sqrt(c))/sqrt(b))*b**4 + 840*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c* 
x) + sqrt(x)*sqrt(c))/sqrt(b))*b**3*c*x - 80*sqrt(c)*sqrt(b + c*x)*a*b**2* 
c - 80*sqrt(c)*sqrt(b + c*x)*a*b*c**2*x + 175*sqrt(c)*sqrt(b + c*x)*b**4 + 
 175*sqrt(c)*sqrt(b + c*x)*b**3*c*x + 480*sqrt(x)*a*b**2*c**2 + 640*sqrt(x 
)*a*b*c**3*x + 96*sqrt(x)*a*c**4*x**2 - 840*sqrt(x)*b**4*c - 1120*sqrt(x)* 
b**3*c**2*x - 168*sqrt(x)*b**2*c**3*x**2 + 48*sqrt(x)*b*c**4*x**3)/(96*sqr 
t(b + c*x)*c**5*(b + c*x))