\(\int \frac {x^4 (A+B x)}{(b x+c x^2)^{3/2}} \, dx\) [153]

Optimal result

Integrand size = 22, antiderivative size = 156 \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 (b B-A c) x^3}{c^2 \sqrt {b x+c x^2}}+\frac {5 b (7 b B-6 A c) \sqrt {b x+c x^2}}{8 c^4}-\frac {5 (7 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^3}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c^2}-\frac {5 b^2 (7 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{9/2}} \] Output:

2*(-A*c+B*b)*x^3/c^2/(c*x^2+b*x)^(1/2)+5/8*b*(-6*A*c+7*B*b)*(c*x^2+b*x)^(1 
/2)/c^4-5/12*(-6*A*c+7*B*b)*x*(c*x^2+b*x)^(1/2)/c^3+1/3*B*x^2*(c*x^2+b*x)^ 
(1/2)/c^2-5/8*b^2*(-6*A*c+7*B*b)*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))/c^(9 
/2)
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {x^{3/2} \left (\sqrt {c} \sqrt {x} (b+c x) \left (105 b^3 B+4 c^3 x^2 (3 A+2 B x)-2 b c^2 x (15 A+7 B x)+b^2 (-90 A c+35 B c x)\right )-30 b^2 (7 b B-6 A c) (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{24 c^{9/2} (x (b+c x))^{3/2}} \] Input:

Integrate[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]
 

Output:

(x^(3/2)*(Sqrt[c]*Sqrt[x]*(b + c*x)*(105*b^3*B + 4*c^3*x^2*(3*A + 2*B*x) - 
 2*b*c^2*x*(15*A + 7*B*x) + b^2*(-90*A*c + 35*B*c*x)) - 30*b^2*(7*b*B - 6* 
A*c)*(b + c*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])] 
))/(24*c^(9/2)*(x*(b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1211, 25, 2192, 27, 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.

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

\(\Big \downarrow \) 1211

\(\displaystyle \frac {\int -\frac {-B c^3 x^3+c^2 (b B-A c) x^2-b c (b B-A c) x+b^2 (b B-A c)}{\sqrt {c x^2+b x}}dx}{c^4}+\frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\int \frac {-B c^3 x^3+c^2 (b B-A c) x^2-b c (b B-A c) x+b^2 (b B-A c)}{\sqrt {c x^2+b x}}dx}{c^4}\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\int \frac {(11 b B-6 A c) x^2 c^3-6 b (b B-A c) x c^2+6 b^2 (b B-A c) c}{2 \sqrt {c x^2+b x}}dx}{3 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\int \frac {(11 b B-6 A c) x^2 c^3-6 b (b B-A c) x c^2+6 b^2 (b B-A c) c}{\sqrt {c x^2+b x}}dx}{6 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\frac {\int \frac {3 b c^2 (8 b (b B-A c)-c (19 b B-14 A c) x)}{2 \sqrt {c x^2+b x}}dx}{2 c}+\frac {1}{2} c^2 x \sqrt {b x+c x^2} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\frac {3}{4} b c \int \frac {8 b (b B-A c)-c (19 b B-14 A c) x}{\sqrt {c x^2+b x}}dx+\frac {1}{2} c^2 x \sqrt {b x+c x^2} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\frac {3}{4} b c \left (\frac {5}{2} b (7 b B-6 A c) \int \frac {1}{\sqrt {c x^2+b x}}dx-\sqrt {b x+c x^2} (19 b B-14 A c)\right )+\frac {1}{2} c^2 x \sqrt {b x+c x^2} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\frac {3}{4} b c \left (5 b (7 b B-6 A c) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}-\sqrt {b x+c x^2} (19 b B-14 A c)\right )+\frac {1}{2} c^2 x \sqrt {b x+c x^2} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 b^2 x (b B-A c)}{c^4 \sqrt {b x+c x^2}}-\frac {\frac {\frac {3}{4} b c \left (\frac {5 b (7 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-\sqrt {b x+c x^2} (19 b B-14 A c)\right )+\frac {1}{2} c^2 x \sqrt {b x+c x^2} (11 b B-6 A c)}{6 c}-\frac {1}{3} B c^2 x^2 \sqrt {b x+c x^2}}{c^4}\)

Input:

Int[(x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x]
 

Output:

(2*b^2*(b*B - A*c)*x)/(c^4*Sqrt[b*x + c*x^2]) - (-1/3*(B*c^2*x^2*Sqrt[b*x 
+ c*x^2]) + ((c^2*(11*b*B - 6*A*c)*x*Sqrt[b*x + c*x^2])/2 + (3*b*c*(-((19* 
b*B - 14*A*c)*Sqrt[b*x + c*x^2]) + (5*b*(7*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x 
)/Sqrt[b*x + c*x^2]])/Sqrt[c]))/4)/(6*c))/c^4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 1211

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( 
e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* 
x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2))   Int[ExpandToSum[((2*c*d - b 
*e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) 
*(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free 
Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] 
&& IGtQ[n, 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 = 1.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {5 \left (-3 \left (A c -\frac {7 B b}{6}\right ) b^{2} \sqrt {x \left (c x +b \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+x \left (3 \left (-\frac {7 B x}{18}+A \right ) b^{2} c^{\frac {3}{2}}+b x \left (\frac {7 B x}{15}+A \right ) c^{\frac {5}{2}}-\frac {2 x^{2} \left (\frac {2 B x}{3}+A \right ) c^{\frac {7}{2}}}{5}-\frac {7 B \sqrt {c}\, b^{3}}{2}\right )\right )}{4 \sqrt {x \left (c x +b \right )}\, c^{\frac {9}{2}}}\)	\(107\)
risch	\(-\frac {\left (-8 B \,c^{2} x^{2}-12 A \,c^{2} x +22 B b c x +42 A b c -57 B \,b^{2}\right ) x \left (c x +b \right )}{24 c^{4} \sqrt {x \left (c x +b \right )}}+\frac {b^{2} \left (30 A \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-\frac {35 B b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-\frac {32 \left (A c -B b \right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{c \left (\frac {b}{c}+x \right )}\right )}{16 c^{4}}\)	\(172\)
default	\(A \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )+B \left (\frac {x^{4}}{3 c \sqrt {c \,x^{2}+b x}}-\frac {7 b \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )}{6 c}\right )\)	\(320\)

Input:

int(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-5/4/(x*(c*x+b))^(1/2)/c^(9/2)*(-3*(A*c-7/6*B*b)*b^2*(x*(c*x+b))^(1/2)*arc 
tanh((x*(c*x+b))^(1/2)/x/c^(1/2))+x*(3*(-7/18*B*x+A)*b^2*c^(3/2)+b*x*(7/15 
*B*x+A)*c^(5/2)-2/5*x^2*(2/3*B*x+A)*c^(7/2)-7/2*B*c^(1/2)*b^3))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.01 \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{4} x^{3} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{2} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, {\left (c^{6} x + b c^{5}\right )}}, \frac {15 \, {\left (7 \, B b^{4} - 6 \, A b^{3} c + {\left (7 \, B b^{3} c - 6 \, A b^{2} c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) + {\left (8 \, B c^{4} x^{3} + 105 \, B b^{3} c - 90 \, A b^{2} c^{2} - 2 \, {\left (7 \, B b c^{3} - 6 \, A c^{4}\right )} x^{2} + 5 \, {\left (7 \, B b^{2} c^{2} - 6 \, A b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, {\left (c^{6} x + b c^{5}\right )}}\right ] \] Input:

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
 

Output:

[-1/48*(15*(7*B*b^4 - 6*A*b^3*c + (7*B*b^3*c - 6*A*b^2*c^2)*x)*sqrt(c)*log 
(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^4*x^3 + 105*B*b^3*c - 
 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4)*x^2 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)* 
x)*sqrt(c*x^2 + b*x))/(c^6*x + b*c^5), 1/24*(15*(7*B*b^4 - 6*A*b^3*c + (7* 
B*b^3*c - 6*A*b^2*c^2)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x 
+ b)) + (8*B*c^4*x^3 + 105*B*b^3*c - 90*A*b^2*c^2 - 2*(7*B*b*c^3 - 6*A*c^4 
)*x^2 + 5*(7*B*b^2*c^2 - 6*A*b*c^3)*x)*sqrt(c*x^2 + b*x))/(c^6*x + b*c^5)]
 

Sympy [F]

\[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{4} \left (A + B x\right )}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**4*(B*x+A)/(c*x**2+b*x)**(3/2),x)
 

Output:

Integral(x**4*(A + B*x)/(x*(b + c*x))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.36 \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {B x^{4}}{3 \, \sqrt {c x^{2} + b x} c} - \frac {7 \, B b x^{3}}{12 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {A x^{3}}{2 \, \sqrt {c x^{2} + b x} c} + \frac {35 \, B b^{2} x^{2}}{24 \, \sqrt {c x^{2} + b x} c^{3}} - \frac {5 \, A b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {35 \, B b^{3} x}{8 \, \sqrt {c x^{2} + b x} c^{4}} - \frac {15 \, A b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} - \frac {35 \, B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {9}{2}}} + \frac {15 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} \] Input:

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
 

Output:

1/3*B*x^4/(sqrt(c*x^2 + b*x)*c) - 7/12*B*b*x^3/(sqrt(c*x^2 + b*x)*c^2) + 1 
/2*A*x^3/(sqrt(c*x^2 + b*x)*c) + 35/24*B*b^2*x^2/(sqrt(c*x^2 + b*x)*c^3) - 
 5/4*A*b*x^2/(sqrt(c*x^2 + b*x)*c^2) + 35/8*B*b^3*x/(sqrt(c*x^2 + b*x)*c^4 
) - 15/4*A*b^2*x/(sqrt(c*x^2 + b*x)*c^3) - 35/16*B*b^3*log(2*c*x + b + 2*s 
qrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) + 15/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^ 
2 + b*x)*sqrt(c))/c^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.05 \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, B x}{c^{2}} - \frac {11 \, B b c^{10} - 6 \, A c^{11}}{c^{13}}\right )} + \frac {3 \, {\left (19 \, B b^{2} c^{9} - 14 \, A b c^{10}\right )}}{c^{13}}\right )} + \frac {5 \, {\left (7 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {9}{2}}} + \frac {2 \, {\left (B b^{4} \sqrt {c} - A b^{3} c^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} c^{5}} \] Input:

integrate(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")
 

Output:

1/24*sqrt(c*x^2 + b*x)*(2*x*(4*B*x/c^2 - (11*B*b*c^10 - 6*A*c^11)/c^13) + 
3*(19*B*b^2*c^9 - 14*A*b*c^10)/c^13) + 5/16*(7*B*b^3 - 6*A*b^2*c)*log(abs( 
2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(9/2) + 2*(B*b^4*sqrt(c) 
 - A*b^3*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)*c^5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \] Input:

int((x^4*(A + B*x))/(b*x + c*x^2)^(3/2),x)
 

Output:

int((x^4*(A + B*x))/(b*x + c*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.16 \[ \int \frac {x^4 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {720 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a \,b^{2} c -840 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{4}-480 \sqrt {c}\, \sqrt {c x +b}\, a \,b^{2} c +525 \sqrt {c}\, \sqrt {c x +b}\, b^{4}-720 \sqrt {x}\, a \,b^{2} c^{2}-240 \sqrt {x}\, a b \,c^{3} x +96 \sqrt {x}\, a \,c^{4} x^{2}+840 \sqrt {x}\, b^{4} c +280 \sqrt {x}\, b^{3} c^{2} x -112 \sqrt {x}\, b^{2} c^{3} x^{2}+64 \sqrt {x}\, b \,c^{4} x^{3}}{192 \sqrt {c x +b}\, c^{5}} \] Input:

int(x^4*(B*x+A)/(c*x^2+b*x)^(3/2),x)
 

Output:

(720*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))* 
a*b**2*c - 840*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c)) 
/sqrt(b))*b**4 - 480*sqrt(c)*sqrt(b + c*x)*a*b**2*c + 525*sqrt(c)*sqrt(b + 
 c*x)*b**4 - 720*sqrt(x)*a*b**2*c**2 - 240*sqrt(x)*a*b*c**3*x + 96*sqrt(x) 
*a*c**4*x**2 + 840*sqrt(x)*b**4*c + 280*sqrt(x)*b**3*c**2*x - 112*sqrt(x)* 
b**2*c**3*x**2 + 64*sqrt(x)*b*c**4*x**3)/(192*sqrt(b + c*x)*c**5)