\(\int \frac {A+B x}{x^4 (b x+c x^2)^{3/2}} \, dx\) [161]

Optimal result

Integrand size = 22, antiderivative size = 192 \[ \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}+\frac {2 (9 b B-10 A c)}{9 b^2 x^3 \sqrt {b x+c x^2}}-\frac {16 (9 b B-10 A c) \sqrt {b x+c x^2}}{63 b^3 x^4}+\frac {32 c (9 b B-10 A c) \sqrt {b x+c x^2}}{105 b^4 x^3}-\frac {128 c^2 (9 b B-10 A c) \sqrt {b x+c x^2}}{315 b^5 x^2}+\frac {256 c^3 (9 b B-10 A c) \sqrt {b x+c x^2}}{315 b^6 x} \] Output:

-2/9*A/b/x^4/(c*x^2+b*x)^(1/2)+2/9*(-10*A*c+9*B*b)/b^2/x^3/(c*x^2+b*x)^(1/ 
2)-16/63*(-10*A*c+9*B*b)*(c*x^2+b*x)^(1/2)/b^3/x^4+32/105*c*(-10*A*c+9*B*b 
)*(c*x^2+b*x)^(1/2)/b^4/x^3-128/315*c^2*(-10*A*c+9*B*b)*(c*x^2+b*x)^(1/2)/ 
b^5/x^2+256/315*c^3*(-10*A*c+9*B*b)*(c*x^2+b*x)^(1/2)/b^6/x
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.64 \[ \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \left (9 b B x \left (5 b^4-8 b^3 c x+16 b^2 c^2 x^2-64 b c^3 x^3-128 c^4 x^4\right )+5 A \left (7 b^5-10 b^4 c x+16 b^3 c^2 x^2-32 b^2 c^3 x^3+128 b c^4 x^4+256 c^5 x^5\right )\right )}{315 b^6 x^4 \sqrt {x (b+c x)}} \] Input:

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

Output:

(-2*(9*b*B*x*(5*b^4 - 8*b^3*c*x + 16*b^2*c^2*x^2 - 64*b*c^3*x^3 - 128*c^4* 
x^4) + 5*A*(7*b^5 - 10*b^4*c*x + 16*b^3*c^2*x^2 - 32*b^2*c^3*x^3 + 128*b*c 
^4*x^4 + 256*c^5*x^5)))/(315*b^6*x^4*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1220, 1129, 1129, 1129, 1088}

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.

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

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(9 b B-10 A c) \int \frac {1}{x^3 \left (c x^2+b x\right )^{3/2}}dx}{9 b}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(9 b B-10 A c) \left (-\frac {8 c \int \frac {1}{x^2 \left (c x^2+b x\right )^{3/2}}dx}{7 b}-\frac {2}{7 b x^3 \sqrt {b x+c x^2}}\right )}{9 b}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(9 b B-10 A c) \left (-\frac {8 c \left (-\frac {6 c \int \frac {1}{x \left (c x^2+b x\right )^{3/2}}dx}{5 b}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}\right )}{7 b}-\frac {2}{7 b x^3 \sqrt {b x+c x^2}}\right )}{9 b}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(9 b B-10 A c) \left (-\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \int \frac {1}{\left (c x^2+b x\right )^{3/2}}dx}{3 b}-\frac {2}{3 b x \sqrt {b x+c x^2}}\right )}{5 b}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}\right )}{7 b}-\frac {2}{7 b x^3 \sqrt {b x+c x^2}}\right )}{9 b}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1088

\(\displaystyle \frac {\left (-\frac {8 c \left (-\frac {6 c \left (\frac {8 c (b+2 c x)}{3 b^3 \sqrt {b x+c x^2}}-\frac {2}{3 b x \sqrt {b x+c x^2}}\right )}{5 b}-\frac {2}{5 b x^2 \sqrt {b x+c x^2}}\right )}{7 b}-\frac {2}{7 b x^3 \sqrt {b x+c x^2}}\right ) (9 b B-10 A c)}{9 b}-\frac {2 A}{9 b x^4 \sqrt {b x+c x^2}}\)

Input:

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

Output:

(-2*A)/(9*b*x^4*Sqrt[b*x + c*x^2]) + ((9*b*B - 10*A*c)*(-2/(7*b*x^3*Sqrt[b 
*x + c*x^2]) - (8*c*(-2/(5*b*x^2*Sqrt[b*x + c*x^2]) - (6*c*(-2/(3*b*x*Sqrt 
[b*x + c*x^2]) + (8*c*(b + 2*c*x))/(3*b^3*Sqrt[b*x + c*x^2])))/(5*b)))/(7* 
b)))/(9*b)
 

Defintions of rubi rules used

rule 1088

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 
2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0]
 

rule 1129

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

rule 1220

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.53

method	result	size
pseudoelliptic	\(\frac {\left (-90 B x -70 A \right ) b^{5}+100 c \left (\frac {36 B x}{25}+A \right ) x \,b^{4}-160 c^{2} x^{2} \left (\frac {9 B x}{5}+A \right ) b^{3}+320 c^{3} x^{3} \left (\frac {18 B x}{5}+A \right ) b^{2}-1280 c^{4} x^{4} \left (-\frac {9 B x}{5}+A \right ) b -2560 A \,c^{5} x^{5}}{315 \sqrt {x \left (c x +b \right )}\, x^{4} b^{6}}\)	\(102\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (1280 A \,c^{5} x^{5}-1152 B b \,c^{4} x^{5}+640 A b \,c^{4} x^{4}-576 B \,b^{2} c^{3} x^{4}-160 A \,b^{2} c^{3} x^{3}+144 B \,b^{3} c^{2} x^{3}+80 A \,b^{3} c^{2} x^{2}-72 B \,b^{4} c \,x^{2}-50 A \,b^{4} c x +45 B \,b^{5} x +35 A \,b^{5}\right )}{315 x^{3} b^{6} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(134\)
orering	\(-\frac {2 \left (c x +b \right ) \left (1280 A \,c^{5} x^{5}-1152 B b \,c^{4} x^{5}+640 A b \,c^{4} x^{4}-576 B \,b^{2} c^{3} x^{4}-160 A \,b^{2} c^{3} x^{3}+144 B \,b^{3} c^{2} x^{3}+80 A \,b^{3} c^{2} x^{2}-72 B \,b^{4} c \,x^{2}-50 A \,b^{4} c x +45 B \,b^{5} x +35 A \,b^{5}\right )}{315 x^{3} b^{6} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(134\)
risch	\(-\frac {2 \left (c x +b \right ) \left (965 A \,c^{4} x^{4}-837 B b \,c^{3} x^{4}-325 A b \,c^{3} x^{3}+261 B \,b^{2} c^{2} x^{3}+165 A \,b^{2} c^{2} x^{2}-117 B \,b^{3} c \,x^{2}-85 A \,b^{3} c x +45 B \,b^{4} x +35 A \,b^{4}\right )}{315 b^{6} x^{4} \sqrt {x \left (c x +b \right )}}-\frac {2 c^{4} \left (A c -B b \right ) x}{\sqrt {x \left (c x +b \right )}\, b^{6}}\)	\(135\)
trager	\(-\frac {2 \left (1280 A \,c^{5} x^{5}-1152 B b \,c^{4} x^{5}+640 A b \,c^{4} x^{4}-576 B \,b^{2} c^{3} x^{4}-160 A \,b^{2} c^{3} x^{3}+144 B \,b^{3} c^{2} x^{3}+80 A \,b^{3} c^{2} x^{2}-72 B \,b^{4} c \,x^{2}-50 A \,b^{4} c x +45 B \,b^{5} x +35 A \,b^{5}\right ) \sqrt {c \,x^{2}+b x}}{315 \left (c x +b \right ) b^{6} x^{5}}\)	\(136\)
default	\(A \left (-\frac {2}{9 b \,x^{4} \sqrt {c \,x^{2}+b x}}-\frac {10 c \left (-\frac {2}{7 b \,x^{3} \sqrt {c \,x^{2}+b x}}-\frac {8 c \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\right )}{9 b}\right )+B \left (-\frac {2}{7 b \,x^{3} \sqrt {c \,x^{2}+b x}}-\frac {8 c \left (-\frac {2}{5 b \,x^{2} \sqrt {c \,x^{2}+b x}}-\frac {6 c \left (-\frac {2}{3 b x \sqrt {c \,x^{2}+b x}}+\frac {8 c \left (2 c x +b \right )}{3 b^{3} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\right )\)	\(222\)

Input:

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

Output:

1/315*((-90*B*x-70*A)*b^5+100*c*(36/25*B*x+A)*x*b^4-160*c^2*x^2*(9/5*B*x+A 
)*b^3+320*c^3*x^3*(18/5*B*x+A)*b^2-1280*c^4*x^4*(-9/5*B*x+A)*b-2560*A*c^5* 
x^5)/(x*(c*x+b))^(1/2)/x^4/b^6
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (35 \, A b^{5} - 128 \, {\left (9 \, B b c^{4} - 10 \, A c^{5}\right )} x^{5} - 64 \, {\left (9 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 16 \, {\left (9 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{3} - 8 \, {\left (9 \, B b^{4} c - 10 \, A b^{3} c^{2}\right )} x^{2} + 5 \, {\left (9 \, B b^{5} - 10 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x}}{315 \, {\left (b^{6} c x^{6} + b^{7} x^{5}\right )}} \] Input:

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

Output:

-2/315*(35*A*b^5 - 128*(9*B*b*c^4 - 10*A*c^5)*x^5 - 64*(9*B*b^2*c^3 - 10*A 
*b*c^4)*x^4 + 16*(9*B*b^3*c^2 - 10*A*b^2*c^3)*x^3 - 8*(9*B*b^4*c - 10*A*b^ 
3*c^2)*x^2 + 5*(9*B*b^5 - 10*A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^6*c*x^6 + b^ 
7*x^5)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {256 \, B c^{4} x}{35 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {512 \, A c^{5} x}{63 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {128 \, B c^{3}}{35 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {256 \, A c^{4}}{63 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {32 \, B c^{2}}{35 \, \sqrt {c x^{2} + b x} b^{3} x} + \frac {64 \, A c^{3}}{63 \, \sqrt {c x^{2} + b x} b^{4} x} + \frac {16 \, B c}{35 \, \sqrt {c x^{2} + b x} b^{2} x^{2}} - \frac {32 \, A c^{2}}{63 \, \sqrt {c x^{2} + b x} b^{3} x^{2}} - \frac {2 \, B}{7 \, \sqrt {c x^{2} + b x} b x^{3}} + \frac {20 \, A c}{63 \, \sqrt {c x^{2} + b x} b^{2} x^{3}} - \frac {2 \, A}{9 \, \sqrt {c x^{2} + b x} b x^{4}} \] Input:

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

Output:

256/35*B*c^4*x/(sqrt(c*x^2 + b*x)*b^5) - 512/63*A*c^5*x/(sqrt(c*x^2 + b*x) 
*b^6) + 128/35*B*c^3/(sqrt(c*x^2 + b*x)*b^4) - 256/63*A*c^4/(sqrt(c*x^2 + 
b*x)*b^5) - 32/35*B*c^2/(sqrt(c*x^2 + b*x)*b^3*x) + 64/63*A*c^3/(sqrt(c*x^ 
2 + b*x)*b^4*x) + 16/35*B*c/(sqrt(c*x^2 + b*x)*b^2*x^2) - 32/63*A*c^2/(sqr 
t(c*x^2 + b*x)*b^3*x^2) - 2/7*B/(sqrt(c*x^2 + b*x)*b*x^3) + 20/63*A*c/(sqr 
t(c*x^2 + b*x)*b^2*x^3) - 2/9*A/(sqrt(c*x^2 + b*x)*b*x^4)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.75 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {1300\,A\,c^5-1044\,B\,b\,c^4}{315\,b^6}-\frac {4\,c^4\,\left (965\,A\,c-837\,B\,b\right )}{315\,b^6}\right )-\frac {2\,c^3\,\left (965\,A\,c-837\,B\,b\right )}{315\,b^5}\right )}{x\,\left (b+c\,x\right )}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{9\,b^2\,x^5}-\frac {\left (18\,B\,b^2-34\,A\,b\,c\right )\,\sqrt {c\,x^2+b\,x}}{63\,b^4\,x^4}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}\,\left (55\,A\,c-39\,B\,b\right )}{105\,b^4\,x^3}+\frac {2\,c^2\,\sqrt {c\,x^2+b\,x}\,\left (325\,A\,c-261\,B\,b\right )}{315\,b^5\,x^2} \] Input:

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

Output:

((b*x + c*x^2)^(1/2)*(x*((1300*A*c^5 - 1044*B*b*c^4)/(315*b^6) - (4*c^4*(9 
65*A*c - 837*B*b))/(315*b^6)) - (2*c^3*(965*A*c - 837*B*b))/(315*b^5)))/(x 
*(b + c*x)) - (2*A*(b*x + c*x^2)^(1/2))/(9*b^2*x^5) - ((18*B*b^2 - 34*A*b* 
c)*(b*x + c*x^2)^(1/2))/(63*b^4*x^4) - (2*c*(b*x + c*x^2)^(1/2)*(55*A*c - 
39*B*b))/(105*b^4*x^3) + (2*c^2*(b*x + c*x^2)^(1/2)*(325*A*c - 261*B*b))/( 
315*b^5*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{x^4 \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {512 \sqrt {c}\, \sqrt {c x +b}\, a \,c^{4} x^{5}}{63}-\frac {256 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{3} x^{5}}{35}-\frac {2 \sqrt {x}\, a \,b^{5}}{9}+\frac {20 \sqrt {x}\, a \,b^{4} c x}{63}-\frac {32 \sqrt {x}\, a \,b^{3} c^{2} x^{2}}{63}+\frac {64 \sqrt {x}\, a \,b^{2} c^{3} x^{3}}{63}-\frac {256 \sqrt {x}\, a b \,c^{4} x^{4}}{63}-\frac {512 \sqrt {x}\, a \,c^{5} x^{5}}{63}-\frac {2 \sqrt {x}\, b^{6} x}{7}+\frac {16 \sqrt {x}\, b^{5} c \,x^{2}}{35}-\frac {32 \sqrt {x}\, b^{4} c^{2} x^{3}}{35}+\frac {128 \sqrt {x}\, b^{3} c^{3} x^{4}}{35}+\frac {256 \sqrt {x}\, b^{2} c^{4} x^{5}}{35}}{\sqrt {c x +b}\, b^{6} x^{5}} \] Input:

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

Output:

(2*(1280*sqrt(c)*sqrt(b + c*x)*a*c**4*x**5 - 1152*sqrt(c)*sqrt(b + c*x)*b* 
*2*c**3*x**5 - 35*sqrt(x)*a*b**5 + 50*sqrt(x)*a*b**4*c*x - 80*sqrt(x)*a*b* 
*3*c**2*x**2 + 160*sqrt(x)*a*b**2*c**3*x**3 - 640*sqrt(x)*a*b*c**4*x**4 - 
1280*sqrt(x)*a*c**5*x**5 - 45*sqrt(x)*b**6*x + 72*sqrt(x)*b**5*c*x**2 - 14 
4*sqrt(x)*b**4*c**2*x**3 + 576*sqrt(x)*b**3*c**3*x**4 + 1152*sqrt(x)*b**2* 
c**4*x**5))/(315*sqrt(b + c*x)*b**6*x**5)