3.1.77 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx\) [77]

3.1.77.1 Optimal result
3.1.77.2 Mathematica [A] (verified)
3.1.77.3 Rubi [A] (verified)
3.1.77.4 Maple [A] (verified)
3.1.77.5 Fricas [A] (verification not implemented)
3.1.77.6 Sympy [F]
3.1.77.7 Maxima [A] (verification not implemented)
3.1.77.8 Giac [B] (verification not implemented)
3.1.77.9 Mupad [B] (verification not implemented)

3.1.77.1 Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}-\frac {2 (11 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}+\frac {4 c (11 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{231 b^3 x^5}-\frac {16 c^2 (11 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{1155 b^4 x^4}+\frac {32 c^3 (11 b B-8 A c) \left (b x+c x^2\right )^{3/2}}{3465 b^5 x^3} \]

output
-2/11*A*(c*x^2+b*x)^(3/2)/b/x^7-2/99*(-8*A*c+11*B*b)*(c*x^2+b*x)^(3/2)/b^2 
/x^6+4/231*c*(-8*A*c+11*B*b)*(c*x^2+b*x)^(3/2)/b^3/x^5-16/1155*c^2*(-8*A*c 
+11*B*b)*(c*x^2+b*x)^(3/2)/b^4/x^4+32/3465*c^3*(-8*A*c+11*B*b)*(c*x^2+b*x) 
^(3/2)/b^5/x^3
 
3.1.77.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.62 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=\frac {2 (x (b+c x))^{3/2} \left (11 b B x \left (-35 b^3+30 b^2 c x-24 b c^2 x^2+16 c^3 x^3\right )+A \left (-315 b^4+280 b^3 c x-240 b^2 c^2 x^2+192 b c^3 x^3-128 c^4 x^4\right )\right )}{3465 b^5 x^7} \]

input
Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^7,x]
 
output
(2*(x*(b + c*x))^(3/2)*(11*b*B*x*(-35*b^3 + 30*b^2*c*x - 24*b*c^2*x^2 + 16 
*c^3*x^3) + A*(-315*b^4 + 280*b^3*c*x - 240*b^2*c^2*x^2 + 192*b*c^3*x^3 - 
128*c^4*x^4)))/(3465*b^5*x^7)
 
3.1.77.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.96, 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, 1123}

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.

\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(11 b B-8 A c) \int \frac {\sqrt {c x^2+b x}}{x^6}dx}{11 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(11 b B-8 A c) \left (-\frac {2 c \int \frac {\sqrt {c x^2+b x}}{x^5}dx}{3 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}\right )}{11 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(11 b B-8 A c) \left (-\frac {2 c \left (-\frac {4 c \int \frac {\sqrt {c x^2+b x}}{x^4}dx}{7 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}\right )}{3 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}\right )}{11 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {(11 b B-8 A c) \left (-\frac {2 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {\sqrt {c x^2+b x}}{x^3}dx}{5 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}\right )}{7 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}\right )}{3 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}\right )}{11 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}\)

\(\Big \downarrow \) 1123

\(\displaystyle \frac {\left (-\frac {2 c \left (-\frac {4 c \left (\frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}\right )}{7 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{7 b x^5}\right )}{3 b}-\frac {2 \left (b x+c x^2\right )^{3/2}}{9 b x^6}\right ) (11 b B-8 A c)}{11 b}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}\)

input
Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^7,x]
 
output
(-2*A*(b*x + c*x^2)^(3/2))/(11*b*x^7) + ((11*b*B - 8*A*c)*((-2*(b*x + c*x^ 
2)^(3/2))/(9*b*x^6) - (2*c*((-2*(b*x + c*x^2)^(3/2))/(7*b*x^5) - (4*c*((-2 
*(b*x + c*x^2)^(3/2))/(5*b*x^4) + (4*c*(b*x + c*x^2)^(3/2))/(15*b^2*x^3))) 
/(7*b)))/(3*b)))/(11*b)
 

3.1.77.3.1 Defintions of rubi rules used

rule 1123
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)/((p + 1)*(2*c*d - b 
*e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 
0] && EqQ[m + 2*p + 2, 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 
]
 
3.1.77.4 Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55

method result size
pseudoelliptic \(-\frac {2 \left (\left (\frac {11 B x}{9}+A \right ) b^{4}-\frac {8 c \left (\frac {33 B x}{28}+A \right ) x \,b^{3}}{9}+\frac {16 c^{2} \left (\frac {11 B x}{10}+A \right ) x^{2} b^{2}}{21}-\frac {64 c^{3} \left (\frac {11 B x}{12}+A \right ) x^{3} b}{105}+\frac {128 A \,c^{4} x^{4}}{315}\right ) \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}{11 x^{6} b^{5}}\) \(88\)
gosper \(-\frac {2 \left (c x +b \right ) \left (128 A \,c^{4} x^{4}-176 B b \,c^{3} x^{4}-192 A b \,c^{3} x^{3}+264 B \,b^{2} c^{2} x^{3}+240 A \,b^{2} c^{2} x^{2}-330 B \,b^{3} c \,x^{2}-280 A \,b^{3} c x +385 b^{4} B x +315 A \,b^{4}\right ) \sqrt {c \,x^{2}+b x}}{3465 x^{6} b^{5}}\) \(110\)
trager \(-\frac {2 \left (128 A \,c^{5} x^{5}-176 B b \,c^{4} x^{5}-64 A b \,c^{4} x^{4}+88 B \,b^{2} c^{3} x^{4}+48 A \,b^{2} c^{3} x^{3}-66 B \,b^{3} c^{2} x^{3}-40 A \,b^{3} c^{2} x^{2}+55 B \,b^{4} c \,x^{2}+35 A \,b^{4} c x +385 B \,b^{5} x +315 A \,b^{5}\right ) \sqrt {c \,x^{2}+b x}}{3465 x^{6} b^{5}}\) \(129\)
risch \(-\frac {2 \left (c x +b \right ) \left (128 A \,c^{5} x^{5}-176 B b \,c^{4} x^{5}-64 A b \,c^{4} x^{4}+88 B \,b^{2} c^{3} x^{4}+48 A \,b^{2} c^{3} x^{3}-66 B \,b^{3} c^{2} x^{3}-40 A \,b^{3} c^{2} x^{2}+55 B \,b^{4} c \,x^{2}+35 A \,b^{4} c x +385 B \,b^{5} x +315 A \,b^{5}\right )}{3465 x^{5} \sqrt {x \left (c x +b \right )}\, b^{5}}\) \(132\)
default \(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{11 b \,x^{7}}-\frac {8 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{9 b \,x^{6}}-\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\right )}{3 b}\right )}{11 b}\right )+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{9 b \,x^{6}}-\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\right )}{3 b}\right )\) \(216\)

input
int((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x,method=_RETURNVERBOSE)
 
output
-2/11*((11/9*B*x+A)*b^4-8/9*c*(33/28*B*x+A)*x*b^3+16/21*c^2*(11/10*B*x+A)* 
x^2*b^2-64/105*c^3*(11/12*B*x+A)*x^3*b+128/315*A*c^4*x^4)*(c*x+b)*(x*(c*x+ 
b))^(1/2)/x^6/b^5
 
3.1.77.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=-\frac {2 \, {\left (315 \, A b^{5} - 16 \, {\left (11 \, B b c^{4} - 8 \, A c^{5}\right )} x^{5} + 8 \, {\left (11 \, B b^{2} c^{3} - 8 \, A b c^{4}\right )} x^{4} - 6 \, {\left (11 \, B b^{3} c^{2} - 8 \, A b^{2} c^{3}\right )} x^{3} + 5 \, {\left (11 \, B b^{4} c - 8 \, A b^{3} c^{2}\right )} x^{2} + 35 \, {\left (11 \, B b^{5} + A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x}}{3465 \, b^{5} x^{6}} \]

input
integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x, algorithm="fricas")
 
output
-2/3465*(315*A*b^5 - 16*(11*B*b*c^4 - 8*A*c^5)*x^5 + 8*(11*B*b^2*c^3 - 8*A 
*b*c^4)*x^4 - 6*(11*B*b^3*c^2 - 8*A*b^2*c^3)*x^3 + 5*(11*B*b^4*c - 8*A*b^3 
*c^2)*x^2 + 35*(11*B*b^5 + A*b^4*c)*x)*sqrt(c*x^2 + b*x)/(b^5*x^6)
 
3.1.77.6 Sympy [F]

\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{7}}\, dx \]

input
integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**7,x)
 
output
Integral(sqrt(x*(b + c*x))*(A + B*x)/x**7, x)
 
3.1.77.7 Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.49 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=\frac {32 \, \sqrt {c x^{2} + b x} B c^{4}}{315 \, b^{4} x} - \frac {256 \, \sqrt {c x^{2} + b x} A c^{5}}{3465 \, b^{5} x} - \frac {16 \, \sqrt {c x^{2} + b x} B c^{3}}{315 \, b^{3} x^{2}} + \frac {128 \, \sqrt {c x^{2} + b x} A c^{4}}{3465 \, b^{4} x^{2}} + \frac {4 \, \sqrt {c x^{2} + b x} B c^{2}}{105 \, b^{2} x^{3}} - \frac {32 \, \sqrt {c x^{2} + b x} A c^{3}}{1155 \, b^{3} x^{3}} - \frac {2 \, \sqrt {c x^{2} + b x} B c}{63 \, b x^{4}} + \frac {16 \, \sqrt {c x^{2} + b x} A c^{2}}{693 \, b^{2} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} B}{9 \, x^{5}} - \frac {2 \, \sqrt {c x^{2} + b x} A c}{99 \, b x^{5}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{11 \, x^{6}} \]

input
integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x, algorithm="maxima")
 
output
32/315*sqrt(c*x^2 + b*x)*B*c^4/(b^4*x) - 256/3465*sqrt(c*x^2 + b*x)*A*c^5/ 
(b^5*x) - 16/315*sqrt(c*x^2 + b*x)*B*c^3/(b^3*x^2) + 128/3465*sqrt(c*x^2 + 
 b*x)*A*c^4/(b^4*x^2) + 4/105*sqrt(c*x^2 + b*x)*B*c^2/(b^2*x^3) - 32/1155* 
sqrt(c*x^2 + b*x)*A*c^3/(b^3*x^3) - 2/63*sqrt(c*x^2 + b*x)*B*c/(b*x^4) + 1 
6/693*sqrt(c*x^2 + b*x)*A*c^2/(b^2*x^4) - 2/9*sqrt(c*x^2 + b*x)*B/x^5 - 2/ 
99*sqrt(c*x^2 + b*x)*A*c/(b*x^5) - 2/11*sqrt(c*x^2 + b*x)*A/x^6
 
3.1.77.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (140) = 280\).

Time = 0.27 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.32 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=\frac {2 \, {\left (6930 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7} B c^{\frac {5}{2}} + 19404 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b c^{2} + 11088 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A c^{3} + 21945 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{2} c^{\frac {3}{2}} + 36960 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b c^{\frac {5}{2}} + 12375 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{3} c + 51480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{2} c^{2} + 3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{4} \sqrt {c} + 38115 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{3} c^{\frac {3}{2}} + 385 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{5} + 15785 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{4} c + 3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{5} \sqrt {c} + 315 \, A b^{6}\right )}}{3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{11}} \]

input
integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x, algorithm="giac")
 
output
2/3465*(6930*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*c^(5/2) + 19404*(sqrt(c)* 
x - sqrt(c*x^2 + b*x))^6*B*b*c^2 + 11088*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6 
*A*c^3 + 21945*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c^(3/2) + 36960*(sq 
rt(c)*x - sqrt(c*x^2 + b*x))^5*A*b*c^(5/2) + 12375*(sqrt(c)*x - sqrt(c*x^2 
 + b*x))^4*B*b^3*c + 51480*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^2*c^2 + 3 
465*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*sqrt(c) + 38115*(sqrt(c)*x - s 
qrt(c*x^2 + b*x))^3*A*b^3*c^(3/2) + 385*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2* 
B*b^5 + 15785*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^4*c + 3465*(sqrt(c)*x 
- sqrt(c*x^2 + b*x))*A*b^5*sqrt(c) + 315*A*b^6)/(sqrt(c)*x - sqrt(c*x^2 + 
b*x))^11
 
3.1.77.9 Mupad [B] (verification not implemented)

Time = 11.11 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.49 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=\frac {16\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{693\,b^2\,x^4}-\frac {2\,B\,\sqrt {c\,x^2+b\,x}}{9\,x^5}-\frac {2\,A\,c\,\sqrt {c\,x^2+b\,x}}{99\,b\,x^5}-\frac {2\,B\,c\,\sqrt {c\,x^2+b\,x}}{63\,b\,x^4}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{11\,x^6}-\frac {32\,A\,c^3\,\sqrt {c\,x^2+b\,x}}{1155\,b^3\,x^3}+\frac {128\,A\,c^4\,\sqrt {c\,x^2+b\,x}}{3465\,b^4\,x^2}-\frac {256\,A\,c^5\,\sqrt {c\,x^2+b\,x}}{3465\,b^5\,x}+\frac {4\,B\,c^2\,\sqrt {c\,x^2+b\,x}}{105\,b^2\,x^3}-\frac {16\,B\,c^3\,\sqrt {c\,x^2+b\,x}}{315\,b^3\,x^2}+\frac {32\,B\,c^4\,\sqrt {c\,x^2+b\,x}}{315\,b^4\,x} \]

input
int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^7,x)
 
output
(16*A*c^2*(b*x + c*x^2)^(1/2))/(693*b^2*x^4) - (2*B*(b*x + c*x^2)^(1/2))/( 
9*x^5) - (2*A*c*(b*x + c*x^2)^(1/2))/(99*b*x^5) - (2*B*c*(b*x + c*x^2)^(1/ 
2))/(63*b*x^4) - (2*A*(b*x + c*x^2)^(1/2))/(11*x^6) - (32*A*c^3*(b*x + c*x 
^2)^(1/2))/(1155*b^3*x^3) + (128*A*c^4*(b*x + c*x^2)^(1/2))/(3465*b^4*x^2) 
 - (256*A*c^5*(b*x + c*x^2)^(1/2))/(3465*b^5*x) + (4*B*c^2*(b*x + c*x^2)^( 
1/2))/(105*b^2*x^3) - (16*B*c^3*(b*x + c*x^2)^(1/2))/(315*b^3*x^2) + (32*B 
*c^4*(b*x + c*x^2)^(1/2))/(315*b^4*x)