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

Optimal result

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

[Out]

-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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {806, 672, 664} \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^7} \, dx=\frac {32 c^3 \left (b x+c x^2\right )^{3/2} (11 b B-8 A c)}{3465 b^5 x^3}-\frac {16 c^2 \left (b x+c x^2\right )^{3/2} (11 b B-8 A c)}{1155 b^4 x^4}+\frac {4 c \left (b x+c x^2\right )^{3/2} (11 b B-8 A c)}{231 b^3 x^5}-\frac {2 \left (b x+c x^2\right )^{3/2} (11 b B-8 A c)}{99 b^2 x^6}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7} \]

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^7,x]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(11*b*x^7) - (2*(11*b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/(99*b^2*x^6) + (4*c*(11*b*B -
8*A*c)*(b*x + c*x^2)^(3/2))/(231*b^3*x^5) - (16*c^2*(11*b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/(1155*b^4*x^4) + (3
2*c^3*(11*b*B - 8*A*c)*(b*x + c*x^2)^(3/2))/(3465*b^5*x^3)

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> 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] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[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] && NeQ[b^2 - 4*a
*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 806

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] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 A \left (b x+c x^2\right )^{3/2}}{11 b x^7}+\frac {\left (2 \left (-7 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx}{11 b} \\ & = -\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 {(2 c (11 b B-8 A c)) \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx}{33 b^2} \\ & = -\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 {\left (8 c^2 (11 b B-8 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx}{231 b^3} \\ & = -\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 {\left (16 c^3 (11 b B-8 A c)\right ) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{1155 b^4} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^7,x]

[Out]

(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)

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	\(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{11 b \,x^{7}}-\frac {8 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{9 b \,x^{6}}-\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\right )}{3 b}\right )}{11 b}\right )+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{9 b \,x^{6}}-\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{7 b \,x^{5}}-\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{5 b \,x^{4}}+\frac {4 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15 b^{2} x^{3}}\right )}{7 b}\right )}{3 b}\right )\)	\(216\)

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

-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

Fricas [A] (verification not implemented)

none

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

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x, algorithm="fricas")

[Out]

-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)

Sympy [F]

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

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**7,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**7, x)

Maxima [A] (verification not implemented)

none

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

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x, algorithm="maxima")

[Out]

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) - 32/115
5*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) + 16/693*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

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

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^7,x, algorithm="giac")

[Out]

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 + 1
1088*(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*(
sqrt(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 + 3465*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*sqrt(c) + 38115*(sqrt(c)*x
- sqrt(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

Mupad [B] (verification not implemented)

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

[In]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^7,x)

[Out]

(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)