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

Optimal. Leaf size=126 \[ -\frac {256 c^4 \left (b x+c x^2\right )^{3/2}}{3465 b^5 x^3}+\frac {128 c^3 \left (b x+c x^2\right )^{3/2}}{1155 b^4 x^4}-\frac {32 c^2 \left (b x+c x^2\right )^{3/2}}{231 b^3 x^5}+\frac {16 c \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7} \]

[Out]

-2/11*(c*x^2+b*x)^(3/2)/b/x^7+16/99*c*(c*x^2+b*x)^(3/2)/b^2/x^6-32/231*c^2*(c*x^2+b*x)^(3/2)/b^3/x^5+128/1155*
c^3*(c*x^2+b*x)^(3/2)/b^4/x^4-256/3465*c^4*(c*x^2+b*x)^(3/2)/b^5/x^3

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Rubi [A]  time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {658, 650} \[ -\frac {256 c^4 \left (b x+c x^2\right )^{3/2}}{3465 b^5 x^3}+\frac {128 c^3 \left (b x+c x^2\right )^{3/2}}{1155 b^4 x^4}-\frac {32 c^2 \left (b x+c x^2\right )^{3/2}}{231 b^3 x^5}+\frac {16 c \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/x^7,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(11*b*x^7) + (16*c*(b*x + c*x^2)^(3/2))/(99*b^2*x^6) - (32*c^2*(b*x + c*x^2)^(3/2))/(
231*b^3*x^5) + (128*c^3*(b*x + c*x^2)^(3/2))/(1155*b^4*x^4) - (256*c^4*(b*x + c*x^2)^(3/2))/(3465*b^5*x^3)

Rule 650

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 658

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {b x+c x^2}}{x^7} \, dx &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7}-\frac {(8 c) \int \frac {\sqrt {b x+c x^2}}{x^6} \, dx}{11 b}\\ &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7}+\frac {16 c \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}+\frac {\left (16 c^2\right ) \int \frac {\sqrt {b x+c x^2}}{x^5} \, dx}{33 b^2}\\ &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7}+\frac {16 c \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{3/2}}{231 b^3 x^5}-\frac {\left (64 c^3\right ) \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx}{231 b^3}\\ &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7}+\frac {16 c \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{3/2}}{231 b^3 x^5}+\frac {128 c^3 \left (b x+c x^2\right )^{3/2}}{1155 b^4 x^4}+\frac {\left (128 c^4\right ) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{1155 b^4}\\ &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{11 b x^7}+\frac {16 c \left (b x+c x^2\right )^{3/2}}{99 b^2 x^6}-\frac {32 c^2 \left (b x+c x^2\right )^{3/2}}{231 b^3 x^5}+\frac {128 c^3 \left (b x+c x^2\right )^{3/2}}{1155 b^4 x^4}-\frac {256 c^4 \left (b x+c x^2\right )^{3/2}}{3465 b^5 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 73, normalized size = 0.58 \[ -\frac {2 \sqrt {x (b+c x)} \left (315 b^5+35 b^4 c x-40 b^3 c^2 x^2+48 b^2 c^3 x^3-64 b c^4 x^4+128 c^5 x^5\right )}{3465 b^5 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2]/x^7,x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(315*b^5 + 35*b^4*c*x - 40*b^3*c^2*x^2 + 48*b^2*c^3*x^3 - 64*b*c^4*x^4 + 128*c^5*x^5))/(
3465*b^5*x^6)

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fricas [A]  time = 0.86, size = 71, normalized size = 0.56 \[ -\frac {2 \, {\left (128 \, c^{5} x^{5} - 64 \, b c^{4} x^{4} + 48 \, b^{2} c^{3} x^{3} - 40 \, b^{3} c^{2} x^{2} + 35 \, b^{4} c x + 315 \, b^{5}\right )} \sqrt {c x^{2} + b x}}{3465 \, b^{5} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(128*c^5*x^5 - 64*b*c^4*x^4 + 48*b^2*c^3*x^3 - 40*b^3*c^2*x^2 + 35*b^4*c*x + 315*b^5)*sqrt(c*x^2 + b*x
)/(b^5*x^6)

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giac [A]  time = 0.18, size = 194, normalized size = 1.54 \[ \frac {2 \, {\left (11088 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} c^{3} + 36960 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} b c^{\frac {5}{2}} + 51480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} b^{2} c^{2} + 38115 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} b^{3} c^{\frac {3}{2}} + 15785 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b^{4} c + 3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{5} \sqrt {c} + 315 \, b^{6}\right )}}{3465 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(11088*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*c^3 + 36960*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*b*c^(5/2) + 5148
0*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*b^2*c^2 + 38115*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*b^3*c^(3/2) + 15785*(sqr
t(c)*x - sqrt(c*x^2 + b*x))^2*b^4*c + 3465*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^5*sqrt(c) + 315*b^6)/(sqrt(c)*x -
sqrt(c*x^2 + b*x))^11

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maple [A]  time = 0.04, size = 66, normalized size = 0.52 \[ -\frac {2 \left (c x +b \right ) \left (128 c^{4} x^{4}-192 x^{3} c^{3} b +240 c^{2} x^{2} b^{2}-280 c x \,b^{3}+315 b^{4}\right ) \sqrt {c \,x^{2}+b x}}{3465 b^{5} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(c*x+b)*(128*c^4*x^4-192*b*c^3*x^3+240*b^2*c^2*x^2-280*b^3*c*x+315*b^4)*(c*x^2+b*x)^(1/2)/b^5/x^6

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maxima [A]  time = 1.41, size = 125, normalized size = 0.99 \[ -\frac {256 \, \sqrt {c x^{2} + b x} c^{5}}{3465 \, b^{5} x} + \frac {128 \, \sqrt {c x^{2} + b x} c^{4}}{3465 \, b^{4} x^{2}} - \frac {32 \, \sqrt {c x^{2} + b x} c^{3}}{1155 \, b^{3} x^{3}} + \frac {16 \, \sqrt {c x^{2} + b x} c^{2}}{693 \, b^{2} x^{4}} - \frac {2 \, \sqrt {c x^{2} + b x} c}{99 \, b x^{5}} - \frac {2 \, \sqrt {c x^{2} + b x}}{11 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-256/3465*sqrt(c*x^2 + b*x)*c^5/(b^5*x) + 128/3465*sqrt(c*x^2 + b*x)*c^4/(b^4*x^2) - 32/1155*sqrt(c*x^2 + b*x)
*c^3/(b^3*x^3) + 16/693*sqrt(c*x^2 + b*x)*c^2/(b^2*x^4) - 2/99*sqrt(c*x^2 + b*x)*c/(b*x^5) - 2/11*sqrt(c*x^2 +
b*x)/x^6

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mupad [B]  time = 0.62, size = 125, normalized size = 0.99 \[ \frac {16\,c^2\,\sqrt {c\,x^2+b\,x}}{693\,b^2\,x^4}-\frac {2\,\sqrt {c\,x^2+b\,x}}{11\,x^6}-\frac {32\,c^3\,\sqrt {c\,x^2+b\,x}}{1155\,b^3\,x^3}+\frac {128\,c^4\,\sqrt {c\,x^2+b\,x}}{3465\,b^4\,x^2}-\frac {256\,c^5\,\sqrt {c\,x^2+b\,x}}{3465\,b^5\,x}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}}{99\,b\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)/x^7,x)

[Out]

(16*c^2*(b*x + c*x^2)^(1/2))/(693*b^2*x^4) - (2*(b*x + c*x^2)^(1/2))/(11*x^6) - (32*c^3*(b*x + c*x^2)^(1/2))/(
1155*b^3*x^3) + (128*c^4*(b*x + c*x^2)^(1/2))/(3465*b^4*x^2) - (256*c^5*(b*x + c*x^2)^(1/2))/(3465*b^5*x) - (2
*c*(b*x + c*x^2)^(1/2))/(99*b*x^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x \left (b + c x\right )}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/x**7, x)

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