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*(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)

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Rubi [A]  time = 0.0527275, antiderivative size = 126, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0141014, size = 73, normalized size = 0.58 \[ -\frac{2 \sqrt{x (b+c x)} \left (-40 b^3 c^2 x^2+48 b^2 c^3 x^3+35 b^4 c x+315 b^5-64 b c^4 x^4+128 c^5 x^5\right )}{3465 b^5 x^6} \]

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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Maple [A]  time = 0.054, size = 66, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 128\,{c}^{4}{x}^{4}-192\,{x}^{3}{c}^{3}b+240\,{c}^{2}{x}^{2}{b}^{2}-280\,cx{b}^{3}+315\,{b}^{4} \right ) }{3465\,{x}^{6}{b}^{5}}\sqrt{c{x}^{2}+bx}} \end{align*}

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)/x^6/b^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.94495, size = 166, normalized size = 1.32 \begin{align*} -\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}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x \left (b + c x\right )}}{x^{7}}\, dx \end{align*}

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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Giac [A]  time = 1.33142, size = 262, normalized size = 2.08 \begin{align*} \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}} \end{align*}

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