3.248 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^{17}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{128 c^4 \left (b x^2+c x^4\right )^{5/2}}{15015 b^5 x^{10}}+\frac{64 c^3 \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}-\frac{16 c^2 \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}+\frac{8 c \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}} \]

[Out]

-(b*x^2 + c*x^4)^(5/2)/(13*b*x^18) + (8*c*(b*x^2 + c*x^4)^(5/2))/(143*b^2*x^16) - (16*c^2*(b*x^2 + c*x^4)^(5/2
))/(429*b^3*x^14) + (64*c^3*(b*x^2 + c*x^4)^(5/2))/(3003*b^4*x^12) - (128*c^4*(b*x^2 + c*x^4)^(5/2))/(15015*b^
5*x^10)

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Rubi [A]  time = 0.262011, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ -\frac{128 c^4 \left (b x^2+c x^4\right )^{5/2}}{15015 b^5 x^{10}}+\frac{64 c^3 \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}-\frac{16 c^2 \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}+\frac{8 c \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^17,x]

[Out]

-(b*x^2 + c*x^4)^(5/2)/(13*b*x^18) + (8*c*(b*x^2 + c*x^4)^(5/2))/(143*b^2*x^16) - (16*c^2*(b*x^2 + c*x^4)^(5/2
))/(429*b^3*x^14) + (64*c^3*(b*x^2 + c*x^4)^(5/2))/(3003*b^4*x^12) - (128*c^4*(b*x^2 + c*x^4)^(5/2))/(15015*b^
5*x^10)

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac{(8 c) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx}{13 b}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}+\frac{8 c \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac{\left (48 c^2\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx}{143 b^2}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}+\frac{8 c \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac{16 c^2 \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}-\frac{\left (64 c^3\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx}{429 b^3}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}+\frac{8 c \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac{16 c^2 \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}+\frac{64 c^3 \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}+\frac{\left (128 c^4\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx}{3003 b^4}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}+\frac{8 c \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}-\frac{16 c^2 \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}+\frac{64 c^3 \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}-\frac{128 c^4 \left (b x^2+c x^4\right )^{5/2}}{15015 b^5 x^{10}}\\ \end{align*}

Mathematica [A]  time = 0.0186833, size = 68, normalized size = 0.5 \[ -\frac{\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (560 b^2 c^2 x^4-840 b^3 c x^2+1155 b^4-320 b c^3 x^6+128 c^4 x^8\right )}{15015 b^5 x^{18}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^17,x]

[Out]

-((x^2*(b + c*x^2))^(5/2)*(1155*b^4 - 840*b^3*c*x^2 + 560*b^2*c^2*x^4 - 320*b*c^3*x^6 + 128*c^4*x^8))/(15015*b
^5*x^18)

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Maple [A]  time = 0.045, size = 72, normalized size = 0.5 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( 128\,{c}^{4}{x}^{8}-320\,{c}^{3}{x}^{6}b+560\,{c}^{2}{x}^{4}{b}^{2}-840\,c{x}^{2}{b}^{3}+1155\,{b}^{4} \right ) }{15015\,{x}^{16}{b}^{5}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^17,x)

[Out]

-1/15015*(c*x^2+b)*(128*c^4*x^8-320*b*c^3*x^6+560*b^2*c^2*x^4-840*b^3*c*x^2+1155*b^4)*(c*x^4+b*x^2)^(3/2)/x^16
/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^4+b*x^2)^(3/2)/x^17,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6755, size = 204, normalized size = 1.5 \begin{align*} -\frac{{\left (128 \, c^{6} x^{12} - 64 \, b c^{5} x^{10} + 48 \, b^{2} c^{4} x^{8} - 40 \, b^{3} c^{3} x^{6} + 35 \, b^{4} c^{2} x^{4} + 1470 \, b^{5} c x^{2} + 1155 \, b^{6}\right )} \sqrt{c x^{4} + b x^{2}}}{15015 \, b^{5} x^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="fricas")

[Out]

-1/15015*(128*c^6*x^12 - 64*b*c^5*x^10 + 48*b^2*c^4*x^8 - 40*b^3*c^3*x^6 + 35*b^4*c^2*x^4 + 1470*b^5*c*x^2 + 1
155*b^6)*sqrt(c*x^4 + b*x^2)/(b^5*x^14)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}{x^{17}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**17,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**17, x)

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Giac [B]  time = 1.26971, size = 356, normalized size = 2.62 \begin{align*} \frac{256 \,{\left (6006 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{16} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) + 12012 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{14} b c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) + 13728 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} b^{2} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) + 4719 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} b^{3} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) + 715 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} b^{4} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) - 286 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} b^{5} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) + 78 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} b^{6} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) - 13 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} b^{7} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right ) + b^{8} c^{\frac{13}{2}} \mathrm{sgn}\left (x\right )\right )}}{15015 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{13}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="giac")

[Out]

256/15015*(6006*(sqrt(c)*x - sqrt(c*x^2 + b))^16*c^(13/2)*sgn(x) + 12012*(sqrt(c)*x - sqrt(c*x^2 + b))^14*b*c^
(13/2)*sgn(x) + 13728*(sqrt(c)*x - sqrt(c*x^2 + b))^12*b^2*c^(13/2)*sgn(x) + 4719*(sqrt(c)*x - sqrt(c*x^2 + b)
)^10*b^3*c^(13/2)*sgn(x) + 715*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b^4*c^(13/2)*sgn(x) - 286*(sqrt(c)*x - sqrt(c*x
^2 + b))^6*b^5*c^(13/2)*sgn(x) + 78*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^6*c^(13/2)*sgn(x) - 13*(sqrt(c)*x - sqrt
(c*x^2 + b))^2*b^7*c^(13/2)*sgn(x) + b^8*c^(13/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^13