3.247 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^{15}} \, dx\)

Optimal. Leaf size=108 \[ \frac{16 c^3 \left (b x^2+c x^4\right )^{5/2}}{1155 b^4 x^{10}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2}}{231 b^3 x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2}}{33 b^2 x^{14}}-\frac{\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}} \]

[Out]

-(b*x^2 + c*x^4)^(5/2)/(11*b*x^16) + (2*c*(b*x^2 + c*x^4)^(5/2))/(33*b^2*x^14) - (8*c^2*(b*x^2 + c*x^4)^(5/2))
/(231*b^3*x^12) + (16*c^3*(b*x^2 + c*x^4)^(5/2))/(1155*b^4*x^10)

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Rubi [A]  time = 0.183943, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ \frac{16 c^3 \left (b x^2+c x^4\right )^{5/2}}{1155 b^4 x^{10}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2}}{231 b^3 x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2}}{33 b^2 x^{14}}-\frac{\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^15,x]

[Out]

-(b*x^2 + c*x^4)^(5/2)/(11*b*x^16) + (2*c*(b*x^2 + c*x^4)^(5/2))/(33*b^2*x^14) - (8*c^2*(b*x^2 + c*x^4)^(5/2))
/(231*b^3*x^12) + (16*c^3*(b*x^2 + c*x^4)^(5/2))/(1155*b^4*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^{15}} \, dx &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac{(6 c) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx}{11 b}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2}}{33 b^2 x^{14}}+\frac{\left (8 c^2\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx}{33 b^2}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2}}{33 b^2 x^{14}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2}}{231 b^3 x^{12}}-\frac{\left (16 c^3\right ) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx}{231 b^3}\\ &=-\frac{\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}+\frac{2 c \left (b x^2+c x^4\right )^{5/2}}{33 b^2 x^{14}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{5/2}}{231 b^3 x^{12}}+\frac{16 c^3 \left (b x^2+c x^4\right )^{5/2}}{1155 b^4 x^{10}}\\ \end{align*}

Mathematica [A]  time = 0.0176025, size = 57, normalized size = 0.53 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (70 b^2 c x^2-105 b^3-40 b c^2 x^4+16 c^3 x^6\right )}{1155 b^4 x^{16}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^15,x]

[Out]

((x^2*(b + c*x^2))^(5/2)*(-105*b^3 + 70*b^2*c*x^2 - 40*b*c^2*x^4 + 16*c^3*x^6))/(1155*b^4*x^16)

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Maple [A]  time = 0.045, size = 61, normalized size = 0.6 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -16\,{c}^{3}{x}^{6}+40\,b{c}^{2}{x}^{4}-70\,{b}^{2}c{x}^{2}+105\,{b}^{3} \right ) }{1155\,{x}^{14}{b}^{4}} \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^15,x)

[Out]

-1/1155*(c*x^2+b)*(-16*c^3*x^6+40*b*c^2*x^4-70*b^2*c*x^2+105*b^3)*(c*x^4+b*x^2)^(3/2)/x^14/b^4

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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^15,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47126, size = 169, normalized size = 1.56 \begin{align*} \frac{{\left (16 \, c^{5} x^{10} - 8 \, b c^{4} x^{8} + 6 \, b^{2} c^{3} x^{6} - 5 \, b^{3} c^{2} x^{4} - 140 \, b^{4} c x^{2} - 105 \, b^{5}\right )} \sqrt{c x^{4} + b x^{2}}}{1155 \, b^{4} x^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^15,x, algorithm="fricas")

[Out]

1/1155*(16*c^5*x^10 - 8*b*c^4*x^8 + 6*b^2*c^3*x^6 - 5*b^3*c^2*x^4 - 140*b^4*c*x^2 - 105*b^5)*sqrt(c*x^4 + b*x^
2)/(b^4*x^12)

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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^{15}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**15,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**15, x)

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Giac [B]  time = 1.26963, size = 319, normalized size = 2.95 \begin{align*} \frac{32 \,{\left (1155 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{14} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 2079 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{12} b c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 2541 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} b^{2} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 825 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} b^{3} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 165 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} b^{4} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) - 55 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} b^{5} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) + 11 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} b^{6} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right ) - b^{7} c^{\frac{11}{2}} \mathrm{sgn}\left (x\right )\right )}}{1155 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^15,x, algorithm="giac")

[Out]

32/1155*(1155*(sqrt(c)*x - sqrt(c*x^2 + b))^14*c^(11/2)*sgn(x) + 2079*(sqrt(c)*x - sqrt(c*x^2 + b))^12*b*c^(11
/2)*sgn(x) + 2541*(sqrt(c)*x - sqrt(c*x^2 + b))^10*b^2*c^(11/2)*sgn(x) + 825*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b
^3*c^(11/2)*sgn(x) + 165*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^4*c^(11/2)*sgn(x) - 55*(sqrt(c)*x - sqrt(c*x^2 + b)
)^4*b^5*c^(11/2)*sgn(x) + 11*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^6*c^(11/2)*sgn(x) - b^7*c^(11/2)*sgn(x))/((sqrt
(c)*x - sqrt(c*x^2 + b))^2 - b)^11