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

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

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Rubi [A]  time = 0.26, antiderivative size = 136, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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])

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*}

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Mathematica [A]  time = 0.02, size = 68, normalized size = 0.50 \begin {gather*} -\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (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\right )}{15015 b^5 x^{18}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15015*((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))/
(b^5*x^18)

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IntegrateAlgebraic [A]  time = 0.28, size = 90, normalized size = 0.66 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-1155 b^6-1470 b^5 c x^2-35 b^4 c^2 x^4+40 b^3 c^3 x^6-48 b^2 c^4 x^8+64 b c^5 x^{10}-128 c^6 x^{12}\right )}{15015 b^5 x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-1155*b^6 - 1470*b^5*c*x^2 - 35*b^4*c^2*x^4 + 40*b^3*c^3*x^6 - 48*b^2*c^4*x^8 + 64*b*c^5
*x^10 - 128*c^6*x^12))/(15015*b^5*x^14)

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fricas [A]  time = 1.14, size = 86, normalized size = 0.63 \begin {gather*} -\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 {gather*}

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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giac [B]  time = 0.29, size = 264, normalized size = 1.94 \begin {gather*} \frac {256 \, {\left (6006 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{16} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 12012 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} b c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 13728 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} b^{2} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 4719 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b^{3} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 715 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{4} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) - 286 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{5} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + 78 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{6} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) - 13 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{7} c^{\frac {13}{2}} \mathrm {sgn}\relax (x) + b^{8} c^{\frac {13}{2}} \mathrm {sgn}\relax (x)\right )}}{15015 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{13}} \end {gather*}

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

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maple [A]  time = 0.01, size = 72, normalized size = 0.53 \begin {gather*} -\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 ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 b^{5} x^{16}} \end {gather*}

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 [A]  time = 1.47, size = 177, normalized size = 1.30 \begin {gather*} -\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{6}}{15015 \, b^{5} x^{2}} + \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{15015 \, b^{4} x^{4}} - \frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{5005 \, b^{3} x^{6}} + \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{3003 \, b^{2} x^{8}} - \frac {\sqrt {c x^{4} + b x^{2}} c^{2}}{429 \, b x^{10}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{1430 \, x^{12}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{130 \, x^{14}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{10 \, x^{16}} \end {gather*}

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]

-128/15015*sqrt(c*x^4 + b*x^2)*c^6/(b^5*x^2) + 64/15015*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^4) - 16/5005*sqrt(c*x^4
+ b*x^2)*c^4/(b^3*x^6) + 8/3003*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^8) - 1/429*sqrt(c*x^4 + b*x^2)*c^2/(b*x^10) +
3/1430*sqrt(c*x^4 + b*x^2)*c/x^12 + 3/130*sqrt(c*x^4 + b*x^2)*b/x^14 - 1/10*(c*x^4 + b*x^2)^(3/2)/x^16

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mupad [B]  time = 5.17, size = 159, normalized size = 1.17 \begin {gather*} \frac {8\,c^3\,\sqrt {c\,x^4+b\,x^2}}{3003\,b^2\,x^8}-\frac {14\,c\,\sqrt {c\,x^4+b\,x^2}}{143\,x^{12}}-\frac {c^2\,\sqrt {c\,x^4+b\,x^2}}{429\,b\,x^{10}}-\frac {b\,\sqrt {c\,x^4+b\,x^2}}{13\,x^{14}}-\frac {16\,c^4\,\sqrt {c\,x^4+b\,x^2}}{5005\,b^3\,x^6}+\frac {64\,c^5\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^4\,x^4}-\frac {128\,c^6\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^17,x)

[Out]

(8*c^3*(b*x^2 + c*x^4)^(1/2))/(3003*b^2*x^8) - (14*c*(b*x^2 + c*x^4)^(1/2))/(143*x^12) - (c^2*(b*x^2 + c*x^4)^
(1/2))/(429*b*x^10) - (b*(b*x^2 + c*x^4)^(1/2))/(13*x^14) - (16*c^4*(b*x^2 + c*x^4)^(1/2))/(5005*b^3*x^6) + (6
4*c^5*(b*x^2 + c*x^4)^(1/2))/(15015*b^4*x^4) - (128*c^6*(b*x^2 + c*x^4)^(1/2))/(15015*b^5*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{17}}\, dx \end {gather*}

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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