3.2.12 \(\int \frac {\sqrt {b x^2+c x^4}}{x^{13}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {128 c^4 \left (b x^2+c x^4\right )^{3/2}}{3465 b^5 x^6}+\frac {64 c^3 \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}-\frac {16 c^2 \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}} \]

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Rubi [A]  time = 0.21, 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 )^{3/2}}{3465 b^5 x^6}+\frac {64 c^3 \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}-\frac {16 c^2 \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^13,x]

[Out]

-(b*x^2 + c*x^4)^(3/2)/(11*b*x^14) + (8*c*(b*x^2 + c*x^4)^(3/2))/(99*b^2*x^12) - (16*c^2*(b*x^2 + c*x^4)^(3/2)
)/(231*b^3*x^10) + (64*c^3*(b*x^2 + c*x^4)^(3/2))/(1155*b^4*x^8) - (128*c^4*(b*x^2 + c*x^4)^(3/2))/(3465*b^5*x
^6)

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 {\sqrt {b x^2+c x^4}}{x^{13}} \, dx &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}-\frac {(8 c) \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx}{11 b}\\ &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}+\frac {\left (16 c^2\right ) \int \frac {\sqrt {b x^2+c x^4}}{x^9} \, dx}{33 b^2}\\ &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {16 c^2 \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}-\frac {\left (64 c^3\right ) \int \frac {\sqrt {b x^2+c x^4}}{x^7} \, dx}{231 b^3}\\ &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {16 c^2 \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}+\frac {64 c^3 \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}+\frac {\left (128 c^4\right ) \int \frac {\sqrt {b x^2+c x^4}}{x^5} \, dx}{1155 b^4}\\ &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{11 b x^{14}}+\frac {8 c \left (b x^2+c x^4\right )^{3/2}}{99 b^2 x^{12}}-\frac {16 c^2 \left (b x^2+c x^4\right )^{3/2}}{231 b^3 x^{10}}+\frac {64 c^3 \left (b x^2+c x^4\right )^{3/2}}{1155 b^4 x^8}-\frac {128 c^4 \left (b x^2+c x^4\right )^{3/2}}{3465 b^5 x^6}\\ \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 )^{3/2} \left (315 b^4-280 b^3 c x^2+240 b^2 c^2 x^4-192 b c^3 x^6+128 c^4 x^8\right )}{3465 b^5 x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^13,x]

[Out]

-1/3465*((x^2*(b + c*x^2))^(3/2)*(315*b^4 - 280*b^3*c*x^2 + 240*b^2*c^2*x^4 - 192*b*c^3*x^6 + 128*c^4*x^8))/(b
^5*x^14)

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IntegrateAlgebraic [A]  time = 0.16, size = 79, normalized size = 0.58 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-315 b^5-35 b^4 c x^2+40 b^3 c^2 x^4-48 b^2 c^3 x^6+64 b c^4 x^8-128 c^5 x^{10}\right )}{3465 b^5 x^{12}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[b*x^2 + c*x^4]/x^13,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-315*b^5 - 35*b^4*c*x^2 + 40*b^3*c^2*x^4 - 48*b^2*c^3*x^6 + 64*b*c^4*x^8 - 128*c^5*x^10)
)/(3465*b^5*x^12)

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fricas [A]  time = 0.94, size = 75, normalized size = 0.55 \begin {gather*} -\frac {{\left (128 \, c^{5} x^{10} - 64 \, b c^{4} x^{8} + 48 \, b^{2} c^{3} x^{6} - 40 \, b^{3} c^{2} x^{4} + 35 \, b^{4} c x^{2} + 315 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, b^{5} x^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="fricas")

[Out]

-1/3465*(128*c^5*x^10 - 64*b*c^4*x^8 + 48*b^2*c^3*x^6 - 40*b^3*c^2*x^4 + 35*b^4*c*x^2 + 315*b^5)*sqrt(c*x^4 +
b*x^2)/(b^5*x^12)

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giac [A]  time = 0.25, size = 206, normalized size = 1.51 \begin {gather*} \frac {256 \, {\left (1386 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 924 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 330 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{2} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{3} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 55 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{4} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 11 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{5} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + b^{6} c^{\frac {11}{2}} \mathrm {sgn}\relax (x)\right )}}{3465 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="giac")

[Out]

256/3465*(1386*(sqrt(c)*x - sqrt(c*x^2 + b))^12*c^(11/2)*sgn(x) + 924*(sqrt(c)*x - sqrt(c*x^2 + b))^10*b*c^(11
/2)*sgn(x) + 330*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b^2*c^(11/2)*sgn(x) - 165*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^3
*c^(11/2)*sgn(x) + 55*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^4*c^(11/2)*sgn(x) - 11*(sqrt(c)*x - sqrt(c*x^2 + b))^2
*b^5*c^(11/2)*sgn(x) + b^6*c^(11/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^11

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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}-192 c^{3} x^{6} b +240 c^{2} x^{4} b^{2}-280 c \,x^{2} b^{3}+315 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 b^{5} x^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(1/2)/x^13,x)

[Out]

-1/3465*(c*x^2+b)*(128*c^4*x^8-192*b*c^3*x^6+240*b^2*c^2*x^4-280*b^3*c*x^2+315*b^4)*(c*x^4+b*x^2)^(1/2)/x^12/b
^5

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maxima [A]  time = 1.53, size = 137, normalized size = 1.01 \begin {gather*} -\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{3465 \, b^{5} x^{2}} + \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{3465 \, b^{4} x^{4}} - \frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{1155 \, b^{3} x^{6}} + \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{693 \, b^{2} x^{8}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{99 \, b x^{10}} - \frac {\sqrt {c x^{4} + b x^{2}}}{11 \, x^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^13,x, algorithm="maxima")

[Out]

-128/3465*sqrt(c*x^4 + b*x^2)*c^5/(b^5*x^2) + 64/3465*sqrt(c*x^4 + b*x^2)*c^4/(b^4*x^4) - 16/1155*sqrt(c*x^4 +
b*x^2)*c^3/(b^3*x^6) + 8/693*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^8) - 1/99*sqrt(c*x^4 + b*x^2)*c/(b*x^10) - 1/11*s
qrt(c*x^4 + b*x^2)/x^12

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mupad [B]  time = 4.62, size = 137, normalized size = 1.01 \begin {gather*} \frac {8\,c^2\,\sqrt {c\,x^4+b\,x^2}}{693\,b^2\,x^8}-\frac {c\,\sqrt {c\,x^4+b\,x^2}}{99\,b\,x^{10}}-\frac {\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {16\,c^3\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^6}+\frac {64\,c^4\,\sqrt {c\,x^4+b\,x^2}}{3465\,b^4\,x^4}-\frac {128\,c^5\,\sqrt {c\,x^4+b\,x^2}}{3465\,b^5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x^13,x)

[Out]

(8*c^2*(b*x^2 + c*x^4)^(1/2))/(693*b^2*x^8) - (c*(b*x^2 + c*x^4)^(1/2))/(99*b*x^10) - (b*x^2 + c*x^4)^(1/2)/(1
1*x^12) - (16*c^3*(b*x^2 + c*x^4)^(1/2))/(1155*b^3*x^6) + (64*c^4*(b*x^2 + c*x^4)^(1/2))/(3465*b^4*x^4) - (128
*c^5*(b*x^2 + c*x^4)^(1/2))/(3465*b^5*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{13}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**13,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**13, x)

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