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]

-1/11*(c*x^4+b*x^2)^(5/2)/b/x^16+2/33*c*(c*x^4+b*x^2)^(5/2)/b^2/x^14-8/231*c^2*(c*x^4+b*x^2)^(5/2)/b^3/x^12+16
/1155*c^3*(c*x^4+b*x^2)^(5/2)/b^4/x^10

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Rubi [A]  time = 0.18, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 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^{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*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.53 \[ \frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (-105 b^3+70 b^2 c x^2-40 b c^2 x^4+16 c^3 x^6\right )}{1155 b^4 x^{16}} \]

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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fricas [A]  time = 1.03, size = 75, normalized size = 0.69 \[ \frac {{\left (16 \, c^{5} x^{10} - 8 \, b c^{4} x^{8} + 6 \, b^{2} c^{3} x^{6} - 5 \, b^{3} c^{2} x^{4} - 140 \, b^{4} c x^{2} - 105 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{1155 \, b^{4} x^{12}} \]

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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giac [B]  time = 0.28, size = 236, normalized size = 2.19 \[ \frac {32 \, {\left (1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 2079 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} b c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 2541 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b^{2} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 825 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{3} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{4} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - 55 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{5} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) + 11 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{6} c^{\frac {11}{2}} \mathrm {sgn}\relax (x) - b^{7} c^{\frac {11}{2}} \mathrm {sgn}\relax (x)\right )}}{1155 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \]

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

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maple [A]  time = 0.00, size = 61, normalized size = 0.56 \[ -\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+40 b \,c^{2} x^{4}-70 b^{2} c \,x^{2}+105 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 b^{4} x^{14}} \]

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 [A]  time = 1.49, size = 153, normalized size = 1.42 \[ \frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{1155 \, b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{1155 \, b^{3} x^{4}} + \frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{385 \, b^{2} x^{6}} - \frac {\sqrt {c x^{4} + b x^{2}} c^{2}}{231 \, b x^{8}} + \frac {\sqrt {c x^{4} + b x^{2}} c}{264 \, x^{10}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{88 \, x^{12}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{8 \, x^{14}} \]

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]

16/1155*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^2) - 8/1155*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^4) + 2/385*sqrt(c*x^4 + b*x^
2)*c^3/(b^2*x^6) - 1/231*sqrt(c*x^4 + b*x^2)*c^2/(b*x^8) + 1/264*sqrt(c*x^4 + b*x^2)*c/x^10 + 3/88*sqrt(c*x^4
+ b*x^2)*b/x^12 - 1/8*(c*x^4 + b*x^2)^(3/2)/x^14

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mupad [B]  time = 4.99, size = 135, normalized size = 1.25 \[ \frac {2\,c^3\,\sqrt {c\,x^4+b\,x^2}}{385\,b^2\,x^6}-\frac {4\,c\,\sqrt {c\,x^4+b\,x^2}}{33\,x^{10}}-\frac {c^2\,\sqrt {c\,x^4+b\,x^2}}{231\,b\,x^8}-\frac {b\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {8\,c^4\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^4}+\frac {16\,c^5\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^4\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^15,x)

[Out]

(2*c^3*(b*x^2 + c*x^4)^(1/2))/(385*b^2*x^6) - (4*c*(b*x^2 + c*x^4)^(1/2))/(33*x^10) - (c^2*(b*x^2 + c*x^4)^(1/
2))/(231*b*x^8) - (b*(b*x^2 + c*x^4)^(1/2))/(11*x^12) - (8*c^4*(b*x^2 + c*x^4)^(1/2))/(1155*b^3*x^4) + (16*c^5
*(b*x^2 + c*x^4)^(1/2))/(1155*b^4*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{15}}\, dx \]

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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