3.2.28 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^{13}} \, dx\)

Optimal. Leaf size=80 \[ -\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2}}{315 b^3 x^{10}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {8 c^2 \left (b x^2+c x^4\right )^{5/2}}{315 b^3 x^{10}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^13,x]

[Out]

-(b*x^2 + c*x^4)^(5/2)/(9*b*x^14) + (4*c*(b*x^2 + c*x^4)^(5/2))/(63*b^2*x^12) - (8*c^2*(b*x^2 + c*x^4)^(5/2))/
(315*b^3*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^{13}} \, dx &=-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}-\frac {(4 c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx}{9 b}\\ &=-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}+\frac {\left (8 c^2\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^9} \, dx}{63 b^2}\\ &=-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}-\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2}}{315 b^3 x^{10}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 46, normalized size = 0.58 \begin {gather*} -\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (35 b^2-20 b c x^2+8 c^2 x^4\right )}{315 b^3 x^{14}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^13,x]

[Out]

-1/315*((x^2*(b + c*x^2))^(5/2)*(35*b^2 - 20*b*c*x^2 + 8*c^2*x^4))/(b^3*x^14)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.25, size = 68, normalized size = 0.85 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-35 b^4-50 b^3 c x^2-3 b^2 c^2 x^4+4 b c^3 x^6-8 c^4 x^8\right )}{315 b^3 x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x^2 + c*x^4)^(3/2)/x^13,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-35*b^4 - 50*b^3*c*x^2 - 3*b^2*c^2*x^4 + 4*b*c^3*x^6 - 8*c^4*x^8))/(315*b^3*x^10)

________________________________________________________________________________________

fricas [A]  time = 1.12, size = 64, normalized size = 0.80 \begin {gather*} -\frac {{\left (8 \, c^{4} x^{8} - 4 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} + 50 \, b^{3} c x^{2} + 35 \, b^{4}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{3} x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="fricas")

[Out]

-1/315*(8*c^4*x^8 - 4*b*c^3*x^6 + 3*b^2*c^2*x^4 + 50*b^3*c*x^2 + 35*b^4)*sqrt(c*x^4 + b*x^2)/(b^3*x^10)

________________________________________________________________________________________

giac [B]  time = 0.27, size = 206, normalized size = 2.58 \begin {gather*} \frac {16 \, {\left (210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 441 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{2} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 126 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{3} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + 36 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{4} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{5} c^{\frac {9}{2}} \mathrm {sgn}\relax (x) + b^{6} c^{\frac {9}{2}} \mathrm {sgn}\relax (x)\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="giac")

[Out]

16/315*(210*(sqrt(c)*x - sqrt(c*x^2 + b))^12*c^(9/2)*sgn(x) + 315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*b*c^(9/2)*s
gn(x) + 441*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b^2*c^(9/2)*sgn(x) + 126*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^3*c^(9/
2)*sgn(x) + 36*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^4*c^(9/2)*sgn(x) - 9*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^5*c^(9
/2)*sgn(x) + b^6*c^(9/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^9

________________________________________________________________________________________

maple [A]  time = 0.00, size = 50, normalized size = 0.62 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-20 b c \,x^{2}+35 b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 b^{3} x^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^13,x)

[Out]

-1/315*(c*x^2+b)*(8*c^2*x^4-20*b*c*x^2+35*b^2)*(c*x^4+b*x^2)^(3/2)/x^12/b^3

________________________________________________________________________________________

maxima [A]  time = 1.52, size = 129, normalized size = 1.61 \begin {gather*} -\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{315 \, b^{3} x^{2}} + \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{315 \, b^{2} x^{4}} - \frac {\sqrt {c x^{4} + b x^{2}} c^{2}}{105 \, b x^{6}} + \frac {\sqrt {c x^{4} + b x^{2}} c}{126 \, x^{8}} + \frac {\sqrt {c x^{4} + b x^{2}} b}{18 \, x^{10}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{6 \, x^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="maxima")

[Out]

-8/315*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^2) + 4/315*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^4) - 1/105*sqrt(c*x^4 + b*x^2)
*c^2/(b*x^6) + 1/126*sqrt(c*x^4 + b*x^2)*c/x^8 + 1/18*sqrt(c*x^4 + b*x^2)*b/x^10 - 1/6*(c*x^4 + b*x^2)^(3/2)/x
^12

________________________________________________________________________________________

mupad [B]  time = 4.72, size = 111, normalized size = 1.39 \begin {gather*} \frac {4\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^2\,x^4}-\frac {10\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,x^8}-\frac {c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b\,x^6}-\frac {b\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {8\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,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^13,x)

[Out]

(4*c^3*(b*x^2 + c*x^4)^(1/2))/(315*b^2*x^4) - (10*c*(b*x^2 + c*x^4)^(1/2))/(63*x^8) - (c^2*(b*x^2 + c*x^4)^(1/
2))/(105*b*x^6) - (b*(b*x^2 + c*x^4)^(1/2))/(9*x^10) - (8*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^3*x^2)

________________________________________________________________________________________

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^{13}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**13,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**13, x)

________________________________________________________________________________________