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

Optimal. Leaf size=167 \[ -\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \begin {gather*} -\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^15,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(14*x^14*(a + b*x^2)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^12
*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*x^10*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])/(8*x^8*(a + b*x^2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{15}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^3}{x^8} \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (\frac {a^3 b^3}{x^8}+\frac {3 a^2 b^4}{x^7}+\frac {3 a b^5}{x^6}+\frac {b^6}{x^5}\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^{12} \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 x^{10} \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (20 a^3+70 a^2 b x^2+84 a b^2 x^4+35 b^3 x^6\right )}{280 x^{14} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/280*(Sqrt[(a + b*x^2)^2]*(20*a^3 + 70*a^2*b*x^2 + 84*a*b^2*x^4 + 35*b^3*x^6))/(x^14*(a + b*x^2))

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IntegrateAlgebraic [B]  time = 1.17, size = 444, normalized size = 2.66 \begin {gather*} \frac {8 b^6 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-20 a^9 b-190 a^8 b^2 x^2-804 a^7 b^3 x^4-1989 a^6 b^4 x^6-3170 a^5 b^5 x^8-3375 a^4 b^6 x^{10}-2400 a^3 b^7 x^{12}-1099 a^2 b^8 x^{14}-294 a b^9 x^{16}-35 b^{10} x^{18}\right )+8 \sqrt {b^2} b^6 \left (20 a^{10}+210 a^9 b x^2+994 a^8 b^2 x^4+2793 a^7 b^3 x^6+5159 a^6 b^4 x^8+6545 a^5 b^5 x^{10}+5775 a^4 b^6 x^{12}+3499 a^3 b^7 x^{14}+1393 a^2 b^8 x^{16}+329 a b^9 x^{18}+35 b^{10} x^{20}\right )}{35 \sqrt {b^2} x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-64 a^6 b^6-384 a^5 b^7 x^2-960 a^4 b^8 x^4-1280 a^3 b^9 x^6-960 a^2 b^{10} x^8-384 a b^{11} x^{10}-64 b^{12} x^{12}\right )+35 x^{14} \left (64 a^7 b^7+448 a^6 b^8 x^2+1344 a^5 b^9 x^4+2240 a^4 b^{10} x^6+2240 a^3 b^{11} x^8+1344 a^2 b^{12} x^{10}+448 a b^{13} x^{12}+64 b^{14} x^{14}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^15,x]

[Out]

(8*b^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-20*a^9*b - 190*a^8*b^2*x^2 - 804*a^7*b^3*x^4 - 1989*a^6*b^4*x^6 - 317
0*a^5*b^5*x^8 - 3375*a^4*b^6*x^10 - 2400*a^3*b^7*x^12 - 1099*a^2*b^8*x^14 - 294*a*b^9*x^16 - 35*b^10*x^18) + 8
*b^6*Sqrt[b^2]*(20*a^10 + 210*a^9*b*x^2 + 994*a^8*b^2*x^4 + 2793*a^7*b^3*x^6 + 5159*a^6*b^4*x^8 + 6545*a^5*b^5
*x^10 + 5775*a^4*b^6*x^12 + 3499*a^3*b^7*x^14 + 1393*a^2*b^8*x^16 + 329*a*b^9*x^18 + 35*b^10*x^20))/(35*Sqrt[b
^2]*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-64*a^6*b^6 - 384*a^5*b^7*x^2 - 960*a^4*b^8*x^4 - 1280*a^3*b^9*x^6 -
960*a^2*b^10*x^8 - 384*a*b^11*x^10 - 64*b^12*x^12) + 35*x^14*(64*a^7*b^7 + 448*a^6*b^8*x^2 + 1344*a^5*b^9*x^4
+ 2240*a^4*b^10*x^6 + 2240*a^3*b^11*x^8 + 1344*a^2*b^12*x^10 + 448*a*b^13*x^12 + 64*b^14*x^14))

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fricas [A]  time = 1.08, size = 37, normalized size = 0.22 \begin {gather*} -\frac {35 \, b^{3} x^{6} + 84 \, a b^{2} x^{4} + 70 \, a^{2} b x^{2} + 20 \, a^{3}}{280 \, x^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/280*(35*b^3*x^6 + 84*a*b^2*x^4 + 70*a^2*b*x^2 + 20*a^3)/x^14

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giac [A]  time = 0.16, size = 69, normalized size = 0.41 \begin {gather*} -\frac {35 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 84 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 70 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{280 \, x^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/280*(35*b^3*x^6*sgn(b*x^2 + a) + 84*a*b^2*x^4*sgn(b*x^2 + a) + 70*a^2*b*x^2*sgn(b*x^2 + a) + 20*a^3*sgn(b*x
^2 + a))/x^14

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \begin {gather*} -\frac {\left (35 b^{3} x^{6}+84 a \,b^{2} x^{4}+70 a^{2} b \,x^{2}+20 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (b \,x^{2}+a \right )^{3} x^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^15,x)

[Out]

-1/280*(35*b^3*x^6+84*a*b^2*x^4+70*a^2*b*x^2+20*a^3)*((b*x^2+a)^2)^(3/2)/x^14/(b*x^2+a)^3

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maxima [A]  time = 1.13, size = 35, normalized size = 0.21 \begin {gather*} -\frac {b^{3}}{8 \, x^{8}} - \frac {3 \, a b^{2}}{10 \, x^{10}} - \frac {a^{2} b}{4 \, x^{12}} - \frac {a^{3}}{14 \, x^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^3/x^8 - 3/10*a*b^2/x^10 - 1/4*a^2*b/x^12 - 1/14*a^3/x^14

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mupad [B]  time = 4.21, size = 151, normalized size = 0.90 \begin {gather*} -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{10\,x^{10}\,\left (b\,x^2+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{4\,x^{12}\,\left (b\,x^2+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^15,x)

[Out]

- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(8*x
^8*(a + b*x^2)) - (3*a*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(10*x^10*(a + b*x^2)) - (a^2*b*(a^2 + b^2*x^4 +
2*a*b*x^2)^(1/2))/(4*x^12*(a + b*x^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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