3.7.1 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^{19}} \, dx\) [601]

Optimal. Leaf size=255 \[ -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 x^{18} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^{14} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^{12} \left (a+b x^2\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^{10} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )} \]

[Out]

-1/18*a^5*((b*x^2+a)^2)^(1/2)/x^18/(b*x^2+a)-5/16*a^4*b*((b*x^2+a)^2)^(1/2)/x^16/(b*x^2+a)-5/7*a^3*b^2*((b*x^2
+a)^2)^(1/2)/x^14/(b*x^2+a)-5/6*a^2*b^3*((b*x^2+a)^2)^(1/2)/x^12/(b*x^2+a)-1/2*a*b^4*((b*x^2+a)^2)^(1/2)/x^10/
(b*x^2+a)-1/8*b^5*((b*x^2+a)^2)^(1/2)/x^8/(b*x^2+a)

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Rubi [A]
time = 0.10, antiderivative size = 255, 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 = {1125, 660, 45} \begin {gather*} -\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^{10} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^{12} \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 x^{18} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^{14} \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)^(5/2)/x^19,x]

[Out]

-1/18*(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^18*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16
*x^16*(a + b*x^2)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*x^14*(a + b*x^2)) - (5*a^2*b^3*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])/(6*x^12*(a + b*x^2)) - (a*b^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^10*(a + b*x^2)) - (b
^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^8*(a + b*x^2))

Rule 45

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 660

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 1125

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 )^{5/2}}{x^{19}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^{10}} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \frac {\left (a b+b^2 x\right )^5}{x^{10}} \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (\frac {a^5 b^5}{x^{10}}+\frac {5 a^4 b^6}{x^9}+\frac {10 a^3 b^7}{x^8}+\frac {10 a^2 b^8}{x^7}+\frac {5 a b^9}{x^6}+\frac {b^{10}}{x^5}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{18 x^{18} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^{14} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^{12} \left (a+b x^2\right )}-\frac {a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^{10} \left (a+b x^2\right )}-\frac {b^5 \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 = 0.01, size = 83, normalized size = 0.33 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (56 a^5+315 a^4 b x^2+720 a^3 b^2 x^4+840 a^2 b^3 x^6+504 a b^4 x^8+126 b^5 x^{10}\right )}{1008 x^{18} \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)^(5/2)/x^19,x]

[Out]

-1/1008*(Sqrt[(a + b*x^2)^2]*(56*a^5 + 315*a^4*b*x^2 + 720*a^3*b^2*x^4 + 840*a^2*b^3*x^6 + 504*a*b^4*x^8 + 126
*b^5*x^10))/(x^18*(a + b*x^2))

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Maple [A]
time = 0.02, size = 80, normalized size = 0.31

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{18} a^{5}-\frac {5}{16} b \,a^{4} x^{2}-\frac {5}{7} b^{2} a^{3} x^{4}-\frac {5}{6} a^{2} b^{3} x^{6}-\frac {1}{2} b^{4} a \,x^{8}-\frac {1}{8} b^{5} x^{10}\right )}{\left (b \,x^{2}+a \right ) x^{18}}\)	\(79\)
gosper	\(-\frac {\left (126 b^{5} x^{10}+504 b^{4} a \,x^{8}+840 a^{2} b^{3} x^{6}+720 b^{2} a^{3} x^{4}+315 b \,a^{4} x^{2}+56 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{1008 x^{18} \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(-\frac {\left (126 b^{5} x^{10}+504 b^{4} a \,x^{8}+840 a^{2} b^{3} x^{6}+720 b^{2} a^{3} x^{4}+315 b \,a^{4} x^{2}+56 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{1008 x^{18} \left (b \,x^{2}+a \right )^{5}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^19,x,method=_RETURNVERBOSE)

[Out]

-1/1008*(126*b^5*x^10+504*a*b^4*x^8+840*a^2*b^3*x^6+720*a^3*b^2*x^4+315*a^4*b*x^2+56*a^5)*((b*x^2+a)^2)^(5/2)/
x^18/(b*x^2+a)^5

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Maxima [A]
time = 0.27, size = 57, normalized size = 0.22 \begin {gather*} -\frac {b^{5}}{8 \, x^{8}} - \frac {a b^{4}}{2 \, x^{10}} - \frac {5 \, a^{2} b^{3}}{6 \, x^{12}} - \frac {5 \, a^{3} b^{2}}{7 \, x^{14}} - \frac {5 \, a^{4} b}{16 \, x^{16}} - \frac {a^{5}}{18 \, x^{18}} \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)^(5/2)/x^19,x, algorithm="maxima")

[Out]

-1/8*b^5/x^8 - 1/2*a*b^4/x^10 - 5/6*a^2*b^3/x^12 - 5/7*a^3*b^2/x^14 - 5/16*a^4*b/x^16 - 1/18*a^5/x^18

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Fricas [A]
time = 0.33, size = 59, normalized size = 0.23 \begin {gather*} -\frac {126 \, b^{5} x^{10} + 504 \, a b^{4} x^{8} + 840 \, a^{2} b^{3} x^{6} + 720 \, a^{3} b^{2} x^{4} + 315 \, a^{4} b x^{2} + 56 \, a^{5}}{1008 \, x^{18}} \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)^(5/2)/x^19,x, algorithm="fricas")

[Out]

-1/1008*(126*b^5*x^10 + 504*a*b^4*x^8 + 840*a^2*b^3*x^6 + 720*a^3*b^2*x^4 + 315*a^4*b*x^2 + 56*a^5)/x^18

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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 {5}{2}}}{x^{19}}\, 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)**(5/2)/x**19,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**19, x)

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Giac [A]
time = 4.97, size = 107, normalized size = 0.42 \begin {gather*} -\frac {126 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 504 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 840 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 720 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 315 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 56 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{1008 \, x^{18}} \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)^(5/2)/x^19,x, algorithm="giac")

[Out]

-1/1008*(126*b^5*x^10*sgn(b*x^2 + a) + 504*a*b^4*x^8*sgn(b*x^2 + a) + 840*a^2*b^3*x^6*sgn(b*x^2 + a) + 720*a^3
*b^2*x^4*sgn(b*x^2 + a) + 315*a^4*b*x^2*sgn(b*x^2 + a) + 56*a^5*sgn(b*x^2 + a))/x^18

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Mupad [B]
time = 4.27, size = 231, normalized size = 0.91 \begin {gather*} -\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{18\,x^{18}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{8\,x^8\,\left (b\,x^2+a\right )}-\frac {a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^{10}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{16\,x^{16}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{6\,x^{12}\,\left (b\,x^2+a\right )}-\frac {5\,a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^{14}\,\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)^(5/2)/x^19,x)

[Out]

- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(18*x^18*(a + b*x^2)) - (b^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(8*x
^8*(a + b*x^2)) - (a*b^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(2*x^10*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2
*a*b*x^2)^(1/2))/(16*x^16*(a + b*x^2)) - (5*a^2*b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(6*x^12*(a + b*x^2)) -
(5*a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(7*x^14*(a + b*x^2))

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