\(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^7} \, dx\) [566]

Optimal result

Integrand size = 26, antiderivative size = 163 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \]

[Out]

-1/6*a^3*((b*x^2+a)^2)^(1/2)/x^6/(b*x^2+a)-3/4*a^2*b*((b*x^2+a)^2)^(1/2)/x^4/(b*x^2+a)-3/2*a*b^2*((b*x^2+a)^2)
^(1/2)/x^2/(b*x^2+a)+b^3*ln(x)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1126, 272, 45} \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^7,x]

[Out]

-1/6*(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^6*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*x^
4*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*(a + b*x^2)) + (b^3*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4]*Log[x])/(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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^3}{x^7} \, dx}{b^2 \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 \frac {\left (a b+b^2 x\right )^3}{x^4} \, 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} \text {Subst}\left (\int \left (\frac {a^3 b^3}{x^4}+\frac {3 a^2 b^4}{x^3}+\frac {3 a b^5}{x^2}+\frac {b^6}{x}\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}}{6 x^6 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 x^4 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {2 a^3 \sqrt {a^2}+9 \left (a^2\right )^{3/2} b x^2+18 a \sqrt {a^2} b^2 x^4-2 a^3 \sqrt {\left (a+b x^2\right )^2}-7 a^2 b x^2 \sqrt {\left (a+b x^2\right )^2}-11 a b^2 x^4 \sqrt {\left (a+b x^2\right )^2}-12 a b^3 x^6 \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-12 \sqrt {a^2} b^3 x^6 \log \left (x^2\right )+6 \sqrt {a^2} b^3 x^6 \log \left (a \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )+6 \sqrt {a^2} b^3 x^6 \log \left (a \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 a x^6} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^7,x]

[Out]

(2*a^3*Sqrt[a^2] + 9*(a^2)^(3/2)*b*x^2 + 18*a*Sqrt[a^2]*b^2*x^4 - 2*a^3*Sqrt[(a + b*x^2)^2] - 7*a^2*b*x^2*Sqrt
[(a + b*x^2)^2] - 11*a*b^2*x^4*Sqrt[(a + b*x^2)^2] - 12*a*b^3*x^6*ArcTanh[(b*x^2)/(Sqrt[a^2] - Sqrt[(a + b*x^2
)^2])] - 12*Sqrt[a^2]*b^3*x^6*Log[x^2] + 6*Sqrt[a^2]*b^3*x^6*Log[a*(Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2])]
+ 6*Sqrt[a^2]*b^3*x^6*Log[a*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/(24*a*x^6)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.31

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (6 b^{3} \ln \left (x^{2}\right ) x^{6}-18 b^{2} x^{4} a -9 a^{2} b \,x^{2}-2 a^{3}\right )}{12 x^{6}}\)	\(50\)
default	\(\frac {{\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (12 \ln \left (x \right ) x^{6} b^{3}-18 b^{2} x^{4} a -9 a^{2} b \,x^{2}-2 a^{3}\right )}{12 \left (b \,x^{2}+a \right )^{3} x^{6}}\)	\(60\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3}{2} b^{2} x^{4} a -\frac {3}{4} a^{2} b \,x^{2}-\frac {1}{6} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{6}}+\frac {b^{3} \ln \left (x \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(76\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

1/12*csgn(b*x^2+a)*(6*b^3*ln(x^2)*x^6-18*b^2*x^4*a-9*a^2*b*x^2-2*a^3)/x^6

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {12 \, b^{3} x^{6} \log \left (x\right ) - 18 \, a b^{2} x^{4} - 9 \, a^{2} b x^{2} - 2 \, a^{3}}{12 \, x^{6}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x, algorithm="fricas")

[Out]

1/12*(12*b^3*x^6*log(x) - 18*a*b^2*x^4 - 9*a^2*b*x^2 - 2*a^3)/x^6

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**7,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**7, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.20 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=b^{3} \log \left (x\right ) - \frac {3 \, a b^{2}}{2 \, x^{2}} - \frac {3 \, a^{2} b}{4 \, x^{4}} - \frac {a^{3}}{6 \, x^{6}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x, algorithm="maxima")

[Out]

b^3*log(x) - 3/2*a*b^2/x^2 - 3/4*a^2*b/x^4 - 1/6*a^3/x^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\frac {1}{2} \, b^{3} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {11 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 18 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{12 \, x^{6}} \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^7,x, algorithm="giac")

[Out]

1/2*b^3*log(x^2)*sgn(b*x^2 + a) - 1/12*(11*b^3*x^6*sgn(b*x^2 + a) + 18*a*b^2*x^4*sgn(b*x^2 + a) + 9*a^2*b*x^2*
sgn(b*x^2 + a) + 2*a^3*sgn(b*x^2 + a))/x^6

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}}{x^7} \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^7,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^7, x)