\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{x^7} \, dx\) [160]

Optimal result

Integrand size = 24, antiderivative size = 151 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {3 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[Out]

-1/6*a^3*((b*x+a)^2)^(1/2)/x^6/(b*x+a)-3/5*a^2*b*((b*x+a)^2)^(1/2)/x^5/(b*x+a)-3/4*a*b^2*((b*x+a)^2)^(1/2)/x^4
/(b*x+a)-1/3*b^3*((b*x+a)^2)^(1/2)/x^3/(b*x+a)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {3 a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)} \]

[In]

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

[Out]

-1/6*(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x^6*(a + b*x)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^5*(a +
b*x)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b*x)) - (b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^
3*(a + b*x))

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]

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.36 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {\sqrt {(a+b x)^2} \left (10 a^3+36 a^2 b x+45 a b^2 x^2+20 b^3 x^3\right )}{60 x^6 (a+b x)} \]

[In]

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

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(10*a^3 + 36*a^2*b*x + 45*a*b^2*x^2 + 20*b^3*x^3))/(x^6*(a + b*x))

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.34

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {1}{3} b^{3} x^{3}-\frac {3}{4} a \,b^{2} x^{2}-\frac {3}{5} a^{2} b x -\frac {1}{6} a^{3}\right )}{\left (b x +a \right ) x^{6}}\)	\(51\)
gosper	\(-\frac {\left (20 b^{3} x^{3}+45 a \,b^{2} x^{2}+36 a^{2} b x +10 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\)	\(52\)
default	\(-\frac {\left (20 b^{3} x^{3}+45 a \,b^{2} x^{2}+36 a^{2} b x +10 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\)	\(52\)

[In]

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

[Out]

((b*x+a)^2)^(1/2)/(b*x+a)/x^6*(-1/3*b^3*x^3-3/4*a*b^2*x^2-3/5*a^2*b*x-1/6*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {20 \, b^{3} x^{3} + 45 \, a b^{2} x^{2} + 36 \, a^{2} b x + 10 \, a^{3}}{60 \, x^{6}} \]

[In]

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

[Out]

-1/60*(20*b^3*x^3 + 45*a*b^2*x^2 + 36*a^2*b*x + 10*a^3)/x^6

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{6}}{4 \, a^{6}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{5}}{4 \, a^{5} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{4}}{4 \, a^{6} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{3}}{4 \, a^{5} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{2}}{4 \, a^{4} x^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b}{30 \, a^{3} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{6 \, a^{2} x^{6}} \]

[In]

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

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^6/a^6 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^5/(a^5*x) - 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*b^4/(a^6*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^3/(a^5*x^3) - 1/4*(b^2*x^2 + 2*a*b
*x + a^2)^(5/2)*b^2/(a^4*x^4) + 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b/(a^3*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a^
2)^(5/2)/(a^2*x^6)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.49 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {b^{6} \mathrm {sgn}\left (b x + a\right )}{60 \, a^{3}} - \frac {20 \, b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 45 \, a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 36 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{3} \mathrm {sgn}\left (b x + a\right )}{60 \, x^{6}} \]

[In]

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

[Out]

-1/60*b^6*sgn(b*x + a)/a^3 - 1/60*(20*b^3*x^3*sgn(b*x + a) + 45*a*b^2*x^2*sgn(b*x + a) + 36*a^2*b*x*sgn(b*x +
a) + 10*a^3*sgn(b*x + a))/x^6

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^3\,\left (a+b\,x\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )} \]

[In]

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

[Out]

- (a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(3*x^3*(a +
b*x)) - (3*a*b^2*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*x^4*(a + b*x)) - (3*a^2*b*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)
)/(5*x^5*(a + b*x))