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

Optimal. Leaf size=151 \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac{3 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0347421, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 43} \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac{3 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^7} \, dx &=\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]  time = 0.0138948, size = 55, normalized size = 0.36 \[ -\frac{\sqrt{(a+b x)^2} \left (36 a^2 b x+10 a^3+45 a b^2 x^2+20 b^3 x^3\right )}{60 x^6 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.176, size = 52, normalized size = 0.3 \begin{align*} -{\frac{20\,{b}^{3}{x}^{3}+45\,a{b}^{2}{x}^{2}+36\,b{a}^{2}x+10\,{a}^{3}}{60\,{x}^{6} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.01139, size = 82, normalized size = 0.54 \begin{align*} -\frac{20 \, b^{3} x^{3} + 45 \, a b^{2} x^{2} + 36 \, a^{2} b x + 10 \, a^{3}}{60 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{x^{7}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]  time = 1.49118, size = 100, normalized size = 0.66 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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