3.678 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=210 \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{6 x^6 (a+b x)}-\frac {3 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 x^5 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{4 x^4 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 76} \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{6 x^6 (a+b x)}-\frac {3 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{5 x^5 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{4 x^4 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^8,x]

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*x^7*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
6*x^6*(a + b*x)) - (3*a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^5*(a + b*x)) - (b^2*(A*b + 3*a*B)*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b*x)) - (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b*x))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^8} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3 (A+B x)}{x^8} \, 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 A b^3}{x^8}+\frac {a^2 b^3 (3 A b+a B)}{x^7}+\frac {3 a b^4 (A b+a B)}{x^6}+\frac {b^5 (A b+3 a B)}{x^5}+\frac {b^6 B}{x^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 87, normalized size = 0.41 \[ -\frac {\sqrt {(a+b x)^2} \left (10 a^3 (6 A+7 B x)+42 a^2 b x (5 A+6 B x)+63 a b^2 x^2 (4 A+5 B x)+35 b^3 x^3 (3 A+4 B x)\right )}{420 x^7 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^8,x]

[Out]

-1/420*(Sqrt[(a + b*x)^2]*(35*b^3*x^3*(3*A + 4*B*x) + 63*a*b^2*x^2*(4*A + 5*B*x) + 42*a^2*b*x*(5*A + 6*B*x) +
10*a^3*(6*A + 7*B*x)))/(x^7*(a + b*x))

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fricas [A]  time = 0.87, size = 73, normalized size = 0.35 \[ -\frac {140 \, B b^{3} x^{4} + 60 \, A a^{3} + 105 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 252 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 70 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{420 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x, algorithm="fricas")

[Out]

-1/420*(140*B*b^3*x^4 + 60*A*a^3 + 105*(3*B*a*b^2 + A*b^3)*x^3 + 252*(B*a^2*b + A*a*b^2)*x^2 + 70*(B*a^3 + 3*A
*a^2*b)*x)/x^7

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giac [A]  time = 0.23, size = 149, normalized size = 0.71 \[ -\frac {{\left (7 \, B a b^{6} - 3 \, A b^{7}\right )} \mathrm {sgn}\left (b x + a\right )}{420 \, a^{4}} - \frac {140 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 315 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 252 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 252 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 70 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 210 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 60 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{420 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x, algorithm="giac")

[Out]

-1/420*(7*B*a*b^6 - 3*A*b^7)*sgn(b*x + a)/a^4 - 1/420*(140*B*b^3*x^4*sgn(b*x + a) + 315*B*a*b^2*x^3*sgn(b*x +
a) + 105*A*b^3*x^3*sgn(b*x + a) + 252*B*a^2*b*x^2*sgn(b*x + a) + 252*A*a*b^2*x^2*sgn(b*x + a) + 70*B*a^3*x*sgn
(b*x + a) + 210*A*a^2*b*x*sgn(b*x + a) + 60*A*a^3*sgn(b*x + a))/x^7

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maple [A]  time = 0.05, size = 92, normalized size = 0.44 \[ -\frac {\left (140 B \,b^{3} x^{4}+105 A \,b^{3} x^{3}+315 B a \,b^{2} x^{3}+252 A a \,b^{2} x^{2}+252 B \,a^{2} b \,x^{2}+210 A \,a^{2} b x +70 B \,a^{3} x +60 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{420 \left (b x +a \right )^{3} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x)

[Out]

-1/420*(140*B*b^3*x^4+105*A*b^3*x^3+315*B*a*b^2*x^3+252*A*a*b^2*x^2+252*B*a^2*b*x^2+210*A*a^2*b*x+70*B*a^3*x+6
0*A*a^3)*((b*x+a)^2)^(3/2)/x^7/(b*x+a)^3

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maxima [B]  time = 0.55, size = 435, normalized size = 2.07 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{6}}{4 \, a^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{7}}{4 \, a^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5}}{4 \, a^{5} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{6}}{4 \, a^{6} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{4 \, a^{6} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{4 \, a^{7} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{4 \, a^{5} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{4 \, a^{6} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{4 \, a^{4} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{4 \, a^{5} x^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{30 \, a^{3} x^{5}} - \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{70 \, a^{4} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{6 \, a^{2} x^{6}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{14 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{7 \, a^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^8,x, algorithm="maxima")

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^6/a^6 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^7/a^7 + 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*B*b^5/(a^5*x) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^6/(a^6*x) - 1/4*(b^2*x^2 + 2*a*b
*x + a^2)^(5/2)*B*b^4/(a^6*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x^2) + 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*B*b^3/(a^5*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^3) - 1/4*(b^2*x^2 + 2*a*b*x +
a^2)^(5/2)*B*b^2/(a^4*x^4) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^4) + 7/30*(b^2*x^2 + 2*a*b*x +
a^2)^(5/2)*B*b/(a^3*x^5) - 17/70*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a^
2)^(5/2)*B/(a^2*x^6) + 3/14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^6) - 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(5/2
)*A/(a^2*x^7)

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mupad [B]  time = 1.19, size = 196, normalized size = 0.93 \[ -\frac {\left (\frac {B\,a^3}{6}+\frac {A\,b\,a^2}{2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^6\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{4}+\frac {3\,B\,a\,b^2}{4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^4\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^3\,\left (a+b\,x\right )}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^8,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{8}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**8,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**8, x)

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