3.675 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=187 \[ -\frac {A (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 a x^4}-\frac {3 a^2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b^2 B \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 B \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {a^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {770, 78, 43} \[ -\frac {A (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 a x^4}-\frac {a^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {3 a^2 b B \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {3 a b^2 B \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^3 B \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^5,x]

[Out]

-(a^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b*x)) - (3*a^2*b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a
+ b*x)) - (3*a*b^2*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x)) - (A*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/(4*a*x^4) + (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x])/(a + b*x)

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])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [A]  time = 0.03, size = 88, normalized size = 0.47 \[ -\frac {\sqrt {(a+b x)^2} \left (a^3 (3 A+4 B x)+6 a^2 b x (2 A+3 B x)+18 a b^2 x^2 (A+2 B x)+12 A b^3 x^3-12 b^3 B x^4 \log (x)\right )}{12 x^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^5,x]

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(12*A*b^3*x^3 + 18*a*b^2*x^2*(A + 2*B*x) + 6*a^2*b*x*(2*A + 3*B*x) + a^3*(3*A + 4*B*x
) - 12*b^3*B*x^4*Log[x]))/(x^4*(a + b*x))

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fricas [A]  time = 0.72, size = 75, normalized size = 0.40 \[ \frac {12 \, B b^{3} x^{4} \log \relax (x) - 3 \, A a^{3} - 12 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 18 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 4 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{12 \, x^{4}} \]

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^5,x, algorithm="fricas")

[Out]

1/12*(12*B*b^3*x^4*log(x) - 3*A*a^3 - 12*(3*B*a*b^2 + A*b^3)*x^3 - 18*(B*a^2*b + A*a*b^2)*x^2 - 4*(B*a^3 + 3*A
*a^2*b)*x)/x^4

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giac [A]  time = 0.18, size = 121, normalized size = 0.65 \[ B b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {3 \, A a^{3} \mathrm {sgn}\left (b x + a\right ) + 12 \, {\left (3 \, B a b^{2} \mathrm {sgn}\left (b x + a\right ) + A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 18 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) + A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 4 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} x}{12 \, x^{4}} \]

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^5,x, algorithm="giac")

[Out]

B*b^3*log(abs(x))*sgn(b*x + a) - 1/12*(3*A*a^3*sgn(b*x + a) + 12*(3*B*a*b^2*sgn(b*x + a) + A*b^3*sgn(b*x + a))
*x^3 + 18*(B*a^2*b*sgn(b*x + a) + A*a*b^2*sgn(b*x + a))*x^2 + 4*(B*a^3*sgn(b*x + a) + 3*A*a^2*b*sgn(b*x + a))*
x)/x^4

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maple [A]  time = 0.06, size = 94, normalized size = 0.50 \[ -\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (-12 B \,b^{3} x^{4} \ln \relax (x )+12 A \,b^{3} x^{3}+36 B a \,b^{2} x^{3}+18 A a \,b^{2} x^{2}+18 B \,a^{2} b \,x^{2}+12 A \,a^{2} b x +4 B \,a^{3} x +3 A \,a^{3}\right )}{12 \left (b x +a \right )^{3} x^{4}} \]

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^5,x)

[Out]

-1/12*((b*x+a)^2)^(3/2)*(-12*b^3*B*ln(x)*x^4+12*A*b^3*x^3+36*B*a*b^2*x^3+18*A*a*b^2*x^2+18*B*a^2*b*x^2+12*A*a^
2*b*x+4*B*a^3*x+3*A*a^3)/(b*x+a)^3/x^4

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maxima [B]  time = 0.76, size = 379, normalized size = 2.03 \[ \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{4} x}{2 \, a^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3}}{2 \, a} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{3}}{6 \, a^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{4}}{4 \, a^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2}}{2 \, a^{2} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{4 \, a^{3} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{4 \, a^{4} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{3 \, a^{2} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{4 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{4 \, a^{2} x^{4}} \]

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^5,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*B*b^3*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*B*b^3*log(2*a*b*x/abs(x) + 2*a^2/ab
s(x)) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^4*x/a^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^3/a - 1/6*(b^2*x
^2 + 2*a*b*x + a^2)^(3/2)*B*b^3/a^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^4/a^4 - 1/2*(b^2*x^2 + 2*a*b*x +
a^2)^(3/2)*B*b^2/(a^2*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/(a^3*x) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^
(5/2)*B*b/(a^3*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2
)*B/(a^2*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2
*x^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^5} \,d x \]

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^5,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{5}}\, dx \]

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**5,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**5, x)

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