3.8.15 \(\int \frac {x^3 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [715]

Optimal. Leaf size=202 \[ -\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (A b-2 a B) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 202, 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 = {784, 78} \begin {gather*} -\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) (A b-2 a B) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x) (A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(3*A*b - 4*a*B))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^3*(A*b - a*B))/(2*b^5*(a + b*x)*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) + ((A*b - 3*a*B)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^2*(a + b*x))/(2*b^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(A*b - 2*a*B)*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 784

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

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Mathematica [A]
time = 0.04, size = 117, normalized size = 0.58 \begin {gather*} \frac {7 a^4 B+4 a b^3 x^2 (A-B x)+b^4 x^3 (2 A+B x)-a^2 b^2 x (4 A+11 B x)+a^3 (-5 A b+2 b B x)+6 a (-A b+2 a B) (a+b x)^2 \log (a+b x)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*B + 4*a*b^3*x^2*(A - B*x) + b^4*x^3*(2*A + B*x) - a^2*b^2*x*(4*A + 11*B*x) + a^3*(-5*A*b + 2*b*B*x) + 6
*a*(-(A*b) + 2*a*B)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.73, size = 191, normalized size = 0.95

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{2} b B \,x^{2}+A b x -3 B a x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 A \,a^{2} b +4 B \,a^{3}\right ) x -\frac {a^{3} \left (5 A b -7 B a \right )}{2 b}\right )}{\left (b x +a \right )^{3} b^{4}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, a \left (A b -2 B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(129\)
default	\(-\frac {\left (-b^{4} B \,x^{4}+6 A \ln \left (b x +a \right ) a \,b^{3} x^{2}-2 A \,b^{4} x^{3}-12 B \ln \left (b x +a \right ) a^{2} b^{2} x^{2}+4 B a \,b^{3} x^{3}+12 A \ln \left (b x +a \right ) a^{2} b^{2} x -4 A a \,b^{3} x^{2}-24 B \ln \left (b x +a \right ) a^{3} b x +11 B \,a^{2} b^{2} x^{2}+6 A \ln \left (b x +a \right ) a^{3} b +4 A \,a^{2} b^{2} x -12 B \ln \left (b x +a \right ) a^{4}-2 B \,a^{3} b x +5 A \,a^{3} b -7 B \,a^{4}\right ) \left (b x +a \right )}{2 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-b^4*B*x^4+6*A*ln(b*x+a)*a*b^3*x^2-2*A*b^4*x^3-12*B*ln(b*x+a)*a^2*b^2*x^2+4*B*a*b^3*x^3+12*A*ln(b*x+a)*a
^2*b^2*x-4*A*a*b^3*x^2-24*B*ln(b*x+a)*a^3*b*x+11*B*a^2*b^2*x^2+6*A*ln(b*x+a)*a^3*b+4*A*a^2*b^2*x-12*B*ln(b*x+a
)*a^4-2*B*a^3*b*x+5*A*a^3*b-7*B*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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Maxima [A]
time = 0.28, size = 242, normalized size = 1.20 \begin {gather*} \frac {B x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, B a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {A x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {6 \, B a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {3 \, A a \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {5 \, B a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {2 \, A a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {12 \, B a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, A a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, B a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, A a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + A*x^2/(sqrt(
b^2*x^2 + 2*a*b*x + a^2)*b^2) + 6*B*a^2*log(x + a/b)/b^5 - 3*A*a*log(x + a/b)/b^4 - 5*B*a^3/(sqrt(b^2*x^2 + 2*
a*b*x + a^2)*b^5) + 2*A*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) + 12*B*a^3*x/(b^6*(x + a/b)^2) - 6*A*a^2*x/(b^
5*(x + a/b)^2) + 23/2*B*a^4/(b^7*(x + a/b)^2) - 11/2*A*a^3/(b^6*(x + a/b)^2)

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Fricas [A]
time = 1.48, size = 171, normalized size = 0.85 \begin {gather*} \frac {B b^{4} x^{4} + 7 \, B a^{4} - 5 \, A a^{3} b - 2 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{3} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x + 6 \, {\left (2 \, B a^{4} - A a^{3} b + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 2 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b^4*x^4 + 7*B*a^4 - 5*A*a^3*b - 2*(2*B*a*b^3 - A*b^4)*x^3 - (11*B*a^2*b^2 - 4*A*a*b^3)*x^2 + 2*(B*a^3*b
- 2*A*a^2*b^2)*x + 6*(2*B*a^4 - A*a^3*b + (2*B*a^2*b^2 - A*a*b^3)*x^2 + 2*(2*B*a^3*b - A*a^2*b^2)*x)*log(b*x
+ a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.45, size = 134, normalized size = 0.66 \begin {gather*} \frac {3 \, {\left (2 \, B a^{2} - A a b\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {B b^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, B a b^{2} x \mathrm {sgn}\left (b x + a\right ) + 2 \, A b^{3} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{6}} + \frac {7 \, B a^{4} - 5 \, A a^{3} b + 2 \, {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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