3.722 \(\int \frac {x^4 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=245 \[ -\frac {a^2 (3 A b-5 a B)}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a (2 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-5 a B) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4 (A b-a B)}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (4 A b-5 a B)}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 245, 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, 77} \[ -\frac {a^4 (A b-a B)}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (4 A b-5 a B)}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2 (3 A b-5 a B)}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a (2 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-5 a B) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

Rule 77

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 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 {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {x^4 (A+B x)}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {B}{b^{10}}-\frac {a^4 (-A b+a B)}{b^{10} (a+b x)^5}+\frac {a^3 (-4 A b+5 a B)}{b^{10} (a+b x)^4}-\frac {2 a^2 (-3 A b+5 a B)}{b^{10} (a+b x)^3}+\frac {2 a (-2 A b+5 a B)}{b^{10} (a+b x)^2}+\frac {A b-5 a B}{b^{10} (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 a (2 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4 (A b-a B)}{4 b^6 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (4 A b-5 a B)}{3 b^6 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2 (3 A b-5 a B)}{b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-5 a B) (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 127, normalized size = 0.52 \[ \frac {-77 a^5 B+a^4 b (25 A-248 B x)+4 a^3 b^2 x (22 A-63 B x)+12 a^2 b^3 x^2 (9 A-4 B x)+48 a b^4 x^3 (A+B x)+12 (a+b x)^4 (A b-5 a B) \log (a+b x)+12 b^5 B x^5}{12 b^6 (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-77*a^5*B + 12*b^5*B*x^5 + a^4*b*(25*A - 248*B*x) + 4*a^3*b^2*x*(22*A - 63*B*x) + 12*a^2*b^3*x^2*(9*A - 4*B*x
) + 48*a*b^4*x^3*(A + B*x) + 12*(A*b - 5*a*B)*(a + b*x)^4*Log[a + b*x])/(12*b^6*(a + b*x)^3*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.89, size = 252, normalized size = 1.03 \[ \frac {12 \, B b^{5} x^{5} + 48 \, B a b^{4} x^{4} - 77 \, B a^{5} + 25 \, A a^{4} b - 48 \, {\left (B a^{2} b^{3} - A a b^{4}\right )} x^{3} - 36 \, {\left (7 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} - 8 \, {\left (31 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x - 12 \, {\left (5 \, B a^{5} - A a^{4} b + {\left (5 \, B a b^{4} - A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} - A a b^{4}\right )} x^{3} + 6 \, {\left (5 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (5 \, B a^{4} b - A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{12 \, {\left (b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/12*(12*B*b^5*x^5 + 48*B*a*b^4*x^4 - 77*B*a^5 + 25*A*a^4*b - 48*(B*a^2*b^3 - A*a*b^4)*x^3 - 36*(7*B*a^3*b^2 -
3*A*a^2*b^3)*x^2 - 8*(31*B*a^4*b - 11*A*a^3*b^2)*x - 12*(5*B*a^5 - A*a^4*b + (5*B*a*b^4 - A*b^5)*x^4 + 4*(5*B
*a^2*b^3 - A*a*b^4)*x^3 + 6*(5*B*a^3*b^2 - A*a^2*b^3)*x^2 + 4*(5*B*a^4*b - A*a^3*b^2)*x)*log(b*x + a))/(b^10*x
^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.07, size = 273, normalized size = 1.11 \[ \frac {\left (12 A \,b^{5} x^{4} \ln \left (b x +a \right )-60 B a \,b^{4} x^{4} \ln \left (b x +a \right )+12 B \,b^{5} x^{5}+48 A a \,b^{4} x^{3} \ln \left (b x +a \right )-240 B \,a^{2} b^{3} x^{3} \ln \left (b x +a \right )+48 B a \,b^{4} x^{4}+72 A \,a^{2} b^{3} x^{2} \ln \left (b x +a \right )+48 A a \,b^{4} x^{3}-360 B \,a^{3} b^{2} x^{2} \ln \left (b x +a \right )-48 B \,a^{2} b^{3} x^{3}+48 A \,a^{3} b^{2} x \ln \left (b x +a \right )+108 A \,a^{2} b^{3} x^{2}-240 B \,a^{4} b x \ln \left (b x +a \right )-252 B \,a^{3} b^{2} x^{2}+12 A \,a^{4} b \ln \left (b x +a \right )+88 A \,a^{3} b^{2} x -60 B \,a^{5} \ln \left (b x +a \right )-248 B \,a^{4} b x +25 A \,a^{4} b -77 B \,a^{5}\right ) \left (b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/12*(12*A*ln(b*x+a)*x^4*b^5-60*B*ln(b*x+a)*x^4*a*b^4+12*B*b^5*x^5+48*A*ln(b*x+a)*x^3*a*b^4-240*B*ln(b*x+a)*x^
3*a^2*b^3+48*B*a*b^4*x^4+72*A*a^2*b^3*x^2*ln(b*x+a)+48*A*a*b^4*x^3-360*B*a^3*b^2*x^2*ln(b*x+a)-48*B*a^2*b^3*x^
3+48*A*a^3*b^2*x*ln(b*x+a)+108*A*a^2*b^3*x^2-240*B*a^4*b*x*ln(b*x+a)-252*B*a^3*b^2*x^2+12*A*a^4*b*ln(b*x+a)+88
*A*a^3*b^2*x-60*B*a^5*ln(b*x+a)-248*B*a^4*b*x+25*A*a^4*b-77*B*a^5)*(b*x+a)/b^6/((b*x+a)^2)^(5/2)

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maxima [A]  time = 0.79, size = 211, normalized size = 0.86 \[ \frac {1}{12} \, B {\left (\frac {12 \, b^{5} x^{5} + 48 \, a b^{4} x^{4} - 48 \, a^{2} b^{3} x^{3} - 252 \, a^{3} b^{2} x^{2} - 248 \, a^{4} b x - 77 \, a^{5}}{b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}} - \frac {60 \, a \log \left (b x + a\right )}{b^{6}}\right )} + \frac {1}{12} \, A {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/12*B*((12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5)/(b^10*x^4 + 4*a*
b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6) - 60*a*log(b*x + a)/b^6) + 1/12*A*((48*a*b^3*x^3 + 108*a^2*b^
2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)
/b^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x**4*(A + B*x)/((a + b*x)**2)**(5/2), x)

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