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

Optimal. Leaf size=188 \[ \frac {x^4 (A b-a B)}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

3*a*B/b^5/((b*x+a)^2)^(1/2)+1/4*(A*b-B*a)*x^4/a/b/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/3*a^3*B/b^5/(b*x+a)^2/((b*x+a)
^2)^(1/2)-3/2*a^2*B/b^5/(b*x+a)/((b*x+a)^2)^(1/2)+B*(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 = 188, 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} \[ \frac {x^4 (A b-a B)}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 {x^3 (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^3 (A+B x)}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 B \left (a b+b^2 x\right )\right ) \int \frac {x^3}{\left (a b+b^2 x\right )^4} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 B \left (a b+b^2 x\right )\right ) \int \left (-\frac {a^3}{b^7 (a+b x)^4}+\frac {3 a^2}{b^7 (a+b x)^3}-\frac {3 a}{b^7 (a+b x)^2}+\frac {1}{b^7 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 a B}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^4}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 B}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a^2 B}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {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 = 0.05, size = 103, normalized size = 0.55 \[ \frac {25 a^4 B+a^3 (88 b B x-3 A b)-12 a^2 b^2 x (A-9 B x)+6 a b^3 x^2 (8 B x-3 A)+12 B (a+b x)^4 \log (a+b x)-12 A b^4 x^3}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*a^4*B - 12*A*b^4*x^3 - 12*a^2*b^2*x*(A - 9*B*x) + 6*a*b^3*x^2*(-3*A + 8*B*x) + a^3*(-3*A*b + 88*b*B*x) + 1
2*B*(a + b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x)^3*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.85, size = 175, normalized size = 0.93 \[ \frac {25 \, B a^{4} - 3 \, A a^{3} b + 12 \, {\left (4 \, B a b^{3} - A b^{4}\right )} x^{3} + 18 \, {\left (6 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 4 \, {\left (22 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x + 12 \, {\left (B b^{4} x^{4} + 4 \, B a b^{3} x^{3} + 6 \, B a^{2} b^{2} x^{2} + 4 \, B a^{3} b x + B a^{4}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

1/12*(25*B*a^4 - 3*A*a^3*b + 12*(4*B*a*b^3 - A*b^4)*x^3 + 18*(6*B*a^2*b^2 - A*a*b^3)*x^2 + 4*(22*B*a^3*b - 3*A
*a^2*b^2)*x + 12*(B*b^4*x^4 + 4*B*a*b^3*x^3 + 6*B*a^2*b^2*x^2 + 4*B*a^3*b*x + B*a^4)*log(b*x + a))/(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)

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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^3*(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 = 0.06, size = 168, normalized size = 0.89 \[ -\frac {\left (-12 B \,b^{4} x^{4} \ln \left (b x +a \right )-48 B a \,b^{3} x^{3} \ln \left (b x +a \right )+12 A \,b^{4} x^{3}-72 B \,a^{2} b^{2} x^{2} \ln \left (b x +a \right )-48 B a \,b^{3} x^{3}+18 A a \,b^{3} x^{2}-48 B \,a^{3} b x \ln \left (b x +a \right )-108 B \,a^{2} b^{2} x^{2}+12 A \,a^{2} b^{2} x -12 B \,a^{4} \ln \left (b x +a \right )-88 B \,a^{3} b x +3 A \,a^{3} b -25 B \,a^{4}\right ) \left (b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{5}} \]

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)^(5/2),x)

[Out]

-1/12*(-12*B*ln(b*x+a)*x^4*b^4-48*B*ln(b*x+a)*x^3*a*b^3+12*A*b^4*x^3-72*B*a^2*b^2*x^2*ln(b*x+a)-48*B*a*b^3*x^3
+18*A*a*b^3*x^2-48*B*a^3*b*x*ln(b*x+a)-108*B*a^2*b^2*x^2+12*A*a^2*b^2*x-12*B*a^4*ln(b*x+a)-88*B*a^3*b*x+3*A*a^
3*b-25*B*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(5/2)

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maxima [A]  time = 0.65, size = 201, normalized size = 1.07 \[ \frac {1}{12} \, B {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{12} \, A {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/12*B*((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) - 1/12*A*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x
^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^3*(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 \[ \int \frac {x^{3} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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