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 \[ \frac{x^4 (A b-a B)}{4 a b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 a B}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 a^2 B}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a^3 B}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[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])

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Rubi [A]  time = 0.0988829, antiderivative size = 188, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [A]  time = 0.0502164, size = 103, normalized size = 0.55 \[ \frac{-12 a^2 b^2 x (A-9 B x)+a^3 (88 b B x-3 A b)+25 a^4 B+6 a b^3 x^2 (8 B x-3 A)+12 B (a+b x)^4 \log (a+b x)-12 A b^4 x^3}{12 b^5 (a+b x)^3 \sqrt{(a+b x)^2}} \]

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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Maple [A]  time = 0.012, size = 168, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -12\,B\ln \left ( bx+a \right ){x}^{4}{b}^{4}-48\,B\ln \left ( bx+a \right ){x}^{3}a{b}^{3}+12\,A{x}^{3}{b}^{4}-72\,B\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}-48\,B{x}^{3}a{b}^{3}+18\,A{x}^{2}a{b}^{3}-48\,B\ln \left ( bx+a \right ) x{a}^{3}b-108\,B{x}^{2}{a}^{2}{b}^{2}+12\,A{a}^{2}{b}^{2}x-12\,B\ln \left ( bx+a \right ){a}^{4}-88\,B{a}^{3}bx+3\,A{a}^{3}b-25\,B{a}^{4} \right ) \left ( bx+a \right ) }{12\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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*x^3*b^4-72*B*ln(b*x+a)*x^2*a^2*b^2-48*B*x^3*a*b^3
+18*A*x^2*a*b^3-48*B*ln(b*x+a)*x*a^3*b-108*B*x^2*a^2*b^2+12*A*a^2*b^2*x-12*B*ln(b*x+a)*a^4-88*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 = 1.06213, size = 315, normalized size = 1.68 \begin{align*} \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{3 \, a^{3} b}{{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} - \frac{8 \, a^{2}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} + \frac{6 \, a}{{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{6 \, a^{3}}{{\left (b^{2}\right )}^{\frac{5}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{4}}\right )} \end{align*}

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) + 3*a^3*b/((b^2)^(9/2)*(x + a/b)^4) - 8*a^2/((b^2)^(7/2)*(x + a/b)^3) + 6*a/((b
^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(x + a/b)^4))

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Fricas [A]  time = 1.55083, size = 370, normalized size = 1.97 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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