3.1596 \(\int \frac{(d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=210 \[ \frac{e^3 x (a+b x) (4 b d-3 a e)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 e^2 (a+b x) (b d-a e)^2 \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 e (b d-a e)^3}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^4}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(-4*e*(b*d - a*e)^3)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^4/(2*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (e^3*(4*b*d - 3*a*e)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*x^2*(a + b*x))/(2*b^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (6*e^2*(b*d - a*e)^2*(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 = 0.13473, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac{e^3 x (a+b x) (4 b d-3 a e)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 e^2 (a+b x) (b d-a e)^2 \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 e (b d-a e)^3}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^4}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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{(d+e x)^4}{\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{(d+e x)^4}{\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{e^3 (4 b d-3 a e)}{b^7}+\frac{e^4 x}{b^6}+\frac{(b d-a e)^4}{b^7 (a+b x)^3}+\frac{4 e (b d-a e)^3}{b^7 (a+b x)^2}+\frac{6 e^2 (b d-a e)^2}{b^7 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 e (b d-a e)^3}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^4}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^3 (4 b d-3 a e) x (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{6 e^2 (b d-a e)^2 (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 = 0.0843384, size = 174, normalized size = 0.83 \[ \frac{a^2 b^2 e^2 \left (18 d^2-16 d e x-11 e^2 x^2\right )+2 a^3 b e^3 (e x-10 d)+7 a^4 e^4-4 a b^3 e \left (-6 d^2 e x+d^3-4 d e^2 x^2+e^3 x^3\right )+12 e^2 (a+b x)^2 (b d-a e)^2 \log (a+b x)+b^4 \left (-8 d^3 e x-d^4+8 d e^3 x^3+e^4 x^4\right )}{2 b^5 (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(7*a^4*e^4 + 2*a^3*b*e^3*(-10*d + e*x) + a^2*b^2*e^2*(18*d^2 - 16*d*e*x - 11*e^2*x^2) - 4*a*b^3*e*(d^3 - 6*d^2
*e*x - 4*d*e^2*x^2 + e^3*x^3) + b^4*(-d^4 - 8*d^3*e*x + 8*d*e^3*x^3 + e^4*x^4) + 12*e^2*(b*d - a*e)^2*(a + b*x
)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.204, size = 341, normalized size = 1.6 \begin{align*}{\frac{ \left ({x}^{4}{b}^{4}{e}^{4}+12\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}-24\,\ln \left ( bx+a \right ){x}^{2}a{b}^{3}d{e}^{3}+12\,\ln \left ( bx+a \right ){x}^{2}{b}^{4}{d}^{2}{e}^{2}-4\,{x}^{3}a{b}^{3}{e}^{4}+8\,{x}^{3}{b}^{4}d{e}^{3}+24\,\ln \left ( bx+a \right ) x{a}^{3}b{e}^{4}-48\,\ln \left ( bx+a \right ) x{a}^{2}{b}^{2}d{e}^{3}+24\,\ln \left ( bx+a \right ) xa{b}^{3}{d}^{2}{e}^{2}-11\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}+16\,{x}^{2}a{b}^{3}d{e}^{3}+12\,\ln \left ( bx+a \right ){a}^{4}{e}^{4}-24\,\ln \left ( bx+a \right ){a}^{3}bd{e}^{3}+12\,\ln \left ( bx+a \right ){a}^{2}{b}^{2}{d}^{2}{e}^{2}+2\,x{a}^{3}b{e}^{4}-16\,x{a}^{2}{b}^{2}d{e}^{3}+24\,xa{b}^{3}{d}^{2}{e}^{2}-8\,x{b}^{4}{d}^{3}e+7\,{a}^{4}{e}^{4}-20\,{a}^{3}bd{e}^{3}+18\,{d}^{2}{e}^{2}{a}^{2}{b}^{2}-4\,a{b}^{3}{d}^{3}e-{b}^{4}{d}^{4} \right ) \left ( bx+a \right ) }{2\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(x^4*b^4*e^4+12*ln(b*x+a)*x^2*a^2*b^2*e^4-24*ln(b*x+a)*x^2*a*b^3*d*e^3+12*ln(b*x+a)*x^2*b^4*d^2*e^2-4*x^3*
a*b^3*e^4+8*x^3*b^4*d*e^3+24*ln(b*x+a)*x*a^3*b*e^4-48*ln(b*x+a)*x*a^2*b^2*d*e^3+24*ln(b*x+a)*x*a*b^3*d^2*e^2-1
1*x^2*a^2*b^2*e^4+16*x^2*a*b^3*d*e^3+12*ln(b*x+a)*a^4*e^4-24*ln(b*x+a)*a^3*b*d*e^3+12*ln(b*x+a)*a^2*b^2*d^2*e^
2+2*x*a^3*b*e^4-16*x*a^2*b^2*d*e^3+24*x*a*b^3*d^2*e^2-8*x*b^4*d^3*e+7*a^4*e^4-20*a^3*b*d*e^3+18*d^2*e^2*a^2*b^
2-4*a*b^3*d^3*e-b^4*d^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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Maxima [B]  time = 1.08357, size = 657, normalized size = 3.13 \begin{align*} \frac{e^{4} x^{3}}{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{4 \, d e^{3} x^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{5 \, a e^{4} x^{2}}{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac{6 \, d^{2} e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{12 \, a d e^{3} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} + \frac{6 \, a^{2} e^{4} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{2}} + \frac{9 \, a^{2} b^{2} d^{2} e^{2}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{18 \, a^{3} b d e^{3}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{9 \, a^{4} e^{4}}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{12 \, a b d^{2} e^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{24 \, a^{2} d e^{3} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{12 \, a^{3} e^{4} x}{{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{4 \, d^{3} e}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{8 \, a^{2} d e^{3}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac{5 \, a^{3} e^{4}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} - \frac{d^{4}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, a d^{3} e}{{\left (b^{2}\right )}^{\frac{3}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} - \frac{4 \, a^{3} d e^{3}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{5 \, a^{4} e^{4}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}} b^{4}{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*e^4*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 4*d*e^3*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*a*e^4*
x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*d^2*e^2*log(x + a/b)/(b^2)^(3/2) - 12*a*d*e^3*log(x + a/b)/((b^2)^
(3/2)*b) + 6*a^2*e^4*log(x + a/b)/((b^2)^(3/2)*b^2) + 9*a^2*b^2*d^2*e^2/((b^2)^(7/2)*(x + a/b)^2) - 18*a^3*b*d
*e^3/((b^2)^(7/2)*(x + a/b)^2) + 9*a^4*e^4/((b^2)^(7/2)*(x + a/b)^2) + 12*a*b*d^2*e^2*x/((b^2)^(5/2)*(x + a/b)
^2) - 24*a^2*d*e^3*x/((b^2)^(5/2)*(x + a/b)^2) + 12*a^3*e^4*x/((b^2)^(5/2)*b*(x + a/b)^2) - 4*d^3*e/(sqrt(b^2*
x^2 + 2*a*b*x + a^2)*b^2) + 8*a^2*d*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 5*a^3*e^4/(sqrt(b^2*x^2 + 2*a*b*
x + a^2)*b^5) - 1/2*d^4/((b^2)^(3/2)*(x + a/b)^2) + 2*a*d^3*e/((b^2)^(3/2)*b*(x + a/b)^2) - 4*a^3*d*e^3/((b^2)
^(3/2)*b^3*(x + a/b)^2) + 5/2*a^4*e^4/((b^2)^(3/2)*b^4*(x + a/b)^2)

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Fricas [A]  time = 1.58009, size = 586, normalized size = 2.79 \begin{align*} \frac{b^{4} e^{4} x^{4} - b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 20 \, a^{3} b d e^{3} + 7 \, a^{4} e^{4} + 4 \,{\left (2 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} +{\left (16 \, a b^{3} d e^{3} - 11 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \,{\left (4 \, b^{4} d^{3} e - 12 \, a b^{3} d^{2} e^{2} + 8 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x + 12 \,{\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} +{\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \,{\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(b^4*e^4*x^4 - b^4*d^4 - 4*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 20*a^3*b*d*e^3 + 7*a^4*e^4 + 4*(2*b^4*d*e^3
- a*b^3*e^4)*x^3 + (16*a*b^3*d*e^3 - 11*a^2*b^2*e^4)*x^2 - 2*(4*b^4*d^3*e - 12*a*b^3*d^2*e^2 + 8*a^2*b^2*d*e^3
- a^3*b*e^4)*x + 12*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*
x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((d + e*x)**4/((a + b*x)**2)**(3/2), 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((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x