3.1606 \(\int \frac{(d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=209 \[ -\frac{4 e^3 (b d-a e)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e^2 (b d-a e)^2}{b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 e (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^4}{4 b^5 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(-4*e^3*(b*d - a*e))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^4/(4*b^5*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (4*e*(b*d - a*e)^3)/(3*b^5*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*e^2*(b*d - a*e)^2)/(
b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(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 = 0.126887, antiderivative size = 209, 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} \[ -\frac{4 e^3 (b d-a e)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e^2 (b d-a e)^2}{b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 e (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^4}{4 b^5 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*e^3*(b*d - a*e))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^4/(4*b^5*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (4*e*(b*d - a*e)^3)/(3*b^5*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*e^2*(b*d - a*e)^2)/(
b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^4}{\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 d-a e)^4}{b^9 (a+b x)^5}+\frac{4 e (b d-a e)^3}{b^9 (a+b x)^4}+\frac{6 e^2 (b d-a e)^2}{b^9 (a+b x)^3}+\frac{4 e^3 (b d-a e)}{b^9 (a+b x)^2}+\frac{e^4}{b^9 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 e^3 (b d-a e)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^4}{4 b^5 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 e (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e^2 (b d-a e)^2}{b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^4 (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 = 0.0764801, size = 138, normalized size = 0.66 \[ \frac{12 e^4 (a+b x)^4 \log (a+b x)-(b d-a e) \left (a^2 b e^2 (13 d+88 e x)+25 a^3 e^3+a b^2 e \left (7 d^2+40 d e x+108 e^2 x^2\right )+b^3 \left (16 d^2 e x+3 d^3+36 d e^2 x^2+48 e^3 x^3\right )\right )}{12 b^5 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*d - a*e)*(25*a^3*e^3 + a^2*b*e^2*(13*d + 88*e*x) + a*b^2*e*(7*d^2 + 40*d*e*x + 108*e^2*x^2) + b^3*(3*d^3
+ 16*d^2*e*x + 36*d*e^2*x^2 + 48*e^3*x^3))) + 12*e^4*(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 = 0.198, size = 267, normalized size = 1.3 \begin{align*}{\frac{ \left ( 12\,\ln \left ( bx+a \right ){x}^{4}{b}^{4}{e}^{4}+48\,\ln \left ( bx+a \right ){x}^{3}a{b}^{3}{e}^{4}+72\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}+48\,{x}^{3}a{b}^{3}{e}^{4}-48\,{x}^{3}{b}^{4}d{e}^{3}+48\,\ln \left ( bx+a \right ) x{a}^{3}b{e}^{4}+108\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}-72\,{x}^{2}a{b}^{3}d{e}^{3}-36\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+12\,\ln \left ( bx+a \right ){a}^{4}{e}^{4}+88\,x{a}^{3}b{e}^{4}-48\,x{a}^{2}{b}^{2}d{e}^{3}-24\,xa{b}^{3}{d}^{2}{e}^{2}-16\,x{b}^{4}{d}^{3}e+25\,{a}^{4}{e}^{4}-12\,{a}^{3}bd{e}^{3}-6\,{d}^{2}{e}^{2}{a}^{2}{b}^{2}-4\,a{b}^{3}{d}^{3}e-3\,{b}^{4}{d}^{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((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/12*(12*ln(b*x+a)*x^4*b^4*e^4+48*ln(b*x+a)*x^3*a*b^3*e^4+72*ln(b*x+a)*x^2*a^2*b^2*e^4+48*x^3*a*b^3*e^4-48*x^3
*b^4*d*e^3+48*ln(b*x+a)*x*a^3*b*e^4+108*x^2*a^2*b^2*e^4-72*x^2*a*b^3*d*e^3-36*x^2*b^4*d^2*e^2+12*ln(b*x+a)*a^4
*e^4+88*x*a^3*b*e^4-48*x*a^2*b^2*d*e^3-24*x*a*b^3*d^2*e^2-16*x*b^4*d^3*e+25*a^4*e^4-12*a^3*b*d*e^3-6*d^2*e^2*a
^2*b^2-4*a*b^3*d^3*e-3*b^4*d^4)*(b*x+a)/b^5/((b*x+a)^2)^(5/2)

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Maxima [B]  time = 1.14381, size = 502, normalized size = 2.4 \begin{align*} \frac{1}{12} \, e^{4}{\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}{3} \, d e^{3}{\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 )} - \frac{1}{3} \, d^{3} e{\left (\frac{4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{3 \, a}{{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{4}}\right )} - \frac{1}{2} \, d^{2} e^{2}{\left (\frac{3 \, a^{2} b^{2}}{{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} - \frac{8 \, a b}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} + \frac{6}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}}\right )} - \frac{d^{4}}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{4}} \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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51676, size = 545, normalized size = 2.61 \begin{align*} -\frac{3 \, b^{4} d^{4} + 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 25 \, a^{4} e^{4} + 48 \,{\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 36 \,{\left (b^{4} d^{2} e^{2} + 2 \, a b^{3} d e^{3} - 3 \, a^{2} b^{2} e^{4}\right )} x^{2} + 8 \,{\left (2 \, b^{4} d^{3} e + 3 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} - 11 \, a^{3} b e^{4}\right )} x - 12 \,{\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{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((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*b^4*d^4 + 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 12*a^3*b*d*e^3 - 25*a^4*e^4 + 48*(b^4*d*e^3 - a*b^3*e^4
)*x^3 + 36*(b^4*d^2*e^2 + 2*a*b^3*d*e^3 - 3*a^2*b^2*e^4)*x^2 + 8*(2*b^4*d^3*e + 3*a*b^3*d^2*e^2 + 6*a^2*b^2*d*
e^3 - 11*a^3*b*e^4)*x - 12*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^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{\left (d + e x\right )^{4}}{\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((e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((d + e*x)**4/((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((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x