3.2033 \(\int \frac{(a+b x) (d+e x)^5}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=252 \[ \frac{5 e^4 x (a+b x) (4 b d-3 a e)}{3 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{20 e^2 (b d-a e)^3}{3 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{10 e^3 (a+b x) (b d-a e)^2 \log (a+b x)}{b^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (b d-a e)^4}{6 b^6 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{5 e^5 x^2 (a+b x)}{6 b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-(d + e*x)^5/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (20*e^2*(b*d - a*e)^3)/(3*b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^
2]) - (5*e*(b*d - a*e)^4)/(6*b^6*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^4*(4*b*d - 3*a*e)*x*(a + b*x)
)/(3*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^5*x^2*(a + b*x))/(6*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (10*e^
3*(b*d - a*e)^2*(a + b*x)*Log[a + b*x])/(b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.192574, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {768, 646, 43} \[ \frac{5 e^4 x (a+b x) (4 b d-3 a e)}{3 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{20 e^2 (b d-a e)^3}{3 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{10 e^3 (a+b x) (b d-a e)^2 \log (a+b x)}{b^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (b d-a e)^4}{6 b^6 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{5 e^5 x^2 (a+b x)}{6 b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d + e*x)^5/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (20*e^2*(b*d - a*e)^3)/(3*b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^
2]) - (5*e*(b*d - a*e)^4)/(6*b^6*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^4*(4*b*d - 3*a*e)*x*(a + b*x)
)/(3*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*e^5*x^2*(a + b*x))/(6*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (10*e^
3*(b*d - a*e)^2*(a + b*x)*Log[a + b*x])/(b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

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{(a+b x) (d+e x)^5}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac{(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{(5 e) \int \frac{(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac{(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{\left (5 b e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^4}{\left (a b+b^2 x\right )^3} \, dx}{3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{\left (5 b e \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}{3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^5}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{20 e^2 (b d-a e)^3}{3 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 e (b d-a e)^4}{6 b^6 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e^4 (4 b d-3 a e) x (a+b x)}{3 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 e^5 x^2 (a+b x)}{6 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{10 e^3 (b d-a e)^2 (a+b x) \log (a+b x)}{b^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.127119, size = 232, normalized size = 0.92 \[ \frac{-a^2 b^3 e^2 \left (-270 d^2 e x+20 d^3+90 d e^2 x^2+63 e^3 x^3\right )+a^3 b^2 e^3 \left (110 d^2-270 d e x-9 e^2 x^2\right )+a^4 b e^4 (81 e x-130 d)+47 a^5 e^5-5 a b^4 e \left (-36 d^2 e^2 x^2+12 d^3 e x+d^4-18 d e^3 x^3+3 e^4 x^4\right )+60 e^3 (a+b x)^3 (b d-a e)^2 \log (a+b x)+b^5 \left (-60 d^3 e^2 x^2-15 d^4 e x-2 d^5+30 d e^4 x^4+3 e^5 x^5\right )}{6 b^6 \left ((a+b x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(47*a^5*e^5 + a^4*b*e^4*(-130*d + 81*e*x) + a^3*b^2*e^3*(110*d^2 - 270*d*e*x - 9*e^2*x^2) - a^2*b^3*e^2*(20*d^
3 - 270*d^2*e*x + 90*d*e^2*x^2 + 63*e^3*x^3) - 5*a*b^4*e*(d^4 + 12*d^3*e*x - 36*d^2*e^2*x^2 - 18*d*e^3*x^3 + 3
*e^4*x^4) + b^5*(-2*d^5 - 15*d^4*e*x - 60*d^3*e^2*x^2 + 30*d*e^4*x^4 + 3*e^5*x^5) + 60*e^3*(b*d - a*e)^2*(a +
b*x)^3*Log[a + b*x])/(6*b^6*((a + b*x)^2)^(3/2))

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Maple [B]  time = 0.016, size = 495, normalized size = 2. \begin{align*}{\frac{ \left ( 3\,{x}^{5}{b}^{5}{e}^{5}+180\,\ln \left ( bx+a \right ){x}^{2}a{b}^{4}{d}^{2}{e}^{3}+47\,{a}^{5}{e}^{5}-2\,{b}^{5}{d}^{5}+60\,\ln \left ( bx+a \right ){a}^{5}{e}^{5}-360\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{3}d{e}^{4}-120\,\ln \left ( bx+a \right ){x}^{3}a{b}^{4}d{e}^{4}-360\,\ln \left ( bx+a \right ) x{a}^{3}{b}^{2}d{e}^{4}+180\,\ln \left ( bx+a \right ) x{a}^{2}{b}^{3}{d}^{2}{e}^{3}-15\,{x}^{4}a{b}^{4}{e}^{5}+30\,{x}^{4}{b}^{5}d{e}^{4}-9\,{x}^{2}{a}^{3}{b}^{2}{e}^{5}-60\,{x}^{2}{b}^{5}{d}^{3}{e}^{2}+81\,x{a}^{4}b{e}^{5}-15\,x{b}^{5}{d}^{4}e-63\,{x}^{3}{a}^{2}{b}^{3}{e}^{5}-5\,a{d}^{4}{b}^{4}e-20\,{a}^{2}{b}^{3}{d}^{3}{e}^{2}+110\,{a}^{3}{d}^{2}{b}^{2}{e}^{3}-130\,{a}^{4}bd{e}^{4}+60\,\ln \left ( bx+a \right ){a}^{3}{b}^{2}{d}^{2}{e}^{3}+60\,\ln \left ( bx+a \right ){x}^{3}{b}^{5}{d}^{2}{e}^{3}+180\,\ln \left ( bx+a \right ){x}^{2}{a}^{3}{b}^{2}{e}^{5}+180\,\ln \left ( bx+a \right ) x{a}^{4}b{e}^{5}+60\,\ln \left ( bx+a \right ){x}^{3}{a}^{2}{b}^{3}{e}^{5}+180\,{x}^{2}a{b}^{4}{d}^{2}{e}^{3}+270\,x{a}^{2}{b}^{3}{d}^{2}{e}^{3}-60\,xa{b}^{4}{d}^{3}{e}^{2}-270\,x{a}^{3}{b}^{2}d{e}^{4}+90\,{x}^{3}a{b}^{4}d{e}^{4}-90\,{x}^{2}{a}^{2}{b}^{3}d{e}^{4}-120\,\ln \left ( bx+a \right ){a}^{4}bd{e}^{4} \right ) \left ( bx+a \right ) ^{2}}{6\,{b}^{6}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/6*(3*x^5*b^5*e^5+180*ln(b*x+a)*x^2*a*b^4*d^2*e^3+47*a^5*e^5-2*b^5*d^5+60*ln(b*x+a)*a^5*e^5-360*ln(b*x+a)*x^2
*a^2*b^3*d*e^4-120*ln(b*x+a)*x^3*a*b^4*d*e^4-360*ln(b*x+a)*x*a^3*b^2*d*e^4+180*ln(b*x+a)*x*a^2*b^3*d^2*e^3-15*
x^4*a*b^4*e^5+30*x^4*b^5*d*e^4-9*x^2*a^3*b^2*e^5-60*x^2*b^5*d^3*e^2+81*x*a^4*b*e^5-15*x*b^5*d^4*e-63*x^3*a^2*b
^3*e^5-5*a*d^4*b^4*e-20*a^2*b^3*d^3*e^2+110*a^3*d^2*b^2*e^3-130*a^4*b*d*e^4+60*ln(b*x+a)*a^3*b^2*d^2*e^3+60*ln
(b*x+a)*x^3*b^5*d^2*e^3+180*ln(b*x+a)*x^2*a^3*b^2*e^5+180*ln(b*x+a)*x*a^4*b*e^5+60*ln(b*x+a)*x^3*a^2*b^3*e^5+1
80*x^2*a*b^4*d^2*e^3+270*x*a^2*b^3*d^2*e^3-60*x*a*b^4*d^3*e^2-270*x*a^3*b^2*d*e^4+90*x^3*a*b^4*d*e^4-90*x^2*a^
2*b^3*d*e^4-120*ln(b*x+a)*a^4*b*d*e^4)*(b*x+a)^2/b^6/((b*x+a)^2)^(5/2)

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Maxima [B]  time = 1.37763, size = 1493, normalized size = 5.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/4*b*e^5*((2*b^6*x^6 - 12*a*b^5*x^5 - 68*a^2*b^4*x^4 - 32*a^3*b^3*x^3 + 132*a^4*b^2*x^2 + 168*a^5*b*x + 57*a^
6)/(b^11*x^4 + 4*a*b^10*x^3 + 6*a^2*b^9*x^2 + 4*a^3*b^8*x + a^4*b^7) + 60*a^2*log(b*x + a)/b^7) + 5/12*b*d*e^4
*((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*e^5*((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) + 5/6*b*d^2*e^3*((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) + 5/12*a*d*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) - 5/6*b*d^3*e^2*(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)) - 5/6*a*d^2*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)) - 1/12*b*d^5*(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)) - 5/12*a*d^4*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)) - 5/12*b*d^4*e*(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)) - 5/6*a*d^3*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*a*d^5/((b^2)^(5/2)*(x + a/b)^4)

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Fricas [B]  time = 1.71039, size = 867, normalized size = 3.44 \begin{align*} \frac{3 \, b^{5} e^{5} x^{5} - 2 \, b^{5} d^{5} - 5 \, a b^{4} d^{4} e - 20 \, a^{2} b^{3} d^{3} e^{2} + 110 \, a^{3} b^{2} d^{2} e^{3} - 130 \, a^{4} b d e^{4} + 47 \, a^{5} e^{5} + 15 \,{\left (2 \, b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 9 \,{\left (10 \, a b^{4} d e^{4} - 7 \, a^{2} b^{3} e^{5}\right )} x^{3} - 3 \,{\left (20 \, b^{5} d^{3} e^{2} - 60 \, a b^{4} d^{2} e^{3} + 30 \, a^{2} b^{3} d e^{4} + 3 \, a^{3} b^{2} e^{5}\right )} x^{2} - 3 \,{\left (5 \, b^{5} d^{4} e + 20 \, a b^{4} d^{3} e^{2} - 90 \, a^{2} b^{3} d^{2} e^{3} + 90 \, a^{3} b^{2} d e^{4} - 27 \, a^{4} b e^{5}\right )} x + 60 \,{\left (a^{3} b^{2} d^{2} e^{3} - 2 \, a^{4} b d e^{4} + a^{5} e^{5} +{\left (b^{5} d^{2} e^{3} - 2 \, a b^{4} d e^{4} + a^{2} b^{3} e^{5}\right )} x^{3} + 3 \,{\left (a b^{4} d^{2} e^{3} - 2 \, a^{2} b^{3} d e^{4} + a^{3} b^{2} e^{5}\right )} x^{2} + 3 \,{\left (a^{2} b^{3} d^{2} e^{3} - 2 \, a^{3} b^{2} d e^{4} + a^{4} b e^{5}\right )} x\right )} \log \left (b x + a\right )}{6 \,{\left (b^{9} x^{3} + 3 \, a b^{8} x^{2} + 3 \, a^{2} b^{7} x + a^{3} b^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^5/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/6*(3*b^5*e^5*x^5 - 2*b^5*d^5 - 5*a*b^4*d^4*e - 20*a^2*b^3*d^3*e^2 + 110*a^3*b^2*d^2*e^3 - 130*a^4*b*d*e^4 +
47*a^5*e^5 + 15*(2*b^5*d*e^4 - a*b^4*e^5)*x^4 + 9*(10*a*b^4*d*e^4 - 7*a^2*b^3*e^5)*x^3 - 3*(20*b^5*d^3*e^2 - 6
0*a*b^4*d^2*e^3 + 30*a^2*b^3*d*e^4 + 3*a^3*b^2*e^5)*x^2 - 3*(5*b^5*d^4*e + 20*a*b^4*d^3*e^2 - 90*a^2*b^3*d^2*e
^3 + 90*a^3*b^2*d*e^4 - 27*a^4*b*e^5)*x + 60*(a^3*b^2*d^2*e^3 - 2*a^4*b*d*e^4 + a^5*e^5 + (b^5*d^2*e^3 - 2*a*b
^4*d*e^4 + a^2*b^3*e^5)*x^3 + 3*(a*b^4*d^2*e^3 - 2*a^2*b^3*d*e^4 + a^3*b^2*e^5)*x^2 + 3*(a^2*b^3*d^2*e^3 - 2*a
^3*b^2*d*e^4 + a^4*b*e^5)*x)*log(b*x + a))/(b^9*x^3 + 3*a*b^8*x^2 + 3*a^2*b^7*x + a^3*b^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right ) \left (d + e x\right )^{5}}{\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((b*x+a)*(e*x+d)**5/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((a + b*x)*(d + e*x)**5/((a + b*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}{\left (e x + d\right )}^{5}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x + a)*(e*x + d)^5/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)