3.2107 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{9/2}} \, dx\)

Optimal. Leaf size=258 \[ \frac{2 b^4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x}}{e^5 (a+b x)}+\frac{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) \sqrt{d+e x}}-\frac{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) (d+e x)^{3/2}}+\frac{8 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^5 (a+b x) (d+e x)^{5/2}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{7 e^5 (a+b x) (d+e x)^{7/2}} \]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)*(d + e*x)^(7/2)) + (8*b*(b*d - a*e)^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)*(d + e*x)^(5/2)) - (4*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/(e^5*(a + b*x)*(d + e*x)^(3/2)) + (8*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d +
e*x]) + (2*b^4*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x))

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Rubi [A]  time = 0.103988, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {770, 21, 43} \[ \frac{2 b^4 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x}}{e^5 (a+b x)}+\frac{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) \sqrt{d+e x}}-\frac{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) (d+e x)^{3/2}}+\frac{8 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^5 (a+b x) (d+e x)^{5/2}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{7 e^5 (a+b x) (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(9/2),x]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)*(d + e*x)^(7/2)) + (8*b*(b*d - a*e)^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)*(d + e*x)^(5/2)) - (4*b^2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/(e^5*(a + b*x)*(d + e*x)^(3/2)) + (8*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d +
e*x]) + (2*b^4*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x))

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{9/2}} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{(a+b x) \left (a b+b^2 x\right )^3}{(d+e x)^{9/2}} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{(a+b x)^4}{(d+e x)^{9/2}} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(-b d+a e)^4}{e^4 (d+e x)^{9/2}}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)^{7/2}}+\frac{6 b^2 (b d-a e)^2}{e^4 (d+e x)^{5/2}}-\frac{4 b^3 (b d-a e)}{e^4 (d+e x)^{3/2}}+\frac{b^4}{e^4 \sqrt{d+e x}}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 (b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^{7/2}}+\frac{8 b (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^{5/2}}-\frac{4 b^2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)^{3/2}}+\frac{8 b^3 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt{d+e x}}+\frac{2 b^4 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0932884, size = 173, normalized size = 0.67 \[ -\frac{2 \sqrt{(a+b x)^2} \left (2 a^2 b^2 e^2 \left (8 d^2+28 d e x+35 e^2 x^2\right )+4 a^3 b e^3 (2 d+7 e x)+5 a^4 e^4+4 a b^3 e \left (56 d^2 e x+16 d^3+70 d e^2 x^2+35 e^3 x^3\right )+b^4 \left (-\left (560 d^2 e^2 x^2+448 d^3 e x+128 d^4+280 d e^3 x^3+35 e^4 x^4\right )\right )\right )}{35 e^5 (a+b x) (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(9/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(5*a^4*e^4 + 4*a^3*b*e^3*(2*d + 7*e*x) + 2*a^2*b^2*e^2*(8*d^2 + 28*d*e*x + 35*e^2*x^2) +
4*a*b^3*e*(16*d^3 + 56*d^2*e*x + 70*d*e^2*x^2 + 35*e^3*x^3) - b^4*(128*d^4 + 448*d^3*e*x + 560*d^2*e^2*x^2 +
280*d*e^3*x^3 + 35*e^4*x^4)))/(35*e^5*(a + b*x)*(d + e*x)^(7/2))

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Maple [A]  time = 0.007, size = 202, normalized size = 0.8 \begin{align*} -{\frac{-70\,{x}^{4}{b}^{4}{e}^{4}+280\,{x}^{3}a{b}^{3}{e}^{4}-560\,{x}^{3}{b}^{4}d{e}^{3}+140\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}+560\,{x}^{2}a{b}^{3}d{e}^{3}-1120\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+56\,x{a}^{3}b{e}^{4}+112\,x{a}^{2}{b}^{2}d{e}^{3}+448\,xa{b}^{3}{d}^{2}{e}^{2}-896\,x{b}^{4}{d}^{3}e+10\,{a}^{4}{e}^{4}+16\,d{e}^{3}{a}^{3}b+32\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}+128\,a{b}^{3}{d}^{3}e-256\,{b}^{4}{d}^{4}}{35\, \left ( bx+a \right ) ^{3}{e}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}} \left ( ex+d \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(9/2),x)

[Out]

-2/35/(e*x+d)^(7/2)*(-35*b^4*e^4*x^4+140*a*b^3*e^4*x^3-280*b^4*d*e^3*x^3+70*a^2*b^2*e^4*x^2+280*a*b^3*d*e^3*x^
2-560*b^4*d^2*e^2*x^2+28*a^3*b*e^4*x+56*a^2*b^2*d*e^3*x+224*a*b^3*d^2*e^2*x-448*b^4*d^3*e*x+5*a^4*e^4+8*a^3*b*
d*e^3+16*a^2*b^2*d^2*e^2+64*a*b^3*d^3*e-128*b^4*d^4)*((b*x+a)^2)^(3/2)/e^5/(b*x+a)^3

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Maxima [A]  time = 1.28321, size = 470, normalized size = 1.82 \begin{align*} -\frac{2 \,{\left (35 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} + 8 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} + 5 \, a^{3} e^{3} + 35 \,{\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 7 \,{\left (8 \, b^{3} d^{2} e + 4 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} a}{35 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )} \sqrt{e x + d}} + \frac{2 \,{\left (35 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} - 48 \, a b^{2} d^{3} e - 8 \, a^{2} b d^{2} e^{2} - 2 \, a^{3} d e^{3} + 35 \,{\left (8 \, b^{3} d e^{3} - 3 \, a b^{2} e^{4}\right )} x^{3} + 35 \,{\left (16 \, b^{3} d^{2} e^{2} - 6 \, a b^{2} d e^{3} - a^{2} b e^{4}\right )} x^{2} + 7 \,{\left (64 \, b^{3} d^{3} e - 24 \, a b^{2} d^{2} e^{2} - 4 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} x\right )} b}{35 \,{\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )} \sqrt{e x + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(9/2),x, algorithm="maxima")

[Out]

-2/35*(35*b^3*e^3*x^3 + 16*b^3*d^3 + 8*a*b^2*d^2*e + 6*a^2*b*d*e^2 + 5*a^3*e^3 + 35*(2*b^3*d*e^2 + a*b^2*e^3)*
x^2 + 7*(8*b^3*d^2*e + 4*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*a/((e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)*sqrt
(e*x + d)) + 2/35*(35*b^3*e^4*x^4 + 128*b^3*d^4 - 48*a*b^2*d^3*e - 8*a^2*b*d^2*e^2 - 2*a^3*d*e^3 + 35*(8*b^3*d
*e^3 - 3*a*b^2*e^4)*x^3 + 35*(16*b^3*d^2*e^2 - 6*a*b^2*d*e^3 - a^2*b*e^4)*x^2 + 7*(64*b^3*d^3*e - 24*a*b^2*d^2
*e^2 - 4*a^2*b*d*e^3 - a^3*e^4)*x)*b/((e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)*sqrt(e*x + d))

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Fricas [A]  time = 1.06598, size = 462, normalized size = 1.79 \begin{align*} \frac{2 \,{\left (35 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} - 64 \, a b^{3} d^{3} e - 16 \, a^{2} b^{2} d^{2} e^{2} - 8 \, a^{3} b d e^{3} - 5 \, a^{4} e^{4} + 140 \,{\left (2 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 70 \,{\left (8 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 28 \,{\left (16 \, b^{4} d^{3} e - 8 \, a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{35 \,{\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(9/2),x, algorithm="fricas")

[Out]

2/35*(35*b^4*e^4*x^4 + 128*b^4*d^4 - 64*a*b^3*d^3*e - 16*a^2*b^2*d^2*e^2 - 8*a^3*b*d*e^3 - 5*a^4*e^4 + 140*(2*
b^4*d*e^3 - a*b^3*e^4)*x^3 + 70*(8*b^4*d^2*e^2 - 4*a*b^3*d*e^3 - a^2*b^2*e^4)*x^2 + 28*(16*b^4*d^3*e - 8*a*b^3
*d^2*e^2 - 2*a^2*b^2*d*e^3 - a^3*b*e^4)*x)*sqrt(e*x + d)/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x
+ d^4*e^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.19406, size = 419, normalized size = 1.62 \begin{align*} 2 \, \sqrt{x e + d} b^{4} e^{\left (-5\right )} \mathrm{sgn}\left (b x + a\right ) + \frac{2 \,{\left (140 \,{\left (x e + d\right )}^{3} b^{4} d \mathrm{sgn}\left (b x + a\right ) - 70 \,{\left (x e + d\right )}^{2} b^{4} d^{2} \mathrm{sgn}\left (b x + a\right ) + 28 \,{\left (x e + d\right )} b^{4} d^{3} \mathrm{sgn}\left (b x + a\right ) - 5 \, b^{4} d^{4} \mathrm{sgn}\left (b x + a\right ) - 140 \,{\left (x e + d\right )}^{3} a b^{3} e \mathrm{sgn}\left (b x + a\right ) + 140 \,{\left (x e + d\right )}^{2} a b^{3} d e \mathrm{sgn}\left (b x + a\right ) - 84 \,{\left (x e + d\right )} a b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 20 \, a b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) - 70 \,{\left (x e + d\right )}^{2} a^{2} b^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 84 \,{\left (x e + d\right )} a^{2} b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) - 30 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 28 \,{\left (x e + d\right )} a^{3} b e^{3} \mathrm{sgn}\left (b x + a\right ) + 20 \, a^{3} b d e^{3} \mathrm{sgn}\left (b x + a\right ) - 5 \, a^{4} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{35 \,{\left (x e + d\right )}^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(9/2),x, algorithm="giac")

[Out]

2*sqrt(x*e + d)*b^4*e^(-5)*sgn(b*x + a) + 2/35*(140*(x*e + d)^3*b^4*d*sgn(b*x + a) - 70*(x*e + d)^2*b^4*d^2*sg
n(b*x + a) + 28*(x*e + d)*b^4*d^3*sgn(b*x + a) - 5*b^4*d^4*sgn(b*x + a) - 140*(x*e + d)^3*a*b^3*e*sgn(b*x + a)
+ 140*(x*e + d)^2*a*b^3*d*e*sgn(b*x + a) - 84*(x*e + d)*a*b^3*d^2*e*sgn(b*x + a) + 20*a*b^3*d^3*e*sgn(b*x + a
) - 70*(x*e + d)^2*a^2*b^2*e^2*sgn(b*x + a) + 84*(x*e + d)*a^2*b^2*d*e^2*sgn(b*x + a) - 30*a^2*b^2*d^2*e^2*sgn
(b*x + a) - 28*(x*e + d)*a^3*b*e^3*sgn(b*x + a) + 20*a^3*b*d*e^3*sgn(b*x + a) - 5*a^4*e^4*sgn(b*x + a))*e^(-5)
/(x*e + d)^(7/2)