3.2105 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=260 \[ \frac{2 b^4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^5 (a+b x)}-\frac{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^5 (a+b x)}+\frac{12 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}{e^5 (a+b x)}+\frac{8 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^5 (a+b x) \sqrt{d+e x}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{3 e^5 (a+b x) (d+e x)^{3/2}} \]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^(3/2)) + (8*b*(b*d - a*e)^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d + e*x]) + (12*b^2*(b*d - a*e)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/(e^5*(a + b*x)) - (8*b^3*(b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a +
b*x)) + (2*b^4*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x))

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Rubi [A]  time = 0.103578, antiderivative size = 260, 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} \[ \frac{2 b^4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^5 (a+b x)}-\frac{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^5 (a+b x)}+\frac{12 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}{e^5 (a+b x)}+\frac{8 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^5 (a+b x) \sqrt{d+e x}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{3 e^5 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^(3/2)) + (8*b*(b*d - a*e)^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d + e*x]) + (12*b^2*(b*d - a*e)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*
x + b^2*x^2])/(e^5*(a + b*x)) - (8*b^3*(b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a +
b*x)) + (2*b^4*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*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)^{5/2}} \, 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)^{5/2}} \, 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)^{5/2}} \, 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)^{5/2}}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)^{3/2}}+\frac{6 b^2 (b d-a e)^2}{e^4 \sqrt{d+e x}}-\frac{4 b^3 (b d-a e) \sqrt{d+e x}}{e^4}+\frac{b^4 (d+e x)^{3/2}}{e^4}\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}}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac{8 b (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt{d+e x}}+\frac{12 b^2 (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac{8 b^3 (b d-a e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}+\frac{2 b^4 (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0999293, size = 171, normalized size = 0.66 \[ \frac{2 \sqrt{(a+b x)^2} \left (30 a^2 b^2 e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )-20 a^3 b e^3 (2 d+3 e x)-5 a^4 e^4+20 a b^3 e \left (-24 d^2 e x-16 d^3-6 d e^2 x^2+e^3 x^3\right )+b^4 \left (48 d^2 e^2 x^2+192 d^3 e x+128 d^4-8 d e^3 x^3+3 e^4 x^4\right )\right )}{15 e^5 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(2*Sqrt[(a + b*x)^2]*(-5*a^4*e^4 - 20*a^3*b*e^3*(2*d + 3*e*x) + 30*a^2*b^2*e^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2)
+ 20*a*b^3*e*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3) + b^4*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*
d*e^3*x^3 + 3*e^4*x^4)))/(15*e^5*(a + b*x)*(d + e*x)^(3/2))

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Maple [A]  time = 0.007, size = 202, normalized size = 0.8 \begin{align*} -{\frac{-6\,{x}^{4}{b}^{4}{e}^{4}-40\,{x}^{3}a{b}^{3}{e}^{4}+16\,{x}^{3}{b}^{4}d{e}^{3}-180\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}+240\,{x}^{2}a{b}^{3}d{e}^{3}-96\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+120\,x{a}^{3}b{e}^{4}-720\,x{a}^{2}{b}^{2}d{e}^{3}+960\,xa{b}^{3}{d}^{2}{e}^{2}-384\,x{b}^{4}{d}^{3}e+10\,{a}^{4}{e}^{4}+80\,d{e}^{3}{a}^{3}b-480\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}+640\,a{b}^{3}{d}^{3}e-256\,{b}^{4}{d}^{4}}{15\, \left ( bx+a \right ) ^{3}{e}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}} \left ( ex+d \right ) ^{-{\frac{3}{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)^(5/2),x)

[Out]

-2/15/(e*x+d)^(3/2)*(-3*b^4*e^4*x^4-20*a*b^3*e^4*x^3+8*b^4*d*e^3*x^3-90*a^2*b^2*e^4*x^2+120*a*b^3*d*e^3*x^2-48
*b^4*d^2*e^2*x^2+60*a^3*b*e^4*x-360*a^2*b^2*d*e^3*x+480*a*b^3*d^2*e^2*x-192*b^4*d^3*e*x+5*a^4*e^4+40*a^3*b*d*e
^3-240*a^2*b^2*d^2*e^2+320*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 = 1.18028, size = 410, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \,{\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \,{\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} a}{3 \,{\left (e^{5} x + d e^{4}\right )} \sqrt{e x + d}} + \frac{2 \,{\left (3 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} - 240 \, a b^{2} d^{3} e + 120 \, a^{2} b d^{2} e^{2} - 10 \, a^{3} d e^{3} -{\left (8 \, b^{3} d e^{3} - 15 \, a b^{2} e^{4}\right )} x^{3} + 3 \,{\left (16 \, b^{3} d^{2} e^{2} - 30 \, a b^{2} d e^{3} + 15 \, a^{2} b e^{4}\right )} x^{2} + 3 \,{\left (64 \, b^{3} d^{3} e - 120 \, a b^{2} d^{2} e^{2} + 60 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x\right )} b}{15 \,{\left (e^{6} x + d 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)^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
 3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*a/((e^5*x + d*e^4)*sqrt(e*x + d)) + 2/15*(3*b^3*e^4*x^4 + 1
28*b^3*d^4 - 240*a*b^2*d^3*e + 120*a^2*b*d^2*e^2 - 10*a^3*d*e^3 - (8*b^3*d*e^3 - 15*a*b^2*e^4)*x^3 + 3*(16*b^3
*d^2*e^2 - 30*a*b^2*d*e^3 + 15*a^2*b*e^4)*x^2 + 3*(64*b^3*d^3*e - 120*a*b^2*d^2*e^2 + 60*a^2*b*d*e^3 - 5*a^3*e
^4)*x)*b/((e^6*x + d*e^5)*sqrt(e*x + d))

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Fricas [A]  time = 0.971286, size = 431, normalized size = 1.66 \begin{align*} \frac{2 \,{\left (3 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} - 320 \, a b^{3} d^{3} e + 240 \, a^{2} b^{2} d^{2} e^{2} - 40 \, a^{3} b d e^{3} - 5 \, a^{4} e^{4} - 4 \,{\left (2 \, b^{4} d e^{3} - 5 \, a b^{3} e^{4}\right )} x^{3} + 6 \,{\left (8 \, b^{4} d^{2} e^{2} - 20 \, a b^{3} d e^{3} + 15 \, a^{2} b^{2} e^{4}\right )} x^{2} + 12 \,{\left (16 \, b^{4} d^{3} e - 40 \, a b^{3} d^{2} e^{2} + 30 \, a^{2} b^{2} d e^{3} - 5 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*b^4*e^4*x^4 + 128*b^4*d^4 - 320*a*b^3*d^3*e + 240*a^2*b^2*d^2*e^2 - 40*a^3*b*d*e^3 - 5*a^4*e^4 - 4*(2*
b^4*d*e^3 - 5*a*b^3*e^4)*x^3 + 6*(8*b^4*d^2*e^2 - 20*a*b^3*d*e^3 + 15*a^2*b^2*e^4)*x^2 + 12*(16*b^4*d^3*e - 40
*a*b^3*d^2*e^2 + 30*a^2*b^2*d*e^3 - 5*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*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)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.18537, size = 431, normalized size = 1.66 \begin{align*} \frac{2}{15} \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} e^{20} \mathrm{sgn}\left (b x + a\right ) - 20 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d e^{20} \mathrm{sgn}\left (b x + a\right ) + 90 \, \sqrt{x e + d} b^{4} d^{2} e^{20} \mathrm{sgn}\left (b x + a\right ) + 20 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} e^{21} \mathrm{sgn}\left (b x + a\right ) - 180 \, \sqrt{x e + d} a b^{3} d e^{21} \mathrm{sgn}\left (b x + a\right ) + 90 \, \sqrt{x e + d} a^{2} b^{2} e^{22} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-25\right )} + \frac{2 \,{\left (12 \,{\left (x e + d\right )} b^{4} d^{3} \mathrm{sgn}\left (b x + a\right ) - b^{4} d^{4} \mathrm{sgn}\left (b x + a\right ) - 36 \,{\left (x e + d\right )} a b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 4 \, a b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 36 \,{\left (x e + d\right )} a^{2} b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) - 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 12 \,{\left (x e + d\right )} a^{3} b e^{3} \mathrm{sgn}\left (b x + a\right ) + 4 \, a^{3} b d e^{3} \mathrm{sgn}\left (b x + a\right ) - a^{4} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{3 \,{\left (x e + d\right )}^{\frac{3}{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)^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(x*e + d)^(5/2)*b^4*e^20*sgn(b*x + a) - 20*(x*e + d)^(3/2)*b^4*d*e^20*sgn(b*x + a) + 90*sqrt(x*e + d)*
b^4*d^2*e^20*sgn(b*x + a) + 20*(x*e + d)^(3/2)*a*b^3*e^21*sgn(b*x + a) - 180*sqrt(x*e + d)*a*b^3*d*e^21*sgn(b*
x + a) + 90*sqrt(x*e + d)*a^2*b^2*e^22*sgn(b*x + a))*e^(-25) + 2/3*(12*(x*e + d)*b^4*d^3*sgn(b*x + a) - b^4*d^
4*sgn(b*x + a) - 36*(x*e + d)*a*b^3*d^2*e*sgn(b*x + a) + 4*a*b^3*d^3*e*sgn(b*x + a) + 36*(x*e + d)*a^2*b^2*d*e
^2*sgn(b*x + a) - 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 12*(x*e + d)*a^3*b*e^3*sgn(b*x + a) + 4*a^3*b*d*e^3*sgn(b*x
 + a) - a^4*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^(3/2)