3.1849 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=304 \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) \sqrt{d+e x}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac{2 b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^5 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.142961, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {770, 77} \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) \sqrt{d+e x}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac{2 b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^5 (a+b x)} \]

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)^3*(B*d - A*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^(3/2)) + (2*(b*d - a*e
)^2*(4*b*B*d - 3*A*b*e - a*B*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d + e*x]) + (6*b*(b*d - a*e
)*(2*b*B*d - A*b*e - a*B*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)) - (2*b^2*(4*b*B*d - A
*b*e - 3*a*B*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^3*B*(d + e*x)^(5/2)*Sq
rt[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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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{\left (a b+b^2 x\right )^3 (A+B x)}{(d+e x)^{5/2}} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (-\frac{b^3 (b d-a e)^3 (-B d+A e)}{e^4 (d+e x)^{5/2}}+\frac{b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^{3/2}}-\frac{3 b^4 (b d-a e) (-2 b B d+A b e+a B e)}{e^4 \sqrt{d+e x}}+\frac{b^5 (-4 b B d+A b e+3 a B e) \sqrt{d+e x}}{e^4}+\frac{b^6 B (d+e x)^{3/2}}{e^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{2 (b d-a e)^3 (B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac{2 (b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt{d+e x}}+\frac{6 b (b d-a e) (2 b B d-A b e-a B e) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac{2 b^2 (4 b B d-A b e-3 a B 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^3 B (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 = 0.189254, size = 241, normalized size = 0.79 \[ \frac{2 \sqrt{(a+b x)^2} \left (15 a^2 b e^2 \left (B \left (8 d^2+12 d e x+3 e^2 x^2\right )-A e (2 d+3 e x)\right )-5 a^3 e^3 (A e+2 B d+3 B e x)+15 a b^2 e \left (A e \left (8 d^2+12 d e x+3 e^2 x^2\right )+B \left (-24 d^2 e x-16 d^3-6 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (5 A e \left (-24 d^2 e x-16 d^3-6 d e^2 x^2+e^3 x^3\right )+B \left (48 d^2 e^2 x^2+192 d^3 e x+128 d^4-8 d e^3 x^3+3 e^4 x^4\right )\right )\right )}{15 e^5 (a+b x) (d+e x)^{3/2}} \]

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^3*e^3*(2*B*d + A*e + 3*B*e*x) + 15*a^2*b*e^2*(-(A*e*(2*d + 3*e*x)) + B*(8*d^2 + 12*
d*e*x + 3*e^2*x^2)) + 15*a*b^2*e*(A*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + B*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 +
e^3*x^3)) + b^3*(5*A*e*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3) + B*(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 = 0.008, size = 317, normalized size = 1. \begin{align*} -{\frac{-6\,B{x}^{4}{b}^{3}{e}^{4}-10\,A{x}^{3}{b}^{3}{e}^{4}-30\,B{x}^{3}a{b}^{2}{e}^{4}+16\,B{x}^{3}{b}^{3}d{e}^{3}-90\,A{x}^{2}a{b}^{2}{e}^{4}+60\,A{x}^{2}{b}^{3}d{e}^{3}-90\,B{x}^{2}{a}^{2}b{e}^{4}+180\,B{x}^{2}a{b}^{2}d{e}^{3}-96\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}+90\,Ax{a}^{2}b{e}^{4}-360\,Axa{b}^{2}d{e}^{3}+240\,Ax{b}^{3}{d}^{2}{e}^{2}+30\,Bx{a}^{3}{e}^{4}-360\,Bx{a}^{2}bd{e}^{3}+720\,Bxa{b}^{2}{d}^{2}{e}^{2}-384\,Bx{b}^{3}{d}^{3}e+10\,A{a}^{3}{e}^{4}+60\,Ad{e}^{3}{a}^{2}b-240\,Aa{b}^{2}{d}^{2}{e}^{2}+160\,A{b}^{3}{d}^{3}e+20\,Bd{e}^{3}{a}^{3}-240\,B{a}^{2}b{d}^{2}{e}^{2}+480\,Ba{b}^{2}{d}^{3}e-256\,B{b}^{3}{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*b^3*e^4*x^4-5*A*b^3*e^4*x^3-15*B*a*b^2*e^4*x^3+8*B*b^3*d*e^3*x^3-45*A*a*b^2*e^4*x^2+
30*A*b^3*d*e^3*x^2-45*B*a^2*b*e^4*x^2+90*B*a*b^2*d*e^3*x^2-48*B*b^3*d^2*e^2*x^2+45*A*a^2*b*e^4*x-180*A*a*b^2*d
*e^3*x+120*A*b^3*d^2*e^2*x+15*B*a^3*e^4*x-180*B*a^2*b*d*e^3*x+360*B*a*b^2*d^2*e^2*x-192*B*b^3*d^3*e*x+5*A*a^3*
e^4+30*A*a^2*b*d*e^3-120*A*a*b^2*d^2*e^2+80*A*b^3*d^3*e+10*B*a^3*d*e^3-120*B*a^2*b*d^2*e^2+240*B*a*b^2*d^3*e-1
28*B*b^3*d^4)*((b*x+a)^2)^(3/2)/e^5/(b*x+a)^3

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Maxima [A]  time = 1.09398, size = 410, normalized size = 1.35 \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 = 1.35636, size = 605, normalized size = 1.99 \begin{align*} \frac{2 \,{\left (3 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} - 5 \, A a^{3} e^{4} - 80 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 120 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 10 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} -{\left (8 \, B b^{3} d e^{3} - 5 \,{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \,{\left (16 \, B b^{3} d^{2} e^{2} - 10 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 15 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \,{\left (64 \, B b^{3} d^{3} e - 40 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 60 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 5 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} 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*b^3*e^4*x^4 + 128*B*b^3*d^4 - 5*A*a^3*e^4 - 80*(3*B*a*b^2 + A*b^3)*d^3*e + 120*(B*a^2*b + A*a*b^2)*d
^2*e^2 - 10*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*d^2*e^2
- 10*(3*B*a*b^2 + A*b^3)*d*e^3 + 15*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 3*(64*B*b^3*d^3*e - 40*(3*B*a*b^2 + A*b^3)*
d^2*e^2 + 60*(B*a^2*b + A*a*b^2)*d*e^3 - 5*(B*a^3 + 3*A*a^2*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 [B]  time = 1.21495, size = 687, normalized size = 2.26 \begin{align*} \frac{2}{15} \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{3} e^{20} \mathrm{sgn}\left (b x + a\right ) - 20 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} d e^{20} \mathrm{sgn}\left (b x + a\right ) + 90 \, \sqrt{x e + d} B b^{3} d^{2} e^{20} \mathrm{sgn}\left (b x + a\right ) + 15 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{2} e^{21} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} e^{21} \mathrm{sgn}\left (b x + a\right ) - 135 \, \sqrt{x e + d} B a b^{2} d e^{21} \mathrm{sgn}\left (b x + a\right ) - 45 \, \sqrt{x e + d} A b^{3} d e^{21} \mathrm{sgn}\left (b x + a\right ) + 45 \, \sqrt{x e + d} B a^{2} b e^{22} \mathrm{sgn}\left (b x + a\right ) + 45 \, \sqrt{x e + d} A a 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 b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - B b^{3} d^{4} \mathrm{sgn}\left (b x + a\right ) - 27 \,{\left (x e + d\right )} B a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) - 9 \,{\left (x e + d\right )} A b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, B a b^{2} d^{3} e \mathrm{sgn}\left (b x + a\right ) + A b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 18 \,{\left (x e + d\right )} B a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) + 18 \,{\left (x e + d\right )} A a b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, B a^{2} b d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, A a b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \,{\left (x e + d\right )} B a^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) - 9 \,{\left (x e + d\right )} A a^{2} b e^{3} \mathrm{sgn}\left (b x + a\right ) + B a^{3} d e^{3} \mathrm{sgn}\left (b x + a\right ) + 3 \, A a^{2} b d e^{3} \mathrm{sgn}\left (b x + a\right ) - A a^{3} 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*b^3*e^20*sgn(b*x + a) - 20*(x*e + d)^(3/2)*B*b^3*d*e^20*sgn(b*x + a) + 90*sqrt(x*e +
d)*B*b^3*d^2*e^20*sgn(b*x + a) + 15*(x*e + d)^(3/2)*B*a*b^2*e^21*sgn(b*x + a) + 5*(x*e + d)^(3/2)*A*b^3*e^21*
sgn(b*x + a) - 135*sqrt(x*e + d)*B*a*b^2*d*e^21*sgn(b*x + a) - 45*sqrt(x*e + d)*A*b^3*d*e^21*sgn(b*x + a) + 45
*sqrt(x*e + d)*B*a^2*b*e^22*sgn(b*x + a) + 45*sqrt(x*e + d)*A*a*b^2*e^22*sgn(b*x + a))*e^(-25) + 2/3*(12*(x*e
+ d)*B*b^3*d^3*sgn(b*x + a) - B*b^3*d^4*sgn(b*x + a) - 27*(x*e + d)*B*a*b^2*d^2*e*sgn(b*x + a) - 9*(x*e + d)*A
*b^3*d^2*e*sgn(b*x + a) + 3*B*a*b^2*d^3*e*sgn(b*x + a) + A*b^3*d^3*e*sgn(b*x + a) + 18*(x*e + d)*B*a^2*b*d*e^2
*sgn(b*x + a) + 18*(x*e + d)*A*a*b^2*d*e^2*sgn(b*x + a) - 3*B*a^2*b*d^2*e^2*sgn(b*x + a) - 3*A*a*b^2*d^2*e^2*s
gn(b*x + a) - 3*(x*e + d)*B*a^3*e^3*sgn(b*x + a) - 9*(x*e + d)*A*a^2*b*e^3*sgn(b*x + a) + B*a^3*d*e^3*sgn(b*x
+ a) + 3*A*a^2*b*d*e^3*sgn(b*x + a) - A*a^3*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^(3/2)