3.1848 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=302 \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-3 a B e-A b e+4 b B d)}{5 e^5 (a+b x)}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{e^5 (a+b x) \sqrt{d+e x}}+\frac{2 b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^5 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.144891, antiderivative size = 302, 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} \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-3 a B e-A b e+4 b B d)}{5 e^5 (a+b x)}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{e^5 (a+b x) \sqrt{d+e x}}+\frac{2 b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^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*d - a*e)^2*
(4*b*B*d - 3*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*(b*d - a*e)*(2
*b*B*d - A*b*e - a*B*e)*(d + e*x)^(3/2)*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)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)) + (2*b^3*B*(d + e*x)^(7/2)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2])/(7*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)^{3/2}} \, 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)^{3/2}} \, 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)^{3/2}}+\frac{b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 \sqrt{d+e x}}-\frac{3 b^4 (b d-a e) (-2 b B d+A b e+a B e) \sqrt{d+e x}}{e^4}+\frac{b^5 (-4 b B d+A b e+3 a B e) (d+e x)^{3/2}}{e^4}+\frac{b^6 B (d+e x)^{5/2}}{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}}{e^5 (a+b x) \sqrt{d+e x}}-\frac{2 (b d-a e)^2 (4 b B d-3 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 (b d-a e) (2 b B d-A b e-a B e) (d+e x)^{3/2} \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)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)}+\frac{2 b^3 B (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.147694, size = 240, normalized size = 0.79 \[ \frac{2 \sqrt{(a+b x)^2} \left (35 a^2 b e^2 \left (3 A e (2 d+e x)+B \left (-8 d^2-4 d e x+e^2 x^2\right )\right )+35 a^3 e^3 (-A e+2 B d+B e x)+7 a b^2 e \left (5 A e \left (-8 d^2-4 d e x+e^2 x^2\right )+3 B \left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (7 A e \left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right )+B \left (16 d^2 e^2 x^2-64 d^3 e x-128 d^4-8 d e^3 x^3+5 e^4 x^4\right )\right )\right )}{35 e^5 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(35*a^3*e^3*(2*B*d - A*e + B*e*x) + 35*a^2*b*e^2*(3*A*e*(2*d + e*x) + B*(-8*d^2 - 4*d*e*x
+ e^2*x^2)) + 7*a*b^2*e*(5*A*e*(-8*d^2 - 4*d*e*x + e^2*x^2) + 3*B*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3
)) + b^3*(7*A*e*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3) + B*(-128*d^4 - 64*d^3*e*x + 16*d^2*e^2*x^2 - 8*d
*e^3*x^3 + 5*e^4*x^4))))/(35*e^5*(a + b*x)*Sqrt[d + e*x])

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Maple [A]  time = 0.007, size = 317, normalized size = 1.1 \begin{align*} -{\frac{-10\,B{x}^{4}{b}^{3}{e}^{4}-14\,A{x}^{3}{b}^{3}{e}^{4}-42\,B{x}^{3}a{b}^{2}{e}^{4}+16\,B{x}^{3}{b}^{3}d{e}^{3}-70\,A{x}^{2}a{b}^{2}{e}^{4}+28\,A{x}^{2}{b}^{3}d{e}^{3}-70\,B{x}^{2}{a}^{2}b{e}^{4}+84\,B{x}^{2}a{b}^{2}d{e}^{3}-32\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}-210\,Ax{a}^{2}b{e}^{4}+280\,Axa{b}^{2}d{e}^{3}-112\,Ax{b}^{3}{d}^{2}{e}^{2}-70\,Bx{a}^{3}{e}^{4}+280\,Bx{a}^{2}bd{e}^{3}-336\,Bxa{b}^{2}{d}^{2}{e}^{2}+128\,Bx{b}^{3}{d}^{3}e+70\,A{a}^{3}{e}^{4}-420\,Ad{e}^{3}{a}^{2}b+560\,Aa{b}^{2}{d}^{2}{e}^{2}-224\,A{b}^{3}{d}^{3}e-140\,Bd{e}^{3}{a}^{3}+560\,B{a}^{2}b{d}^{2}{e}^{2}-672\,Ba{b}^{2}{d}^{3}e+256\,B{b}^{3}{d}^{4}}{35\, \left ( bx+a \right ) ^{3}{e}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{ex+d}}}} \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)^(3/2),x)

[Out]

-2/35/(e*x+d)^(1/2)*(-5*B*b^3*e^4*x^4-7*A*b^3*e^4*x^3-21*B*a*b^2*e^4*x^3+8*B*b^3*d*e^3*x^3-35*A*a*b^2*e^4*x^2+
14*A*b^3*d*e^3*x^2-35*B*a^2*b*e^4*x^2+42*B*a*b^2*d*e^3*x^2-16*B*b^3*d^2*e^2*x^2-105*A*a^2*b*e^4*x+140*A*a*b^2*
d*e^3*x-56*A*b^3*d^2*e^2*x-35*B*a^3*e^4*x+140*B*a^2*b*d*e^3*x-168*B*a*b^2*d^2*e^2*x+64*B*b^3*d^3*e*x+35*A*a^3*
e^4-210*A*a^2*b*d*e^3+280*A*a*b^2*d^2*e^2-112*A*b^3*d^3*e-70*B*a^3*d*e^3+280*B*a^2*b*d^2*e^2-336*B*a*b^2*d^3*e
+128*B*b^3*d^4)*((b*x+a)^2)^(3/2)/e^5/(b*x+a)^3

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Maxima [A]  time = 1.05141, size = 381, normalized size = 1.26 \begin{align*} \frac{2 \,{\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} -{\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} +{\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} A}{5 \, \sqrt{e x + d} e^{4}} + \frac{2 \,{\left (5 \, b^{3} e^{4} x^{4} - 128 \, b^{3} d^{4} + 336 \, a b^{2} d^{3} e - 280 \, a^{2} b d^{2} e^{2} + 70 \, a^{3} d e^{3} -{\left (8 \, b^{3} d e^{3} - 21 \, a b^{2} e^{4}\right )} x^{3} +{\left (16 \, b^{3} d^{2} e^{2} - 42 \, a b^{2} d e^{3} + 35 \, a^{2} b e^{4}\right )} x^{2} -{\left (64 \, b^{3} d^{3} e - 168 \, a b^{2} d^{2} e^{2} + 140 \, a^{2} b d e^{3} - 35 \, a^{3} e^{4}\right )} x\right )} B}{35 \, \sqrt{e x + d} e^{5}} \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)^(3/2),x, algorithm="maxima")

[Out]

2/5*(b^3*e^3*x^3 + 16*b^3*d^3 - 40*a*b^2*d^2*e + 30*a^2*b*d*e^2 - 5*a^3*e^3 - (2*b^3*d*e^2 - 5*a*b^2*e^3)*x^2
+ (8*b^3*d^2*e - 20*a*b^2*d*e^2 + 15*a^2*b*e^3)*x)*A/(sqrt(e*x + d)*e^4) + 2/35*(5*b^3*e^4*x^4 - 128*b^3*d^4 +
336*a*b^2*d^3*e - 280*a^2*b*d^2*e^2 + 70*a^3*d*e^3 - (8*b^3*d*e^3 - 21*a*b^2*e^4)*x^3 + (16*b^3*d^2*e^2 - 42*
a*b^2*d*e^3 + 35*a^2*b*e^4)*x^2 - (64*b^3*d^3*e - 168*a*b^2*d^2*e^2 + 140*a^2*b*d*e^3 - 35*a^3*e^4)*x)*B/(sqrt
(e*x + d)*e^5)

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Fricas [A]  time = 1.35509, size = 583, normalized size = 1.93 \begin{align*} \frac{2 \,{\left (5 \, B b^{3} e^{4} x^{4} - 128 \, B b^{3} d^{4} - 35 \, A a^{3} e^{4} + 112 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e - 280 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 70 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} -{\left (8 \, B b^{3} d e^{3} - 7 \,{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} +{\left (16 \, B b^{3} d^{2} e^{2} - 14 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 35 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} -{\left (64 \, B b^{3} d^{3} e - 56 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 140 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 35 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt{e x + d}}{35 \,{\left (e^{6} x + d 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)^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*B*b^3*e^4*x^4 - 128*B*b^3*d^4 - 35*A*a^3*e^4 + 112*(3*B*a*b^2 + A*b^3)*d^3*e - 280*(B*a^2*b + A*a*b^2)
*d^2*e^2 + 70*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 7*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + (16*B*b^3*d^2*e^2
- 14*(3*B*a*b^2 + A*b^3)*d*e^3 + 35*(B*a^2*b + A*a*b^2)*e^4)*x^2 - (64*B*b^3*d^3*e - 56*(3*B*a*b^2 + A*b^3)*d^
2*e^2 + 140*(B*a^2*b + A*a*b^2)*d*e^3 - 35*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^6*x + d*e^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{\left (d + e x\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/(d + e*x)**(3/2), x)

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Giac [B]  time = 1.17814, size = 709, normalized size = 2.35 \begin{align*} \frac{2}{35} \,{\left (5 \,{\left (x e + d\right )}^{\frac{7}{2}} B b^{3} e^{30} \mathrm{sgn}\left (b x + a\right ) - 28 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{3} d e^{30} \mathrm{sgn}\left (b x + a\right ) + 70 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} d^{2} e^{30} \mathrm{sgn}\left (b x + a\right ) - 140 \, \sqrt{x e + d} B b^{3} d^{3} e^{30} \mathrm{sgn}\left (b x + a\right ) + 21 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{2} e^{31} \mathrm{sgn}\left (b x + a\right ) + 7 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{3} e^{31} \mathrm{sgn}\left (b x + a\right ) - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{2} d e^{31} \mathrm{sgn}\left (b x + a\right ) - 35 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} d e^{31} \mathrm{sgn}\left (b x + a\right ) + 315 \, \sqrt{x e + d} B a b^{2} d^{2} e^{31} \mathrm{sgn}\left (b x + a\right ) + 105 \, \sqrt{x e + d} A b^{3} d^{2} e^{31} \mathrm{sgn}\left (b x + a\right ) + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b e^{32} \mathrm{sgn}\left (b x + a\right ) + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{2} e^{32} \mathrm{sgn}\left (b x + a\right ) - 210 \, \sqrt{x e + d} B a^{2} b d e^{32} \mathrm{sgn}\left (b x + a\right ) - 210 \, \sqrt{x e + d} A a b^{2} d e^{32} \mathrm{sgn}\left (b x + a\right ) + 35 \, \sqrt{x e + d} B a^{3} e^{33} \mathrm{sgn}\left (b x + a\right ) + 105 \, \sqrt{x e + d} A a^{2} b e^{33} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-35\right )} - \frac{2 \,{\left (B b^{3} d^{4} \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 ) + 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 ) - 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 )}}{\sqrt{x e + 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)^(3/2),x, algorithm="giac")

[Out]

2/35*(5*(x*e + d)^(7/2)*B*b^3*e^30*sgn(b*x + a) - 28*(x*e + d)^(5/2)*B*b^3*d*e^30*sgn(b*x + a) + 70*(x*e + d)^
(3/2)*B*b^3*d^2*e^30*sgn(b*x + a) - 140*sqrt(x*e + d)*B*b^3*d^3*e^30*sgn(b*x + a) + 21*(x*e + d)^(5/2)*B*a*b^2
*e^31*sgn(b*x + a) + 7*(x*e + d)^(5/2)*A*b^3*e^31*sgn(b*x + a) - 105*(x*e + d)^(3/2)*B*a*b^2*d*e^31*sgn(b*x +
a) - 35*(x*e + d)^(3/2)*A*b^3*d*e^31*sgn(b*x + a) + 315*sqrt(x*e + d)*B*a*b^2*d^2*e^31*sgn(b*x + a) + 105*sqrt
(x*e + d)*A*b^3*d^2*e^31*sgn(b*x + a) + 35*(x*e + d)^(3/2)*B*a^2*b*e^32*sgn(b*x + a) + 35*(x*e + d)^(3/2)*A*a*
b^2*e^32*sgn(b*x + a) - 210*sqrt(x*e + d)*B*a^2*b*d*e^32*sgn(b*x + a) - 210*sqrt(x*e + d)*A*a*b^2*d*e^32*sgn(b
*x + a) + 35*sqrt(x*e + d)*B*a^3*e^33*sgn(b*x + a) + 105*sqrt(x*e + d)*A*a^2*b*e^33*sgn(b*x + a))*e^(-35) - 2*
(B*b^3*d^4*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) + 3*B*a^2*b*d^2*e^2*sgn(b*x
+ a) + 3*A*a*b^2*d^2*e^2*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*sg
n(b*x + a))*e^(-5)/sqrt(x*e + d)