3.1847 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=306 \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (-3 a B e-A b e+4 b B d)}{7 e^5 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e) (-a B e-A b e+2 b B d)}{5 e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{3 e^5 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3 (B d-A e)}{e^5 (a+b x)}+\frac{2 b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^5 (a+b x)} \]

[Out]

(2*(b*d - a*e)^3*(B*d - A*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)) - (2*(b*d - a*e)^2*(
4*b*B*d - 3*A*b*e - 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)) + (6*b*(b*d - a*e)
*(2*b*B*d - A*b*e - 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^2*(4*b*B*d
- A*b*e - 3*a*B*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)) + (2*b^3*B*(d + e*x)^(9/2)
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*e^5*(a + b*x))

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Rubi [A]  time = 0.147518, antiderivative size = 306, 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} \[ -\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (-3 a B e-A b e+4 b B d)}{7 e^5 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e) (-a B e-A b e+2 b B d)}{5 e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{3 e^5 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3 (B d-A e)}{e^5 (a+b x)}+\frac{2 b^3 B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)^3*(B*d - A*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)) - (2*(b*d - a*e)^2*(
4*b*B*d - 3*A*b*e - 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)) + (6*b*(b*d - a*e)
*(2*b*B*d - A*b*e - 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^2*(4*b*B*d
- A*b*e - 3*a*B*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)) + (2*b^3*B*(d + e*x)^(9/2)
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*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}}{\sqrt{d+e x}} \, 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)}{\sqrt{d+e x}} \, 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 \sqrt{d+e x}}+\frac{b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e) \sqrt{d+e x}}{e^4}-\frac{3 b^4 (b d-a e) (-2 b B d+A b e+a B e) (d+e x)^{3/2}}{e^4}+\frac{b^5 (-4 b B d+A b e+3 a B e) (d+e x)^{5/2}}{e^4}+\frac{b^6 B (d+e x)^{7/2}}{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{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac{2 (b d-a e)^2 (4 b B d-3 A b e-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{6 b (b d-a e) (2 b B d-A b e-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^2 (4 b B d-A b e-3 a B e) (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x)}+\frac{2 b^3 B (d+e x)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{9 e^5 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.118301, size = 163, normalized size = 0.53 \[ \frac{2 \left ((a+b x)^2\right )^{3/2} \sqrt{d+e x} \left (-45 b^2 (d+e x)^3 (-3 a B e-A b e+4 b B d)+189 b (d+e x)^2 (b d-a e) (-a B e-A b e+2 b B d)-105 (d+e x) (b d-a e)^2 (-a B e-3 A b e+4 b B d)+315 (b d-a e)^3 (B d-A e)+35 b^3 B (d+e x)^4\right )}{315 e^5 (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a + b*x)^2)^(3/2)*Sqrt[d + e*x]*(315*(b*d - a*e)^3*(B*d - A*e) - 105*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a
*B*e)*(d + e*x) + 189*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2 - 45*b^2*(4*b*B*d - A*b*e - 3*a*B*e)
*(d + e*x)^3 + 35*b^3*B*(d + e*x)^4))/(315*e^5*(a + b*x)^3)

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Maple [A]  time = 0.007, size = 317, normalized size = 1. \begin{align*}{\frac{70\,B{x}^{4}{b}^{3}{e}^{4}+90\,A{x}^{3}{b}^{3}{e}^{4}+270\,B{x}^{3}a{b}^{2}{e}^{4}-80\,B{x}^{3}{b}^{3}d{e}^{3}+378\,A{x}^{2}a{b}^{2}{e}^{4}-108\,A{x}^{2}{b}^{3}d{e}^{3}+378\,B{x}^{2}{a}^{2}b{e}^{4}-324\,B{x}^{2}a{b}^{2}d{e}^{3}+96\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}+630\,Ax{a}^{2}b{e}^{4}-504\,Axa{b}^{2}d{e}^{3}+144\,Ax{b}^{3}{d}^{2}{e}^{2}+210\,Bx{a}^{3}{e}^{4}-504\,Bx{a}^{2}bd{e}^{3}+432\,Bxa{b}^{2}{d}^{2}{e}^{2}-128\,Bx{b}^{3}{d}^{3}e+630\,A{a}^{3}{e}^{4}-1260\,Ad{e}^{3}{a}^{2}b+1008\,Aa{b}^{2}{d}^{2}{e}^{2}-288\,A{b}^{3}{d}^{3}e-420\,Bd{e}^{3}{a}^{3}+1008\,B{a}^{2}b{d}^{2}{e}^{2}-864\,Ba{b}^{2}{d}^{3}e+256\,B{b}^{3}{d}^{4}}{315\, \left ( bx+a \right ) ^{3}{e}^{5}}\sqrt{ex+d} \left ( \left ( bx+a \right ) ^{2} \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)^(1/2),x)

[Out]

2/315*(e*x+d)^(1/2)*(35*B*b^3*e^4*x^4+45*A*b^3*e^4*x^3+135*B*a*b^2*e^4*x^3-40*B*b^3*d*e^3*x^3+189*A*a*b^2*e^4*
x^2-54*A*b^3*d*e^3*x^2+189*B*a^2*b*e^4*x^2-162*B*a*b^2*d*e^3*x^2+48*B*b^3*d^2*e^2*x^2+315*A*a^2*b*e^4*x-252*A*
a*b^2*d*e^3*x+72*A*b^3*d^2*e^2*x+105*B*a^3*e^4*x-252*B*a^2*b*d*e^3*x+216*B*a*b^2*d^2*e^2*x-64*B*b^3*d^3*e*x+31
5*A*a^3*e^4-630*A*a^2*b*d*e^3+504*A*a*b^2*d^2*e^2-144*A*b^3*d^3*e-210*B*a^3*d*e^3+504*B*a^2*b*d^2*e^2-432*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 = 1.03012, size = 516, normalized size = 1.69 \begin{align*} \frac{2 \,{\left (5 \, b^{3} e^{4} x^{4} - 16 \, b^{3} d^{4} + 56 \, a b^{2} d^{3} e - 70 \, a^{2} b d^{2} e^{2} + 35 \, a^{3} d e^{3} -{\left (b^{3} d e^{3} - 21 \, a b^{2} e^{4}\right )} x^{3} +{\left (2 \, b^{3} d^{2} e^{2} - 7 \, a b^{2} d e^{3} + 35 \, a^{2} b e^{4}\right )} x^{2} -{\left (8 \, b^{3} d^{3} e - 28 \, a b^{2} d^{2} e^{2} + 35 \, a^{2} b d e^{3} - 35 \, a^{3} e^{4}\right )} x\right )} A}{35 \, \sqrt{e x + d} e^{4}} + \frac{2 \,{\left (35 \, b^{3} e^{5} x^{5} + 128 \, b^{3} d^{5} - 432 \, a b^{2} d^{4} e + 504 \, a^{2} b d^{3} e^{2} - 210 \, a^{3} d^{2} e^{3} - 5 \,{\left (b^{3} d e^{4} - 27 \, a b^{2} e^{5}\right )} x^{4} +{\left (8 \, b^{3} d^{2} e^{3} - 27 \, a b^{2} d e^{4} + 189 \, a^{2} b e^{5}\right )} x^{3} -{\left (16 \, b^{3} d^{3} e^{2} - 54 \, a b^{2} d^{2} e^{3} + 63 \, a^{2} b d e^{4} - 105 \, a^{3} e^{5}\right )} x^{2} +{\left (64 \, b^{3} d^{4} e - 216 \, a b^{2} d^{3} e^{2} + 252 \, a^{2} b d^{2} e^{3} - 105 \, a^{3} d e^{4}\right )} x\right )} B}{315 \, \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)^(1/2),x, algorithm="maxima")

[Out]

2/35*(5*b^3*e^4*x^4 - 16*b^3*d^4 + 56*a*b^2*d^3*e - 70*a^2*b*d^2*e^2 + 35*a^3*d*e^3 - (b^3*d*e^3 - 21*a*b^2*e^
4)*x^3 + (2*b^3*d^2*e^2 - 7*a*b^2*d*e^3 + 35*a^2*b*e^4)*x^2 - (8*b^3*d^3*e - 28*a*b^2*d^2*e^2 + 35*a^2*b*d*e^3
- 35*a^3*e^4)*x)*A/(sqrt(e*x + d)*e^4) + 2/315*(35*b^3*e^5*x^5 + 128*b^3*d^5 - 432*a*b^2*d^4*e + 504*a^2*b*d^
3*e^2 - 210*a^3*d^2*e^3 - 5*(b^3*d*e^4 - 27*a*b^2*e^5)*x^4 + (8*b^3*d^2*e^3 - 27*a*b^2*d*e^4 + 189*a^2*b*e^5)*
x^3 - (16*b^3*d^3*e^2 - 54*a*b^2*d^2*e^3 + 63*a^2*b*d*e^4 - 105*a^3*e^5)*x^2 + (64*b^3*d^4*e - 216*a*b^2*d^3*e
^2 + 252*a^2*b*d^2*e^3 - 105*a^3*d*e^4)*x)*B/(sqrt(e*x + d)*e^5)

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Fricas [A]  time = 1.40732, size = 579, normalized size = 1.89 \begin{align*} \frac{2 \,{\left (35 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} + 315 \, A a^{3} e^{4} - 144 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 504 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 210 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 5 \,{\left (8 \, B b^{3} d e^{3} - 9 \,{\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} - 18 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 63 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} -{\left (64 \, B b^{3} d^{3} e - 72 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 252 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 105 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt{e x + d}}{315 \, 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)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*B*b^3*e^4*x^4 + 128*B*b^3*d^4 + 315*A*a^3*e^4 - 144*(3*B*a*b^2 + A*b^3)*d^3*e + 504*(B*a^2*b + A*a*b
^2)*d^2*e^2 - 210*(B*a^3 + 3*A*a^2*b)*d*e^3 - 5*(8*B*b^3*d*e^3 - 9*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*
d^2*e^2 - 18*(3*B*a*b^2 + A*b^3)*d*e^3 + 63*(B*a^2*b + A*a*b^2)*e^4)*x^2 - (64*B*b^3*d^3*e - 72*(3*B*a*b^2 + A
*b^3)*d^2*e^2 + 252*(B*a^2*b + A*a*b^2)*d*e^3 - 105*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/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)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.13258, size = 531, normalized size = 1.74 \begin{align*} \frac{2}{315} \,{\left (105 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} B a^{3} e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 315 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} A a^{2} b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 63 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} d + 15 \, \sqrt{x e + d} d^{2}\right )} B a^{2} b e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 63 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} d + 15 \, \sqrt{x e + d} d^{2}\right )} A a b^{2} e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 27 \,{\left (5 \,{\left (x e + d\right )}^{\frac{7}{2}} - 21 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2} - 35 \, \sqrt{x e + d} d^{3}\right )} B a b^{2} e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) + 9 \,{\left (5 \,{\left (x e + d\right )}^{\frac{7}{2}} - 21 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2} - 35 \, \sqrt{x e + d} d^{3}\right )} A b^{3} e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) +{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 180 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 378 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 420 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3} + 315 \, \sqrt{x e + d} d^{4}\right )} B b^{3} e^{\left (-4\right )} \mathrm{sgn}\left (b x + a\right ) + 315 \, \sqrt{x e + d} A a^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\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)^(1/2),x, algorithm="giac")

[Out]

2/315*(105*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*B*a^3*e^(-1)*sgn(b*x + a) + 315*((x*e + d)^(3/2) - 3*sqrt(x*e
+ d)*d)*A*a^2*b*e^(-1)*sgn(b*x + a) + 63*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*B*
a^2*b*e^(-2)*sgn(b*x + a) + 63*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*A*a*b^2*e^(-2
)*sgn(b*x + a) + 27*(5*(x*e + d)^(7/2) - 21*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)
*B*a*b^2*e^(-3)*sgn(b*x + a) + 9*(5*(x*e + d)^(7/2) - 21*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(
x*e + d)*d^3)*A*b^3*e^(-3)*sgn(b*x + a) + (35*(x*e + d)^(9/2) - 180*(x*e + d)^(7/2)*d + 378*(x*e + d)^(5/2)*d^
2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*e + d)*d^4)*B*b^3*e^(-4)*sgn(b*x + a) + 315*sqrt(x*e + d)*A*a^3*sgn(b
*x + a))*e^(-1)