3.1687 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=202 \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^4 (a+b x)}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{e^4 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}{e^4 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^4 (a+b x) \sqrt{d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.0645532, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {646, 43} \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^4 (a+b x)}-\frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{e^4 (a+b x)}+\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}{e^4 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^4 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

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

Mathematica [A]  time = 0.0595021, size = 119, normalized size = 0.59 \[ -\frac{2 \sqrt{(a+b x)^2} \left (-15 a^2 b e^2 (2 d+e x)+5 a^3 e^3+5 a b^2 e \left (8 d^2+4 d e x-e^2 x^2\right )+b^3 \left (-\left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right )\right )\right )}{5 e^4 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(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]*(5*a^3*e^3 - 15*a^2*b*e^2*(2*d + e*x) + 5*a*b^2*e*(8*d^2 + 4*d*e*x - e^2*x^2) - b^3*(16*
d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3)))/(5*e^4*(a + b*x)*Sqrt[d + e*x])

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Maple [A]  time = 0.154, size = 132, normalized size = 0.7 \begin{align*} -{\frac{-2\,{x}^{3}{b}^{3}{e}^{3}-10\,{x}^{2}a{b}^{2}{e}^{3}+4\,{x}^{2}{b}^{3}d{e}^{2}-30\,x{a}^{2}b{e}^{3}+40\,xa{b}^{2}d{e}^{2}-16\,x{b}^{3}{d}^{2}e+10\,{a}^{3}{e}^{3}-60\,d{e}^{2}{a}^{2}b+80\,a{b}^{2}{d}^{2}e-32\,{b}^{3}{d}^{3}}{5\, \left ( bx+a \right ) ^{3}{e}^{4}} \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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(3/2),x)

[Out]

-2/5/(e*x+d)^(1/2)*(-b^3*e^3*x^3-5*a*b^2*e^3*x^2+2*b^3*d*e^2*x^2-15*a^2*b*e^3*x+20*a*b^2*d*e^2*x-8*b^3*d^2*e*x
+5*a^3*e^3-30*a^2*b*d*e^2+40*a*b^2*d^2*e-16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3

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Maxima [A]  time = 1.16769, size = 154, normalized size = 0.76 \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 )}}{5 \, \sqrt{e x + d} e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)/(sqrt(e*x + d)*e^4)

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Fricas [A]  time = 1.5908, size = 259, normalized size = 1.28 \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 )} \sqrt{e x + d}}{5 \,{\left (e^{5} x + d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*sqrt(e*x + d)/(e^5*x + d*e^4)

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

[Out]

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

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Giac [A]  time = 1.19567, size = 284, normalized size = 1.41 \begin{align*} \frac{2}{5} \,{\left ({\left (x e + d\right )}^{\frac{5}{2}} b^{3} e^{16} \mathrm{sgn}\left (b x + a\right ) - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d e^{16} \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} b^{3} d^{2} e^{16} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} e^{17} \mathrm{sgn}\left (b x + a\right ) - 30 \, \sqrt{x e + d} a b^{2} d e^{17} \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} a^{2} b e^{18} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-20\right )} + \frac{2 \,{\left (b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{\sqrt{x e + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*((x*e + d)^(5/2)*b^3*e^16*sgn(b*x + a) - 5*(x*e + d)^(3/2)*b^3*d*e^16*sgn(b*x + a) + 15*sqrt(x*e + d)*b^3*
d^2*e^16*sgn(b*x + a) + 5*(x*e + d)^(3/2)*a*b^2*e^17*sgn(b*x + a) - 30*sqrt(x*e + d)*a*b^2*d*e^17*sgn(b*x + a)
+ 15*sqrt(x*e + d)*a^2*b*e^18*sgn(b*x + a))*e^(-20) + 2*(b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) +
3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*e^(-4)/sqrt(x*e + d)