3.1688 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=204 \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^4 (a+b x)}-\frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)}{e^4 (a+b x)}-\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) \sqrt{d+e x}}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^4 (a+b x) (d+e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0677803, antiderivative size = 204, 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} \[ \frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^4 (a+b x)}-\frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)}{e^4 (a+b x)}-\frac{6 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) \sqrt{d+e x}}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^4 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

Mathematica [A]  time = 0.0629554, size = 119, normalized size = 0.58 \[ -\frac{2 \sqrt{(a+b x)^2} \left (3 a^2 b e^2 (2 d+3 e x)+a^3 e^3-3 a b^2 e \left (8 d^2+12 d e x+3 e^2 x^2\right )+b^3 \left (24 d^2 e x+16 d^3+6 d e^2 x^2-e^3 x^3\right )\right )}{3 e^4 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(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]*(a^3*e^3 + 3*a^2*b*e^2*(2*d + 3*e*x) - 3*a*b^2*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + b^3*(1
6*d^3 + 24*d^2*e*x + 6*d*e^2*x^2 - e^3*x^3)))/(3*e^4*(a + b*x)*(d + e*x)^(3/2))

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Maple [A]  time = 0.153, size = 131, normalized size = 0.6 \begin{align*} -{\frac{-2\,{x}^{3}{b}^{3}{e}^{3}-18\,{x}^{2}a{b}^{2}{e}^{3}+12\,{x}^{2}{b}^{3}d{e}^{2}+18\,x{a}^{2}b{e}^{3}-72\,xa{b}^{2}d{e}^{2}+48\,x{b}^{3}{d}^{2}e+2\,{a}^{3}{e}^{3}+12\,d{e}^{2}{a}^{2}b-48\,a{b}^{2}{d}^{2}e+32\,{b}^{3}{d}^{3}}{3\, \left ( bx+a \right ) ^{3}{e}^{4}} \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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(5/2),x)

[Out]

-2/3/(e*x+d)^(3/2)*(-b^3*e^3*x^3-9*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+9*a^2*b*e^3*x-36*a*b^2*d*e^2*x+24*b^3*d^2*e*x
+a^3*e^3+6*a^2*b*d*e^2-24*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 = 1.10806, size = 169, normalized size = 0.83 \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 )}}{3 \,{\left (e^{5} x + d e^{4}\right )} \sqrt{e x + 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)^(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)/((e^5*x + d*e^4)*sqrt(e*x + d))

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Fricas [A]  time = 1.51978, size = 281, normalized size = 1.38 \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 )} \sqrt{e x + d}}{3 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^(5/2),x, algorithm="fricas")

[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)*sqrt(e*x + d)/(e^6*x^2 + 2*d*e^5*x + d^2*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{5}{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)**(5/2),x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**(5/2), x)

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Giac [A]  time = 1.19761, size = 273, normalized size = 1.34 \begin{align*} \frac{2}{3} \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{3} e^{8} \mathrm{sgn}\left (b x + a\right ) - 9 \, \sqrt{x e + d} b^{3} d e^{8} \mathrm{sgn}\left (b x + a\right ) + 9 \, \sqrt{x e + d} a b^{2} e^{9} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-12\right )} - \frac{2 \,{\left (9 \,{\left (x e + d\right )} b^{3} d^{2} \mathrm{sgn}\left (b x + a\right ) - b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - 18 \,{\left (x e + d\right )} a b^{2} d e \mathrm{sgn}\left (b x + a\right ) + 3 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 9 \,{\left (x e + d\right )} a^{2} b e^{2} \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 )}}{3 \,{\left (x e + d\right )}^{\frac{3}{2}}} \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)^(5/2),x, algorithm="giac")

[Out]

2/3*((x*e + d)^(3/2)*b^3*e^8*sgn(b*x + a) - 9*sqrt(x*e + d)*b^3*d*e^8*sgn(b*x + a) + 9*sqrt(x*e + d)*a*b^2*e^9
*sgn(b*x + a))*e^(-12) - 2/3*(9*(x*e + d)*b^3*d^2*sgn(b*x + a) - b^3*d^3*sgn(b*x + a) - 18*(x*e + d)*a*b^2*d*e
*sgn(b*x + a) + 3*a*b^2*d^2*e*sgn(b*x + a) + 9*(x*e + d)*a^2*b*e^2*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)/(x*e + d)^(3/2)