### 3.1686 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{\sqrt{d+e x}} \, dx$$

Optimal. Leaf size=204 $\frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^4 (a+b x)}-\frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^4 (a+b x)}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}{e^4 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}{e^4 (a+b x)}$

[Out]

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

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Rubi [A]  time = 0.0635751, 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} $\frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^4 (a+b x)}-\frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^4 (a+b x)}+\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}{e^4 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}{e^4 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0652632, size = 119, normalized size = 0.58 $\frac{2 \sqrt{(a+b x)^2} \sqrt{d+e x} \left (35 a^2 b e^2 (e x-2 d)+35 a^3 e^3+7 a b^2 e \left (8 d^2-4 d e x+3 e^2 x^2\right )+b^3 \left (8 d^2 e x-16 d^3-6 d e^2 x^2+5 e^3 x^3\right )\right )}{35 e^4 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(35*a^3*e^3 + 35*a^2*b*e^2*(-2*d + e*x) + 7*a*b^2*e*(8*d^2 - 4*d*e*x + 3*e^
2*x^2) + b^3*(-16*d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3*x^3)))/(35*e^4*(a + b*x))

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Maple [A]  time = 0.152, size = 132, normalized size = 0.7 \begin{align*}{\frac{10\,{x}^{3}{b}^{3}{e}^{3}+42\,{x}^{2}a{b}^{2}{e}^{3}-12\,{x}^{2}{b}^{3}d{e}^{2}+70\,x{a}^{2}b{e}^{3}-56\,xa{b}^{2}d{e}^{2}+16\,x{b}^{3}{d}^{2}e+70\,{a}^{3}{e}^{3}-140\,d{e}^{2}{a}^{2}b+112\,a{b}^{2}{d}^{2}e-32\,{b}^{3}{d}^{3}}{35\, \left ( bx+a \right ) ^{3}{e}^{4}}\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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x)

[Out]

2/35*(e*x+d)^(1/2)*(5*b^3*e^3*x^3+21*a*b^2*e^3*x^2-6*b^3*d*e^2*x^2+35*a^2*b*e^3*x-28*a*b^2*d*e^2*x+8*b^3*d^2*e
*x+35*a^3*e^3-70*a^2*b*d*e^2+56*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 = 1.07642, size = 221, normalized size = 1.08 \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 )}}{35 \, \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)^(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)/(sqrt(e*x + d)*e^4)

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Fricas [A]  time = 1.51798, size = 251, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (5 \, b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 56 \, a b^{2} d^{2} e - 70 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} - 3 \,{\left (2 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} +{\left (8 \, b^{3} d^{2} e - 28 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{35 \, 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)^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*b^3*e^3*x^3 - 16*b^3*d^3 + 56*a*b^2*d^2*e - 70*a^2*b*d*e^2 + 35*a^3*e^3 - 3*(2*b^3*d*e^2 - 7*a*b^2*e^3
)*x^2 + (8*b^3*d^2*e - 28*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d)/e^4

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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**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 = 1.22602, size = 225, normalized size = 1.1 \begin{align*} \frac{2}{35} \,{\left (35 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} a^{2} b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 7 \,{\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 b^{2} e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) +{\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^{3} e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) + 35 \, \sqrt{x e + d} 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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/35*(35*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a^2*b*e^(-1)*sgn(b*x + a) + 7*(3*(x*e + d)^(5/2) - 10*(x*e + d)
^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a*b^2*e^(-2)*sgn(b*x + a) + (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^3*e^(-3)*sgn(b*x + a) + 35*sqrt(x*e + d)*a^3*sgn(b*x + a))*e^(-1)