3.1556 \(\int (d+e x)^3 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=172 \[ \frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)}{2 b^4}+\frac{3 e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^2}{5 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)^3}{4 b^4}+\frac{e^3 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4} \]

[Out]

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

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Rubi [A]  time = 0.131752, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)}{2 b^4}+\frac{3 e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^2}{5 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)^3}{4 b^4}+\frac{e^3 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0619149, size = 171, normalized size = 0.99 \[ \frac{x \sqrt{(a+b x)^2} \left (21 a^2 b x \left (20 d^2 e x+10 d^3+15 d e^2 x^2+4 e^3 x^3\right )+35 a^3 \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+7 a b^2 x^2 \left (45 d^2 e x+20 d^3+36 d e^2 x^2+10 e^3 x^3\right )+b^3 x^3 \left (84 d^2 e x+35 d^3+70 d e^2 x^2+20 e^3 x^3\right )\right )}{140 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(35*a^3*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 21*a^2*b*x*(10*d^3 + 20*d^2*e*x + 1
5*d*e^2*x^2 + 4*e^3*x^3) + 7*a*b^2*x^2*(20*d^3 + 45*d^2*e*x + 36*d*e^2*x^2 + 10*e^3*x^3) + b^3*x^3*(35*d^3 + 8
4*d^2*e*x + 70*d*e^2*x^2 + 20*e^3*x^3)))/(140*(a + b*x))

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Maple [A]  time = 0.155, size = 206, normalized size = 1.2 \begin{align*}{\frac{x \left ( 20\,{b}^{3}{e}^{3}{x}^{6}+70\,{x}^{5}{b}^{2}a{e}^{3}+70\,{x}^{5}{b}^{3}d{e}^{2}+84\,{x}^{4}b{a}^{2}{e}^{3}+252\,{x}^{4}{b}^{2}ad{e}^{2}+84\,{x}^{4}{b}^{3}{d}^{2}e+35\,{x}^{3}{a}^{3}{e}^{3}+315\,{x}^{3}b{a}^{2}d{e}^{2}+315\,{x}^{3}a{b}^{2}{d}^{2}e+35\,{x}^{3}{b}^{3}{d}^{3}+140\,{a}^{3}d{e}^{2}{x}^{2}+420\,{a}^{2}b{d}^{2}e{x}^{2}+140\,a{b}^{2}{d}^{3}{x}^{2}+210\,x{a}^{3}{d}^{2}e+210\,xb{a}^{2}{d}^{3}+140\,{a}^{3}{d}^{3} \right ) }{140\, \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*x*(20*b^3*e^3*x^6+70*a*b^2*e^3*x^5+70*b^3*d*e^2*x^5+84*a^2*b*e^3*x^4+252*a*b^2*d*e^2*x^4+84*b^3*d^2*e*x^
4+35*a^3*e^3*x^3+315*a^2*b*d*e^2*x^3+315*a*b^2*d^2*e*x^3+35*b^3*d^3*x^3+140*a^3*d*e^2*x^2+420*a^2*b*d^2*e*x^2+
140*a*b^2*d^3*x^2+210*a^3*d^2*e*x+210*a^2*b*d^3*x+140*a^3*d^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5251, size = 344, normalized size = 2. \begin{align*} \frac{1}{7} \, b^{3} e^{3} x^{7} + a^{3} d^{3} x + \frac{1}{2} \,{\left (b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{6} + \frac{3}{5} \,{\left (b^{3} d^{2} e + 3 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{5} + \frac{1}{4} \,{\left (b^{3} d^{3} + 9 \, a b^{2} d^{2} e + 9 \, a^{2} b d e^{2} + a^{3} e^{3}\right )} x^{4} +{\left (a b^{2} d^{3} + 3 \, a^{2} b d^{2} e + a^{3} d e^{2}\right )} x^{3} + \frac{3}{2} \,{\left (a^{2} b d^{3} + a^{3} d^{2} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*e^3*x^7 + a^3*d^3*x + 1/2*(b^3*d*e^2 + a*b^2*e^3)*x^6 + 3/5*(b^3*d^2*e + 3*a*b^2*d*e^2 + a^2*b*e^3)*x^
5 + 1/4*(b^3*d^3 + 9*a*b^2*d^2*e + 9*a^2*b*d*e^2 + a^3*e^3)*x^4 + (a*b^2*d^3 + 3*a^2*b*d^2*e + a^3*d*e^2)*x^3
+ 3/2*(a^2*b*d^3 + a^3*d^2*e)*x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15484, size = 378, normalized size = 2.2 \begin{align*} \frac{1}{7} \, b^{3} x^{7} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, b^{3} d x^{6} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, b^{3} d^{2} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, b^{3} d^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a b^{2} x^{6} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{9}{5} \, a b^{2} d x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{9}{4} \, a b^{2} d^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + a b^{2} d^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, a^{2} b x^{5} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{9}{4} \, a^{2} b d x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b d^{2} x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a^{2} b d^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, a^{3} x^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + a^{3} d x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a^{3} d^{2} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{3} d^{3} x \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7*e^3*sgn(b*x + a) + 1/2*b^3*d*x^6*e^2*sgn(b*x + a) + 3/5*b^3*d^2*x^5*e*sgn(b*x + a) + 1/4*b^3*d^3*x
^4*sgn(b*x + a) + 1/2*a*b^2*x^6*e^3*sgn(b*x + a) + 9/5*a*b^2*d*x^5*e^2*sgn(b*x + a) + 9/4*a*b^2*d^2*x^4*e*sgn(
b*x + a) + a*b^2*d^3*x^3*sgn(b*x + a) + 3/5*a^2*b*x^5*e^3*sgn(b*x + a) + 9/4*a^2*b*d*x^4*e^2*sgn(b*x + a) + 3*
a^2*b*d^2*x^3*e*sgn(b*x + a) + 3/2*a^2*b*d^3*x^2*sgn(b*x + a) + 1/4*a^3*x^4*e^3*sgn(b*x + a) + a^3*d*x^3*e^2*s
gn(b*x + a) + 3/2*a^3*d^2*x^2*e*sgn(b*x + a) + a^3*d^3*x*sgn(b*x + a)