3.1555 $$\int (d+e x)^4 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx$$

Optimal. Leaf size=200 $\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^8}{8 e^4 (a+b x)}-\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^7 (b d-a e)}{7 e^4 (a+b x)}+\frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6 (b d-a e)^2}{2 e^4 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)^3}{5 e^4 (a+b x)}$

[Out]

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

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Rubi [A]  time = 0.159973, antiderivative size = 200, 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} $\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^8}{8 e^4 (a+b x)}-\frac{3 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^7 (b d-a e)}{7 e^4 (a+b x)}+\frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6 (b d-a e)^2}{2 e^4 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)^3}{5 e^4 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^4*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

-((b*d - a*e)^3*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^4*(a + b*x)) + (b*(b*d - a*e)^2*(d + e*x)^6*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^4*(a + b*x)) - (3*b^2*(b*d - a*e)*(d + e*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(7*e^4*(a + b*x)) + (b^3*(d + e*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*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 (d+e x)^4 \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)^4 \, 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 (d+e x)^4}{e^3}+\frac{3 b^4 (b d-a e)^2 (d+e x)^5}{e^3}-\frac{3 b^5 (b d-a e) (d+e x)^6}{e^3}+\frac{b^6 (d+e x)^7}{e^3}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{(b d-a e)^3 (d+e x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x)}+\frac{b (b d-a e)^2 (d+e x)^6 \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^4 (a+b x)}-\frac{3 b^2 (b d-a e) (d+e x)^7 \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x)}+\frac{b^3 (d+e x)^8 \sqrt{a^2+2 a b x+b^2 x^2}}{8 e^4 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.076116, size = 215, normalized size = 1.08 $\frac{x \sqrt{(a+b x)^2} \left (28 a^2 b x \left (45 d^2 e^2 x^2+40 d^3 e x+15 d^4+24 d e^3 x^3+5 e^4 x^4\right )+56 a^3 \left (10 d^2 e^2 x^2+10 d^3 e x+5 d^4+5 d e^3 x^3+e^4 x^4\right )+8 a b^2 x^2 \left (126 d^2 e^2 x^2+105 d^3 e x+35 d^4+70 d e^3 x^3+15 e^4 x^4\right )+b^3 x^3 \left (280 d^2 e^2 x^2+224 d^3 e x+70 d^4+160 d e^3 x^3+35 e^4 x^4\right )\right )}{280 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^4*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(56*a^3*(5*d^4 + 10*d^3*e*x + 10*d^2*e^2*x^2 + 5*d*e^3*x^3 + e^4*x^4) + 28*a^2*b*x*(15*d^
4 + 40*d^3*e*x + 45*d^2*e^2*x^2 + 24*d*e^3*x^3 + 5*e^4*x^4) + 8*a*b^2*x^2*(35*d^4 + 105*d^3*e*x + 126*d^2*e^2*
x^2 + 70*d*e^3*x^3 + 15*e^4*x^4) + b^3*x^3*(70*d^4 + 224*d^3*e*x + 280*d^2*e^2*x^2 + 160*d*e^3*x^3 + 35*e^4*x^
4)))/(280*(a + b*x))

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Maple [A]  time = 0.154, size = 264, normalized size = 1.3 \begin{align*}{\frac{x \left ( 35\,{b}^{3}{e}^{4}{x}^{7}+120\,{x}^{6}{b}^{2}a{e}^{4}+160\,{x}^{6}{b}^{3}d{e}^{3}+140\,{x}^{5}b{a}^{2}{e}^{4}+560\,{x}^{5}{b}^{2}ad{e}^{3}+280\,{x}^{5}{b}^{3}{d}^{2}{e}^{2}+56\,{x}^{4}{a}^{3}{e}^{4}+672\,{x}^{4}b{a}^{2}d{e}^{3}+1008\,{x}^{4}{b}^{2}a{d}^{2}{e}^{2}+224\,{x}^{4}{b}^{3}{d}^{3}e+280\,{x}^{3}{a}^{3}d{e}^{3}+1260\,{x}^{3}b{a}^{2}{d}^{2}{e}^{2}+840\,{x}^{3}{b}^{2}a{d}^{3}e+70\,{x}^{3}{b}^{3}{d}^{4}+560\,{a}^{3}{d}^{2}{e}^{2}{x}^{2}+1120\,{a}^{2}b{d}^{3}e{x}^{2}+280\,a{b}^{2}{d}^{4}{x}^{2}+560\,x{a}^{3}{d}^{3}e+420\,xb{a}^{2}{d}^{4}+280\,{a}^{3}{d}^{4} \right ) }{280\, \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)^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/280*x*(35*b^3*e^4*x^7+120*a*b^2*e^4*x^6+160*b^3*d*e^3*x^6+140*a^2*b*e^4*x^5+560*a*b^2*d*e^3*x^5+280*b^3*d^2*
e^2*x^5+56*a^3*e^4*x^4+672*a^2*b*d*e^3*x^4+1008*a*b^2*d^2*e^2*x^4+224*b^3*d^3*e*x^4+280*a^3*d*e^3*x^3+1260*a^2
*b*d^2*e^2*x^3+840*a*b^2*d^3*e*x^3+70*b^3*d^4*x^3+560*a^3*d^2*e^2*x^2+1120*a^2*b*d^3*e*x^2+280*a*b^2*d^4*x^2+5
60*a^3*d^3*e*x+420*a^2*b*d^4*x+280*a^3*d^4)*((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)^4*(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 = 1.57092, size = 471, normalized size = 2.36 \begin{align*} \frac{1}{8} \, b^{3} e^{4} x^{8} + a^{3} d^{4} x + \frac{1}{7} \,{\left (4 \, b^{3} d e^{3} + 3 \, a b^{2} e^{4}\right )} x^{7} + \frac{1}{2} \,{\left (2 \, b^{3} d^{2} e^{2} + 4 \, a b^{2} d e^{3} + a^{2} b e^{4}\right )} x^{6} + \frac{1}{5} \,{\left (4 \, b^{3} d^{3} e + 18 \, a b^{2} d^{2} e^{2} + 12 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} x^{5} + \frac{1}{4} \,{\left (b^{3} d^{4} + 12 \, a b^{2} d^{3} e + 18 \, a^{2} b d^{2} e^{2} + 4 \, a^{3} d e^{3}\right )} x^{4} +{\left (a b^{2} d^{4} + 4 \, a^{2} b d^{3} e + 2 \, a^{3} d^{2} e^{2}\right )} x^{3} + \frac{1}{2} \,{\left (3 \, a^{2} b d^{4} + 4 \, a^{3} d^{3} e\right )} x^{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/8*b^3*e^4*x^8 + a^3*d^4*x + 1/7*(4*b^3*d*e^3 + 3*a*b^2*e^4)*x^7 + 1/2*(2*b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b
*e^4)*x^6 + 1/5*(4*b^3*d^3*e + 18*a*b^2*d^2*e^2 + 12*a^2*b*d*e^3 + a^3*e^4)*x^5 + 1/4*(b^3*d^4 + 12*a*b^2*d^3*
e + 18*a^2*b*d^2*e^2 + 4*a^3*d*e^3)*x^4 + (a*b^2*d^4 + 4*a^2*b*d^3*e + 2*a^3*d^2*e^2)*x^3 + 1/2*(3*a^2*b*d^4 +
4*a^3*d^3*e)*x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{4} \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)**4*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((d + e*x)**4*((a + b*x)**2)**(3/2), x)

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Giac [B]  time = 1.17076, size = 482, normalized size = 2.41 \begin{align*} \frac{1}{8} \, b^{3} x^{8} e^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{4}{7} \, b^{3} d x^{7} e^{3} \mathrm{sgn}\left (b x + a\right ) + b^{3} d^{2} x^{6} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{4}{5} \, b^{3} d^{3} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, b^{3} d^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{7} \, a b^{2} x^{7} e^{4} \mathrm{sgn}\left (b x + a\right ) + 2 \, a b^{2} d x^{6} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{18}{5} \, a b^{2} d^{2} x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a b^{2} d^{3} x^{4} e \mathrm{sgn}\left (b x + a\right ) + a b^{2} d^{4} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a^{2} b x^{6} e^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{12}{5} \, a^{2} b d x^{5} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{9}{2} \, a^{2} b d^{2} x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + 4 \, a^{2} b d^{3} x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a^{2} b d^{4} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, a^{3} x^{5} e^{4} \mathrm{sgn}\left (b x + a\right ) + a^{3} d x^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} d^{2} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} d^{3} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{3} d^{4} 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)^4*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/8*b^3*x^8*e^4*sgn(b*x + a) + 4/7*b^3*d*x^7*e^3*sgn(b*x + a) + b^3*d^2*x^6*e^2*sgn(b*x + a) + 4/5*b^3*d^3*x^5
*e*sgn(b*x + a) + 1/4*b^3*d^4*x^4*sgn(b*x + a) + 3/7*a*b^2*x^7*e^4*sgn(b*x + a) + 2*a*b^2*d*x^6*e^3*sgn(b*x +
a) + 18/5*a*b^2*d^2*x^5*e^2*sgn(b*x + a) + 3*a*b^2*d^3*x^4*e*sgn(b*x + a) + a*b^2*d^4*x^3*sgn(b*x + a) + 1/2*a
^2*b*x^6*e^4*sgn(b*x + a) + 12/5*a^2*b*d*x^5*e^3*sgn(b*x + a) + 9/2*a^2*b*d^2*x^4*e^2*sgn(b*x + a) + 4*a^2*b*d
^3*x^3*e*sgn(b*x + a) + 3/2*a^2*b*d^4*x^2*sgn(b*x + a) + 1/5*a^3*x^5*e^4*sgn(b*x + a) + a^3*d*x^4*e^3*sgn(b*x
+ a) + 2*a^3*d^2*x^3*e^2*sgn(b*x + a) + 2*a^3*d^3*x^2*e*sgn(b*x + a) + a^3*d^4*x*sgn(b*x + a)