3.1572 \(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^2}{6 b^3}+\frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

[Out]

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

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Rubi [A]  time = 0.139914, antiderivative size = 125, 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} \[ \frac{2 e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^2}{6 b^3}+\frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0676425, size = 187, normalized size = 1.5 \[ \frac{x \sqrt{(a+b x)^2} \left (56 a^3 b^2 x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right )+28 a^2 b^3 x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right )+70 a^4 b x \left (6 d^2+8 d e x+3 e^2 x^2\right )+56 a^5 \left (3 d^2+3 d e x+e^2 x^2\right )+8 a b^4 x^4 \left (21 d^2+35 d e x+15 e^2 x^2\right )+b^5 x^5 \left (28 d^2+48 d e x+21 e^2 x^2\right )\right )}{168 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(56*a^5*(3*d^2 + 3*d*e*x + e^2*x^2) + 70*a^4*b*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 56*a^3*b
^2*x^2*(10*d^2 + 15*d*e*x + 6*e^2*x^2) + 28*a^2*b^3*x^3*(15*d^2 + 24*d*e*x + 10*e^2*x^2) + 8*a*b^4*x^4*(21*d^2
+ 35*d*e*x + 15*e^2*x^2) + b^5*x^5*(28*d^2 + 48*d*e*x + 21*e^2*x^2)))/(168*(a + b*x))

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Maple [B]  time = 0.153, size = 230, normalized size = 1.8 \begin{align*}{\frac{x \left ( 21\,{e}^{2}{b}^{5}{x}^{7}+120\,{x}^{6}{e}^{2}a{b}^{4}+48\,{x}^{6}de{b}^{5}+280\,{x}^{5}{e}^{2}{a}^{2}{b}^{3}+280\,{x}^{5}dea{b}^{4}+28\,{x}^{5}{d}^{2}{b}^{5}+336\,{a}^{3}{b}^{2}{e}^{2}{x}^{4}+672\,{a}^{2}{b}^{3}de{x}^{4}+168\,a{b}^{4}{d}^{2}{x}^{4}+210\,{x}^{3}{e}^{2}{a}^{4}b+840\,{x}^{3}de{a}^{3}{b}^{2}+420\,{x}^{3}{d}^{2}{a}^{2}{b}^{3}+56\,{x}^{2}{e}^{2}{a}^{5}+560\,{x}^{2}de{a}^{4}b+560\,{x}^{2}{d}^{2}{a}^{3}{b}^{2}+168\,xde{a}^{5}+420\,x{d}^{2}{a}^{4}b+168\,{d}^{2}{a}^{5} \right ) }{168\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/168*x*(21*b^5*e^2*x^7+120*a*b^4*e^2*x^6+48*b^5*d*e*x^6+280*a^2*b^3*e^2*x^5+280*a*b^4*d*e*x^5+28*b^5*d^2*x^5+
336*a^3*b^2*e^2*x^4+672*a^2*b^3*d*e*x^4+168*a*b^4*d^2*x^4+210*a^4*b*e^2*x^3+840*a^3*b^2*d*e*x^3+420*a^2*b^3*d^
2*x^3+56*a^5*e^2*x^2+560*a^4*b*d*e*x^2+560*a^3*b^2*d^2*x^2+168*a^5*d*e*x+420*a^4*b*d^2*x+168*a^5*d^2)*((b*x+a)
^2)^(5/2)/(b*x+a)^5

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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)^2*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.59418, size = 417, normalized size = 3.34 \begin{align*} \frac{1}{8} \, b^{5} e^{2} x^{8} + a^{5} d^{2} x + \frac{1}{7} \,{\left (2 \, b^{5} d e + 5 \, a b^{4} e^{2}\right )} x^{7} + \frac{1}{6} \,{\left (b^{5} d^{2} + 10 \, a b^{4} d e + 10 \, a^{2} b^{3} e^{2}\right )} x^{6} +{\left (a b^{4} d^{2} + 4 \, a^{2} b^{3} d e + 2 \, a^{3} b^{2} e^{2}\right )} x^{5} + \frac{5}{4} \,{\left (2 \, a^{2} b^{3} d^{2} + 4 \, a^{3} b^{2} d e + a^{4} b e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (10 \, a^{3} b^{2} d^{2} + 10 \, a^{4} b d e + a^{5} e^{2}\right )} x^{3} + \frac{1}{2} \,{\left (5 \, a^{4} b d^{2} + 2 \, a^{5} d e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/8*b^5*e^2*x^8 + a^5*d^2*x + 1/7*(2*b^5*d*e + 5*a*b^4*e^2)*x^7 + 1/6*(b^5*d^2 + 10*a*b^4*d*e + 10*a^2*b^3*e^2
)*x^6 + (a*b^4*d^2 + 4*a^2*b^3*d*e + 2*a^3*b^2*e^2)*x^5 + 5/4*(2*a^2*b^3*d^2 + 4*a^3*b^2*d*e + a^4*b*e^2)*x^4
+ 1/3*(10*a^3*b^2*d^2 + 10*a^4*b*d*e + a^5*e^2)*x^3 + 1/2*(5*a^4*b*d^2 + 2*a^5*d*e)*x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.14295, size = 432, normalized size = 3.46 \begin{align*} \frac{1}{8} \, b^{5} x^{8} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{7} \, b^{5} d x^{7} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{6} \, b^{5} d^{2} x^{6} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{7} \, a b^{4} x^{7} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, a b^{4} d x^{6} e \mathrm{sgn}\left (b x + a\right ) + a b^{4} d^{2} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, a^{2} b^{3} x^{6} e^{2} \mathrm{sgn}\left (b x + a\right ) + 4 \, a^{2} b^{3} d x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{2} b^{3} d^{2} x^{4} \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} b^{2} x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, a^{3} b^{2} d x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, a^{3} b^{2} d^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{4} \, a^{4} b x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, a^{4} b d x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, a^{4} b d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, a^{5} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + a^{5} d x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{5} d^{2} 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)^2*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/8*b^5*x^8*e^2*sgn(b*x + a) + 2/7*b^5*d*x^7*e*sgn(b*x + a) + 1/6*b^5*d^2*x^6*sgn(b*x + a) + 5/7*a*b^4*x^7*e^2
*sgn(b*x + a) + 5/3*a*b^4*d*x^6*e*sgn(b*x + a) + a*b^4*d^2*x^5*sgn(b*x + a) + 5/3*a^2*b^3*x^6*e^2*sgn(b*x + a)
+ 4*a^2*b^3*d*x^5*e*sgn(b*x + a) + 5/2*a^2*b^3*d^2*x^4*sgn(b*x + a) + 2*a^3*b^2*x^5*e^2*sgn(b*x + a) + 5*a^3*
b^2*d*x^4*e*sgn(b*x + a) + 10/3*a^3*b^2*d^2*x^3*sgn(b*x + a) + 5/4*a^4*b*x^4*e^2*sgn(b*x + a) + 10/3*a^4*b*d*x
^3*e*sgn(b*x + a) + 5/2*a^4*b*d^2*x^2*sgn(b*x + a) + 1/3*a^5*x^3*e^2*sgn(b*x + a) + a^5*d*x^2*e*sgn(b*x + a) +
a^5*d^2*x*sgn(b*x + a)