3.1973 \(\int (a+b x) (d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=125 \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)}{3 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^2}{5 b^3}+\frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^3} \]

[Out]

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

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Rubi [A]  time = 0.134894, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)}{3 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^2}{5 b^3}+\frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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

Mathematica [A]  time = 0.0565029, size = 157, normalized size = 1.26 \[ \frac{x \sqrt{(a+b x)^2} \left (21 a^2 b^2 x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right )+35 a^3 b x \left (6 d^2+8 d e x+3 e^2 x^2\right )+35 a^4 \left (3 d^2+3 d e x+e^2 x^2\right )+7 a b^3 x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right )+b^4 x^4 \left (21 d^2+35 d e x+15 e^2 x^2\right )\right )}{105 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(35*a^4*(3*d^2 + 3*d*e*x + e^2*x^2) + 35*a^3*b*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 21*a^2*b
^2*x^2*(10*d^2 + 15*d*e*x + 6*e^2*x^2) + 7*a*b^3*x^3*(15*d^2 + 24*d*e*x + 10*e^2*x^2) + b^4*x^4*(21*d^2 + 35*d
*e*x + 15*e^2*x^2)))/(105*(a + b*x))

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Maple [B]  time = 0.006, size = 189, normalized size = 1.5 \begin{align*}{\frac{x \left ( 15\,{e}^{2}{b}^{4}{x}^{6}+70\,{x}^{5}{e}^{2}a{b}^{3}+35\,{x}^{5}de{b}^{4}+126\,{x}^{4}{e}^{2}{a}^{2}{b}^{2}+168\,{x}^{4}dea{b}^{3}+21\,{x}^{4}{d}^{2}{b}^{4}+105\,{a}^{3}b{e}^{2}{x}^{3}+315\,{a}^{2}{b}^{2}de{x}^{3}+105\,a{b}^{3}{d}^{2}{x}^{3}+35\,{x}^{2}{e}^{2}{a}^{4}+280\,{x}^{2}de{a}^{3}b+210\,{x}^{2}{d}^{2}{a}^{2}{b}^{2}+105\,{a}^{4}dex+210\,{a}^{3}b{d}^{2}x+105\,{d}^{2}{a}^{4} \right ) }{105\, \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((b*x+a)*(e*x+d)^2*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/105*x*(15*b^4*e^2*x^6+70*a*b^3*e^2*x^5+35*b^4*d*e*x^5+126*a^2*b^2*e^2*x^4+168*a*b^3*d*e*x^4+21*b^4*d^2*x^4+1
05*a^3*b*e^2*x^3+315*a^2*b^2*d*e*x^3+105*a*b^3*d^2*x^3+35*a^4*e^2*x^2+280*a^3*b*d*e*x^2+210*a^2*b^2*d^2*x^2+10
5*a^4*d*e*x+210*a^3*b*d^2*x+105*a^4*d^2)*((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((b*x+a)*(e*x+d)^2*(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 = 1.51265, size = 323, normalized size = 2.58 \begin{align*} \frac{1}{7} \, b^{4} e^{2} x^{7} + a^{4} d^{2} x + \frac{1}{3} \,{\left (b^{4} d e + 2 \, a b^{3} e^{2}\right )} x^{6} + \frac{1}{5} \,{\left (b^{4} d^{2} + 8 \, a b^{3} d e + 6 \, a^{2} b^{2} e^{2}\right )} x^{5} +{\left (a b^{3} d^{2} + 3 \, a^{2} b^{2} d e + a^{3} b e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (6 \, a^{2} b^{2} d^{2} + 8 \, a^{3} b d e + a^{4} e^{2}\right )} x^{3} +{\left (2 \, a^{3} b d^{2} + a^{4} d e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.13244, size = 351, normalized size = 2.81 \begin{align*} \frac{1}{7} \, b^{4} x^{7} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, b^{4} d x^{6} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, b^{4} d^{2} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{3} \, a b^{3} x^{6} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{8}{5} \, a b^{3} d x^{5} e \mathrm{sgn}\left (b x + a\right ) + a b^{3} d^{2} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{6}{5} \, a^{2} b^{2} x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d x^{4} e \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{2} b^{2} d^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + a^{3} b x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{8}{3} \, a^{3} b d x^{3} e \mathrm{sgn}\left (b x + a\right ) + 2 \, a^{3} b d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, a^{4} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + a^{4} d x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{4} d^{2} x \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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