3.1743 \(\int (A+B x) (d+e x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=135 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (-2 a B e+A b e+b B d)}{7 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B) (b d-a e)}{6 b^3}+\frac{B e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

[Out]

((A*b - a*B)*(b*d - a*e)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3) + ((b*B*d + A*b*e - 2*a*B*e)*(a +
b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^3) + (B*e*(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 = 0.203986, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {770, 77} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (-2 a B e+A b e+b B d)}{7 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B) (b d-a e)}{6 b^3}+\frac{B e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*(b*d - a*e)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3) + ((b*B*d + A*b*e - 2*a*B*e)*(a +
b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^3) + (B*e*(a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.11434, size = 214, normalized size = 1.59 \[ \frac{x \sqrt{(a+b x)^2} \left (28 a^3 b^2 x^2 (5 A (4 d+3 e x)+3 B x (5 d+4 e x))+28 a^2 b^3 x^3 (3 A (5 d+4 e x)+2 B x (6 d+5 e x))+70 a^4 b x (A (6 d+4 e x)+B x (4 d+3 e x))+28 a^5 (3 A (2 d+e x)+B x (3 d+2 e x))+4 a b^4 x^4 (7 A (6 d+5 e x)+5 B x (7 d+6 e x))+b^5 x^5 (4 A (7 d+6 e x)+3 B x (8 d+7 e x))\right )}{168 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(28*a^5*(3*A*(2*d + e*x) + B*x*(3*d + 2*e*x)) + 28*a^3*b^2*x^2*(5*A*(4*d + 3*e*x) + 3*B*x
*(5*d + 4*e*x)) + 70*a^4*b*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x)) + 28*a^2*b^3*x^3*(3*A*(5*d + 4*e*x) + 2*B*x
*(6*d + 5*e*x)) + 4*a*b^4*x^4*(7*A*(6*d + 5*e*x) + 5*B*x*(7*d + 6*e*x)) + b^5*x^5*(4*A*(7*d + 6*e*x) + 3*B*x*(
8*d + 7*e*x))))/(168*(a + b*x))

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Maple [B]  time = 0.007, size = 284, normalized size = 2.1 \begin{align*}{\frac{x \left ( 21\,Be{b}^{5}{x}^{7}+24\,{x}^{6}A{b}^{5}e+120\,{x}^{6}Bea{b}^{4}+24\,{x}^{6}B{b}^{5}d+140\,{x}^{5}Aa{b}^{4}e+28\,{x}^{5}Ad{b}^{5}+280\,{x}^{5}Be{a}^{2}{b}^{3}+140\,{x}^{5}Ba{b}^{4}d+336\,A{a}^{2}{b}^{3}e{x}^{4}+168\,Aa{b}^{4}d{x}^{4}+336\,B{a}^{3}{b}^{2}e{x}^{4}+336\,B{a}^{2}{b}^{3}d{x}^{4}+420\,{x}^{3}A{a}^{3}{b}^{2}e+420\,{x}^{3}Ad{a}^{2}{b}^{3}+210\,{x}^{3}Be{a}^{4}b+420\,{x}^{3}B{a}^{3}{b}^{2}d+280\,{x}^{2}A{a}^{4}be+560\,{x}^{2}Ad{a}^{3}{b}^{2}+56\,{x}^{2}Be{a}^{5}+280\,{x}^{2}B{a}^{4}bd+84\,xA{a}^{5}e+420\,xAd{a}^{4}b+84\,xB{a}^{5}d+168\,Ad{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((B*x+A)*(e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

1/168*x*(21*B*b^5*e*x^7+24*A*b^5*e*x^6+120*B*a*b^4*e*x^6+24*B*b^5*d*x^6+140*A*a*b^4*e*x^5+28*A*b^5*d*x^5+280*B
*a^2*b^3*e*x^5+140*B*a*b^4*d*x^5+336*A*a^2*b^3*e*x^4+168*A*a*b^4*d*x^4+336*B*a^3*b^2*e*x^4+336*B*a^2*b^3*d*x^4
+420*A*a^3*b^2*e*x^3+420*A*a^2*b^3*d*x^3+210*B*a^4*b*e*x^3+420*B*a^3*b^2*d*x^3+280*A*a^4*b*e*x^2+560*A*a^3*b^2
*d*x^2+56*B*a^5*e*x^2+280*B*a^4*b*d*x^2+84*A*a^5*e*x+420*A*a^4*b*d*x+84*B*a^5*d*x+168*A*a^5*d)*((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((B*x+A)*(e*x+d)*(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 = 1.54318, size = 509, normalized size = 3.77 \begin{align*} \frac{1}{8} \, B b^{5} e x^{8} + A a^{5} d x + \frac{1}{7} \,{\left (B b^{5} d +{\left (5 \, B a b^{4} + A b^{5}\right )} e\right )} x^{7} + \frac{1}{6} \,{\left ({\left (5 \, B a b^{4} + A b^{5}\right )} d + 5 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} e\right )} x^{6} +{\left ({\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} d + 2 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} e\right )} x^{5} + \frac{5}{4} \,{\left (2 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} d +{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} e\right )} x^{4} + \frac{1}{3} \,{\left (5 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} d +{\left (B a^{5} + 5 \, A a^{4} b\right )} e\right )} x^{3} + \frac{1}{2} \,{\left (A a^{5} e +{\left (B a^{5} + 5 \, A a^{4} b\right )} d\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Giac [B]  time = 1.14919, size = 574, normalized size = 4.25 \begin{align*} \frac{1}{8} \, B b^{5} x^{8} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{7} \, B b^{5} d x^{7} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{7} \, B a b^{4} x^{7} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{7} \, A b^{5} x^{7} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{6} \, B a b^{4} d x^{6} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{6} \, A b^{5} d x^{6} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, B a^{2} b^{3} x^{6} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{6} \, A a b^{4} x^{6} e \mathrm{sgn}\left (b x + a\right ) + 2 \, B a^{2} b^{3} d x^{5} \mathrm{sgn}\left (b x + a\right ) + A a b^{4} d x^{5} \mathrm{sgn}\left (b x + a\right ) + 2 \, B a^{3} b^{2} x^{5} e \mathrm{sgn}\left (b x + a\right ) + 2 \, A a^{2} b^{3} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, B a^{3} b^{2} d x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, A a^{2} b^{3} d x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{4} \, B a^{4} b x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, A a^{3} b^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, B a^{4} b d x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, A a^{3} b^{2} d x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B a^{5} x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, A a^{4} b x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a^{5} d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, A a^{4} b d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A a^{5} x^{2} e \mathrm{sgn}\left (b x + a\right ) + A a^{5} d 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)*(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

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