3.1731 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=106 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 (d+e x)^4 (b d-a e)^2}+\frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (B d-A e)}{5 (d+e x)^5 (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.0594985, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {769, 646, 37} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B)}{4 (d+e x)^4 (b d-a e)^2}+\frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (B d-A e)}{5 (d+e x)^5 (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^6,x]

[Out]

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

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.117581, size = 229, normalized size = 2.16 \[ -\frac{\sqrt{(a+b x)^2} \left (a^2 b e^2 \left (3 A e (d+5 e x)+2 B \left (d^2+5 d e x+10 e^2 x^2\right )\right )+a^3 e^3 (4 A e+B (d+5 e x))+a b^2 e \left (2 A e \left (d^2+5 d e x+10 e^2 x^2\right )+3 B \left (5 d^2 e x+d^3+10 d e^2 x^2+10 e^3 x^3\right )\right )+b^3 \left (A e \left (5 d^2 e x+d^3+10 d e^2 x^2+10 e^3 x^3\right )+4 B \left (10 d^2 e^2 x^2+5 d^3 e x+d^4+10 d e^3 x^3+5 e^4 x^4\right )\right )\right )}{20 e^5 (a+b x) (d+e x)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^6,x]

[Out]

-(Sqrt[(a + b*x)^2]*(a^3*e^3*(4*A*e + B*(d + 5*e*x)) + a^2*b*e^2*(3*A*e*(d + 5*e*x) + 2*B*(d^2 + 5*d*e*x + 10*
e^2*x^2)) + a*b^2*e*(2*A*e*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*B*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3)) +
b^3*(A*e*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + 4*B*(d^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3
+ 5*e^4*x^4))))/(20*e^5*(a + b*x)*(d + e*x)^5)

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Maple [B]  time = 0.007, size = 315, normalized size = 3. \begin{align*} -{\frac{20\,B{x}^{4}{b}^{3}{e}^{4}+10\,A{x}^{3}{b}^{3}{e}^{4}+30\,B{x}^{3}a{b}^{2}{e}^{4}+40\,B{x}^{3}{b}^{3}d{e}^{3}+20\,A{x}^{2}a{b}^{2}{e}^{4}+10\,A{x}^{2}{b}^{3}d{e}^{3}+20\,B{x}^{2}{a}^{2}b{e}^{4}+30\,B{x}^{2}a{b}^{2}d{e}^{3}+40\,B{x}^{2}{b}^{3}{d}^{2}{e}^{2}+15\,Ax{a}^{2}b{e}^{4}+10\,Axa{b}^{2}d{e}^{3}+5\,Ax{b}^{3}{d}^{2}{e}^{2}+5\,Bx{a}^{3}{e}^{4}+10\,Bx{a}^{2}bd{e}^{3}+15\,Bxa{b}^{2}{d}^{2}{e}^{2}+20\,Bx{b}^{3}{d}^{3}e+4\,A{a}^{3}{e}^{4}+3\,Ad{e}^{3}{a}^{2}b+2\,Aa{b}^{2}{d}^{2}{e}^{2}+A{b}^{3}{d}^{3}e+Bd{e}^{3}{a}^{3}+2\,B{a}^{2}b{d}^{2}{e}^{2}+3\,Ba{b}^{2}{d}^{3}e+4\,B{b}^{3}{d}^{4}}{20\, \left ( ex+d \right ) ^{5}{e}^{5} \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)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x)

[Out]

-1/20*(20*B*b^3*e^4*x^4+10*A*b^3*e^4*x^3+30*B*a*b^2*e^4*x^3+40*B*b^3*d*e^3*x^3+20*A*a*b^2*e^4*x^2+10*A*b^3*d*e
^3*x^2+20*B*a^2*b*e^4*x^2+30*B*a*b^2*d*e^3*x^2+40*B*b^3*d^2*e^2*x^2+15*A*a^2*b*e^4*x+10*A*a*b^2*d*e^3*x+5*A*b^
3*d^2*e^2*x+5*B*a^3*e^4*x+10*B*a^2*b*d*e^3*x+15*B*a*b^2*d^2*e^2*x+20*B*b^3*d^3*e*x+4*A*a^3*e^4+3*A*a^2*b*d*e^3
+2*A*a*b^2*d^2*e^2+A*b^3*d^3*e+B*a^3*d*e^3+2*B*a^2*b*d^2*e^2+3*B*a*b^2*d^3*e+4*B*b^3*d^4)*((b*x+a)^2)^(3/2)/(e
*x+d)^5/e^5/(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)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.88718, size = 630, normalized size = 5.94 \begin{align*} -\frac{20 \, B b^{3} e^{4} x^{4} + 4 \, B b^{3} d^{4} + 4 \, A a^{3} e^{4} +{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 2 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 10 \,{\left (4 \, B b^{3} d e^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 10 \,{\left (4 \, B b^{3} d^{2} e^{2} +{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 2 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 5 \,{\left (4 \, B b^{3} d^{3} e +{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 2 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} +{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{20 \,{\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/20*(20*B*b^3*e^4*x^4 + 4*B*b^3*d^4 + 4*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 2*(B*a^2*b + A*a*b^2)*d^2*e^
2 + (B*a^3 + 3*A*a^2*b)*d*e^3 + 10*(4*B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 10*(4*B*b^3*d^2*e^2 + (3*B*
a*b^2 + A*b^3)*d*e^3 + 2*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 5*(4*B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 2*(B*
a^2*b + A*a*b^2)*d*e^3 + (B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^10*x^5 + 5*d*e^9*x^4 + 10*d^2*e^8*x^3 + 10*d^3*e^7*x^2
+ 5*d^4*e^6*x + d^5*e^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.13934, size = 574, normalized size = 5.42 \begin{align*} -\frac{{\left (20 \, B b^{3} x^{4} e^{4} \mathrm{sgn}\left (b x + a\right ) + 40 \, B b^{3} d x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 40 \, B b^{3} d^{2} x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 20 \, B b^{3} d^{3} x e \mathrm{sgn}\left (b x + a\right ) + 4 \, B b^{3} d^{4} \mathrm{sgn}\left (b x + a\right ) + 30 \, B a b^{2} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 10 \, A b^{3} x^{3} e^{4} \mathrm{sgn}\left (b x + a\right ) + 30 \, B a b^{2} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 10 \, A b^{3} d x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) + 15 \, B a b^{2} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, A b^{3} d^{2} x e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, B a b^{2} d^{3} e \mathrm{sgn}\left (b x + a\right ) + A b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 20 \, B a^{2} b x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) + 20 \, A a b^{2} x^{2} e^{4} \mathrm{sgn}\left (b x + a\right ) + 10 \, B a^{2} b d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 10 \, A a b^{2} d x e^{3} \mathrm{sgn}\left (b x + a\right ) + 2 \, B a^{2} b d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \, A a b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, B a^{3} x e^{4} \mathrm{sgn}\left (b x + a\right ) + 15 \, A a^{2} b x e^{4} \mathrm{sgn}\left (b x + a\right ) + B a^{3} d e^{3} \mathrm{sgn}\left (b x + a\right ) + 3 \, A a^{2} b d e^{3} \mathrm{sgn}\left (b x + a\right ) + 4 \, A a^{3} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{20 \,{\left (x e + d\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/20*(20*B*b^3*x^4*e^4*sgn(b*x + a) + 40*B*b^3*d*x^3*e^3*sgn(b*x + a) + 40*B*b^3*d^2*x^2*e^2*sgn(b*x + a) + 2
0*B*b^3*d^3*x*e*sgn(b*x + a) + 4*B*b^3*d^4*sgn(b*x + a) + 30*B*a*b^2*x^3*e^4*sgn(b*x + a) + 10*A*b^3*x^3*e^4*s
gn(b*x + a) + 30*B*a*b^2*d*x^2*e^3*sgn(b*x + a) + 10*A*b^3*d*x^2*e^3*sgn(b*x + a) + 15*B*a*b^2*d^2*x*e^2*sgn(b
*x + a) + 5*A*b^3*d^2*x*e^2*sgn(b*x + a) + 3*B*a*b^2*d^3*e*sgn(b*x + a) + A*b^3*d^3*e*sgn(b*x + a) + 20*B*a^2*
b*x^2*e^4*sgn(b*x + a) + 20*A*a*b^2*x^2*e^4*sgn(b*x + a) + 10*B*a^2*b*d*x*e^3*sgn(b*x + a) + 10*A*a*b^2*d*x*e^
3*sgn(b*x + a) + 2*B*a^2*b*d^2*e^2*sgn(b*x + a) + 2*A*a*b^2*d^2*e^2*sgn(b*x + a) + 5*B*a^3*x*e^4*sgn(b*x + a)
+ 15*A*a^2*b*x*e^4*sgn(b*x + a) + B*a^3*d*e^3*sgn(b*x + a) + 3*A*a^2*b*d*e^3*sgn(b*x + a) + 4*A*a^3*e^4*sgn(b*
x + a))*e^(-5)/(x*e + d)^5