3.1981 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=41 \[ \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 (d+e x)^5 (b d-a e)} \]

[Out]

1/5*(b^2*x^2+2*a*b*x+a^2)^(5/2)/(-a*e+b*d)/(e*x+d)^5

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {767} \[ \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 (d+e x)^5 (b d-a e)} \]

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

Rule 767

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

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 {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 (b d-a e) (d+e x)^5}\\ \end {align*}

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Mathematica [B]  time = 0.06, size = 158, normalized size = 3.85 \[ -\frac {\sqrt {(a+b x)^2} \left (a^4 e^4+a^3 b e^3 (d+5 e x)+a^2 b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+a b^3 e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+b^4 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )\right )}{5 e^5 (a+b x) (d+e x)^5} \]

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]

-1/5*(Sqrt[(a + b*x)^2]*(a^4*e^4 + a^3*b*e^3*(d + 5*e*x) + a^2*b^2*e^2*(d^2 + 5*d*e*x + 10*e^2*x^2) + a*b^3*e*
(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + b^4*(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)))/(e^5*(a + b*x)*(d + e*x)^5)

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fricas [B]  time = 1.09, size = 215, normalized size = 5.24 \[ -\frac {5 \, b^{4} e^{4} x^{4} + b^{4} d^{4} + a b^{3} d^{3} e + a^{2} b^{2} d^{2} e^{2} + a^{3} b d e^{3} + a^{4} e^{4} + 10 \, {\left (b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{3} + 10 \, {\left (b^{4} d^{2} e^{2} + a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 5 \, {\left (b^{4} d^{3} e + a b^{3} d^{2} e^{2} + a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x}{5 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \]

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]

-1/5*(5*b^4*e^4*x^4 + b^4*d^4 + a*b^3*d^3*e + a^2*b^2*d^2*e^2 + a^3*b*d*e^3 + a^4*e^4 + 10*(b^4*d*e^3 + a*b^3*
e^4)*x^3 + 10*(b^4*d^2*e^2 + a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 5*(b^4*d^3*e + a*b^3*d^2*e^2 + a^2*b^2*d*e^3 + a
^3*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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giac [B]  time = 0.21, size = 260, normalized size = 6.34 \[ -\frac {{\left (5 \, b^{4} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, b^{4} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, b^{4} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, b^{4} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{2} b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{5 \, {\left (x e + d\right )}^{5}} \]

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]

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

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maple [B]  time = 0.06, size = 197, normalized size = 4.80 \[ -\frac {\left (5 b^{4} e^{4} x^{4}+10 a \,b^{3} e^{4} x^{3}+10 b^{4} d \,e^{3} x^{3}+10 a^{2} b^{2} e^{4} x^{2}+10 a \,b^{3} d \,e^{3} x^{2}+10 b^{4} d^{2} e^{2} x^{2}+5 a^{3} b \,e^{4} x +5 a^{2} b^{2} d \,e^{3} x +5 a \,b^{3} d^{2} e^{2} x +5 b^{4} d^{3} e x +a^{4} e^{4}+a^{3} b d \,e^{3}+a^{2} b^{2} d^{2} e^{2}+a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 \left (e x +d \right )^{5} \left (b x +a \right )^{3} e^{5}} \]

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]

-1/5*(5*b^4*e^4*x^4+10*a*b^3*e^4*x^3+10*b^4*d*e^3*x^3+10*a^2*b^2*e^4*x^2+10*a*b^3*d*e^3*x^2+10*b^4*d^2*e^2*x^2
+5*a^3*b*e^4*x+5*a^2*b^2*d*e^3*x+5*a*b^3*d^2*e^2*x+5*b^4*d^3*e*x+a^4*e^4+a^3*b*d*e^3+a^2*b^2*d^2*e^2+a*b^3*d^3
*e+b^4*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.15, size = 449, normalized size = 10.95 \[ \frac {\left (\frac {-4\,a^3\,b\,e^3+6\,a^2\,b^2\,d\,e^2-4\,a\,b^3\,d^2\,e+b^4\,d^3}{4\,e^5}+\frac {d\,\left (\frac {d\,\left (\frac {b^4\,d}{4\,e^3}-\frac {b^3\,\left (4\,a\,e-b\,d\right )}{4\,e^3}\right )}{e}+\frac {b^2\,\left (6\,a^2\,e^2-4\,a\,b\,d\,e+b^2\,d^2\right )}{4\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {a^4}{5\,e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {4\,a\,b^3}{5\,e}-\frac {b^4\,d}{5\,e^2}\right )}{e}-\frac {6\,a^2\,b^2}{5\,e}\right )}{e}+\frac {4\,a^3\,b}{5\,e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {6\,a^2\,b^2\,e^2-8\,a\,b^3\,d\,e+3\,b^4\,d^2}{3\,e^5}+\frac {d\,\left (\frac {b^4\,d}{3\,e^4}-\frac {2\,b^3\,\left (2\,a\,e-b\,d\right )}{3\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}+\frac {\left (\frac {3\,b^4\,d-4\,a\,b^3\,e}{2\,e^5}+\frac {b^4\,d}{2\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2}-\frac {b^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{e^5\,\left (a+b\,x\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^6,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{6}}\, dx \]

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]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**6, x)

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