3.1564 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=48 \[ \frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (d+e x)^4 (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 37} \[ \frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (d+e x)^4 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^5,x]

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [B]  time = 0.04, size = 109, normalized size = 2.27 \[ -\frac {\sqrt {(a+b x)^2} \left (a^3 e^3+a^2 b e^2 (d+4 e x)+a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )+b^3 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )}{4 e^4 (a+b x) (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^5,x]

[Out]

-1/4*(Sqrt[(a + b*x)^2]*(a^3*e^3 + a^2*b*e^2*(d + 4*e*x) + a*b^2*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + b^3*(d^3 + 4*
d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^4)

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fricas [B]  time = 0.91, size = 143, normalized size = 2.98 \[ -\frac {4 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + a b^{2} d^{2} e + a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 4 \, {\left (b^{3} d^{2} e + a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{4 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.19, size = 166, normalized size = 3.46 \[ -\frac {{\left (4 \, b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, a b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{4 \, {\left (x e + d\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 128, normalized size = 2.67 \[ -\frac {\left (4 b^{3} e^{3} x^{3}+6 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+4 a^{2} b \,e^{3} x +4 a \,b^{2} d \,e^{2} x +4 b^{3} d^{2} e x +a^{3} e^{3}+a^{2} b d \,e^{2}+a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{4 \left (e x +d \right )^{4} \left (b x +a \right )^{3} e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x)

[Out]

-1/4*(4*b^3*e^3*x^3+6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+4*a^2*b*e^3*x+4*a*b^2*d*e^2*x+4*b^3*d^2*e*x+a^3*e^3+a^2*b*
d*e^2+a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^4/e^4/(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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,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 = 0.63, size = 284, normalized size = 5.92 \[ \frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{2\,e^4}+\frac {b^3\,d}{2\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{3\,e^4}+\frac {d\,\left (\frac {b^3\,d}{3\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{3\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}-\frac {\left (\frac {a^3}{4\,e}-\frac {d\,\left (\frac {3\,a^2\,b}{4\,e}-\frac {d\,\left (\frac {3\,a\,b^2}{4\,e}-\frac {b^3\,d}{4\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{e^4\,\left (a+b\,x\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/(d + e*x)^5,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**5,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**5, x)

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