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]

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

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Rubi [A]  time = 0.0190633, antiderivative size = 48, normalized size of antiderivative = 1., 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 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{\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*}

Mathematica [B]  time = 0.0424282, size = 109, normalized size = 2.27 $-\frac{\sqrt{(a+b x)^2} \left (a^2 b e^2 (d+4 e x)+a^3 e^3+a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )+b^3 \left (4 d^2 e x+d^3+6 d e^2 x^2+4 e^3 x^3\right )\right )}{4 e^4 (a+b x) (d+e x)^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(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)))/(4*e^4*(a + b*x)*(d + e*x)^4)

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Maple [B]  time = 0.157, size = 128, normalized size = 2.7 \begin{align*} -{\frac{4\,{x}^{3}{b}^{3}{e}^{3}+6\,{x}^{2}a{b}^{2}{e}^{3}+6\,{x}^{2}{b}^{3}d{e}^{2}+4\,x{a}^{2}b{e}^{3}+4\,xa{b}^{2}d{e}^{2}+4\,x{b}^{3}{d}^{2}e+{a}^{3}{e}^{3}+d{e}^{2}{a}^{2}b+a{b}^{2}{d}^{2}e+{b}^{3}{d}^{3}}{4\, \left ( ex+d \right ) ^{4}{e}^{4} \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^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., 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^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.53743, size = 284, normalized size = 5.92 \begin{align*} -\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 )}} \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 1.18077, size = 224, normalized size = 4.67 \begin{align*} -\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}} \end{align*}

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