### 3.1079 $$\int \frac{(d+e x)^6}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=39 $\frac{(d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e}$

[Out]

((d + e*x)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])/(2*c^3*e)

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Rubi [A]  time = 0.0225454, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.062, Rules used = {642, 609} $\frac{(d+e x) \sqrt{c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^6/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

((d + e*x)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])/(2*c^3*e)

Rule 642

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

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

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

Mathematica [A]  time = 0.0032829, size = 33, normalized size = 0.85 $\frac{x (d+e x) (2 d+e x)}{2 c^2 \sqrt{c (d+e x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^6/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

(x*(d + e*x)*(2*d + e*x))/(2*c^2*Sqrt[c*(d + e*x)^2])

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Maple [A]  time = 0.04, size = 40, normalized size = 1. \begin{align*}{\frac{x \left ( ex+2\,d \right ) \left ( ex+d \right ) ^{5}}{2} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x)

[Out]

1/2*x*(e*x+2*d)*(e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)

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Maxima [B]  time = 1.27596, size = 313, normalized size = 8.03 \begin{align*} \frac{e^{4} x^{5}}{2 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c} + \frac{5 \, d e^{3} x^{4}}{2 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c} - \frac{25 \, c^{2} d^{6} e^{4}}{4 \, \left (c e^{2}\right )^{\frac{9}{2}}{\left (x + \frac{d}{e}\right )}^{4}} - \frac{10 \, d^{3} e x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c} + \frac{50 \, c d^{5} e^{3}}{3 \, \left (c e^{2}\right )^{\frac{7}{2}}{\left (x + \frac{d}{e}\right )}^{3}} - \frac{26 \, d^{5}}{3 \,{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac{3}{2}} c e} - \frac{25 \, d^{4} e^{2}}{2 \, \left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{2}} + \frac{25 \, d^{6}}{4 \, \left (c e^{2}\right )^{\frac{5}{2}}{\left (x + \frac{d}{e}\right )}^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

1/2*e^4*x^5/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 5/2*d*e^3*x^4/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c
) - 25/4*c^2*d^6*e^4/((c*e^2)^(9/2)*(x + d/e)^4) - 10*d^3*e*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 50
/3*c*d^5*e^3/((c*e^2)^(7/2)*(x + d/e)^3) - 26/3*d^5/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e) - 25/2*d^4*e^2
/((c*e^2)^(5/2)*(x + d/e)^2) + 25/4*d^6/((c*e^2)^(5/2)*(x + d/e)^4)

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Fricas [A]  time = 2.30854, size = 101, normalized size = 2.59 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}{\left (e x^{2} + 2 \, d x\right )}}{2 \,{\left (c^{3} e x + c^{3} d\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

1/2*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*(e*x^2 + 2*d*x)/(c^3*e*x + c^3*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{6}}{\left (c \left (d + e x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**6/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

Integral((d + e*x)**6/(c*(d + e*x)**2)**(5/2), x)

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Giac [B]  time = 1.48774, size = 146, normalized size = 3.74 \begin{align*} -\frac{\frac{9 \, d^{5} e^{\left (-1\right )}}{c} - 4 \, C_{0} d^{3} e^{\left (-3\right )} -{\left (12 \, C_{0} d^{2} e^{\left (-2\right )} - \frac{25 \, d^{4}}{c} -{\left (\frac{20 \, d^{3} e}{c} - 12 \, C_{0} d e^{\left (-1\right )} -{\left (x{\left (\frac{x e^{4}}{c} + \frac{5 \, d e^{3}}{c}\right )} + 4 \, C_{0}\right )} x\right )} x\right )} x}{2 \,{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

-1/2*(9*d^5*e^(-1)/c - 4*C_0*d^3*e^(-3) - (12*C_0*d^2*e^(-2) - 25*d^4/c - (20*d^3*e/c - 12*C_0*d*e^(-1) - (x*(
x*e^4/c + 5*d*e^3/c) + 4*C_0)*x)*x)*x)/(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)