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

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {(d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e} \end {gather*}

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 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)]

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]

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*}

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Mathematica [A]  time = 0.00, size = 33, normalized size = 0.85 \begin {gather*} \frac {x (d+e x) (2 d+e x)}{2 c^2 \sqrt {c (d+e x)^2}} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.47, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 48, normalized size = 1.23 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.66, size = 108, normalized size = 2.77 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.05, size = 40, normalized size = 1.03 \begin {gather*} \frac {\left (e x +2 d \right ) \left (e x +d \right )^{5} x}{2 \left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}} \end {gather*}

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 = 1.73, size = 175, normalized size = 4.49 \begin {gather*} \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 {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 {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}}{2 \, c^{\frac {5}{2}} e^{3} {\left (x + \frac {d}{e}\right )}^{2}} + \frac {50 \, d^{5}}{3 \, c^{\frac {5}{2}} e^{4} {\left (x + \frac {d}{e}\right )}^{3}} \end {gather*}

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
) - 10*d^3*e*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) - 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/(c^(5/2)*e^3*(x + d/e)^2) + 50/3*d^5/(c^(5/2)*e^4*(x + d/e)^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^6}{{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{6}}{\left (c \left (d + e x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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