3.1983 \(\int \frac {a d e+(c d^2+a e^2) x+c d e x^2}{(d+e x)^{11/2}} \, dx\)

Optimal. Leaf size=43 \[ -\frac {2 \left (a-\frac {c d^2}{e^2}\right )}{7 (d+e x)^{7/2}}-\frac {2 c d}{5 e^2 (d+e x)^{5/2}} \]

[Out]

-2/7*(a-c*d^2/e^2)/(e*x+d)^(7/2)-2/5*c*d/e^2/(e*x+d)^(5/2)

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {24, 43} \[ -\frac {2 \left (a-\frac {c d^2}{e^2}\right )}{7 (d+e x)^{7/2}}-\frac {2 c d}{5 e^2 (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)/(d + e*x)^(11/2),x]

[Out]

(-2*(a - (c*d^2)/e^2))/(7*(d + e*x)^(7/2)) - (2*c*d)/(5*e^2*(d + e*x)^(5/2))

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^{11/2}} \, dx &=\frac {\int \frac {a e^3+c d e^2 x}{(d+e x)^{9/2}} \, dx}{e^2}\\ &=\frac {\int \left (\frac {-c d^2 e+a e^3}{(d+e x)^{9/2}}+\frac {c d e}{(d+e x)^{7/2}}\right ) \, dx}{e^2}\\ &=-\frac {2 \left (a-\frac {c d^2}{e^2}\right )}{7 (d+e x)^{7/2}}-\frac {2 c d}{5 e^2 (d+e x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.79 \[ -\frac {2 \left (5 a e^2+c d (2 d+7 e x)\right )}{35 e^2 (d+e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)/(d + e*x)^(11/2),x]

[Out]

(-2*(5*a*e^2 + c*d*(2*d + 7*e*x)))/(35*e^2*(d + e*x)^(7/2))

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fricas [B]  time = 0.92, size = 74, normalized size = 1.72 \[ -\frac {2 \, {\left (7 \, c d e x + 2 \, c d^{2} + 5 \, a e^{2}\right )} \sqrt {e x + d}}{35 \, {\left (e^{6} x^{4} + 4 \, d e^{5} x^{3} + 6 \, d^{2} e^{4} x^{2} + 4 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)/(e*x+d)^(11/2),x, algorithm="fricas")

[Out]

-2/35*(7*c*d*e*x + 2*c*d^2 + 5*a*e^2)*sqrt(e*x + d)/(e^6*x^4 + 4*d*e^5*x^3 + 6*d^2*e^4*x^2 + 4*d^3*e^3*x + d^4
*e^2)

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giac [A]  time = 0.24, size = 48, normalized size = 1.12 \[ -\frac {2 \, {\left (7 \, {\left (x e + d\right )}^{2} c d - 5 \, {\left (x e + d\right )} c d^{2} + 5 \, {\left (x e + d\right )} a e^{2}\right )} e^{\left (-2\right )}}{35 \, {\left (x e + d\right )}^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)/(e*x+d)^(11/2),x, algorithm="giac")

[Out]

-2/35*(7*(x*e + d)^2*c*d - 5*(x*e + d)*c*d^2 + 5*(x*e + d)*a*e^2)*e^(-2)/(x*e + d)^(9/2)

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maple [A]  time = 0.05, size = 32, normalized size = 0.74 \[ -\frac {2 \left (7 c d e x +5 a \,e^{2}+2 c \,d^{2}\right )}{35 \left (e x +d \right )^{\frac {7}{2}} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)/(e*x+d)^(11/2),x)

[Out]

-2/35/(e*x+d)^(7/2)*(7*c*d*e*x+5*a*e^2+2*c*d^2)/e^2

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maxima [A]  time = 1.18, size = 34, normalized size = 0.79 \[ -\frac {2 \, {\left (7 \, {\left (e x + d\right )} c d - 5 \, c d^{2} + 5 \, a e^{2}\right )}}{35 \, {\left (e x + d\right )}^{\frac {7}{2}} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)/(e*x+d)^(11/2),x, algorithm="maxima")

[Out]

-2/35*(7*(e*x + d)*c*d - 5*c*d^2 + 5*a*e^2)/((e*x + d)^(7/2)*e^2)

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mupad [B]  time = 0.59, size = 31, normalized size = 0.72 \[ -\frac {4\,c\,d^2+14\,c\,x\,d\,e+10\,a\,e^2}{35\,e^2\,{\left (d+e\,x\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)/(d + e*x)^(11/2),x)

[Out]

-(10*a*e^2 + 4*c*d^2 + 14*c*d*e*x)/(35*e^2*(d + e*x)^(7/2))

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sympy [A]  time = 13.21, size = 248, normalized size = 5.77 \[ \begin {cases} - \frac {10 a e^{2}}{35 d^{3} e^{2} \sqrt {d + e x} + 105 d^{2} e^{3} x \sqrt {d + e x} + 105 d e^{4} x^{2} \sqrt {d + e x} + 35 e^{5} x^{3} \sqrt {d + e x}} - \frac {4 c d^{2}}{35 d^{3} e^{2} \sqrt {d + e x} + 105 d^{2} e^{3} x \sqrt {d + e x} + 105 d e^{4} x^{2} \sqrt {d + e x} + 35 e^{5} x^{3} \sqrt {d + e x}} - \frac {14 c d e x}{35 d^{3} e^{2} \sqrt {d + e x} + 105 d^{2} e^{3} x \sqrt {d + e x} + 105 d e^{4} x^{2} \sqrt {d + e x} + 35 e^{5} x^{3} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {c x^{2}}{2 d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)/(e*x+d)**(11/2),x)

[Out]

Piecewise((-10*a*e**2/(35*d**3*e**2*sqrt(d + e*x) + 105*d**2*e**3*x*sqrt(d + e*x) + 105*d*e**4*x**2*sqrt(d + e
*x) + 35*e**5*x**3*sqrt(d + e*x)) - 4*c*d**2/(35*d**3*e**2*sqrt(d + e*x) + 105*d**2*e**3*x*sqrt(d + e*x) + 105
*d*e**4*x**2*sqrt(d + e*x) + 35*e**5*x**3*sqrt(d + e*x)) - 14*c*d*e*x/(35*d**3*e**2*sqrt(d + e*x) + 105*d**2*e
**3*x*sqrt(d + e*x) + 105*d*e**4*x**2*sqrt(d + e*x) + 35*e**5*x**3*sqrt(d + e*x)), Ne(e, 0)), (c*x**2/(2*d**(7
/2)), True))

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