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

Optimal. Leaf size=54 \[ \frac {\left (c d^2-a e^2\right ) (a e+c d x)^3}{3 c^2 d^2}+\frac {e (a e+c d x)^4}{4 c^2 d^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 54, 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 = {626, 43} \begin {gather*} \frac {\left (c d^2-a e^2\right ) (a e+c d x)^3}{3 c^2 d^2}+\frac {e (a e+c d x)^4}{4 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 626

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 54, normalized size = 1.00 \begin {gather*} \frac {1}{12} x \left (6 a^2 e^2 (2 d+e x)+4 a c d e x (3 d+2 e x)+c^2 d^2 x^2 (4 d+3 e x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(6*a^2*e^2*(2*d + e*x) + 4*a*c*d*e*x*(3*d + 2*e*x) + c^2*d^2*x^2*(4*d + 3*e*x)))/12

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IntegrateAlgebraic [A]  time = 0.12, size = 100, normalized size = 1.85 \begin {gather*} \frac {(d+e x)^4 \left (\frac {6 a^2 e^4}{(d+e x)^2}-\frac {12 a c d^2 e^2}{(d+e x)^2}+\frac {8 a c d e^2}{d+e x}+\frac {6 c^2 d^4}{(d+e x)^2}-\frac {8 c^2 d^3}{d+e x}+3 c^2 d^2\right )}{12 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^4*(3*c^2*d^2 + (6*c^2*d^4)/(d + e*x)^2 - (12*a*c*d^2*e^2)/(d + e*x)^2 + (6*a^2*e^4)/(d + e*x)^2 - (
8*c^2*d^3)/(d + e*x) + (8*a*c*d*e^2)/(d + e*x)))/(12*e^3)

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fricas [A]  time = 0.38, size = 64, normalized size = 1.19 \begin {gather*} \frac {1}{4} \, c^{2} d^{2} e x^{4} + a^{2} d e^{2} x + \frac {1}{3} \, {\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a c d^{2} e + a^{2} e^{3}\right )} x^{2} \end {gather*}

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)^2/(e*x+d),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 72, normalized size = 1.33 \begin {gather*} \frac {1}{12} \, {\left (3 \, c^{2} d^{2} x^{4} e^{5} + 4 \, c^{2} d^{3} x^{3} e^{4} + 8 \, a c d x^{3} e^{6} + 12 \, a c d^{2} x^{2} e^{5} + 6 \, a^{2} x^{2} e^{7} + 12 \, a^{2} d x e^{6}\right )} e^{\left (-4\right )} \end {gather*}

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)^2/(e*x+d),x, algorithm="giac")

[Out]

1/12*(3*c^2*d^2*x^4*e^5 + 4*c^2*d^3*x^3*e^4 + 8*a*c*d*x^3*e^6 + 12*a*c*d^2*x^2*e^5 + 6*a^2*x^2*e^7 + 12*a^2*d*
x*e^6)*e^(-4)

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maple [A]  time = 0.05, size = 77, normalized size = 1.43 \begin {gather*} \frac {c^{2} d^{2} e \,x^{4}}{4}+a^{2} d \,e^{2} x +\frac {\left (a c d \,e^{2}+\left (a \,e^{2}+c \,d^{2}\right ) c d \right ) x^{3}}{3}+\frac {\left (a c \,d^{2} e +\left (a \,e^{2}+c \,d^{2}\right ) a e \right ) x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 64, normalized size = 1.19 \begin {gather*} \frac {1}{4} \, c^{2} d^{2} e x^{4} + a^{2} d e^{2} x + \frac {1}{3} \, {\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a c d^{2} e + a^{2} e^{3}\right )} x^{2} \end {gather*}

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)^2/(e*x+d),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.58, size = 63, normalized size = 1.17 \begin {gather*} x^2\,\left (\frac {a^2\,e^3}{2}+c\,a\,d^2\,e\right )+x^3\,\left (\frac {c^2\,d^3}{3}+\frac {2\,a\,c\,d\,e^2}{3}\right )+\frac {c^2\,d^2\,e\,x^4}{4}+a^2\,d\,e^2\,x \end {gather*}

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)^2/(d + e*x),x)

[Out]

x^2*((a^2*e^3)/2 + a*c*d^2*e) + x^3*((c^2*d^3)/3 + (2*a*c*d*e^2)/3) + (c^2*d^2*e*x^4)/4 + a^2*d*e^2*x

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sympy [A]  time = 0.13, size = 66, normalized size = 1.22 \begin {gather*} a^{2} d e^{2} x + \frac {c^{2} d^{2} e x^{4}}{4} + x^{3} \left (\frac {2 a c d e^{2}}{3} + \frac {c^{2} d^{3}}{3}\right ) + x^{2} \left (\frac {a^{2} e^{3}}{2} + a c d^{2} e\right ) \end {gather*}

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)**2/(e*x+d),x)

[Out]

a**2*d*e**2*x + c**2*d**2*e*x**4/4 + x**3*(2*a*c*d*e**2/3 + c**2*d**3/3) + x**2*(a**2*e**3/2 + a*c*d**2*e)

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