### 3.1994 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^{3/2}} \, dx$$

Optimal. Leaf size=119 $-\frac{2 c^2 d^2 (d+e x)^{9/2} \left (c d^2-a e^2\right )}{3 e^4}+\frac{6 c d (d+e x)^{7/2} \left (c d^2-a e^2\right )^2}{7 e^4}-\frac{2 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3}{5 e^4}+\frac{2 c^3 d^3 (d+e x)^{11/2}}{11 e^4}$

[Out]

(-2*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2))/(5*e^4) + (6*c*d*(c*d^2 - a*e^2)^2*(d + e*x)^(7/2))/(7*e^4) - (2*c^2*d^
2*(c*d^2 - a*e^2)*(d + e*x)^(9/2))/(3*e^4) + (2*c^3*d^3*(d + e*x)^(11/2))/(11*e^4)

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Rubi [A]  time = 0.0576717, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.054, Rules used = {626, 43} $-\frac{2 c^2 d^2 (d+e x)^{9/2} \left (c d^2-a e^2\right )}{3 e^4}+\frac{6 c d (d+e x)^{7/2} \left (c d^2-a e^2\right )^2}{7 e^4}-\frac{2 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3}{5 e^4}+\frac{2 c^3 d^3 (d+e x)^{11/2}}{11 e^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^(3/2),x]

[Out]

(-2*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2))/(5*e^4) + (6*c*d*(c*d^2 - a*e^2)^2*(d + e*x)^(7/2))/(7*e^4) - (2*c^2*d^
2*(c*d^2 - a*e^2)*(d + e*x)^(9/2))/(3*e^4) + (2*c^3*d^3*(d + e*x)^(11/2))/(11*e^4)

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]

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{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{3/2}} \, dx &=\int (a e+c d x)^3 (d+e x)^{3/2} \, dx\\ &=\int \left (\frac{\left (-c d^2+a e^2\right )^3 (d+e x)^{3/2}}{e^3}+\frac{3 c d \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}{e^3}-\frac{3 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{7/2}}{e^3}+\frac{c^3 d^3 (d+e x)^{9/2}}{e^3}\right ) \, dx\\ &=-\frac{2 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}{5 e^4}+\frac{6 c d \left (c d^2-a e^2\right )^2 (d+e x)^{7/2}}{7 e^4}-\frac{2 c^2 d^2 \left (c d^2-a e^2\right ) (d+e x)^{9/2}}{3 e^4}+\frac{2 c^3 d^3 (d+e x)^{11/2}}{11 e^4}\\ \end{align*}

Mathematica [A]  time = 0.0712888, size = 98, normalized size = 0.82 $\frac{2 (d+e x)^{5/2} \left (-385 c^2 d^2 (d+e x)^2 \left (c d^2-a e^2\right )+495 c d (d+e x) \left (c d^2-a e^2\right )^2-231 \left (c d^2-a e^2\right )^3+105 c^3 d^3 (d+e x)^3\right )}{1155 e^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(d + e*x)^(5/2)*(-231*(c*d^2 - a*e^2)^3 + 495*c*d*(c*d^2 - a*e^2)^2*(d + e*x) - 385*c^2*d^2*(c*d^2 - a*e^2)
*(d + e*x)^2 + 105*c^3*d^3*(d + e*x)^3))/(1155*e^4)

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Maple [A]  time = 0.043, size = 131, normalized size = 1.1 \begin{align*}{\frac{210\,{x}^{3}{c}^{3}{d}^{3}{e}^{3}+770\,a{c}^{2}{d}^{2}{e}^{4}{x}^{2}-140\,{c}^{3}{d}^{4}{e}^{2}{x}^{2}+990\,{a}^{2}cd{e}^{5}x-440\,a{c}^{2}{d}^{3}{e}^{3}x+80\,{c}^{3}{d}^{5}ex+462\,{a}^{3}{e}^{6}-396\,{a}^{2}c{d}^{2}{e}^{4}+176\,a{c}^{2}{d}^{4}{e}^{2}-32\,{c}^{3}{d}^{6}}{1155\,{e}^{4}} \left ( ex+d \right ) ^{{\frac{5}{2}}}} \end{align*}

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)^3/(e*x+d)^(3/2),x)

[Out]

2/1155*(e*x+d)^(5/2)*(105*c^3*d^3*e^3*x^3+385*a*c^2*d^2*e^4*x^2-70*c^3*d^4*e^2*x^2+495*a^2*c*d*e^5*x-220*a*c^2
*d^3*e^3*x+40*c^3*d^5*e*x+231*a^3*e^6-198*a^2*c*d^2*e^4+88*a*c^2*d^4*e^2-16*c^3*d^6)/e^4

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Maxima [A]  time = 1.04634, size = 185, normalized size = 1.55 \begin{align*} \frac{2 \,{\left (105 \,{\left (e x + d\right )}^{\frac{11}{2}} c^{3} d^{3} - 385 \,{\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )}{\left (e x + d\right )}^{\frac{9}{2}} + 495 \,{\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )}{\left (e x + d\right )}^{\frac{7}{2}} - 231 \,{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )}{\left (e x + d\right )}^{\frac{5}{2}}\right )}}{1155 \, e^{4}} \end{align*}

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)^3/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

2/1155*(105*(e*x + d)^(11/2)*c^3*d^3 - 385*(c^3*d^4 - a*c^2*d^2*e^2)*(e*x + d)^(9/2) + 495*(c^3*d^5 - 2*a*c^2*
d^3*e^2 + a^2*c*d*e^4)*(e*x + d)^(7/2) - 231*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*(e*x + d)
^(5/2))/e^4

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Fricas [B]  time = 1.83046, size = 498, normalized size = 4.18 \begin{align*} \frac{2 \,{\left (105 \, c^{3} d^{3} e^{5} x^{5} - 16 \, c^{3} d^{8} + 88 \, a c^{2} d^{6} e^{2} - 198 \, a^{2} c d^{4} e^{4} + 231 \, a^{3} d^{2} e^{6} + 35 \,{\left (4 \, c^{3} d^{4} e^{4} + 11 \, a c^{2} d^{2} e^{6}\right )} x^{4} + 5 \,{\left (c^{3} d^{5} e^{3} + 110 \, a c^{2} d^{3} e^{5} + 99 \, a^{2} c d e^{7}\right )} x^{3} - 3 \,{\left (2 \, c^{3} d^{6} e^{2} - 11 \, a c^{2} d^{4} e^{4} - 264 \, a^{2} c d^{2} e^{6} - 77 \, a^{3} e^{8}\right )} x^{2} +{\left (8 \, c^{3} d^{7} e - 44 \, a c^{2} d^{5} e^{3} + 99 \, a^{2} c d^{3} e^{5} + 462 \, a^{3} d e^{7}\right )} x\right )} \sqrt{e x + d}}{1155 \, e^{4}} \end{align*}

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)^3/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/1155*(105*c^3*d^3*e^5*x^5 - 16*c^3*d^8 + 88*a*c^2*d^6*e^2 - 198*a^2*c*d^4*e^4 + 231*a^3*d^2*e^6 + 35*(4*c^3*
d^4*e^4 + 11*a*c^2*d^2*e^6)*x^4 + 5*(c^3*d^5*e^3 + 110*a*c^2*d^3*e^5 + 99*a^2*c*d*e^7)*x^3 - 3*(2*c^3*d^6*e^2
- 11*a*c^2*d^4*e^4 - 264*a^2*c*d^2*e^6 - 77*a^3*e^8)*x^2 + (8*c^3*d^7*e - 44*a*c^2*d^5*e^3 + 99*a^2*c*d^3*e^5
+ 462*a^3*d*e^7)*x)*sqrt(e*x + d)/e^4

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Sympy [A]  time = 90.1039, size = 971, normalized size = 8.16 \begin{align*} \text{result too large to display} \end{align*}

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)**3/(e*x+d)**(3/2),x)

[Out]

Piecewise((-(2*a**3*d**3*e**3/sqrt(d + e*x) + 6*a**3*d**2*e**3*(-d/sqrt(d + e*x) - sqrt(d + e*x)) + 6*a**3*d*e
**3*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3) + 2*a**3*e**3*(-d**3/sqrt(d + e*x) - 3*d**2*
sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5) + 6*a**2*c*d**4*e*(-d/sqrt(d + e*x) - sqrt(d + e*x))
+ 18*a**2*c*d**3*e*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3) + 18*a**2*c*d**2*e*(-d**3/sqr
t(d + e*x) - 3*d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5) + 6*a**2*c*d*e*(d**4/sqrt(d + e*x
) + 4*d**3*sqrt(d + e*x) - 2*d**2*(d + e*x)**(3/2) + 4*d*(d + e*x)**(5/2)/5 - (d + e*x)**(7/2)/7) + 6*a*c**2*d
**5*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e + 18*a*c**2*d**4*(-d**3/sqrt(d + e*x) - 3*
d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e + 18*a*c**2*d**3*(d**4/sqrt(d + e*x) + 4*d**3*
sqrt(d + e*x) - 2*d**2*(d + e*x)**(3/2) + 4*d*(d + e*x)**(5/2)/5 - (d + e*x)**(7/2)/7)/e + 6*a*c**2*d**2*(-d**
5/sqrt(d + e*x) - 5*d**4*sqrt(d + e*x) + 10*d**3*(d + e*x)**(3/2)/3 - 2*d**2*(d + e*x)**(5/2) + 5*d*(d + e*x)*
*(7/2)/7 - (d + e*x)**(9/2)/9)/e + 2*c**3*d**6*(-d**3/sqrt(d + e*x) - 3*d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2
) - (d + e*x)**(5/2)/5)/e**3 + 6*c**3*d**5*(d**4/sqrt(d + e*x) + 4*d**3*sqrt(d + e*x) - 2*d**2*(d + e*x)**(3/2
) + 4*d*(d + e*x)**(5/2)/5 - (d + e*x)**(7/2)/7)/e**3 + 6*c**3*d**4*(-d**5/sqrt(d + e*x) - 5*d**4*sqrt(d + e*x
) + 10*d**3*(d + e*x)**(3/2)/3 - 2*d**2*(d + e*x)**(5/2) + 5*d*(d + e*x)**(7/2)/7 - (d + e*x)**(9/2)/9)/e**3 +
2*c**3*d**3*(d**6/sqrt(d + e*x) + 6*d**5*sqrt(d + e*x) - 5*d**4*(d + e*x)**(3/2) + 4*d**3*(d + e*x)**(5/2) -
15*d**2*(d + e*x)**(7/2)/7 + 2*d*(d + e*x)**(9/2)/3 - (d + e*x)**(11/2)/11)/e**3)/e, Ne(e, 0)), (c**3*d**(9/2)
*x**4/4, True))

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Giac [A]  time = 1.18415, size = 250, normalized size = 2.1 \begin{align*} \frac{2}{1155} \,{\left (105 \,{\left (x e + d\right )}^{\frac{11}{2}} c^{3} d^{3} e^{40} - 385 \,{\left (x e + d\right )}^{\frac{9}{2}} c^{3} d^{4} e^{40} + 495 \,{\left (x e + d\right )}^{\frac{7}{2}} c^{3} d^{5} e^{40} - 231 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{3} d^{6} e^{40} + 385 \,{\left (x e + d\right )}^{\frac{9}{2}} a c^{2} d^{2} e^{42} - 990 \,{\left (x e + d\right )}^{\frac{7}{2}} a c^{2} d^{3} e^{42} + 693 \,{\left (x e + d\right )}^{\frac{5}{2}} a c^{2} d^{4} e^{42} + 495 \,{\left (x e + d\right )}^{\frac{7}{2}} a^{2} c d e^{44} - 693 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{2} c d^{2} e^{44} + 231 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{3} e^{46}\right )} e^{\left (-44\right )} \end{align*}

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)^3/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2/1155*(105*(x*e + d)^(11/2)*c^3*d^3*e^40 - 385*(x*e + d)^(9/2)*c^3*d^4*e^40 + 495*(x*e + d)^(7/2)*c^3*d^5*e^4
0 - 231*(x*e + d)^(5/2)*c^3*d^6*e^40 + 385*(x*e + d)^(9/2)*a*c^2*d^2*e^42 - 990*(x*e + d)^(7/2)*a*c^2*d^3*e^42
+ 693*(x*e + d)^(5/2)*a*c^2*d^4*e^42 + 495*(x*e + d)^(7/2)*a^2*c*d*e^44 - 693*(x*e + d)^(5/2)*a^2*c*d^2*e^44
+ 231*(x*e + d)^(5/2)*a^3*e^46)*e^(-44)