### 3.1977 $$\int \sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0193963, antiderivative size = 43, normalized size of antiderivative = 1., 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} $\frac{2}{5} (d+e x)^{5/2} \left (a-\frac{c d^2}{e^2}\right )+\frac{2 c d (d+e x)^{7/2}}{7 e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0249812, size = 34, normalized size = 0.79 $\frac{2 (d+e x)^{5/2} \left (7 a e^2+c d (5 e x-2 d)\right )}{35 e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 32, normalized size = 0.7 \begin{align*}{\frac{10\,cdex+14\,a{e}^{2}-4\,c{d}^{2}}{35\,{e}^{2}} \left ( ex+d \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x)

[Out]

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

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Maxima [A]  time = 0.976588, size = 51, normalized size = 1.19 \begin{align*} \frac{2 \,{\left (5 \,{\left (e x + d\right )}^{\frac{7}{2}} c d - 7 \,{\left (c d^{2} - a e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}}\right )}}{35 \, e^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.27846, size = 163, normalized size = 3.79 \begin{align*} \frac{2 \,{\left (5 \, c d e^{3} x^{3} - 2 \, c d^{4} + 7 \, a d^{2} e^{2} +{\left (8 \, c d^{2} e^{2} + 7 \, a e^{4}\right )} x^{2} +{\left (c d^{3} e + 14 \, a d e^{3}\right )} x\right )} \sqrt{e x + d}}{35 \, e^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

2/35*(5*c*d*e^3*x^3 - 2*c*d^4 + 7*a*d^2*e^2 + (8*c*d^2*e^2 + 7*a*e^4)*x^2 + (c*d^3*e + 14*a*d*e^3)*x)*sqrt(e*x
+ d)/e^2

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Sympy [A]  time = 2.84052, size = 41, normalized size = 0.95 \begin{align*} \frac{2 \left (\frac{c d \left (d + e x\right )^{\frac{7}{2}}}{7 e} + \frac{\left (d + e x\right )^{\frac{5}{2}} \left (a e^{2} - c d^{2}\right )}{5 e}\right )}{e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)

[Out]

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

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Giac [B]  time = 1.13804, size = 157, normalized size = 3.65 \begin{align*} \frac{2}{105} \,{\left (7 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} c d^{2} e^{\left (-1\right )} + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} a d e +{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} c d e^{\left (-1\right )} + 7 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} a e\right )} e^{\left (-1\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")

[Out]

2/105*(7*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*c*d^2*e^(-1) + 35*(x*e + d)^(3/2)*a*d*e + (15*(x*e + d)^(7/
2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*c*d*e^(-1) + 7*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a
*e)*e^(-1)