### 3.2030 $$\int \sqrt{d+e x} \sqrt{a d e+(c d^2+a e^2) x+c d e x^2} \, dx$$

Optimal. Leaf size=109 $\frac{4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 c^2 d^2 (d+e x)^{3/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d \sqrt{d+e x}}$

[Out]

(4*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(15*c^2*d^2*(d + e*x)^(3/2)) + (2*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(5*c*d*Sqrt[d + e*x])

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Rubi [A]  time = 0.0602184, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.051, Rules used = {656, 648} $\frac{4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 c^2 d^2 (d+e x)^{3/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(15*c^2*d^2*(d + e*x)^(3/2)) + (2*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(5*c*d*Sqrt[d + e*x])

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps

\begin{align*} \int \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d \sqrt{d+e x}}+\frac{\left (2 \left (d^2-\frac{a e^2}{c}\right )\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}} \, dx}{5 d}\\ &=\frac{4 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 c^2 d^2 (d+e x)^{3/2}}+\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.0479747, size = 55, normalized size = 0.5 $\frac{2 ((d+e x) (a e+c d x))^{3/2} \left (c d (5 d+3 e x)-2 a e^2\right )}{15 c^2 d^2 (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(-2*a*e^2 + c*d*(5*d + 3*e*x)))/(15*c^2*d^2*(d + e*x)^(3/2))

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Maple [A]  time = 0.042, size = 69, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( -3\,cdex+2\,a{e}^{2}-5\,c{d}^{2} \right ) }{15\,{c}^{2}{d}^{2}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade}{\frac{1}{\sqrt{ex+d}}}} \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)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.08072, size = 112, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (3 \, c^{2} d^{2} e x^{2} + 5 \, a c d^{2} e - 2 \, a^{2} e^{3} +{\left (5 \, c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt{c d x + a e}{\left (e x + d\right )}}{15 \,{\left (c^{2} d^{2} e x + c^{2} d^{3}\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)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*c^2*d^2*e*x^2 + 5*a*c*d^2*e - 2*a^2*e^3 + (5*c^2*d^3 + a*c*d*e^2)*x)*sqrt(c*d*x + a*e)*(e*x + d)/(c^2*
d^2*e*x + c^2*d^3)

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Fricas [A]  time = 1.80272, size = 216, normalized size = 1.98 \begin{align*} \frac{2 \,{\left (3 \, c^{2} d^{2} e x^{2} + 5 \, a c d^{2} e - 2 \, a^{2} e^{3} +{\left (5 \, c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{15 \,{\left (c^{2} d^{2} e x + c^{2} d^{3}\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)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*c^2*d^2*e*x^2 + 5*a*c*d^2*e - 2*a^2*e^3 + (5*c^2*d^3 + a*c*d*e^2)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 +
a*e^2)*x)*sqrt(e*x + d)/(c^2*d^2*e*x + c^2*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (d + e x\right ) \left (a e + c d x\right )} \sqrt{d + e x}\, dx \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)**(1/2),x)

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))*sqrt(d + e*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}\,{d x} \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)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d), x)