### 3.2033 $$\int \frac{\sqrt{a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^{5/2}} \, dx$$

Optimal. Leaf size=129 $\frac{c d \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{e^{3/2} \sqrt{c d^2-a e^2}}-\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x)^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0732543, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {662, 660, 205} $\frac{c d \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{e^{3/2} \sqrt{c d^2-a e^2}}-\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(d + e*x)^(5/2),x]

[Out]

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

Rule 662

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

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{5/2}} \, dx &=-\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+\frac{(c d) \int \frac{1}{\sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 e}\\ &=-\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+(c d) \operatorname{Subst}\left (\int \frac{1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}}\right )\\ &=-\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+\frac{c d \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{c d^2-a e^2} \sqrt{d+e x}}\right )}{e^{3/2} \sqrt{c d^2-a e^2}}\\ \end{align*}

Mathematica [A]  time = 0.159372, size = 112, normalized size = 0.87 $\frac{\sqrt{(d+e x) (a e+c d x)} \left (\frac{c d (d+e x) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d^2-a e^2}}\right )}{\sqrt{c d^2-a e^2} \sqrt{a e+c d x}}-\sqrt{e}\right )}{e^{3/2} (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(d + e*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.244, size = 163, normalized size = 1.3 \begin{align*}{\frac{1}{e} \left ( -{\it Artanh} \left ({e\sqrt{cdx+ae}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \right ) xcde-{\it Artanh} \left ({e\sqrt{cdx+ae}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \right ) c{d}^{2}-\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e} \right ) \sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( ex+d \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cdx+ae}}}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \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)^(1/2)/(e*x+d)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{{\left (e x + d\right )}^{\frac{5}{2}}}\,{d x} \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)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)/(e*x + d)^(5/2), x)

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Fricas [B]  time = 1.87339, size = 1004, normalized size = 7.78 \begin{align*} \left [-\frac{{\left (c d e^{2} x^{2} + 2 \, c d^{2} e x + c d^{3}\right )} \sqrt{-c d^{2} e + a e^{3}} \log \left (-\frac{c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{-c d^{2} e + a e^{3}} \sqrt{e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d^{2} e - a e^{3}\right )} \sqrt{e x + d}}{2 \,{\left (c d^{4} e^{2} - a d^{2} e^{4} +{\left (c d^{2} e^{4} - a e^{6}\right )} x^{2} + 2 \,{\left (c d^{3} e^{3} - a d e^{5}\right )} x\right )}}, -\frac{{\left (c d e^{2} x^{2} + 2 \, c d^{2} e x + c d^{3}\right )} \sqrt{c d^{2} e - a e^{3}} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{c d^{2} e - a e^{3}} \sqrt{e x + d}}{c d e^{2} x^{2} + a d e^{2} +{\left (c d^{2} e + a e^{3}\right )} x}\right ) + \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d^{2} e - a e^{3}\right )} \sqrt{e x + d}}{c d^{4} e^{2} - a d^{2} e^{4} +{\left (c d^{2} e^{4} - a e^{6}\right )} x^{2} + 2 \,{\left (c d^{3} e^{3} - a d e^{5}\right )} x}\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)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/2*((c*d*e^2*x^2 + 2*c*d^2*e*x + c*d^3)*sqrt(-c*d^2*e + a*e^3)*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*
d*e^2 - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(-c*d^2*e + a*e^3)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x
+ d^2)) + 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2*e - a*e^3)*sqrt(e*x + d))/(c*d^4*e^2 - a*d^2*e
^4 + (c*d^2*e^4 - a*e^6)*x^2 + 2*(c*d^3*e^3 - a*d*e^5)*x), -((c*d*e^2*x^2 + 2*c*d^2*e*x + c*d^3)*sqrt(c*d^2*e
- a*e^3)*arctan(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d^2*e - a*e^3)*sqrt(e*x + d)/(c*d*e^2*x^2 +
a*d*e^2 + (c*d^2*e + a*e^3)*x)) + sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2*e - a*e^3)*sqrt(e*x + d)
)/(c*d^4*e^2 - a*d^2*e^4 + (c*d^2*e^4 - a*e^6)*x^2 + 2*(c*d^3*e^3 - a*d*e^5)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(1/2)/(e*x+d)**(5/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

Timed out