### 3.2076 $$\int \frac{\sqrt{d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx$$

Optimal. Leaf size=257 $\frac{5 c d e \sqrt{d+e x}}{\left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{5 e}{3 \sqrt{d+e x} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{2 \sqrt{d+e x}}{3 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}+\frac{5 c d e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}$

[Out]

(-2*Sqrt[d + e*x])/(3*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (5*e)/(3*(c*d^2 - a*e^2
)^2*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*c*d*e*Sqrt[d + e*x])/((c*d^2 - a*e^2)^3*Sq
rt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*c*d*e^(3/2)*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])])/(c*d^2 - a*e^2)^(7/2)

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Rubi [A]  time = 0.209254, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {666, 672, 660, 205} $\frac{5 c d e \sqrt{d+e x}}{\left (c d^2-a e^2\right )^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{5 e}{3 \sqrt{d+e x} \left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{2 \sqrt{d+e x}}{3 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}+\frac{5 c d e^{3/2} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}$

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)^(5/2),x]

[Out]

(-2*Sqrt[d + e*x])/(3*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (5*e)/(3*(c*d^2 - a*e^2
)^2*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*c*d*e*Sqrt[d + e*x])/((c*d^2 - a*e^2)^3*Sq
rt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (5*c*d*e^(3/2)*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])])/(c*d^2 - a*e^2)^(7/2)

Rule 666

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

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 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{d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac{2 \sqrt{d+e x}}{3 \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}}-\frac{(5 e) \int \frac{1}{\sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 \left (c d^2-a e^2\right )}\\ &=-\frac{2 \sqrt{d+e x}}{3 \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}}-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{(5 c d e) \int \frac{\sqrt{d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{2 \left (c d^2-a e^2\right )^2}\\ &=-\frac{2 \sqrt{d+e x}}{3 \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}}-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d e \sqrt{d+e x}}{\left (c d^2-a e^2\right )^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (5 c d e^2\right ) \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 \left (c d^2-a e^2\right )^3}\\ &=-\frac{2 \sqrt{d+e x}}{3 \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}}-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d e \sqrt{d+e x}}{\left (c d^2-a e^2\right )^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (5 c d e^3\right ) \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 )}{\left (c d^2-a e^2\right )^3}\\ &=-\frac{2 \sqrt{d+e x}}{3 \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}}-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d e \sqrt{d+e x}}{\left (c d^2-a e^2\right )^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{5 c d e^{3/2} \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 )}{\left (c d^2-a e^2\right )^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0312925, size = 79, normalized size = 0.31 $-\frac{2 c d (d+e x)^{3/2} \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{3 \left (c d^2-a e^2\right )^2 ((d+e x) (a e+c d x))^{3/2}}$

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)^(5/2),x]

[Out]

(-2*c*d*(d + e*x)^(3/2)*Hypergeometric2F1[-3/2, 2, -1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(3*(c*d^2 - a*
e^2)^2*((a*e + c*d*x)*(d + e*x))^(3/2))

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Maple [A]  time = 0.248, size = 433, normalized size = 1.7 \begin{align*}{\frac{1}{3\, \left ( cdx+ae \right ) ^{2} \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}{x}^{2}{c}^{2}{d}^{2}{e}^{3}+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) xacd{e}^{4}\sqrt{cdx+ae}+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}x{c}^{2}{d}^{3}{e}^{2}+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) ac{d}^{2}{e}^{3}\sqrt{cdx+ae}-15\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{x}^{2}{c}^{2}{d}^{2}{e}^{2}-20\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}xacd{e}^{3}-10\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}x{c}^{2}{d}^{3}e-3\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{a}^{2}{e}^{4}-14\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}ac{d}^{2}{e}^{2}+2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}{c}^{2}{d}^{4} \right ) \left ( ex+d \right ) ^{-{\frac{3}{2}}}{\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((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x)

[Out]

1/3*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*
e)^(1/2)*x^2*c^2*d^2*e^3+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x*a*c*d*e^4*(c*d*x+a*e)^(1/2)
+15*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*x*c^2*d^3*e^2+15*arctanh(e*(c*d*x+a
*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*a*c*d^2*e^3*(c*d*x+a*e)^(1/2)-15*((a*e^2-c*d^2)*e)^(1/2)*x^2*c^2*d^2*e^2-20
*((a*e^2-c*d^2)*e)^(1/2)*x*a*c*d*e^3-10*((a*e^2-c*d^2)*e)^(1/2)*x*c^2*d^3*e-3*((a*e^2-c*d^2)*e)^(1/2)*a^2*e^4-
14*((a*e^2-c*d^2)*e)^(1/2)*a*c*d^2*e^2+2*((a*e^2-c*d^2)*e)^(1/2)*c^2*d^4)/(e*x+d)^(3/2)/(c*d*x+a*e)^2/(a*e^2-c
*d^2)^3/((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{e x + d}}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.15481, size = 2452, normalized size = 9.54 \begin{align*} \text{result too large to display} \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)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(15*(c^3*d^3*e^3*x^4 + a^2*c*d^3*e^3 + 2*(c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^3 + (c^3*d^5*e + 4*a*c^2*d^3*e^
3 + a^2*c*d*e^5)*x^2 + 2*(a*c^2*d^4*e^2 + a^2*c*d^2*e^4)*x)*sqrt(-e/(c*d^2 - a*e^2))*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)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-e
/(c*d^2 - a*e^2)))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*(15*c^2*d^2*e^2*x^2 - 2*c^2*d^4 + 14*a*c*d^2*e^2 + 3*a^2*e^4
+ 10*(c^2*d^3*e + 2*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a^2*c^3*d^8*e^2
- 3*a^3*c^2*d^6*e^4 + 3*a^4*c*d^4*e^6 - a^5*d^2*e^8 + (c^5*d^8*e^2 - 3*a*c^4*d^6*e^4 + 3*a^2*c^3*d^4*e^6 - a^
3*c^2*d^2*e^8)*x^4 + 2*(c^5*d^9*e - 2*a*c^4*d^7*e^3 + 2*a^3*c^2*d^3*e^7 - a^4*c*d*e^9)*x^3 + (c^5*d^10 + a*c^4
*d^8*e^2 - 8*a^2*c^3*d^6*e^4 + 8*a^3*c^2*d^4*e^6 - a^4*c*d^2*e^8 - a^5*e^10)*x^2 + 2*(a*c^4*d^9*e - 2*a^2*c^3*
d^7*e^3 + 2*a^4*c*d^3*e^7 - a^5*d*e^9)*x), 1/3*(15*(c^3*d^3*e^3*x^4 + a^2*c*d^3*e^3 + 2*(c^3*d^4*e^2 + a*c^2*d
^2*e^4)*x^3 + (c^3*d^5*e + 4*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 + 2*(a*c^2*d^4*e^2 + a^2*c*d^2*e^4)*x)*sqrt(e/(c
*d^2 - a*e^2))*arctan(-sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(e/(c*d^2
- a*e^2))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) + (15*c^2*d^2*e^2*x^2 - 2*c^2*d^4 + 14*a*c*d^2*e^2 +
3*a^2*e^4 + 10*(c^2*d^3*e + 2*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(a^2*c
^3*d^8*e^2 - 3*a^3*c^2*d^6*e^4 + 3*a^4*c*d^4*e^6 - a^5*d^2*e^8 + (c^5*d^8*e^2 - 3*a*c^4*d^6*e^4 + 3*a^2*c^3*d^
4*e^6 - a^3*c^2*d^2*e^8)*x^4 + 2*(c^5*d^9*e - 2*a*c^4*d^7*e^3 + 2*a^3*c^2*d^3*e^7 - a^4*c*d*e^9)*x^3 + (c^5*d^
10 + a*c^4*d^8*e^2 - 8*a^2*c^3*d^6*e^4 + 8*a^3*c^2*d^4*e^6 - a^4*c*d^2*e^8 - a^5*e^10)*x^2 + 2*(a*c^4*d^9*e -
2*a^2*c^3*d^7*e^3 + 2*a^4*c*d^3*e^7 - a^5*d*e^9)*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((e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} 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)^(5/2),x, algorithm="giac")

[Out]

sage0*x