### 3.2063 $$\int \frac{1}{(d+e x)^{5/2} \sqrt{a d e+(c d^2+a e^2) x+c d e x^2}} \, dx$$

Optimal. Leaf size=207 $\frac{3 c^2 d^2 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 \sqrt{e} \left (c d^2-a e^2\right )^{5/2}}+\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}+\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right )}$

[Out]

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

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Rubi [A]  time = 0.113309, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {672, 660, 205} $\frac{3 c^2 d^2 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{4 \sqrt{e} \left (c d^2-a e^2\right )^{5/2}}+\frac{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}+\frac{\sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

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

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{1}{(d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{(3 c d) \int \frac{1}{(d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{\left (3 c^2 d^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}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{\left (3 c^2 d^2 e\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 )}{4 \left (c d^2-a e^2\right )^2}\\ &=\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{3 c^2 d^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 )}{4 \sqrt{e} \left (c d^2-a e^2\right )^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0316992, size = 81, normalized size = 0.39 $\frac{2 c^2 d^2 \sqrt{(d+e x) (a e+c d x)} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{\sqrt{d+e x} \left (c d^2-a e^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^(5/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

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

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Maple [A]  time = 0.243, size = 292, normalized size = 1.4 \begin{align*} -{\frac{1}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{2}{c}^{2}{d}^{2}{e}^{2}+6\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) x{c}^{2}{d}^{3}e+3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{2}{d}^{4}-3\,xcde\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}a{e}^{2}-5\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}c{d}^{2} \right ) \left ( ex+d \right ) ^{-{\frac{5}{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(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

[Out]

-1/4*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x^2*c^2*d
^2*e^2+6*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*x*c^2*d^3*e+3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^
2-c*d^2)*e)^(1/2))*c^2*d^4-3*x*c*d*e*(c*d*x+a*e)^(1/2)*((a*e^2-c*d^2)*e)^(1/2)+2*((a*e^2-c*d^2)*e)^(1/2)*(c*d*
x+a*e)^(1/2)*a*e^2-5*((a*e^2-c*d^2)*e)^(1/2)*(c*d*x+a*e)^(1/2)*c*d^2)/(e*x+d)^(5/2)/(c*d*x+a*e)^(1/2)/(a*e^2-c
*d^2)^2/((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{1}{\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(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(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 = 2.28048, size = 1758, normalized size = 8.49 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(3*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*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)*sqr
t(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*(5*c^2*d^4*e - 7*a*c*d^2*e^3 + 2*a^2*e^5 + 3*(c^2*d^3*e^2 - a*c*d*e
^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c^3*d^9*e - 3*a*c^2*d^7*e^3 + 3*a^2*c*d^5*e
^5 - a^3*d^3*e^7 + (c^3*d^6*e^4 - 3*a*c^2*d^4*e^6 + 3*a^2*c*d^2*e^8 - a^3*e^10)*x^3 + 3*(c^3*d^7*e^3 - 3*a*c^2
*d^5*e^5 + 3*a^2*c*d^3*e^7 - a^3*d*e^9)*x^2 + 3*(c^3*d^8*e^2 - 3*a*c^2*d^6*e^4 + 3*a^2*c*d^4*e^6 - a^3*d^2*e^8
)*x), -1/4*(3*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*sqrt(c*d^2*e - a*e^3)*arctan(sqr
t(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)) - (5*c^2*d^4*e - 7*a*c*d^2*e^3 + 2*a^2*e^5 + 3*(c^2*d^3*e^2 - a*c*d*e^4)*x)*sqrt(c*d*e*x^2 + a*d
*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(c^3*d^9*e - 3*a*c^2*d^7*e^3 + 3*a^2*c*d^5*e^5 - a^3*d^3*e^7 + (c^3*d^6
*e^4 - 3*a*c^2*d^4*e^6 + 3*a^2*c*d^2*e^8 - a^3*e^10)*x^3 + 3*(c^3*d^7*e^3 - 3*a*c^2*d^5*e^5 + 3*a^2*c*d^3*e^7
- a^3*d*e^9)*x^2 + 3*(c^3*d^8*e^2 - 3*a*c^2*d^6*e^4 + 3*a^2*c*d^4*e^6 - a^3*d^2*e^8)*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(1/(e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),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(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

sage0*x