### 3.2013 $$\int \frac{(d+e x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx$$

Optimal. Leaf size=94 $-\frac{e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{3/2} d^{3/2} \sqrt{c d^2-a e^2}}-\frac{\sqrt{d+e x}}{c d (a e+c d x)}$

[Out]

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

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Rubi [A]  time = 0.0571539, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {626, 47, 63, 208} $-\frac{e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{c^{3/2} d^{3/2} \sqrt{c d^2-a e^2}}-\frac{\sqrt{d+e x}}{c d (a e+c d x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sqrt[d + e*x]/(c*d*(a*e + c*d*x))) - (e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c^(3/
2)*d^(3/2)*Sqrt[c*d^2 - a*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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.109552, size = 93, normalized size = 0.99 $\frac{e \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{a e^2-c d^2}}\right )}{c^{3/2} d^{3/2} \sqrt{a e^2-c d^2}}-\frac{\sqrt{d+e x}}{a c d e+c^2 d^2 x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.201, size = 84, normalized size = 0.9 \begin{align*} -{\frac{e}{cd \left ( cdex+a{e}^{2} \right ) }\sqrt{ex+d}}+{\frac{e}{cd}\arctan \left ({cd\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \right ){\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99687, size = 625, normalized size = 6.65 \begin{align*} \left [\frac{\sqrt{c^{2} d^{3} - a c d e^{2}}{\left (c d e x + a e^{2}\right )} \log \left (\frac{c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt{c^{2} d^{3} - a c d e^{2}} \sqrt{e x + d}}{c d x + a e}\right ) - 2 \,{\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt{e x + d}}{2 \,{\left (a c^{3} d^{4} e - a^{2} c^{2} d^{2} e^{3} +{\left (c^{4} d^{5} - a c^{3} d^{3} e^{2}\right )} x\right )}}, \frac{\sqrt{-c^{2} d^{3} + a c d e^{2}}{\left (c d e x + a e^{2}\right )} \arctan \left (\frac{\sqrt{-c^{2} d^{3} + a c d e^{2}} \sqrt{e x + d}}{c d e x + c d^{2}}\right ) -{\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt{e x + d}}{a c^{3} d^{4} e - a^{2} c^{2} d^{2} e^{3} +{\left (c^{4} d^{5} - a c^{3} d^{3} e^{2}\right )} x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(c^2*d^3 - a*c*d*e^2)*(c*d*e*x + a*e^2)*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*sqrt(c^2*d^3 - a*c*d*e^2)
*sqrt(e*x + d))/(c*d*x + a*e)) - 2*(c^2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(a*c^3*d^4*e - a^2*c^2*d^2*e^3 + (c^4*
d^5 - a*c^3*d^3*e^2)*x), (sqrt(-c^2*d^3 + a*c*d*e^2)*(c*d*e*x + a*e^2)*arctan(sqrt(-c^2*d^3 + a*c*d*e^2)*sqrt(
e*x + d)/(c*d*e*x + c*d^2)) - (c^2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(a*c^3*d^4*e - a^2*c^2*d^2*e^3 + (c^4*d^5 -
a*c^3*d^3*e^2)*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)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**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((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

Timed out