### 3.2024 $$\int \frac{(d+e x)^{3/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx$$

Optimal. Leaf size=176 $\frac{15 e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^3}+\frac{5 e}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 \sqrt{d+e x} \left (c d^2-a e^2\right ) (a e+c d x)^2}-\frac{15 \sqrt{c} \sqrt{d} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{7/2}}$

[Out]

(15*e^2)/(4*(c*d^2 - a*e^2)^3*Sqrt[d + e*x]) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*Sqrt[d + e*x]) + (5*e)/(4*
(c*d^2 - a*e^2)^2*(a*e + c*d*x)*Sqrt[d + e*x]) - (15*Sqrt[c]*Sqrt[d]*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x
])/Sqrt[c*d^2 - a*e^2]])/(4*(c*d^2 - a*e^2)^(7/2))

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Rubi [A]  time = 0.119063, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {626, 51, 63, 208} $\frac{15 e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^3}+\frac{5 e}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 \sqrt{d+e x} \left (c d^2-a e^2\right ) (a e+c d x)^2}-\frac{15 \sqrt{c} \sqrt{d} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(15*e^2)/(4*(c*d^2 - a*e^2)^3*Sqrt[d + e*x]) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*Sqrt[d + e*x]) + (5*e)/(4*
(c*d^2 - a*e^2)^2*(a*e + c*d*x)*Sqrt[d + e*x]) - (15*Sqrt[c]*Sqrt[d]*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x
])/Sqrt[c*d^2 - a*e^2]])/(4*(c*d^2 - a*e^2)^(7/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)^{3/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac{1}{(a e+c d x)^3 (d+e x)^{3/2}} \, dx\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 \sqrt{d+e x}}-\frac{(5 e) \int \frac{1}{(a e+c d x)^2 (d+e x)^{3/2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 \sqrt{d+e x}}+\frac{5 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt{d+e x}}+\frac{\left (15 e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac{15 e^2}{4 \left (c d^2-a e^2\right )^3 \sqrt{d+e x}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 \sqrt{d+e x}}+\frac{5 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt{d+e x}}+\frac{\left (15 c d e^2\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 \left (c d^2-a e^2\right )^3}\\ &=\frac{15 e^2}{4 \left (c d^2-a e^2\right )^3 \sqrt{d+e x}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 \sqrt{d+e x}}+\frac{5 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt{d+e x}}+\frac{(15 c d e) \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 )}{4 \left (c d^2-a e^2\right )^3}\\ &=\frac{15 e^2}{4 \left (c d^2-a e^2\right )^3 \sqrt{d+e x}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 \sqrt{d+e x}}+\frac{5 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt{d+e x}}-\frac{15 \sqrt{c} \sqrt{d} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0154581, size = 59, normalized size = 0.34 $-\frac{2 e^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{\sqrt{d+e x} \left (a e^2-c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*e^2*Hypergeometric2F1[-1/2, 3, 1/2, -((c*d*(d + e*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 = 0.207, size = 226, normalized size = 1.3 \begin{align*} -2\,{\frac{{e}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}\sqrt{ex+d}}}-{\frac{7\,{c}^{2}{d}^{2}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3} \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{9\,{e}^{4}cda}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{9\,{c}^{2}{d}^{3}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{15\,{e}^{2}cd}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}}\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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

-2*e^2/(a*e^2-c*d^2)^3/(e*x+d)^(1/2)-7/4*e^2/(a*e^2-c*d^2)^3*c^2*d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)-9/4*e^4/(
a*e^2-c*d^2)^3*c*d/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+9/4*e^2/(a*e^2-c*d^2)^3*c^2*d^3/(c*d*e*x+a*e^2)^2*(e*x+d)
^(1/2)-15/4*e^2/(a*e^2-c*d^2)^3*c*d/((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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.13636, size = 1767, normalized size = 10.04 \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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(c^2*d^2*e^3*x^3 + a^2*d*e^4 + (c^2*d^3*e^2 + 2*a*c*d*e^4)*x^2 + (2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(c*
d/(c*d^2 - a*e^2))*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(c*d/(c*d^2 - a*e^2)))
/(c*d*x + a*e)) + 2*(15*c^2*d^2*e^2*x^2 - 2*c^2*d^4 + 9*a*c*d^2*e^2 + 8*a^2*e^4 + 5*(c^2*d^3*e + 5*a*c*d*e^3)*
x)*sqrt(e*x + d))/(a^2*c^3*d^7*e^2 - 3*a^3*c^2*d^5*e^4 + 3*a^4*c*d^3*e^6 - a^5*d*e^8 + (c^5*d^8*e - 3*a*c^4*d^
6*e^3 + 3*a^2*c^3*d^4*e^5 - a^3*c^2*d^2*e^7)*x^3 + (c^5*d^9 - a*c^4*d^7*e^2 - 3*a^2*c^3*d^5*e^4 + 5*a^3*c^2*d^
3*e^6 - 2*a^4*c*d*e^8)*x^2 + (2*a*c^4*d^8*e - 5*a^2*c^3*d^6*e^3 + 3*a^3*c^2*d^4*e^5 + a^4*c*d^2*e^7 - a^5*e^9)
*x), -1/4*(15*(c^2*d^2*e^3*x^3 + a^2*d*e^4 + (c^2*d^3*e^2 + 2*a*c*d*e^4)*x^2 + (2*a*c*d^2*e^3 + a^2*e^5)*x)*sq
rt(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2)) -
(15*c^2*d^2*e^2*x^2 - 2*c^2*d^4 + 9*a*c*d^2*e^2 + 8*a^2*e^4 + 5*(c^2*d^3*e + 5*a*c*d*e^3)*x)*sqrt(e*x + d))/(
a^2*c^3*d^7*e^2 - 3*a^3*c^2*d^5*e^4 + 3*a^4*c*d^3*e^6 - a^5*d*e^8 + (c^5*d^8*e - 3*a*c^4*d^6*e^3 + 3*a^2*c^3*d
^4*e^5 - a^3*c^2*d^2*e^7)*x^3 + (c^5*d^9 - a*c^4*d^7*e^2 - 3*a^2*c^3*d^5*e^4 + 5*a^3*c^2*d^3*e^6 - 2*a^4*c*d*e
^8)*x^2 + (2*a*c^4*d^8*e - 5*a^2*c^3*d^6*e^3 + 3*a^3*c^2*d^4*e^5 + a^4*c*d^2*e^7 - a^5*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)**(3/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out