3.2016 \(\int \frac{1}{\sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=158 \[ \frac{5 c^{3/2} d^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}-\frac{5 c d e}{\sqrt{d+e x} \left (c d^2-a e^2\right )^3}-\frac{1}{(d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{5 e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0927356, antiderivative size = 158, 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} \[ \frac{5 c^{3/2} d^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}-\frac{5 c d e}{\sqrt{d+e x} \left (c d^2-a e^2\right )^3}-\frac{1}{(d+e x)^{3/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{5 e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e)/(3*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2)) - 1/((c*d^2 - a*e^2)*(a*e + c*d*x)*(d + e*x)^(3/2)) - (5*c*d*e)/(
(c*d^2 - a*e^2)^3*Sqrt[d + e*x]) + (5*c^(3/2)*d^(3/2)*e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a
*e^2]])/(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{1}{\sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac{1}{(a e+c d x)^2 (d+e x)^{5/2}} \, dx\\ &=-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{3/2}}-\frac{(5 e) \int \frac{1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{2 \left (c d^2-a e^2\right )}\\ &=-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{3/2}}-\frac{(5 c d e) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{2 \left (c d^2-a e^2\right )^2}\\ &=-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{3/2}}-\frac{5 c d e}{\left (c d^2-a e^2\right )^3 \sqrt{d+e x}}-\frac{\left (5 c^2 d^2 e\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{2 \left (c d^2-a e^2\right )^3}\\ &=-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{3/2}}-\frac{5 c d e}{\left (c d^2-a e^2\right )^3 \sqrt{d+e x}}-\frac{\left (5 c^2 d^2\right ) \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 )}{\left (c d^2-a e^2\right )^3}\\ &=-\frac{5 e}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{3/2}}-\frac{5 c d e}{\left (c d^2-a e^2\right )^3 \sqrt{d+e x}}+\frac{5 c^{3/2} d^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0172592, size = 59, normalized size = 0.37 \[ -\frac{2 e \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{3 (d+e x)^{3/2} \left (a e^2-c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.204, size = 162, normalized size = 1. \begin{align*} -{\frac{2\,e}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}+4\,{\frac{dec}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}\sqrt{ex+d}}}+{\frac{{c}^{2}e{d}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3} \left ( cdex+a{e}^{2} \right ) }\sqrt{ex+d}}+5\,{\frac{{c}^{2}e{d}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}\arctan \left ({\frac{\sqrt{ex+d}cd}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x)

[Out]

-2/3*e/(a*e^2-c*d^2)^2/(e*x+d)^(3/2)+4*e/(a*e^2-c*d^2)^3*c*d/(e*x+d)^(1/2)+e*c^2*d^2/(a*e^2-c*d^2)^3*(e*x+d)^(
1/2)/(c*d*e*x+a*e^2)+5*e*c^2*d^2/(a*e^2-c*d^2)^3/((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(1/(e*x+d)^(1/2)/(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 [B]  time = 2.11454, size = 1751, normalized size = 11.08 \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)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

[-1/6*(15*(c^2*d^2*e^3*x^3 + a*c*d^3*e^2 + (2*c^2*d^3*e^2 + a*c*d*e^4)*x^2 + (c^2*d^4*e + 2*a*c*d^2*e^3)*x)*sq
rt(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 + 3*c^2*d^4 + 14*a*c*d^2*e^2 - 2*a^2*e^4 + 10*(2*c^2*d^3*e + a*c*
d*e^3)*x)*sqrt(e*x + d))/(a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2 - 3*a
*c^3*d^5*e^4 + 3*a^2*c^2*d^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5 + a^3
*c*d^2*e^7 - a^4*e^9)*x^2 + (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*e^8)*x),
1/3*(15*(c^2*d^2*e^3*x^3 + a*c*d^3*e^2 + (2*c^2*d^3*e^2 + a*c*d*e^4)*x^2 + (c^2*d^4*e + 2*a*c*d^2*e^3)*x)*sqrt
(-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 + 3*c^2*d^4 + 14*a*c*d^2*e^2 - 2*a^2*e^4 + 10*(2*c^2*d^3*e + a*c*d*e^3)*x)*sqrt(e*x + d))/(
a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2 - 3*a*c^3*d^5*e^4 + 3*a^2*c^2*d
^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5 + a^3*c*d^2*e^7 - a^4*e^9)*x^2
+ (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*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)**(1/2)/(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(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

Timed out