3.2017 \(\int \frac{1}{(d+e x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=192 \[ -\frac{7 c^2 d^2 e}{\sqrt{d+e x} \left (c d^2-a e^2\right )^4}+\frac{7 c^{5/2} d^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}}-\frac{7 c d e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}-\frac{1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{7 e}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2} \]

[Out]

(-7*e)/(5*(c*d^2 - a*e^2)^2*(d + e*x)^(5/2)) - 1/((c*d^2 - a*e^2)*(a*e + c*d*x)*(d + e*x)^(5/2)) - (7*c*d*e)/(
3*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - (7*c^2*d^2*e)/((c*d^2 - a*e^2)^4*Sqrt[d + e*x]) + (7*c^(5/2)*d^(5/2)*e*
ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(9/2)

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Rubi [A]  time = 0.128608, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \[ -\frac{7 c^2 d^2 e}{\sqrt{d+e x} \left (c d^2-a e^2\right )^4}+\frac{7 c^{5/2} d^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}}-\frac{7 c d e}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3}-\frac{1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}-\frac{7 e}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*e)/(5*(c*d^2 - a*e^2)^2*(d + e*x)^(5/2)) - 1/((c*d^2 - a*e^2)*(a*e + c*d*x)*(d + e*x)^(5/2)) - (7*c*d*e)/(
3*(c*d^2 - a*e^2)^3*(d + e*x)^(3/2)) - (7*c^2*d^2*e)/((c*d^2 - a*e^2)^4*Sqrt[d + e*x]) + (7*c^(5/2)*d^(5/2)*e*
ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(9/2)

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}{(d+e x)^{3/2} \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)^{7/2}} \, dx\\ &=-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac{(7 e) \int \frac{1}{(a e+c d x) (d+e x)^{7/2}} \, dx}{2 \left (c d^2-a e^2\right )}\\ &=-\frac{7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac{(7 c d e) \int \frac{1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{2 \left (c d^2-a e^2\right )^2}\\ &=-\frac{7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac{7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{\left (7 c^2 d^2 e\right ) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{2 \left (c d^2-a e^2\right )^3}\\ &=-\frac{7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac{7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt{d+e x}}-\frac{\left (7 c^3 d^3 e\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{2 \left (c d^2-a e^2\right )^4}\\ &=-\frac{7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac{7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt{d+e x}}-\frac{\left (7 c^3 d^3\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 )^4}\\ &=-\frac{7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac{1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac{7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac{7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt{d+e x}}+\frac{7 c^{5/2} d^{5/2} 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 )^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.018472, size = 59, normalized size = 0.31 \[ -\frac{2 e \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{5 (d+e x)^{5/2} \left (a e^2-c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*Hypergeometric2F1[-5/2, 2, -3/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(5*(-(c*d^2) + a*e^2)^2*(d + e*
x)^(5/2))

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Maple [A]  time = 0.204, size = 193, normalized size = 1. \begin{align*} -{\frac{2\,e}{5\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-6\,{\frac{{c}^{2}e{d}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}\sqrt{ex+d}}}+{\frac{4\,dec}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{{c}^{3}e{d}^{3}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4} \left ( cdex+a{e}^{2} \right ) }\sqrt{ex+d}}-7\,{\frac{{c}^{3}e{d}^{3}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}\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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x)

[Out]

-2/5*e/(a*e^2-c*d^2)^2/(e*x+d)^(5/2)-6*e/(a*e^2-c*d^2)^4*c^2*d^2/(e*x+d)^(1/2)+4/3*e/(a*e^2-c*d^2)^3*c*d/(e*x+
d)^(3/2)-e*c^3*d^3/(a*e^2-c*d^2)^4*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)-7*e*c^3*d^3/(a*e^2-c*d^2)^4/((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)^(3/2)/(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 = 2.13286, size = 2678, normalized size = 13.95 \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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

[1/30*(105*(c^3*d^3*e^4*x^4 + a*c^2*d^5*e^2 + (3*c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + 3*(c^3*d^5*e^2 + a*c^2*d^3
*e^4)*x^2 + (c^3*d^6*e + 3*a*c^2*d^4*e^3)*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*(105*c^3*d^3*e^3*x^3 + 15*c^3*d^6 + 11
6*a*c^2*d^4*e^2 - 32*a^2*c*d^2*e^4 + 6*a^3*e^6 + 35*(7*c^3*d^4*e^2 + 2*a*c^2*d^2*e^4)*x^2 + 7*(23*c^3*d^5*e +
24*a*c^2*d^3*e^3 - 2*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a*c^4*d^11*e - 4*a^2*c^3*d^9*e^3 + 6*a^3*c^2*d^7*e^5 - 4*
a^4*c*d^5*e^7 + a^5*d^3*e^9 + (c^5*d^9*e^3 - 4*a*c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d
*e^11)*x^4 + (3*c^5*d^10*e^2 - 11*a*c^4*d^8*e^4 + 14*a^2*c^3*d^6*e^6 - 6*a^3*c^2*d^4*e^8 - a^4*c*d^2*e^10 + a^
5*e^12)*x^3 + 3*(c^5*d^11*e - 3*a*c^4*d^9*e^3 + 2*a^2*c^3*d^7*e^5 + 2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*
d*e^11)*x^2 + (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 - 11*a^4*c*d^4*e^8 + 3*a^5*d
^2*e^10)*x), 1/15*(105*(c^3*d^3*e^4*x^4 + a*c^2*d^5*e^2 + (3*c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + 3*(c^3*d^5*e^2
+ a*c^2*d^3*e^4)*x^2 + (c^3*d^6*e + 3*a*c^2*d^4*e^3)*x)*sqrt(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sq
rt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2)) - (105*c^3*d^3*e^3*x^3 + 15*c^3*d^6 + 116*a*c^2*d^4*
e^2 - 32*a^2*c*d^2*e^4 + 6*a^3*e^6 + 35*(7*c^3*d^4*e^2 + 2*a*c^2*d^2*e^4)*x^2 + 7*(23*c^3*d^5*e + 24*a*c^2*d^3
*e^3 - 2*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a*c^4*d^11*e - 4*a^2*c^3*d^9*e^3 + 6*a^3*c^2*d^7*e^5 - 4*a^4*c*d^5*e^
7 + a^5*d^3*e^9 + (c^5*d^9*e^3 - 4*a*c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d*e^11)*x^4 +
(3*c^5*d^10*e^2 - 11*a*c^4*d^8*e^4 + 14*a^2*c^3*d^6*e^6 - 6*a^3*c^2*d^4*e^8 - a^4*c*d^2*e^10 + a^5*e^12)*x^3
+ 3*(c^5*d^11*e - 3*a*c^4*d^9*e^3 + 2*a^2*c^3*d^7*e^5 + 2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*d*e^11)*x^2
+ (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 - 11*a^4*c*d^4*e^8 + 3*a^5*d^2*e^10)*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)**(3/2)/(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)^(3/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

Timed out