### 3.2007 $$\int \frac{1}{(d+e x)^{5/2} (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx$$

Optimal. Leaf size=153 $\frac{2 c^2 d^2}{\sqrt{d+e x} \left (c d^2-a e^2\right )^3}-\frac{2 c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}+\frac{2 c d}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}+\frac{2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}$

[Out]

2/(5*(c*d^2 - a*e^2)*(d + e*x)^(5/2)) + (2*c*d)/(3*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2)) + (2*c^2*d^2)/((c*d^2 -
a*e^2)^3*Sqrt[d + e*x]) - (2*c^(5/2)*d^(5/2)*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 = 0.129555, antiderivative size = 153, 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} $\frac{2 c^2 d^2}{\sqrt{d+e x} \left (c d^2-a e^2\right )^3}-\frac{2 c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}}+\frac{2 c d}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}+\frac{2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

2/(5*(c*d^2 - a*e^2)*(d + e*x)^(5/2)) + (2*c*d)/(3*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2)) + (2*c^2*d^2)/((c*d^2 -
a*e^2)^3*Sqrt[d + e*x]) - (2*c^(5/2)*d^(5/2)*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}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx &=\int \frac{1}{(a e+c d x) (d+e x)^{7/2}} \, dx\\ &=\frac{2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{(c d) \int \frac{1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{c d^2-a e^2}\\ &=\frac{2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{\left (c^2 d^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{\left (c d^2-a e^2\right )^2}\\ &=\frac{2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt{d+e x}}+\frac{\left (c^3 d^3\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{\left (c d^2-a e^2\right )^3}\\ &=\frac{2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt{d+e x}}+\frac{\left (2 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 )}{e \left (c d^2-a e^2\right )^3}\\ &=\frac{2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac{2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac{2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt{d+e x}}-\frac{2 c^{5/2} d^{5/2} \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 = 0.0126572, size = 57, normalized size = 0.37 $\frac{2 \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};\frac{c d (d+e x)}{c d^2-a e^2}\right )}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Hypergeometric2F1[-5/2, 1, -3/2, (c*d*(d + e*x))/(c*d^2 - a*e^2)])/(5*(c*d^2 - a*e^2)*(d + e*x)^(5/2))

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Maple [A]  time = 0.198, size = 146, normalized size = 1. \begin{align*} -{\frac{2}{5\,a{e}^{2}-5\,c{d}^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-2\,{\frac{{c}^{2}{d}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}\sqrt{ex+d}}}+{\frac{2\,cd}{3\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{c}^{3}{d}^{3}}{ \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)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x)

[Out]

-2/5/(a*e^2-c*d^2)/(e*x+d)^(5/2)-2*c^2*d^2/(a*e^2-c*d^2)^3/(e*x+d)^(1/2)+2/3*c*d/(a*e^2-c*d^2)^2/(e*x+d)^(3/2)
-2*c^3*d^3/(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)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.05397, size = 1555, normalized size = 10.16 \begin{align*} \left [-\frac{15 \,{\left (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}\right )} \sqrt{\frac{c d}{c d^{2} - a e^{2}}} \log \left (\frac{c d e x + 2 \, c d^{2} - a e^{2} + 2 \,{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{\frac{c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \,{\left (15 \, c^{2} d^{2} e^{2} x^{2} + 23 \, c^{2} d^{4} - 11 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 5 \,{\left (7 \, c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (c^{3} d^{9} - 3 \, a c^{2} d^{7} e^{2} + 3 \, a^{2} c d^{5} e^{4} - a^{3} d^{3} e^{6} +{\left (c^{3} d^{6} e^{3} - 3 \, a c^{2} d^{4} e^{5} + 3 \, a^{2} c d^{2} e^{7} - a^{3} e^{9}\right )} x^{3} + 3 \,{\left (c^{3} d^{7} e^{2} - 3 \, a c^{2} d^{5} e^{4} + 3 \, a^{2} c d^{3} e^{6} - a^{3} d e^{8}\right )} x^{2} + 3 \,{\left (c^{3} d^{8} e - 3 \, a c^{2} d^{6} e^{3} + 3 \, a^{2} c d^{4} e^{5} - a^{3} d^{2} e^{7}\right )} x\right )}}, -\frac{2 \,{\left (15 \,{\left (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}\right )} \sqrt{-\frac{c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac{{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{-\frac{c d}{c d^{2} - a e^{2}}}}{c d e x + c d^{2}}\right ) -{\left (15 \, c^{2} d^{2} e^{2} x^{2} + 23 \, c^{2} d^{4} - 11 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 5 \,{\left (7 \, c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt{e x + d}\right )}}{15 \,{\left (c^{3} d^{9} - 3 \, a c^{2} d^{7} e^{2} + 3 \, a^{2} c d^{5} e^{4} - a^{3} d^{3} e^{6} +{\left (c^{3} d^{6} e^{3} - 3 \, a c^{2} d^{4} e^{5} + 3 \, a^{2} c d^{2} e^{7} - a^{3} e^{9}\right )} x^{3} + 3 \,{\left (c^{3} d^{7} e^{2} - 3 \, a c^{2} d^{5} e^{4} + 3 \, a^{2} c d^{3} e^{6} - a^{3} d e^{8}\right )} x^{2} + 3 \,{\left (c^{3} d^{8} e - 3 \, a c^{2} d^{6} e^{3} + 3 \, a^{2} c d^{4} e^{5} - a^{3} d^{2} e^{7}\right )} x\right )}}\right ] \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),x, algorithm="fricas")

[Out]

[-1/15*(15*(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/(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 + 23*c^2*d^4 - 11*a*c*d^2*e^2 + 3*a^2*e^4 + 5*(7*c^2*d^3*e - a*c*d*e^3)*x)*sqrt(e*x + d))/(c^3*d^9
- 3*a*c^2*d^7*e^2 + 3*a^2*c*d^5*e^4 - a^3*d^3*e^6 + (c^3*d^6*e^3 - 3*a*c^2*d^4*e^5 + 3*a^2*c*d^2*e^7 - a^3*e^
9)*x^3 + 3*(c^3*d^7*e^2 - 3*a*c^2*d^5*e^4 + 3*a^2*c*d^3*e^6 - a^3*d*e^8)*x^2 + 3*(c^3*d^8*e - 3*a*c^2*d^6*e^3
+ 3*a^2*c*d^4*e^5 - a^3*d^2*e^7)*x), -2/15*(15*(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/(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 + 23*c^2*d^4 - 11*a*c*d^2*e^2 + 3*a^2*e^4 + 5*(7*c^2*d^3*e - a*c*d*e^3)*x)*sqrt(e*x +
d))/(c^3*d^9 - 3*a*c^2*d^7*e^2 + 3*a^2*c*d^5*e^4 - a^3*d^3*e^6 + (c^3*d^6*e^3 - 3*a*c^2*d^4*e^5 + 3*a^2*c*d^2*
e^7 - a^3*e^9)*x^3 + 3*(c^3*d^7*e^2 - 3*a*c^2*d^5*e^4 + 3*a^2*c*d^3*e^6 - a^3*d*e^8)*x^2 + 3*(c^3*d^8*e - 3*a*
c^2*d^6*e^3 + 3*a^2*c*d^4*e^5 - a^3*d^2*e^7)*x)]

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Sympy [A]  time = 40.8523, size = 141, normalized size = 0.92 \begin{align*} - \frac{2 c^{2} d^{2}}{\sqrt{d + e x} \left (a e^{2} - c d^{2}\right )^{3}} - \frac{2 c^{2} d^{2} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{a e^{2} - c d^{2}}{c d}}} \right )}}{\sqrt{\frac{a e^{2} - c d^{2}}{c d}} \left (a e^{2} - c d^{2}\right )^{3}} + \frac{2 c d}{3 \left (d + e x\right )^{\frac{3}{2}} \left (a e^{2} - c d^{2}\right )^{2}} - \frac{2}{5 \left (d + e x\right )^{\frac{5}{2}} \left (a e^{2} - c d^{2}\right )} \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),x)

[Out]

-2*c**2*d**2/(sqrt(d + e*x)*(a*e**2 - c*d**2)**3) - 2*c**2*d**2*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d**2)/(c*d
)))/(sqrt((a*e**2 - c*d**2)/(c*d))*(a*e**2 - c*d**2)**3) + 2*c*d/(3*(d + e*x)**(3/2)*(a*e**2 - c*d**2)**2) - 2
/(5*(d + e*x)**(5/2)*(a*e**2 - c*d**2))

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

[Out]

Timed out