### 3.1941 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^6} \, dx$$

Optimal. Leaf size=218 $-\frac{2 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3 (d+e x)}+\frac{c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{7/2}}-\frac{2 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}$

[Out]

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

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Rubi [A]  time = 0.132549, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.081, Rules used = {662, 621, 206} $-\frac{2 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3 (d+e x)}+\frac{c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{7/2}}-\frac{2 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^6,x]

[Out]

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

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^6} \, dx &=-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac{(c d) \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^4} \, dx}{e}\\ &=-\frac{2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac{\left (c^2 d^2\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^2} \, dx}{e^2}\\ &=-\frac{2 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 (d+e x)}-\frac{2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac{\left (c^3 d^3\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{e^3}\\ &=-\frac{2 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 (d+e x)}-\frac{2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac{\left (2 c^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{4 c d e-x^2} \, dx,x,\frac{c d^2+a e^2+2 c d e x}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{e^3}\\ &=-\frac{2 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 (d+e x)}-\frac{2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac{c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac{c d^2+a e^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{e^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.11802, size = 217, normalized size = 1. $\frac{2 \sqrt{(d+e x) (a e+c d x)} \left (\frac{15 c^{5/2} d^{5/2} \sqrt{c d} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )}{\sqrt{c d^2-a e^2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}}-\frac{\sqrt{e} \left (3 a^2 e^4+a c d e^2 (5 d+11 e x)+c^2 d^2 \left (15 d^2+35 d e x+23 e^2 x^2\right )\right )}{(d+e x)^3}\right )}{15 e^{7/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^6,x]

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-((Sqrt[e]*(3*a^2*e^4 + a*c*d*e^2*(5*d + 11*e*x) + c^2*d^2*(15*d^2 + 35*d*e*
x + 23*e^2*x^2)))/(d + e*x)^3) + (15*c^(5/2)*d^(5/2)*Sqrt[c*d]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d
*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[c*d^2 - a*e^2]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*
e^2)])))/(15*e^(7/2))

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Maple [B]  time = 0.052, size = 1764, normalized size = 8.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,x)

[Out]

-2/5/e^6/(a*e^2-c*d^2)/(d/e+x)^6*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)-4/15/e^5*d*c/(a*e^2-c*d^2)^2/(d
/e+x)^5*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)-16/15/e^4*d^2*c^2/(a*e^2-c*d^2)^3/(d/e+x)^4*(c*d*e*(d/e+
x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)+32/5/e^3*d^3*c^3/(a*e^2-c*d^2)^4/(d/e+x)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d
/e+x))^(7/2)-256/15/e^2*d^4*c^4/(a*e^2-c*d^2)^5/(d/e+x)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)+256/15
/e*d^5*c^5/(a*e^2-c*d^2)^5*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(5/2)+32/3*e*d^5*c^5/(a*e^2-c*d^2)^5*a*(c*d
*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x+4*e^3*d^5*c^4/(a*e^2-c*d^2)^5*a^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(
d/e+x))^(1/2)-12*d^8*c^6/(a*e^2-c*d^2)^5*a*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x+e^7*d^3*c^3/(a*e^2-
c*d^2)^5*a^5*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2
))/(d*e*c)^(1/2)-10*e*d^9*c^6/(a*e^2-c*d^2)^5*a^2*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*
(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+5/e*d^11*c^7/(a*e^2-c*d^2)^5*a*ln((1/2*a*e^2-1/2*c*d^2+(
d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)-32/3/e*d^7*c^6/(a*e^2
-c*d^2)^5*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x-16/3/e^2*d^8*c^6/(a*e^2-c*d^2)^5*(c*d*e*(d/e+x)^2+(a
*e^2-c*d^2)*(d/e+x))^(3/2)+2/e^3*d^11*c^7/(a*e^2-c*d^2)^5*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+4/e^2*
d^10*c^7/(a*e^2-c*d^2)^5*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-1/e^3*d^13*c^8/(a*e^2-c*d^2)^5*ln((1/
2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)+16
/3*e^2*d^4*c^4/(a*e^2-c*d^2)^5*a^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+12*e^2*d^6*c^5/(a*e^2-c*d^2)^
5*a^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-2*e^5*d^3*c^3/(a*e^2-c*d^2)^5*a^4*(c*d*e*(d/e+x)^2+(a*e^
2-c*d^2)*(d/e+x))^(1/2)-4/e*d^9*c^6/(a*e^2-c*d^2)^5*a*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-5*e^5*d^5*
c^4/(a*e^2-c*d^2)^5*a^4*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d
/e+x))^(1/2))/(d*e*c)^(1/2)+10*e^3*d^7*c^5/(a*e^2-c*d^2)^5*a^3*ln((1/2*a*e^2-1/2*c*d^2+(d/e+x)*c*d*e)/(d*e*c)^
(1/2)+(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2))/(d*e*c)^(1/2)-4*e^4*d^4*c^4/(a*e^2-c*d^2)^5*a^3*(c*d*e*(d
/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 13.5571, size = 1210, normalized size = 5.55 \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}{e}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \,{\left (2 \, c d e^{2} x + c d^{2} e + a e^{3}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{\frac{c d}{e}} + 8 \,{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \,{\left (23 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} + 5 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} +{\left (35 \, c^{2} d^{3} e + 11 \, a c d e^{3}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{30 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}, -\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}{e}} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt{-\frac{c d}{e}}}{2 \,{\left (c^{2} d^{2} e x^{2} + a c d^{2} e +{\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )}}\right ) + 2 \,{\left (23 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} + 5 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} +{\left (35 \, c^{2} d^{3} e + 11 \, a c d e^{3}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{15 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

[1/30*(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/e)*log(8*c^2*d^2*e^2*x^2 +
c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*(2*c*d*e^2*x + c*d^2*e + a*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)
*x)*sqrt(c*d/e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) - 4*(23*c^2*d^2*e^2*x^2 + 15*c^2*d^4 + 5*a*c*d^2*e^2 + 3*a^2*e^
4 + (35*c^2*d^3*e + 11*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d
^2*e^4*x + d^3*e^3), -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/e)*ar
ctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d/e)/(c^2*d^2*e*x^2 +
a*c*d^2*e + (c^2*d^3 + a*c*d*e^2)*x)) + 2*(23*c^2*d^2*e^2*x^2 + 15*c^2*d^4 + 5*a*c*d^2*e^2 + 3*a^2*e^4 + (35*
c^2*d^3*e + 11*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x
+ d^3*e^3)]

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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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**6,x)

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

Exception raised: TypeError