### 3.1939 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^4} \, dx$$

Optimal. Leaf size=235 $-\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}+\frac{15 c d \left (a-\frac{c d^2}{e^2}\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e}+\frac{15 \sqrt{c} \sqrt{d} \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 e^{7/2}}$

[Out]

(15*c*d*(a - (c*d^2)/e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e) + (5*c*d*(a*d*e + (c*d^2 + a*e^2)
*x + c*d*e*x^2)^(3/2))/(2*e^2*(d + e*x)) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(e*(d + e*x)^3) +
(15*Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^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])])/(8*e^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.211624, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.108, Rules used = {662, 664, 621, 206} $-\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}+\frac{15 c d \left (a-\frac{c d^2}{e^2}\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e}+\frac{15 \sqrt{c} \sqrt{d} \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 e^{7/2}}$

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)^4,x]

[Out]

(15*c*d*(a - (c*d^2)/e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(4*e) + (5*c*d*(a*d*e + (c*d^2 + a*e^2)
*x + c*d*e*x^2)^(3/2))/(2*e^2*(d + e*x)) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(e*(d + e*x)^3) +
(15*Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^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])])/(8*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 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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)^4} \, dx &=-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac{(5 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)^2} \, dx}{e}\\ &=\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}-\frac{\left (15 c d \left (c d^2-a e^2\right )\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{d+e x} \, dx}{4 e^2}\\ &=-\frac{15 c d \left (c d^2-a e^2\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3}+\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac{\left (15 c d \left (c d^2-a e^2\right )^2\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 e^3}\\ &=-\frac{15 c d \left (c d^2-a e^2\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3}+\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac{\left (15 c d \left (c d^2-a e^2\right )^2\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 )}{4 e^3}\\ &=-\frac{15 c d \left (c d^2-a e^2\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e^3}+\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{2 e^2 (d+e x)}-\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^3}+\frac{15 \sqrt{c} \sqrt{d} \left (c d^2-a e^2\right )^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 )}{8 e^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0779383, size = 108, normalized size = 0.46 $\frac{2 c d (a e+c d x)^3 \sqrt{(d+e x) (a e+c d x)} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{7 \left (c d^2-a e^2\right )^2 \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}}}$

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)^4,x]

[Out]

(2*c*d*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*Hypergeometric2F1[3/2, 7/2, 9/2, (e*(a*e + c*d*x))/(-(c*d
^2) + a*e^2)])/(7*(c*d^2 - a*e^2)^2*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)])

________________________________________________________________________________________

Maple [B]  time = 0.05, size = 1617, normalized size = 6.9 \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)^4,x)

[Out]

-2/e^4/(a*e^2-c*d^2)/(d/e+x)^4*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)+12/e^3*d*c/(a*e^2-c*d^2)^2/(d/e+x
)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)-32/e^2*d^2*c^2/(a*e^2-c*d^2)^3/(d/e+x)^2*(c*d*e*(d/e+x)^2+(a
*e^2-c*d^2)*(d/e+x))^(7/2)+32/e*d^3*c^3/(a*e^2-c*d^2)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(5/2)+15/2*e^3
*d^3*c^2/(a*e^2-c*d^2)^3*a^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-20/e*d^5*c^4/(a*e^2-c*d^2)^3*(c*d*e
*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x-10/e^2*d^6*c^4/(a*e^2-c*d^2)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x
))^(3/2)+15/4/e^3*d^9*c^5/(a*e^2-c*d^2)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-15/2*e^4*d^2*c^2/(a*e^
2-c*d^2)^3*a^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x+20*e*d^3*c^3/(a*e^2-c*d^2)^3*a*(c*d*e*(d/e+x)^2
+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x-15/8/e^3*d^11*c^6/(a*e^2-c*d^2)^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)-75/8*e^5*d^3*c^2/(a*e^2-c*d^2)^3*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)
+75/4*e^3*d^5*c^3/(a*e^2-c*d^2)^3*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)+15/2/e^2*d^8*c^5/(a*e^2-c*d^2)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e
+x))^(1/2)*x+45/2*e^2*d^4*c^3/(a*e^2-c*d^2)^3*a^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-15/4*e^5*d*c
/(a*e^2-c*d^2)^3*a^4*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-15/2/e*d^7*c^4/(a*e^2-c*d^2)^3*a*(c*d*e*(d/
e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)+10*e^2*d^2*c^2/(a*e^2-c*d^2)^3*a^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))
^(3/2)-45/2*d^6*c^4/(a*e^2-c*d^2)^3*a*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x+15/8*e^7*d*c/(a*e^2-c*d^
2)^3*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)-75/4*e*d^7*c^4/(a*e^2-c*d^2)^3*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)+75/8/e*d^9*c^5/(a*e^2-c*d^2)^3*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)

________________________________________________________________________________________

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)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 3.8389, size = 1160, normalized size = 4.94 \begin{align*} \left [\frac{15 \,{\left (c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\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 (2 \, c^{2} d^{2} e^{2} x^{2} - 15 \, c^{2} d^{4} + 25 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4} -{\left (5 \, c^{2} d^{3} e - 9 \, 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}}{16 \,{\left (e^{4} x + d e^{3}\right )}}, -\frac{15 \,{\left (c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x\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 (2 \, c^{2} d^{2} e^{2} x^{2} - 15 \, c^{2} d^{4} + 25 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4} -{\left (5 \, c^{2} d^{3} e - 9 \, 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}}{8 \,{\left (e^{4} x + d 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)^4,x, algorithm="fricas")

[Out]

[1/16*(15*(c^2*d^5 - 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x)*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*(2*c^2*d^2*e^2*x^2 - 15*c^2*d^4 + 25*a*c*d^
2*e^2 - 8*a^2*e^4 - (5*c^2*d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^4*x + d*e^3
), -1/8*(15*(c^2*d^5 - 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x)*sqrt(-c*d/e)*arcta
n(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*(2*c^2*d^2*e^2*x^2 - 15*c^2*d^4 + 25*a*c*d^2*e^2 - 8*a^2*e^4 - (5*c^2*
d^3*e - 9*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^4*x + d*e^3)]

________________________________________________________________________________________

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)**4,x)

[Out]

Timed out

________________________________________________________________________________________

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)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError