### 3.1938 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^3} \, dx$$

Optimal. Leaf size=244 $\frac{5 \left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2}-\frac{5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 \sqrt{c} \sqrt{d} e^{7/2}}$

[Out]

(5*(c*d^2 - a*e^2)*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*e^3) - (5*c*d*(
a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(3*e^2) + (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(e*(
d + e*x)^2) - (5*(c*d^2 - a*e^2)^3*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])])/(16*Sqrt[c]*Sqrt[d]*e^(7/2))

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Rubi [A]  time = 0.20238, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.135, Rules used = {662, 664, 612, 621, 206} $\frac{5 \left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2}-\frac{5 \left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 \sqrt{c} \sqrt{d} e^{7/2}}$

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)^3,x]

[Out]

(5*(c*d^2 - a*e^2)*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*e^3) - (5*c*d*(
a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(3*e^2) + (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(e*(
d + e*x)^2) - (5*(c*d^2 - a*e^2)^3*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])])/(16*Sqrt[c]*Sqrt[d]*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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*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)^3} \, 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)^2}-\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} \, 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}}{3 e^2}+\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)^2}+\frac{\left (5 c d \left (c d^2-a e^2\right )\right ) \int \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{2 e^2}\\ &=\frac{5 \left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 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}}{3 e^2}+\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)^2}-\frac{\left (5 \left (c d^2-a e^2\right )^3\right ) \int \frac{1}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 e^3}\\ &=\frac{5 \left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 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}}{3 e^2}+\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)^2}-\frac{\left (5 \left (c d^2-a e^2\right )^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 )}{8 e^3}\\ &=\frac{5 \left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 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}}{3 e^2}+\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)^2}-\frac{5 \left (c d^2-a e^2\right )^3 \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 )}{16 \sqrt{c} \sqrt{d} e^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.524665, size = 252, normalized size = 1.03 $\frac{\sqrt{e} (c d)^{3/2} (d+e x) \left (a^2 c d e^3 (59 e x-40 d)+33 a^3 e^5+a c^2 d^2 e \left (15 d^2-50 d e x+34 e^2 x^2\right )+c^3 d^3 x \left (15 d^2-10 d e x+8 e^2 x^2\right )\right )-15 \sqrt{c} \sqrt{d} \left (c d^2-a e^2\right )^{7/2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )}{24 e^{7/2} (c d)^{3/2} \sqrt{(d+e x) (a e+c d x)}}$

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)^3,x]

[Out]

((c*d)^(3/2)*Sqrt[e]*(d + e*x)*(33*a^3*e^5 + a^2*c*d*e^3*(-40*d + 59*e*x) + c^3*d^3*x*(15*d^2 - 10*d*e*x + 8*e
^2*x^2) + a*c^2*d^2*e*(15*d^2 - 50*d*e*x + 34*e^2*x^2)) - 15*Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^(7/2)*Sqrt[a*e +
c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sq
rt[c*d^2 - a*e^2])])/(24*(c*d)^(3/2)*e^(7/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [B]  time = 0.051, size = 1531, normalized size = 6.3 \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)^3,x)

[Out]

2/e^3/(a*e^2-c*d^2)/(d/e+x)^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)-16/3/e^2*d*c/(a*e^2-c*d^2)^2/(d/e+
x)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(7/2)+16/3/e*d^2*c^2/(a*e^2-c*d^2)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^
2)*(d/e+x))^(5/2)+5/4*e^3*d^2*c/(a*e^2-c*d^2)^2*a^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-10/3/e*d^4*c
^3/(a*e^2-c*d^2)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x-5/3/e^2*d^5*c^3/(a*e^2-c*d^2)^2*(c*d*e*(d/e
+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+5/8/e^3*d^8*c^4/(a*e^2-c*d^2)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/
2)+10/3*e*d^2*c^2/(a*e^2-c*d^2)^2*a*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)*x-5/4*e^4*d*c/(a*e^2-c*d^2)^
2*a^3*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-15/4*d^5*c^3/(a*e^2-c*d^2)^2*a*(c*d*e*(d/e+x)^2+(a*e^2-c
*d^2)*(d/e+x))^(1/2)*x-5/16/e^3*d^10*c^5/(a*e^2-c*d^2)^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)+25/8*e^3*d^4*c^2/(a*e^2-c*d^2)^2*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)+5/4/e^2*
d^7*c^4/(a*e^2-c*d^2)^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x+15/4*e^2*d^3*c^2/(a*e^2-c*d^2)^2*a^2*(
c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)*x-5/8*e^5/(a*e^2-c*d^2)^2*a^4*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e
+x))^(1/2)-5/4/e*d^6*c^3/(a*e^2-c*d^2)^2*a*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(1/2)-25/16*e^5*d^2*c/(a*e^
2-c*d^2)^2*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)+5/3*e^2*d*c/(a*e^2-c*d^2)^2*a^2*(c*d*e*(d/e+x)^2+(a*e^2-c*d^2)*(d/e+x))^(3/2)+5/16*e^7/(a*e
^2-c*d^2)^2*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)-25/8*e*d^6*c^3/(a*e^2-c*d^2)^2*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)+25/16/e*d^8*c^4/(a*e^2-c*d^2)^2*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)

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82404, size = 1130, normalized size = 4.63 \begin{align*} \left [\frac{15 \,{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt{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 \, \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{c d e} + 8 \,{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \,{\left (8 \, c^{3} d^{3} e^{3} x^{2} + 15 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 33 \, a^{2} c d e^{5} - 2 \,{\left (5 \, c^{3} d^{4} e^{2} - 13 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{96 \, c d e^{4}}, \frac{15 \,{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt{-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{-c d e}}{2 \,{\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} +{\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) + 2 \,{\left (8 \, c^{3} d^{3} e^{3} x^{2} + 15 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 33 \, a^{2} c d e^{5} - 2 \,{\left (5 \, c^{3} d^{4} e^{2} - 13 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{48 \, c d e^{4}}\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)^3,x, algorithm="fricas")

[Out]

[1/96*(15*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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*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) + 8*(c^2*d^3*e + a*c*d*e^3)*x) + 4*(8*c^3*d^3*e^3*x^2 + 15*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 33*a^2*c*d*e^5 -
2*(5*c^3*d^4*e^2 - 13*a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c*d*e^4), 1/48*(15*(c^3
*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-c*d*e)*arctan(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^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)
) + 2*(8*c^3*d^3*e^3*x^2 + 15*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 33*a^2*c*d*e^5 - 2*(5*c^3*d^4*e^2 - 13*a*c^2*d^2*
e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c*d*e^4)]

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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)**3,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)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError