### 3.2018 $$\int \frac{(d+e x)^{15/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx$$

Optimal. Leaf size=222 $\frac{21 e^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{4 c^4 d^4}+\frac{63 e^2 \sqrt{d+e x} \left (c d^2-a e^2\right )^2}{4 c^5 d^5}-\frac{63 e^2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}$

[Out]

(63*e^2*(c*d^2 - a*e^2)^2*Sqrt[d + e*x])/(4*c^5*d^5) + (21*e^2*(c*d^2 - a*e^2)*(d + e*x)^(3/2))/(4*c^4*d^4) +
(63*e^2*(d + e*x)^(5/2))/(20*c^3*d^3) - (9*e*(d + e*x)^(7/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(9/2)/(2*c
*d*(a*e + c*d*x)^2) - (63*e^2*(c*d^2 - a*e^2)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2
]])/(4*c^(11/2)*d^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.185903, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.135, Rules used = {626, 47, 50, 63, 208} $\frac{21 e^2 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{4 c^4 d^4}+\frac{63 e^2 \sqrt{d+e x} \left (c d^2-a e^2\right )^2}{4 c^5 d^5}-\frac{63 e^2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(63*e^2*(c*d^2 - a*e^2)^2*Sqrt[d + e*x])/(4*c^5*d^5) + (21*e^2*(c*d^2 - a*e^2)*(d + e*x)^(3/2))/(4*c^4*d^4) +
(63*e^2*(d + e*x)^(5/2))/(20*c^3*d^3) - (9*e*(d + e*x)^(7/2))/(4*c^2*d^2*(a*e + c*d*x)) - (d + e*x)^(9/2)/(2*c
*d*(a*e + c*d*x)^2) - (63*e^2*(c*d^2 - a*e^2)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2
]])/(4*c^(11/2)*d^(11/2))

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac{(d+e x)^{9/2}}{(a e+c d x)^3} \, dx\\ &=-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{(9 e) \int \frac{(d+e x)^{7/2}}{(a e+c d x)^2} \, dx}{4 c d}\\ &=-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{\left (63 e^2\right ) \int \frac{(d+e x)^{5/2}}{a e+c d x} \, dx}{8 c^2 d^2}\\ &=\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{\left (63 e^2 \left (c d^2-a e^2\right )\right ) \int \frac{(d+e x)^{3/2}}{a e+c d x} \, dx}{8 c^3 d^3}\\ &=\frac{21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{\left (63 e^2 \left (c d^2-a e^2\right )^2\right ) \int \frac{\sqrt{d+e x}}{a e+c d x} \, dx}{8 c^4 d^4}\\ &=\frac{63 e^2 \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}{4 c^5 d^5}+\frac{21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{\left (63 e^2 \left (c d^2-a e^2\right )^3\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 c^5 d^5}\\ &=\frac{63 e^2 \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}{4 c^5 d^5}+\frac{21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}+\frac{\left (63 e \left (c d^2-a e^2\right )^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 )}{4 c^5 d^5}\\ &=\frac{63 e^2 \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}{4 c^5 d^5}+\frac{21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac{63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac{9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac{(d+e x)^{9/2}}{2 c d (a e+c d x)^2}-\frac{63 e^2 \left (c d^2-a e^2\right )^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0255233, size = 61, normalized size = 0.27 $\frac{2 e^2 (d+e x)^{11/2} \, _2F_1\left (3,\frac{11}{2};\frac{13}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{11 \left (a e^2-c d^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]

[Out]

(2*e^2*(d + e*x)^(11/2)*Hypergeometric2F1[3, 11/2, 13/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(11*(-(c*d^2)
+ a*e^2)^3)

________________________________________________________________________________________

Maple [B]  time = 0.239, size = 635, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)

[Out]

2/5*e^2*(e*x+d)^(5/2)/c^3/d^3-2*e^4/c^4/d^4*(e*x+d)^(3/2)*a+2*e^2/c^3/d^2*(e*x+d)^(3/2)+12*e^6/c^5/d^5*a^2*(e*
x+d)^(1/2)-24*e^4/c^4/d^3*a*(e*x+d)^(1/2)+12*e^2/c^3/d*(e*x+d)^(1/2)+17/4*e^8/c^4/d^4/(c*d*e*x+a*e^2)^2*(e*x+d
)^(3/2)*a^3-51/4*e^6/c^3/d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*a^2+51/4*e^4/c^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)*
a-17/4*e^2/c*d^2/(c*d*e*x+a*e^2)^2*(e*x+d)^(3/2)+15/4*e^10/c^5/d^5/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^4-15*e^8/
c^4/d^3/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^3+45/2*e^6/c^3/d/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a^2-15*e^4/c^2*d/(c
*d*e*x+a*e^2)^2*(e*x+d)^(1/2)*a+15/4*e^2/c*d^3/(c*d*e*x+a*e^2)^2*(e*x+d)^(1/2)-63/4*e^8/c^5/d^5/((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))*a^3+189/4*e^6/c^4/d^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))*a^2-189/4*e^4/c^3/d/((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))*a+63/4*e^2/c^2*d/((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))

________________________________________________________________________________________

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((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.03045, size = 1748, normalized size = 7.87 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/40*(315*(a^2*c^2*d^4*e^4 - 2*a^3*c*d^2*e^6 + a^4*e^8 + (c^4*d^6*e^2 - 2*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^
2 + 2*(a*c^3*d^5*e^3 - 2*a^2*c^2*d^3*e^5 + a^3*c*d*e^7)*x)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2
- a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(8*c^4*d^4*e^4*x^4 - 10*c^4*d^8
- 45*a*c^3*d^6*e^2 + 483*a^2*c^2*d^4*e^4 - 735*a^3*c*d^2*e^6 + 315*a^4*e^8 + 8*(7*c^4*d^5*e^3 - 3*a*c^3*d^3*e^
5)*x^3 + 24*(12*c^4*d^6*e^2 - 17*a*c^3*d^4*e^4 + 7*a^2*c^2*d^2*e^6)*x^2 - (85*c^4*d^7*e - 831*a*c^3*d^5*e^3 +
1239*a^2*c^2*d^3*e^5 - 525*a^3*c*d*e^7)*x)*sqrt(e*x + d))/(c^7*d^7*x^2 + 2*a*c^6*d^6*e*x + a^2*c^5*d^5*e^2), -
1/20*(315*(a^2*c^2*d^4*e^4 - 2*a^3*c*d^2*e^6 + a^4*e^8 + (c^4*d^6*e^2 - 2*a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^2
+ 2*(a*c^3*d^5*e^3 - 2*a^2*c^2*d^3*e^5 + a^3*c*d*e^7)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c
*d*sqrt(-(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (8*c^4*d^4*e^4*x^4 - 10*c^4*d^8 - 45*a*c^3*d^6*e^2 + 483*a^
2*c^2*d^4*e^4 - 735*a^3*c*d^2*e^6 + 315*a^4*e^8 + 8*(7*c^4*d^5*e^3 - 3*a*c^3*d^3*e^5)*x^3 + 24*(12*c^4*d^6*e^2
- 17*a*c^3*d^4*e^4 + 7*a^2*c^2*d^2*e^6)*x^2 - (85*c^4*d^7*e - 831*a*c^3*d^5*e^3 + 1239*a^2*c^2*d^3*e^5 - 525*
a^3*c*d*e^7)*x)*sqrt(e*x + d))/(c^7*d^7*x^2 + 2*a*c^6*d^6*e*x + a^2*c^5*d^5*e^2)]

________________________________________________________________________________________

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((e*x+d)**(15/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________

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((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

Timed out