### 3.1654 $$\int \frac{(d+e x)^{11/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx$$

Optimal. Leaf size=201 $-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}+\frac{77 e^3 (d+e x)^{3/2} (b d-a e)}{8 b^5}+\frac{231 e^3 \sqrt{d+e x} (b d-a e)^2}{8 b^6}-\frac{231 e^3 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{13/2}}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}$

[Out]

(231*e^3*(b*d - a*e)^2*Sqrt[d + e*x])/(8*b^6) + (77*e^3*(b*d - a*e)*(d + e*x)^(3/2))/(8*b^5) + (231*e^3*(d + e
*x)^(5/2))/(40*b^4) - (33*e^2*(d + e*x)^(7/2))/(8*b^3*(a + b*x)) - (11*e*(d + e*x)^(9/2))/(12*b^2*(a + b*x)^2)
- (d + e*x)^(11/2)/(3*b*(a + b*x)^3) - (231*e^3*(b*d - a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d -
a*e]])/(8*b^(13/2))

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Rubi [A]  time = 0.130738, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.179, Rules used = {27, 47, 50, 63, 208} $-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}+\frac{77 e^3 (d+e x)^{3/2} (b d-a e)}{8 b^5}+\frac{231 e^3 \sqrt{d+e x} (b d-a e)^2}{8 b^6}-\frac{231 e^3 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{13/2}}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(11/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(231*e^3*(b*d - a*e)^2*Sqrt[d + e*x])/(8*b^6) + (77*e^3*(b*d - a*e)*(d + e*x)^(3/2))/(8*b^5) + (231*e^3*(d + e
*x)^(5/2))/(40*b^4) - (33*e^2*(d + e*x)^(7/2))/(8*b^3*(a + b*x)) - (11*e*(d + e*x)^(9/2))/(12*b^2*(a + b*x)^2)
- (d + e*x)^(11/2)/(3*b*(a + b*x)^3) - (231*e^3*(b*d - a*e)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d -
a*e]])/(8*b^(13/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 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)^{11/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^{11/2}}{(a+b x)^4} \, dx\\ &=-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{(11 e) \int \frac{(d+e x)^{9/2}}{(a+b x)^3} \, dx}{6 b}\\ &=-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{\left (33 e^2\right ) \int \frac{(d+e x)^{7/2}}{(a+b x)^2} \, dx}{8 b^2}\\ &=-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{\left (231 e^3\right ) \int \frac{(d+e x)^{5/2}}{a+b x} \, dx}{16 b^3}\\ &=\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{\left (231 e^3 (b d-a e)\right ) \int \frac{(d+e x)^{3/2}}{a+b x} \, dx}{16 b^4}\\ &=\frac{77 e^3 (b d-a e) (d+e x)^{3/2}}{8 b^5}+\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{\left (231 e^3 (b d-a e)^2\right ) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{16 b^5}\\ &=\frac{231 e^3 (b d-a e)^2 \sqrt{d+e x}}{8 b^6}+\frac{77 e^3 (b d-a e) (d+e x)^{3/2}}{8 b^5}+\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{\left (231 e^3 (b d-a e)^3\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b^6}\\ &=\frac{231 e^3 (b d-a e)^2 \sqrt{d+e x}}{8 b^6}+\frac{77 e^3 (b d-a e) (d+e x)^{3/2}}{8 b^5}+\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}+\frac{\left (231 e^2 (b d-a e)^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{8 b^6}\\ &=\frac{231 e^3 (b d-a e)^2 \sqrt{d+e x}}{8 b^6}+\frac{77 e^3 (b d-a e) (d+e x)^{3/2}}{8 b^5}+\frac{231 e^3 (d+e x)^{5/2}}{40 b^4}-\frac{33 e^2 (d+e x)^{7/2}}{8 b^3 (a+b x)}-\frac{11 e (d+e x)^{9/2}}{12 b^2 (a+b x)^2}-\frac{(d+e x)^{11/2}}{3 b (a+b x)^3}-\frac{231 e^3 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{13/2}}\\ \end{align*}

Mathematica [C]  time = 0.0252909, size = 52, normalized size = 0.26 $\frac{2 e^3 (d+e x)^{13/2} \, _2F_1\left (4,\frac{13}{2};\frac{15}{2};-\frac{b (d+e x)}{a e-b d}\right )}{13 (a e-b d)^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(11/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(2*e^3*(d + e*x)^(13/2)*Hypergeometric2F1[4, 13/2, 15/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(13*(-(b*d) + a*e)^
4)

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Maple [B]  time = 0.209, size = 719, normalized size = 3.6 \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)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

-355/8*e^7/b^5/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d*a^4-693/8*e^4/b^4/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*
e-b*d)*b)^(1/2))*a*d^2-267/8*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(5/2)*a^2*d+2/5*e^3*(e*x+d)^(5/2)/b^4+231/8*e^3/b^3
/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*d^3-40*e^4/b^5*a*d*(e*x+d)^(1/2)+89/8*e^6/b^4
/(b*e*x+a*e)^3*(e*x+d)^(5/2)*a^3+59/3*e^7/b^5/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a^4+71/8*e^8/b^6/(b*e*x+a*e)^3*(e*x+
d)^(1/2)*a^5-231/8*e^6/b^6/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3-89/8*e^3/b/(b*e
*x+a*e)^3*(e*x+d)^(5/2)*d^3+59/3*e^3/b/(b*e*x+a*e)^3*(e*x+d)^(3/2)*d^4-8/3*e^4/b^5*(e*x+d)^(3/2)*a+20*e^5/b^6*
a^2*(e*x+d)^(1/2)+8/3*e^3/b^4*(e*x+d)^(3/2)*d+20*e^3/b^4*d^2*(e*x+d)^(1/2)+693/8*e^5/b^5/((a*e-b*d)*b)^(1/2)*a
rctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*d*a^2-236/3*e^6/b^4/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a^3*d+118*e^5/b^3/(
b*e*x+a*e)^3*(e*x+d)^(3/2)*a^2*d^2-236/3*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a*d^3-71/8*e^3/b/(b*e*x+a*e)^3*(e
*x+d)^(1/2)*d^5+267/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(5/2)*a*d^2+355/4*e^6/b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^3*
d^2-355/4*e^5/b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^2*d^3+355/8*e^4/b^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a*d^4

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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((e*x+d)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.0346, size = 2140, normalized size = 10.65 \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)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/240*(3465*(a^3*b^2*d^2*e^3 - 2*a^4*b*d*e^4 + a^5*e^5 + (b^5*d^2*e^3 - 2*a*b^4*d*e^4 + a^2*b^3*e^5)*x^3 + 3*
(a*b^4*d^2*e^3 - 2*a^2*b^3*d*e^4 + a^3*b^2*e^5)*x^2 + 3*(a^2*b^3*d^2*e^3 - 2*a^3*b^2*d*e^4 + a^4*b*e^5)*x)*sqr
t((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(48*b^5*e^5*
x^5 - 40*b^5*d^5 - 110*a*b^4*d^4*e - 495*a^2*b^3*d^3*e^2 + 5313*a^3*b^2*d^2*e^3 - 8085*a^4*b*d*e^4 + 3465*a^5*
e^5 + 16*(26*b^5*d*e^4 - 11*a*b^4*e^5)*x^4 + 16*(173*b^5*d^2*e^3 - 242*a*b^4*d*e^4 + 99*a^2*b^3*e^5)*x^3 - 3*(
445*b^5*d^3*e^2 - 4103*a*b^4*d^2*e^3 + 6039*a^2*b^3*d*e^4 - 2541*a^3*b^2*e^5)*x^2 - 2*(155*b^5*d^4*e + 715*a*b
^4*d^3*e^2 - 7227*a^2*b^3*d^2*e^3 + 10857*a^3*b^2*d*e^4 - 4620*a^4*b*e^5)*x)*sqrt(e*x + d))/(b^9*x^3 + 3*a*b^8
*x^2 + 3*a^2*b^7*x + a^3*b^6), -1/120*(3465*(a^3*b^2*d^2*e^3 - 2*a^4*b*d*e^4 + a^5*e^5 + (b^5*d^2*e^3 - 2*a*b^
4*d*e^4 + a^2*b^3*e^5)*x^3 + 3*(a*b^4*d^2*e^3 - 2*a^2*b^3*d*e^4 + a^3*b^2*e^5)*x^2 + 3*(a^2*b^3*d^2*e^3 - 2*a^
3*b^2*d*e^4 + a^4*b*e^5)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (
48*b^5*e^5*x^5 - 40*b^5*d^5 - 110*a*b^4*d^4*e - 495*a^2*b^3*d^3*e^2 + 5313*a^3*b^2*d^2*e^3 - 8085*a^4*b*d*e^4
+ 3465*a^5*e^5 + 16*(26*b^5*d*e^4 - 11*a*b^4*e^5)*x^4 + 16*(173*b^5*d^2*e^3 - 242*a*b^4*d*e^4 + 99*a^2*b^3*e^5
)*x^3 - 3*(445*b^5*d^3*e^2 - 4103*a*b^4*d^2*e^3 + 6039*a^2*b^3*d*e^4 - 2541*a^3*b^2*e^5)*x^2 - 2*(155*b^5*d^4*
e + 715*a*b^4*d^3*e^2 - 7227*a^2*b^3*d^2*e^3 + 10857*a^3*b^2*d*e^4 - 4620*a^4*b*e^5)*x)*sqrt(e*x + d))/(b^9*x^
3 + 3*a*b^8*x^2 + 3*a^2*b^7*x + a^3*b^6)]

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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((e*x+d)**(11/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.22319, size = 663, normalized size = 3.3 \begin{align*} \frac{231 \,{\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{8 \, \sqrt{-b^{2} d + a b e} b^{6}} - \frac{267 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{5} d^{3} e^{3} - 472 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{5} d^{4} e^{3} + 213 \, \sqrt{x e + d} b^{5} d^{5} e^{3} - 801 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{4} d^{2} e^{4} + 1888 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{4} d^{3} e^{4} - 1065 \, \sqrt{x e + d} a b^{4} d^{4} e^{4} + 801 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{2} b^{3} d e^{5} - 2832 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{3} d^{2} e^{5} + 2130 \, \sqrt{x e + d} a^{2} b^{3} d^{3} e^{5} - 267 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{3} b^{2} e^{6} + 1888 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} b^{2} d e^{6} - 2130 \, \sqrt{x e + d} a^{3} b^{2} d^{2} e^{6} - 472 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{4} b e^{7} + 1065 \, \sqrt{x e + d} a^{4} b d e^{7} - 213 \, \sqrt{x e + d} a^{5} e^{8}}{24 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{6}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{16} e^{3} + 20 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{16} d e^{3} + 150 \, \sqrt{x e + d} b^{16} d^{2} e^{3} - 20 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{15} e^{4} - 300 \, \sqrt{x e + d} a b^{15} d e^{4} + 150 \, \sqrt{x e + d} a^{2} b^{14} e^{5}\right )}}{15 \, b^{20}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(11/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

231/8*(b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(
sqrt(-b^2*d + a*b*e)*b^6) - 1/24*(267*(x*e + d)^(5/2)*b^5*d^3*e^3 - 472*(x*e + d)^(3/2)*b^5*d^4*e^3 + 213*sqrt
(x*e + d)*b^5*d^5*e^3 - 801*(x*e + d)^(5/2)*a*b^4*d^2*e^4 + 1888*(x*e + d)^(3/2)*a*b^4*d^3*e^4 - 1065*sqrt(x*e
+ d)*a*b^4*d^4*e^4 + 801*(x*e + d)^(5/2)*a^2*b^3*d*e^5 - 2832*(x*e + d)^(3/2)*a^2*b^3*d^2*e^5 + 2130*sqrt(x*e
+ d)*a^2*b^3*d^3*e^5 - 267*(x*e + d)^(5/2)*a^3*b^2*e^6 + 1888*(x*e + d)^(3/2)*a^3*b^2*d*e^6 - 2130*sqrt(x*e +
d)*a^3*b^2*d^2*e^6 - 472*(x*e + d)^(3/2)*a^4*b*e^7 + 1065*sqrt(x*e + d)*a^4*b*d*e^7 - 213*sqrt(x*e + d)*a^5*e
^8)/(((x*e + d)*b - b*d + a*e)^3*b^6) + 2/15*(3*(x*e + d)^(5/2)*b^16*e^3 + 20*(x*e + d)^(3/2)*b^16*d*e^3 + 150
*sqrt(x*e + d)*b^16*d^2*e^3 - 20*(x*e + d)^(3/2)*a*b^15*e^4 - 300*sqrt(x*e + d)*a*b^15*d*e^4 + 150*sqrt(x*e +
d)*a^2*b^14*e^5)/b^20