### 3.1721 $$\int \frac{(d+e x)^{11/2}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=346 $-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{385 e^4 (a+b x) (d+e x)^{3/2}}{64 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{1155 e^4 (a+b x) \sqrt{d+e x} (b d-a e)}{64 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1155 e^4 (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

(1155*e^4*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x])/(64*b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (385*e^4*(a + b*x)*(d
+ e*x)^(3/2))/(64*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (231*e^3*(d + e*x)^(5/2))/(64*b^4*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) - (33*e^2*(d + e*x)^(7/2))/(32*b^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (11*e*(d + e*x)^(9/2))
/(24*b^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(11/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) - (1155*e^4*(b*d - a*e)^(3/2)*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(13/2
)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.188314, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {646, 47, 50, 63, 208} $-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{385 e^4 (a+b x) (d+e x)^{3/2}}{64 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{1155 e^4 (a+b x) \sqrt{d+e x} (b d-a e)}{64 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1155 e^4 (a+b x) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(11/2)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(1155*e^4*(b*d - a*e)*(a + b*x)*Sqrt[d + e*x])/(64*b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (385*e^4*(a + b*x)*(d
+ e*x)^(3/2))/(64*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (231*e^3*(d + e*x)^(5/2))/(64*b^4*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) - (33*e^2*(d + e*x)^(7/2))/(32*b^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (11*e*(d + e*x)^(9/2))
/(24*b^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(11/2)/(4*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) - (1155*e^4*(b*d - a*e)^(3/2)*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*b^(13/2
)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 646

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

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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{11/2}}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (11 b^2 e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{9/2}}{\left (a b+b^2 x\right )^4} \, dx}{8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (33 e^2 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{7/2}}{\left (a b+b^2 x\right )^3} \, dx}{16 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (231 e^3 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{\left (a b+b^2 x\right )^2} \, dx}{64 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (1155 e^4 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{128 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{385 e^4 (a+b x) (d+e x)^{3/2}}{64 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (1155 e^4 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{128 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{1155 e^4 (b d-a e) (a+b x) \sqrt{d+e x}}{64 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{385 e^4 (a+b x) (d+e x)^{3/2}}{64 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (1155 e^4 \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{128 b^8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{1155 e^4 (b d-a e) (a+b x) \sqrt{d+e x}}{64 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{385 e^4 (a+b x) (d+e x)^{3/2}}{64 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (1155 e^3 \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{64 b^8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{1155 e^4 (b d-a e) (a+b x) \sqrt{d+e x}}{64 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{385 e^4 (a+b x) (d+e x)^{3/2}}{64 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 e^3 (d+e x)^{5/2}}{64 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{33 e^2 (d+e x)^{7/2}}{32 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{11 e (d+e x)^{9/2}}{24 b^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{11/2}}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1155 e^4 (b d-a e)^{3/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0491772, size = 67, normalized size = 0.19 $-\frac{2 e^4 (a+b x) (d+e x)^{13/2} \, _2F_1\left (5,\frac{13}{2};\frac{15}{2};\frac{b (d+e x)}{b d-a e}\right )}{13 \sqrt{(a+b x)^2} (b d-a e)^5}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(11/2)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-2*e^4*(a + b*x)*(d + e*x)^(13/2)*Hypergeometric2F1[5, 13/2, 15/2, (b*(d + e*x))/(b*d - a*e)])/(13*(b*d - a*e
)^5*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.288, size = 1471, normalized size = 4.3 \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)^(5/2),x)

[Out]

-1/192*(-3465*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^4*a^2*b^4*e^6-3465*arctan(b*(e*x+d)^(1/2)/((a*e-b*
d)*b)^(1/2))*x^4*b^6*d^2*e^4-13860*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^3*a^3*b^3*e^6+2295*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(7/2)*a^2*b^3*e^2-20790*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*a^4*b^2*e^6+5855*((a
*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a^3*b^2*e^3-13860*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^5*b*e^6+5025*
((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^4*b*e^4+6930*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^5*b*d*e^5-3465*
arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^4*b^2*d^2*e^4-128*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x^4*b^5*e^4+
2295*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*b^5*d^2-5855*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^5*d^3+5153*((a*e-b*d)*
b)^(1/2)*(e*x+d)^(3/2)*b^5*d^4+3465*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^5*e^5-1545*((a*e-b*d)*b)^(1/2)*(e*x+d)
^(1/2)*b^5*d^5-3465*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^6*e^6-7680*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)
*x^3*a*b^4*d*e^4-11520*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*a^2*b^3*d*e^4-7680*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1
/2)*x*a^3*b^2*d*e^4-4590*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*a*b^4*d*e-768*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x^2
*a^2*b^3*e^4+7680*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^3*a^2*b^3*e^5+41580*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)
^(1/2))*x^2*a^3*b^3*d*e^5-20790*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*a^2*b^4*d^2*e^4-17565*((a*e-b*
d)*b)^(1/2)*(e*x+d)^(5/2)*a^2*b^3*d*e^2+17565*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a*b^4*d^2*e-512*((a*e-b*d)*b)^
(1/2)*(e*x+d)^(3/2)*x*a^3*b^2*e^4+6930*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^4*a*b^5*d*e^5-512*((a*e-b
*d)*b)^(1/2)*(e*x+d)^(3/2)*x^3*a*b^4*e^4+1920*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^4*a*b^4*e^5+11520*((a*e-b*d)
*b)^(1/2)*(e*x+d)^(1/2)*x^2*a^3*b^2*e^5+27720*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^4*b^2*d*e^5-1386
0*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^3*b^3*d^2*e^4-20612*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^3*b^
2*d*e^3+30918*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^2*b^3*d^2*e^2-20612*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b^4*
d^3*e+7680*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a^4*b*e^5-9645*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^4*b*d*e^4+15
450*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*b^2*d^2*e^3-15450*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b^3*d^3*e^2+
7725*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^4*d^4*e-1920*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^4*b^5*d*e^4+27720*
arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^3*a^2*b^4*d*e^5-13860*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)
)*x^3*a*b^5*d^2*e^4)*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^6/((b*x+a)^2)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{11}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \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)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(11/2)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)

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Fricas [B]  time = 1.70008, size = 2079, normalized size = 6.01 \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)^(5/2),x, algorithm="fricas")

[Out]

[-1/384*(3465*(a^4*b*d*e^4 - a^5*e^5 + (b^5*d*e^4 - a*b^4*e^5)*x^4 + 4*(a*b^4*d*e^4 - a^2*b^3*e^5)*x^3 + 6*(a^
2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 4*(a^3*b^2*d*e^4 - a^4*b*e^5)*x)*sqrt((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*(128*b^5*e^5*x^5 - 48*b^5*d^5 - 88*a*b^4*d^4*e - 198*
a^2*b^3*d^3*e^2 - 693*a^3*b^2*d^2*e^3 + 4620*a^4*b*d*e^4 - 3465*a^5*e^5 + 128*(16*b^5*d*e^4 - 11*a*b^4*e^5)*x^
4 - (2295*b^5*d^2*e^3 - 12782*a*b^4*d*e^4 + 9207*a^2*b^3*e^5)*x^3 - (1030*b^5*d^3*e^2 + 3795*a*b^4*d^2*e^3 - 2
2968*a^2*b^3*d*e^4 + 16863*a^3*b^2*e^5)*x^2 - (328*b^5*d^4*e + 748*a*b^4*d^3*e^2 + 2673*a^2*b^3*d^2*e^3 - 1709
4*a^3*b^2*d*e^4 + 12705*a^4*b*e^5)*x)*sqrt(e*x + d))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a
^4*b^6), -1/192*(3465*(a^4*b*d*e^4 - a^5*e^5 + (b^5*d*e^4 - a*b^4*e^5)*x^4 + 4*(a*b^4*d*e^4 - a^2*b^3*e^5)*x^3
+ 6*(a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 4*(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)) - (128*b^5*e^5*x^5 - 48*b^5*d^5 - 88*a*b^4*d^4*e - 198*a^2*b^3*d^3*
e^2 - 693*a^3*b^2*d^2*e^3 + 4620*a^4*b*d*e^4 - 3465*a^5*e^5 + 128*(16*b^5*d*e^4 - 11*a*b^4*e^5)*x^4 - (2295*b^
5*d^2*e^3 - 12782*a*b^4*d*e^4 + 9207*a^2*b^3*e^5)*x^3 - (1030*b^5*d^3*e^2 + 3795*a*b^4*d^2*e^3 - 22968*a^2*b^3
*d*e^4 + 16863*a^3*b^2*e^5)*x^2 - (328*b^5*d^4*e + 748*a*b^4*d^3*e^2 + 2673*a^2*b^3*d^2*e^3 - 17094*a^3*b^2*d*
e^4 + 12705*a^4*b*e^5)*x)*sqrt(e*x + d))/(b^10*x^4 + 4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.33739, size = 740, normalized size = 2.14 \begin{align*} \frac{1155 \,{\left (b^{2} d^{2} e^{4} - 2 \, a b d e^{5} + a^{2} e^{6}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{64 \, \sqrt{-b^{2} d + a b e} b^{6} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{2295 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{5} d^{2} e^{4} - 5855 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{5} d^{3} e^{4} + 5153 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{5} d^{4} e^{4} - 1545 \, \sqrt{x e + d} b^{5} d^{5} e^{4} - 4590 \,{\left (x e + d\right )}^{\frac{7}{2}} a b^{4} d e^{5} + 17565 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{4} d^{2} e^{5} - 20612 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{4} d^{3} e^{5} + 7725 \, \sqrt{x e + d} a b^{4} d^{4} e^{5} + 2295 \,{\left (x e + d\right )}^{\frac{7}{2}} a^{2} b^{3} e^{6} - 17565 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{2} b^{3} d e^{6} + 30918 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{3} d^{2} e^{6} - 15450 \, \sqrt{x e + d} a^{2} b^{3} d^{3} e^{6} + 5855 \,{\left (x e + d\right )}^{\frac{5}{2}} a^{3} b^{2} e^{7} - 20612 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} b^{2} d e^{7} + 15450 \, \sqrt{x e + d} a^{3} b^{2} d^{2} e^{7} + 5153 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{4} b e^{8} - 7725 \, \sqrt{x e + d} a^{4} b d e^{8} + 1545 \, \sqrt{x e + d} a^{5} e^{9}}{192 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4} b^{6} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac{2 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{10} e^{4} + 15 \, \sqrt{x e + d} b^{10} d e^{4} - 15 \, \sqrt{x e + d} a b^{9} e^{5}\right )}}{3 \, b^{15} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \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)^(5/2),x, algorithm="giac")

[Out]

1155/64*(b^2*d^2*e^4 - 2*a*b*d*e^5 + a^2*e^6)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*
e)*b^6*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1/192*(2295*(x*e + d)^(7/2)*b^5*d^2*e^4 - 5855*(x*e + d)^(5/2)*b^
5*d^3*e^4 + 5153*(x*e + d)^(3/2)*b^5*d^4*e^4 - 1545*sqrt(x*e + d)*b^5*d^5*e^4 - 4590*(x*e + d)^(7/2)*a*b^4*d*e
^5 + 17565*(x*e + d)^(5/2)*a*b^4*d^2*e^5 - 20612*(x*e + d)^(3/2)*a*b^4*d^3*e^5 + 7725*sqrt(x*e + d)*a*b^4*d^4*
e^5 + 2295*(x*e + d)^(7/2)*a^2*b^3*e^6 - 17565*(x*e + d)^(5/2)*a^2*b^3*d*e^6 + 30918*(x*e + d)^(3/2)*a^2*b^3*d
^2*e^6 - 15450*sqrt(x*e + d)*a^2*b^3*d^3*e^6 + 5855*(x*e + d)^(5/2)*a^3*b^2*e^7 - 20612*(x*e + d)^(3/2)*a^3*b^
2*d*e^7 + 15450*sqrt(x*e + d)*a^3*b^2*d^2*e^7 + 5153*(x*e + d)^(3/2)*a^4*b*e^8 - 7725*sqrt(x*e + d)*a^4*b*d*e^
8 + 1545*sqrt(x*e + d)*a^5*e^9)/(((x*e + d)*b - b*d + a*e)^4*b^6*sgn((x*e + d)*b*e - b*d*e + a*e^2)) + 2/3*((x
*e + d)^(3/2)*b^10*e^4 + 15*sqrt(x*e + d)*b^10*d*e^4 - 15*sqrt(x*e + d)*a*b^9*e^5)/(b^15*sgn((x*e + d)*b*e - b
*d*e + a*e^2))