3.1711 \(\int \frac{(d+e x)^{9/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=308 \[ \frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 e^2 (a+b x) (d+e x)^{3/2} (b d-a e)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{63 e^2 (a+b x) \sqrt{d+e x} (b d-a e)^2}{4 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{63 e^2 (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.174307, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 e^2 (a+b x) (d+e x)^{3/2} (b d-a e)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{63 e^2 (a+b x) \sqrt{d+e x} (b d-a e)^2}{4 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{63 e^2 (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(9/2)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(63*e^2*(b*d - a*e)^2*(a + b*x)*Sqrt[d + e*x])/(4*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (21*e^2*(b*d - a*e)*(a
+ b*x)*(d + e*x)^(3/2))/(4*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (63*e^2*(a + b*x)*(d + e*x)^(5/2))/(20*b^3*Sqr
t[a^2 + 2*a*b*x + b^2*x^2]) - (9*e*(d + e*x)^(7/2))/(4*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(9/2)/(2
*b*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (63*e^2*(b*d - a*e)^(5/2)*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*
x])/Sqrt[b*d - a*e]])/(4*b^(11/2)*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)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{9/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (9 e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{7/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (63 e^2 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{a b+b^2 x} \, dx}{8 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (63 e^2 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{21 e^2 (b d-a e) (a+b x) (d+e x)^{3/2}}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (63 e^2 \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{8 b^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{63 e^2 (b d-a e)^2 (a+b x) \sqrt{d+e x}}{4 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 e^2 (b d-a e) (a+b x) (d+e x)^{3/2}}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (63 e^2 \left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{8 b^8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{63 e^2 (b d-a e)^2 (a+b x) \sqrt{d+e x}}{4 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 e^2 (b d-a e) (a+b x) (d+e x)^{3/2}}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (63 e \left (b^2 d-a b e\right )^3 \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 )}{4 b^8 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{63 e^2 (b d-a e)^2 (a+b x) \sqrt{d+e x}}{4 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 e^2 (b d-a e) (a+b x) (d+e x)^{3/2}}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{63 e^2 (a+b x) (d+e x)^{5/2}}{20 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{9 e (d+e x)^{7/2}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{9/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{63 e^2 (b d-a e)^{5/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{11/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0461776, size = 67, normalized size = 0.22 \[ -\frac{2 e^2 (a+b x) (d+e x)^{11/2} \, _2F_1\left (3,\frac{11}{2};\frac{13}{2};\frac{b (d+e x)}{b d-a e}\right )}{11 \sqrt{(a+b x)^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(9/2)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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Maple [B]  time = 0.282, size = 1115, 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)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/20*(-480*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*a*b^3*d*e^3-960*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a^2*b^2*d
*e^3+480*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a*b^3*d^2*e^2+80*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x*a*b^3*d*e^2+
1890*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^3*b^2*d*e^4-1890*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/
2))*x*a^2*b^3*d^2*e^3+630*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a*b^4*d^3*e^2-215*((a*e-b*d)*b)^(1/2)*
(e*x+d)^(3/2)*a^2*b^2*d*e^2+255*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b^3*d^2*e+16*((a*e-b*d)*b)^(1/2)*(e*x+d)^(
5/2)*x*a*b^3*e^2-40*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x^2*a*b^3*e^3+315*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^4*
e^4+75*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^4*d^4-85*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^4*d^3+240*((a*e-b*d)*b
)^(1/2)*(e*x+d)^(1/2)*x^2*a^2*b^2*e^4+240*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x^2*b^4*d^2*e^2-315*arctan(b*(e*x+
d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^5*e^5-630*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a^4*b*e^5+45*((a*e-b*d
)*b)^(1/2)*(e*x+d)^(3/2)*a^3*b*e^3+945*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^4*b*d*e^4-945*arctan(b*(e
*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3*b^2*d^2*e^3+315*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*b^3*d^3*e
^2-315*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*a^3*b^2*e^5+315*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1
/2))*x^2*b^5*d^3*e^2+8*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a^2*b^2*e^2+8*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*x^2*b
^4*e^2+40*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*x^2*b^4*d*e^2+945*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*
a^2*b^3*d*e^4-945*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*a*b^4*d^2*e^3-80*((a*e-b*d)*b)^(1/2)*(e*x+d)
^(3/2)*x*a^2*b^2*e^3+480*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*a^3*b*e^4-780*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a
^3*b*d*e^3+690*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b^2*d^2*e^2-300*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^3*d
^3*e)*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^5/((b*x+a)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{9}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(9/2)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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Fricas [A]  time = 1.69455, size = 1535, normalized size = 4.98 \begin{align*} \left [\frac{315 \,{\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} +{\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \,{\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \,{\left (8 \, b^{4} e^{4} x^{4} - 10 \, b^{4} d^{4} - 45 \, a b^{3} d^{3} e + 483 \, a^{2} b^{2} d^{2} e^{2} - 735 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} + 8 \,{\left (7 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 24 \,{\left (12 \, b^{4} d^{2} e^{2} - 17 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} -{\left (85 \, b^{4} d^{3} e - 831 \, a b^{3} d^{2} e^{2} + 1239 \, a^{2} b^{2} d e^{3} - 525 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{40 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}, -\frac{315 \,{\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} +{\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \,{\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (8 \, b^{4} e^{4} x^{4} - 10 \, b^{4} d^{4} - 45 \, a b^{3} d^{3} e + 483 \, a^{2} b^{2} d^{2} e^{2} - 735 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} + 8 \,{\left (7 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 24 \,{\left (12 \, b^{4} d^{2} e^{2} - 17 \, a b^{3} d e^{3} + 7 \, a^{2} b^{2} e^{4}\right )} x^{2} -{\left (85 \, b^{4} d^{3} e - 831 \, a b^{3} d^{2} e^{2} + 1239 \, a^{2} b^{2} d e^{3} - 525 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{20 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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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)**(9/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.288, size = 602, normalized size = 1.95 \begin{align*} \frac{63 \,{\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \, \sqrt{-b^{2} d + a b e} b^{5} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{17 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d^{3} e^{2} - 15 \, \sqrt{x e + d} b^{4} d^{4} e^{2} - 51 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} d^{2} e^{3} + 60 \, \sqrt{x e + d} a b^{3} d^{3} e^{3} + 51 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{2} d e^{4} - 90 \, \sqrt{x e + d} a^{2} b^{2} d^{2} e^{4} - 17 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{3} b e^{5} + 60 \, \sqrt{x e + d} a^{3} b d e^{5} - 15 \, \sqrt{x e + d} a^{4} e^{6}}{4 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{5} \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{5}{2}} b^{12} e^{2} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{12} d e^{2} + 30 \, \sqrt{x e + d} b^{12} d^{2} e^{2} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{11} e^{3} - 60 \, \sqrt{x e + d} a b^{11} d e^{3} + 30 \, \sqrt{x e + d} a^{2} b^{10} e^{4}\right )}}{5 \, 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)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

63/4*(b^3*d^3*e^2 - 3*a*b^2*d^2*e^3 + 3*a^2*b*d*e^4 - a^3*e^5)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(s
qrt(-b^2*d + a*b*e)*b^5*sgn((x*e + d)*b*e - b*d*e + a*e^2)) - 1/4*(17*(x*e + d)^(3/2)*b^4*d^3*e^2 - 15*sqrt(x*
e + d)*b^4*d^4*e^2 - 51*(x*e + d)^(3/2)*a*b^3*d^2*e^3 + 60*sqrt(x*e + d)*a*b^3*d^3*e^3 + 51*(x*e + d)^(3/2)*a^
2*b^2*d*e^4 - 90*sqrt(x*e + d)*a^2*b^2*d^2*e^4 - 17*(x*e + d)^(3/2)*a^3*b*e^5 + 60*sqrt(x*e + d)*a^3*b*d*e^5 -
15*sqrt(x*e + d)*a^4*e^6)/(((x*e + d)*b - b*d + a*e)^2*b^5*sgn((x*e + d)*b*e - b*d*e + a*e^2)) + 2/5*((x*e +
d)^(5/2)*b^12*e^2 + 5*(x*e + d)^(3/2)*b^12*d*e^2 + 30*sqrt(x*e + d)*b^12*d^2*e^2 - 5*(x*e + d)^(3/2)*a*b^11*e^
3 - 60*sqrt(x*e + d)*a*b^11*d*e^3 + 30*sqrt(x*e + d)*a^2*b^10*e^4)/(b^15*sgn((x*e + d)*b*e - b*d*e + a*e^2))