3.14.65 \(\int \frac {1}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=229 \[ \frac {231 b^{5/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{13/2}}-\frac {231 b^2 e^3}{8 \sqrt {d+e x} (b d-a e)^6}-\frac {77 b e^3}{8 (d+e x)^{3/2} (b d-a e)^5}-\frac {231 e^3}{40 (d+e x)^{5/2} (b d-a e)^4}-\frac {33 e^2}{8 (a+b x) (d+e x)^{5/2} (b d-a e)^3}+\frac {11 e}{12 (a+b x)^2 (d+e x)^{5/2} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 (d+e x)^{5/2} (b d-a e)} \]

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Rubi [A]  time = 0.18, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 51, 63, 208} \begin {gather*} -\frac {231 b^2 e^3}{8 \sqrt {d+e x} (b d-a e)^6}+\frac {231 b^{5/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{13/2}}-\frac {77 b e^3}{8 (d+e x)^{3/2} (b d-a e)^5}-\frac {231 e^3}{40 (d+e x)^{5/2} (b d-a e)^4}-\frac {33 e^2}{8 (a+b x) (d+e x)^{5/2} (b d-a e)^3}+\frac {11 e}{12 (a+b x)^2 (d+e x)^{5/2} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 (d+e x)^{5/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

(-231*e^3)/(40*(b*d - a*e)^4*(d + e*x)^(5/2)) - 1/(3*(b*d - a*e)*(a + b*x)^3*(d + e*x)^(5/2)) + (11*e)/(12*(b*
d - a*e)^2*(a + b*x)^2*(d + e*x)^(5/2)) - (33*e^2)/(8*(b*d - a*e)^3*(a + b*x)*(d + e*x)^(5/2)) - (77*b*e^3)/(8
*(b*d - a*e)^5*(d + e*x)^(3/2)) - (231*b^2*e^3)/(8*(b*d - a*e)^6*Sqrt[d + e*x]) + (231*b^(5/2)*e^3*ArcTanh[(Sq
rt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)^(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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^4 (d+e x)^{7/2}} \, dx\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}-\frac {(11 e) \int \frac {1}{(a+b x)^3 (d+e x)^{7/2}} \, dx}{6 (b d-a e)}\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}+\frac {\left (33 e^2\right ) \int \frac {1}{(a+b x)^2 (d+e x)^{7/2}} \, dx}{8 (b d-a e)^2}\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}-\frac {33 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{5/2}}-\frac {\left (231 e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{7/2}} \, dx}{16 (b d-a e)^3}\\ &=-\frac {231 e^3}{40 (b d-a e)^4 (d+e x)^{5/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}-\frac {33 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{5/2}}-\frac {\left (231 b e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{16 (b d-a e)^4}\\ &=-\frac {231 e^3}{40 (b d-a e)^4 (d+e x)^{5/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}-\frac {33 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{5/2}}-\frac {77 b e^3}{8 (b d-a e)^5 (d+e x)^{3/2}}-\frac {\left (231 b^2 e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{16 (b d-a e)^5}\\ &=-\frac {231 e^3}{40 (b d-a e)^4 (d+e x)^{5/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}-\frac {33 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{5/2}}-\frac {77 b e^3}{8 (b d-a e)^5 (d+e x)^{3/2}}-\frac {231 b^2 e^3}{8 (b d-a e)^6 \sqrt {d+e x}}-\frac {\left (231 b^3 e^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 (b d-a e)^6}\\ &=-\frac {231 e^3}{40 (b d-a e)^4 (d+e x)^{5/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}-\frac {33 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{5/2}}-\frac {77 b e^3}{8 (b d-a e)^5 (d+e x)^{3/2}}-\frac {231 b^2 e^3}{8 (b d-a e)^6 \sqrt {d+e x}}-\frac {\left (231 b^3 e^2\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 d-a e)^6}\\ &=-\frac {231 e^3}{40 (b d-a e)^4 (d+e x)^{5/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{5/2}}+\frac {11 e}{12 (b d-a e)^2 (a+b x)^2 (d+e x)^{5/2}}-\frac {33 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{5/2}}-\frac {77 b e^3}{8 (b d-a e)^5 (d+e x)^{3/2}}-\frac {231 b^2 e^3}{8 (b d-a e)^6 \sqrt {d+e x}}+\frac {231 b^{5/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{13/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.23 \begin {gather*} -\frac {2 e^3 \, _2F_1\left (-\frac {5}{2},4;-\frac {3}{2};-\frac {b (d+e x)}{a e-b d}\right )}{5 (d+e x)^{5/2} (a e-b d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

(-2*e^3*Hypergeometric2F1[-5/2, 4, -3/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^4*(d + e*x)^(5/2)
)

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IntegrateAlgebraic [A]  time = 1.16, size = 416, normalized size = 1.82 \begin {gather*} \frac {e^3 \left (48 a^5 e^5-176 a^4 b e^4 (d+e x)-240 a^4 b d e^4+480 a^3 b^2 d^2 e^3+1584 a^3 b^2 e^3 (d+e x)^2+704 a^3 b^2 d e^3 (d+e x)-480 a^2 b^3 d^3 e^2-1056 a^2 b^3 d^2 e^2 (d+e x)+7623 a^2 b^3 e^2 (d+e x)^3-4752 a^2 b^3 d e^2 (d+e x)^2+240 a b^4 d^4 e+704 a b^4 d^3 e (d+e x)+4752 a b^4 d^2 e (d+e x)^2+9240 a b^4 e (d+e x)^4-15246 a b^4 d e (d+e x)^3-48 b^5 d^5-176 b^5 d^4 (d+e x)-1584 b^5 d^3 (d+e x)^2+7623 b^5 d^2 (d+e x)^3+3465 b^5 (d+e x)^5-9240 b^5 d (d+e x)^4\right )}{120 (d+e x)^{5/2} (b d-a e)^6 (-a e-b (d+e x)+b d)^3}+\frac {231 b^{5/2} e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 (b d-a e)^6 \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

(e^3*(-48*b^5*d^5 + 240*a*b^4*d^4*e - 480*a^2*b^3*d^3*e^2 + 480*a^3*b^2*d^2*e^3 - 240*a^4*b*d*e^4 + 48*a^5*e^5
- 176*b^5*d^4*(d + e*x) + 704*a*b^4*d^3*e*(d + e*x) - 1056*a^2*b^3*d^2*e^2*(d + e*x) + 704*a^3*b^2*d*e^3*(d +
e*x) - 176*a^4*b*e^4*(d + e*x) - 1584*b^5*d^3*(d + e*x)^2 + 4752*a*b^4*d^2*e*(d + e*x)^2 - 4752*a^2*b^3*d*e^2
*(d + e*x)^2 + 1584*a^3*b^2*e^3*(d + e*x)^2 + 7623*b^5*d^2*(d + e*x)^3 - 15246*a*b^4*d*e*(d + e*x)^3 + 7623*a^
2*b^3*e^2*(d + e*x)^3 - 9240*b^5*d*(d + e*x)^4 + 9240*a*b^4*e*(d + e*x)^4 + 3465*b^5*(d + e*x)^5))/(120*(b*d -
a*e)^6*(d + e*x)^(5/2)*(b*d - a*e - b*(d + e*x))^3) + (231*b^(5/2)*e^3*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqr
t[d + e*x])/(b*d - a*e)])/(8*(b*d - a*e)^6*Sqrt[-(b*d) + a*e])

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fricas [B]  time = 0.47, size = 2550, normalized size = 11.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/240*(3465*(b^5*e^6*x^6 + a^3*b^2*d^3*e^3 + 3*(b^5*d*e^5 + a*b^4*e^6)*x^5 + 3*(b^5*d^2*e^4 + 3*a*b^4*d*e^5 +
a^2*b^3*e^6)*x^4 + (b^5*d^3*e^3 + 9*a*b^4*d^2*e^4 + 9*a^2*b^3*d*e^5 + a^3*b^2*e^6)*x^3 + 3*(a*b^4*d^3*e^3 + 3
*a^2*b^3*d^2*e^4 + a^3*b^2*d*e^5)*x^2 + 3*(a^2*b^3*d^3*e^3 + a^3*b^2*d^2*e^4)*x)*sqrt(b/(b*d - a*e))*log((b*e*
x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(3465*b^5*e^5*x^5 + 40*b^5*d
^5 - 310*a*b^4*d^4*e + 1335*a^2*b^3*d^3*e^2 + 2768*a^3*b^2*d^2*e^3 - 416*a^4*b*d*e^4 + 48*a^5*e^5 + 1155*(7*b^
5*d*e^4 + 8*a*b^4*e^5)*x^4 + 231*(23*b^5*d^2*e^3 + 94*a*b^4*d*e^4 + 33*a^2*b^3*e^5)*x^3 + 99*(5*b^5*d^3*e^2 +
146*a*b^4*d^2*e^3 + 183*a^2*b^3*d*e^4 + 16*a^3*b^2*e^5)*x^2 - 11*(10*b^5*d^4*e - 130*a*b^4*d^3*e^2 - 1119*a^2*
b^3*d^2*e^3 - 352*a^3*b^2*d*e^4 + 16*a^4*b*e^5)*x)*sqrt(e*x + d))/(a^3*b^6*d^9 - 6*a^4*b^5*d^8*e + 15*a^5*b^4*
d^7*e^2 - 20*a^6*b^3*d^6*e^3 + 15*a^7*b^2*d^5*e^4 - 6*a^8*b*d^4*e^5 + a^9*d^3*e^6 + (b^9*d^6*e^3 - 6*a*b^8*d^5
*e^4 + 15*a^2*b^7*d^4*e^5 - 20*a^3*b^6*d^3*e^6 + 15*a^4*b^5*d^2*e^7 - 6*a^5*b^4*d*e^8 + a^6*b^3*e^9)*x^6 + 3*(
b^9*d^7*e^2 - 5*a*b^8*d^6*e^3 + 9*a^2*b^7*d^5*e^4 - 5*a^3*b^6*d^4*e^5 - 5*a^4*b^5*d^3*e^6 + 9*a^5*b^4*d^2*e^7
- 5*a^6*b^3*d*e^8 + a^7*b^2*e^9)*x^5 + 3*(b^9*d^8*e - 3*a*b^8*d^7*e^2 - 2*a^2*b^7*d^6*e^3 + 19*a^3*b^6*d^5*e^4
- 30*a^4*b^5*d^4*e^5 + 19*a^5*b^4*d^3*e^6 - 2*a^6*b^3*d^2*e^7 - 3*a^7*b^2*d*e^8 + a^8*b*e^9)*x^4 + (b^9*d^9 +
3*a*b^8*d^8*e - 30*a^2*b^7*d^7*e^2 + 62*a^3*b^6*d^6*e^3 - 36*a^4*b^5*d^5*e^4 - 36*a^5*b^4*d^4*e^5 + 62*a^6*b^
3*d^3*e^6 - 30*a^7*b^2*d^2*e^7 + 3*a^8*b*d*e^8 + a^9*e^9)*x^3 + 3*(a*b^8*d^9 - 3*a^2*b^7*d^8*e - 2*a^3*b^6*d^7
*e^2 + 19*a^4*b^5*d^6*e^3 - 30*a^5*b^4*d^5*e^4 + 19*a^6*b^3*d^4*e^5 - 2*a^7*b^2*d^3*e^6 - 3*a^8*b*d^2*e^7 + a^
9*d*e^8)*x^2 + 3*(a^2*b^7*d^9 - 5*a^3*b^6*d^8*e + 9*a^4*b^5*d^7*e^2 - 5*a^5*b^4*d^6*e^3 - 5*a^6*b^3*d^5*e^4 +
9*a^7*b^2*d^4*e^5 - 5*a^8*b*d^3*e^6 + a^9*d^2*e^7)*x), 1/120*(3465*(b^5*e^6*x^6 + a^3*b^2*d^3*e^3 + 3*(b^5*d*e
^5 + a*b^4*e^6)*x^5 + 3*(b^5*d^2*e^4 + 3*a*b^4*d*e^5 + a^2*b^3*e^6)*x^4 + (b^5*d^3*e^3 + 9*a*b^4*d^2*e^4 + 9*a
^2*b^3*d*e^5 + a^3*b^2*e^6)*x^3 + 3*(a*b^4*d^3*e^3 + 3*a^2*b^3*d^2*e^4 + a^3*b^2*d*e^5)*x^2 + 3*(a^2*b^3*d^3*e
^3 + a^3*b^2*d^2*e^4)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x +
b*d)) - (3465*b^5*e^5*x^5 + 40*b^5*d^5 - 310*a*b^4*d^4*e + 1335*a^2*b^3*d^3*e^2 + 2768*a^3*b^2*d^2*e^3 - 416*a
^4*b*d*e^4 + 48*a^5*e^5 + 1155*(7*b^5*d*e^4 + 8*a*b^4*e^5)*x^4 + 231*(23*b^5*d^2*e^3 + 94*a*b^4*d*e^4 + 33*a^2
*b^3*e^5)*x^3 + 99*(5*b^5*d^3*e^2 + 146*a*b^4*d^2*e^3 + 183*a^2*b^3*d*e^4 + 16*a^3*b^2*e^5)*x^2 - 11*(10*b^5*d
^4*e - 130*a*b^4*d^3*e^2 - 1119*a^2*b^3*d^2*e^3 - 352*a^3*b^2*d*e^4 + 16*a^4*b*e^5)*x)*sqrt(e*x + d))/(a^3*b^6
*d^9 - 6*a^4*b^5*d^8*e + 15*a^5*b^4*d^7*e^2 - 20*a^6*b^3*d^6*e^3 + 15*a^7*b^2*d^5*e^4 - 6*a^8*b*d^4*e^5 + a^9*
d^3*e^6 + (b^9*d^6*e^3 - 6*a*b^8*d^5*e^4 + 15*a^2*b^7*d^4*e^5 - 20*a^3*b^6*d^3*e^6 + 15*a^4*b^5*d^2*e^7 - 6*a^
5*b^4*d*e^8 + a^6*b^3*e^9)*x^6 + 3*(b^9*d^7*e^2 - 5*a*b^8*d^6*e^3 + 9*a^2*b^7*d^5*e^4 - 5*a^3*b^6*d^4*e^5 - 5*
a^4*b^5*d^3*e^6 + 9*a^5*b^4*d^2*e^7 - 5*a^6*b^3*d*e^8 + a^7*b^2*e^9)*x^5 + 3*(b^9*d^8*e - 3*a*b^8*d^7*e^2 - 2*
a^2*b^7*d^6*e^3 + 19*a^3*b^6*d^5*e^4 - 30*a^4*b^5*d^4*e^5 + 19*a^5*b^4*d^3*e^6 - 2*a^6*b^3*d^2*e^7 - 3*a^7*b^2
*d*e^8 + a^8*b*e^9)*x^4 + (b^9*d^9 + 3*a*b^8*d^8*e - 30*a^2*b^7*d^7*e^2 + 62*a^3*b^6*d^6*e^3 - 36*a^4*b^5*d^5*
e^4 - 36*a^5*b^4*d^4*e^5 + 62*a^6*b^3*d^3*e^6 - 30*a^7*b^2*d^2*e^7 + 3*a^8*b*d*e^8 + a^9*e^9)*x^3 + 3*(a*b^8*d
^9 - 3*a^2*b^7*d^8*e - 2*a^3*b^6*d^7*e^2 + 19*a^4*b^5*d^6*e^3 - 30*a^5*b^4*d^5*e^4 + 19*a^6*b^3*d^4*e^5 - 2*a^
7*b^2*d^3*e^6 - 3*a^8*b*d^2*e^7 + a^9*d*e^8)*x^2 + 3*(a^2*b^7*d^9 - 5*a^3*b^6*d^8*e + 9*a^4*b^5*d^7*e^2 - 5*a^
5*b^4*d^6*e^3 - 5*a^6*b^3*d^5*e^4 + 9*a^7*b^2*d^4*e^5 - 5*a^8*b*d^3*e^6 + a^9*d^2*e^7)*x)]

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giac [B]  time = 0.25, size = 470, normalized size = 2.05 \begin {gather*} -\frac {231 \, b^{3} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, {\left (b^{6} d^{6} - 6 \, a b^{5} d^{5} e + 15 \, a^{2} b^{4} d^{4} e^{2} - 20 \, a^{3} b^{3} d^{3} e^{3} + 15 \, a^{4} b^{2} d^{2} e^{4} - 6 \, a^{5} b d e^{5} + a^{6} e^{6}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (150 \, {\left (x e + d\right )}^{2} b^{2} e^{3} + 20 \, {\left (x e + d\right )} b^{2} d e^{3} + 3 \, b^{2} d^{2} e^{3} - 20 \, {\left (x e + d\right )} a b e^{4} - 6 \, a b d e^{4} + 3 \, a^{2} e^{5}\right )}}{15 \, {\left (b^{6} d^{6} - 6 \, a b^{5} d^{5} e + 15 \, a^{2} b^{4} d^{4} e^{2} - 20 \, a^{3} b^{3} d^{3} e^{3} + 15 \, a^{4} b^{2} d^{2} e^{4} - 6 \, a^{5} b d e^{5} + a^{6} e^{6}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} - \frac {213 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{5} e^{3} - 472 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{5} d e^{3} + 267 \, \sqrt {x e + d} b^{5} d^{2} e^{3} + 472 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{4} e^{4} - 534 \, \sqrt {x e + d} a b^{4} d e^{4} + 267 \, \sqrt {x e + d} a^{2} b^{3} e^{5}}{24 \, {\left (b^{6} d^{6} - 6 \, a b^{5} d^{5} e + 15 \, a^{2} b^{4} d^{4} e^{2} - 20 \, a^{3} b^{3} d^{3} e^{3} + 15 \, a^{4} b^{2} d^{2} e^{4} - 6 \, a^{5} b d e^{5} + a^{6} e^{6}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

-231/8*b^3*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^6*d^6 - 6*a*b^5*d^5*e + 15*a^2*b^4*d^4*e^2 - 2
0*a^3*b^3*d^3*e^3 + 15*a^4*b^2*d^2*e^4 - 6*a^5*b*d*e^5 + a^6*e^6)*sqrt(-b^2*d + a*b*e)) - 2/15*(150*(x*e + d)^
2*b^2*e^3 + 20*(x*e + d)*b^2*d*e^3 + 3*b^2*d^2*e^3 - 20*(x*e + d)*a*b*e^4 - 6*a*b*d*e^4 + 3*a^2*e^5)/((b^6*d^6
- 6*a*b^5*d^5*e + 15*a^2*b^4*d^4*e^2 - 20*a^3*b^3*d^3*e^3 + 15*a^4*b^2*d^2*e^4 - 6*a^5*b*d*e^5 + a^6*e^6)*(x*
e + d)^(5/2)) - 1/24*(213*(x*e + d)^(5/2)*b^5*e^3 - 472*(x*e + d)^(3/2)*b^5*d*e^3 + 267*sqrt(x*e + d)*b^5*d^2*
e^3 + 472*(x*e + d)^(3/2)*a*b^4*e^4 - 534*sqrt(x*e + d)*a*b^4*d*e^4 + 267*sqrt(x*e + d)*a^2*b^3*e^5)/((b^6*d^6
- 6*a*b^5*d^5*e + 15*a^2*b^4*d^4*e^2 - 20*a^3*b^3*d^3*e^3 + 15*a^4*b^2*d^2*e^4 - 6*a^5*b*d*e^5 + a^6*e^6)*((x
*e + d)*b - b*d + a*e)^3)

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maple [A]  time = 0.07, size = 344, normalized size = 1.50 \begin {gather*} -\frac {89 \sqrt {e x +d}\, a^{2} b^{3} e^{5}}{8 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{3}}+\frac {89 \sqrt {e x +d}\, a \,b^{4} d \,e^{4}}{4 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{3}}-\frac {89 \sqrt {e x +d}\, b^{5} d^{2} e^{3}}{8 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{3}}-\frac {59 \left (e x +d \right )^{\frac {3}{2}} a \,b^{4} e^{4}}{3 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{3}}+\frac {59 \left (e x +d \right )^{\frac {3}{2}} b^{5} d \,e^{3}}{3 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{3}}-\frac {71 \left (e x +d \right )^{\frac {5}{2}} b^{5} e^{3}}{8 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{3}}-\frac {231 b^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right )^{6} \sqrt {\left (a e -b d \right ) b}}-\frac {20 b^{2} e^{3}}{\left (a e -b d \right )^{6} \sqrt {e x +d}}+\frac {8 b \,e^{3}}{3 \left (a e -b d \right )^{5} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 e^{3}}{5 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

-71/8*e^3/(a*e-b*d)^6*b^5/(b*e*x+a*e)^3*(e*x+d)^(5/2)-59/3*e^4/(a*e-b*d)^6*b^4/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a+5
9/3*e^3/(a*e-b*d)^6*b^5/(b*e*x+a*e)^3*(e*x+d)^(3/2)*d-89/8*e^5/(a*e-b*d)^6*b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^2
+89/4*e^4/(a*e-b*d)^6*b^4/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a*d-89/8*e^3/(a*e-b*d)^6*b^5/(b*e*x+a*e)^3*(e*x+d)^(1/2)
*d^2-231/8*e^3/(a*e-b*d)^6*b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-2/5*e^3/(a*e-b*
d)^4/(e*x+d)^(5/2)-20*e^3/(a*e-b*d)^6*b^2/(e*x+d)^(1/2)+8/3*e^3/(a*e-b*d)^5*b/(e*x+d)^(3/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 0.97, size = 373, normalized size = 1.63 \begin {gather*} -\frac {\frac {2\,e^3}{5\,\left (a\,e-b\,d\right )}+\frac {66\,b^2\,e^3\,{\left (d+e\,x\right )}^2}{5\,{\left (a\,e-b\,d\right )}^3}+\frac {2541\,b^3\,e^3\,{\left (d+e\,x\right )}^3}{40\,{\left (a\,e-b\,d\right )}^4}+\frac {77\,b^4\,e^3\,{\left (d+e\,x\right )}^4}{{\left (a\,e-b\,d\right )}^5}+\frac {231\,b^5\,e^3\,{\left (d+e\,x\right )}^5}{8\,{\left (a\,e-b\,d\right )}^6}-\frac {22\,b\,e^3\,\left (d+e\,x\right )}{15\,{\left (a\,e-b\,d\right )}^2}}{{\left (d+e\,x\right )}^{5/2}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )+b^3\,{\left (d+e\,x\right )}^{11/2}-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{9/2}+{\left (d+e\,x\right )}^{7/2}\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}-\frac {231\,b^{5/2}\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^6\,e^6-6\,a^5\,b\,d\,e^5+15\,a^4\,b^2\,d^2\,e^4-20\,a^3\,b^3\,d^3\,e^3+15\,a^2\,b^4\,d^4\,e^2-6\,a\,b^5\,d^5\,e+b^6\,d^6\right )}{{\left (a\,e-b\,d\right )}^{13/2}}\right )}{8\,{\left (a\,e-b\,d\right )}^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^(7/2)*(a^2 + b^2*x^2 + 2*a*b*x)^2),x)

[Out]

- ((2*e^3)/(5*(a*e - b*d)) + (66*b^2*e^3*(d + e*x)^2)/(5*(a*e - b*d)^3) + (2541*b^3*e^3*(d + e*x)^3)/(40*(a*e
- b*d)^4) + (77*b^4*e^3*(d + e*x)^4)/(a*e - b*d)^5 + (231*b^5*e^3*(d + e*x)^5)/(8*(a*e - b*d)^6) - (22*b*e^3*(
d + e*x))/(15*(a*e - b*d)^2))/((d + e*x)^(5/2)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2) + b^3*(d +
e*x)^(11/2) - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^(9/2) + (d + e*x)^(7/2)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e))
- (231*b^(5/2)*e^3*atan((b^(1/2)*(d + e*x)^(1/2)*(a^6*e^6 + b^6*d^6 + 15*a^2*b^4*d^4*e^2 - 20*a^3*b^3*d^3*e^3
+ 15*a^4*b^2*d^2*e^4 - 6*a*b^5*d^5*e - 6*a^5*b*d*e^5))/(a*e - b*d)^(13/2)))/(8*(a*e - b*d)^(13/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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