3.14.71 \(\int \frac {(d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=198 \[ -\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{5/2}}+\frac {3 e^4 \sqrt {d+e x}}{128 b^3 (a+b x) (b d-a e)^2}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (a+b x)^2 (b d-a e)}-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5} \]

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Rubi [A]  time = 0.10, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 47, 51, 63, 208} \begin {gather*} \frac {3 e^4 \sqrt {d+e x}}{128 b^3 (a+b x) (b d-a e)^2}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (a+b x)^2 (b d-a e)}-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{5/2}}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^2*Sqrt[d + e*x])/(16*b^3*(a + b*x)^3) - (e^3*Sqrt[d + e*x])/(64*b^3*(b*d - a*e)*(a + b*x)^2) + (3*e^4*Sqrt
[d + e*x])/(128*b^3*(b*d - a*e)^2*(a + b*x)) - (e*(d + e*x)^(3/2))/(8*b^2*(a + b*x)^4) - (d + e*x)^(5/2)/(5*b*
(a + b*x)^5) - (3*e^5*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(128*b^(7/2)*(b*d - a*e)^(5/2))

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 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 {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^{5/2}}{(a+b x)^6} \, dx\\ &=-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}+\frac {e \int \frac {(d+e x)^{3/2}}{(a+b x)^5} \, dx}{2 b}\\ &=-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}+\frac {\left (3 e^2\right ) \int \frac {\sqrt {d+e x}}{(a+b x)^4} \, dx}{16 b^2}\\ &=-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}+\frac {e^3 \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{32 b^3}\\ &=-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)^2}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}-\frac {\left (3 e^4\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{128 b^3 (b d-a e)}\\ &=-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)^2}+\frac {3 e^4 \sqrt {d+e x}}{128 b^3 (b d-a e)^2 (a+b x)}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}+\frac {\left (3 e^5\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{256 b^3 (b d-a e)^2}\\ &=-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)^2}+\frac {3 e^4 \sqrt {d+e x}}{128 b^3 (b d-a e)^2 (a+b x)}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}+\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{128 b^3 (b d-a e)^2}\\ &=-\frac {e^2 \sqrt {d+e x}}{16 b^3 (a+b x)^3}-\frac {e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)^2}+\frac {3 e^4 \sqrt {d+e x}}{128 b^3 (b d-a e)^2 (a+b x)}-\frac {e (d+e x)^{3/2}}{8 b^2 (a+b x)^4}-\frac {(d+e x)^{5/2}}{5 b (a+b x)^5}-\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.26 \begin {gather*} \frac {2 e^5 (d+e x)^{7/2} \, _2F_1\left (\frac {7}{2},6;\frac {9}{2};-\frac {b (d+e x)}{a e-b d}\right )}{7 (a e-b d)^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^5*(d + e*x)^(7/2)*Hypergeometric2F1[7/2, 6, 9/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(7*(-(b*d) + a*e)^6)

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IntegrateAlgebraic [A]  time = 1.70, size = 307, normalized size = 1.55 \begin {gather*} \frac {e^5 \sqrt {d+e x} \left (15 a^4 e^4+70 a^3 b e^3 (d+e x)-60 a^3 b d e^3+90 a^2 b^2 d^2 e^2+128 a^2 b^2 e^2 (d+e x)^2-210 a^2 b^2 d e^2 (d+e x)-60 a b^3 d^3 e+210 a b^3 d^2 e (d+e x)-70 a b^3 e (d+e x)^3-256 a b^3 d e (d+e x)^2+15 b^4 d^4-70 b^4 d^3 (d+e x)+128 b^4 d^2 (d+e x)^2-15 b^4 (d+e x)^4+70 b^4 d (d+e x)^3\right )}{640 b^3 (b d-a e)^2 (-a e-b (d+e x)+b d)^5}-\frac {3 e^5 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{128 b^{7/2} (a e-b d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^5*Sqrt[d + e*x]*(15*b^4*d^4 - 60*a*b^3*d^3*e + 90*a^2*b^2*d^2*e^2 - 60*a^3*b*d*e^3 + 15*a^4*e^4 - 70*b^4*d^
3*(d + e*x) + 210*a*b^3*d^2*e*(d + e*x) - 210*a^2*b^2*d*e^2*(d + e*x) + 70*a^3*b*e^3*(d + e*x) + 128*b^4*d^2*(
d + e*x)^2 - 256*a*b^3*d*e*(d + e*x)^2 + 128*a^2*b^2*e^2*(d + e*x)^2 + 70*b^4*d*(d + e*x)^3 - 70*a*b^3*e*(d +
e*x)^3 - 15*b^4*(d + e*x)^4))/(640*b^3*(b*d - a*e)^2*(b*d - a*e - b*(d + e*x))^5) - (3*e^5*ArcTan[(Sqrt[b]*Sqr
t[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(128*b^(7/2)*(-(b*d) + a*e)^(5/2))

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fricas [B]  time = 0.44, size = 1337, normalized size = 6.75

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[1/1280*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^5
)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(128*b^6*
d^5 - 304*a*b^5*d^4*e + 184*a^2*b^4*d^3*e^2 + 2*a^3*b^3*d^2*e^3 + 5*a^4*b^2*d*e^4 - 15*a^5*b*e^5 - 15*(b^6*d*e
^4 - a*b^5*e^5)*x^4 + 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(124*b^6*d^3*e^2 - 357*a*b^5*d^
2*e^3 + 297*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(168*b^6*d^4*e - 424*a*b^5*d^3*e^2 + 279*a^2*b^4*d^2*e^3 +
12*a^3*b^3*d*e^4 - 35*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^3 - 3*a^6*b^6*d^2*e + 3*a^7*b^5*d*e^2 - a^8*b
^4*e^3 + (b^12*d^3 - 3*a*b^11*d^2*e + 3*a^2*b^10*d*e^2 - a^3*b^9*e^3)*x^5 + 5*(a*b^11*d^3 - 3*a^2*b^10*d^2*e +
3*a^3*b^9*d*e^2 - a^4*b^8*e^3)*x^4 + 10*(a^2*b^10*d^3 - 3*a^3*b^9*d^2*e + 3*a^4*b^8*d*e^2 - a^5*b^7*e^3)*x^3
+ 10*(a^3*b^9*d^3 - 3*a^4*b^8*d^2*e + 3*a^5*b^7*d*e^2 - a^6*b^6*e^3)*x^2 + 5*(a^4*b^8*d^3 - 3*a^5*b^7*d^2*e +
3*a^6*b^6*d*e^2 - a^7*b^5*e^3)*x), 1/640*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b^2*
e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^5)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d
)) - (128*b^6*d^5 - 304*a*b^5*d^4*e + 184*a^2*b^4*d^3*e^2 + 2*a^3*b^3*d^2*e^3 + 5*a^4*b^2*d*e^4 - 15*a^5*b*e^5
- 15*(b^6*d*e^4 - a*b^5*e^5)*x^4 + 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(124*b^6*d^3*e^2
- 357*a*b^5*d^2*e^3 + 297*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(168*b^6*d^4*e - 424*a*b^5*d^3*e^2 + 279*a^2
*b^4*d^2*e^3 + 12*a^3*b^3*d*e^4 - 35*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^3 - 3*a^6*b^6*d^2*e + 3*a^7*b^5
*d*e^2 - a^8*b^4*e^3 + (b^12*d^3 - 3*a*b^11*d^2*e + 3*a^2*b^10*d*e^2 - a^3*b^9*e^3)*x^5 + 5*(a*b^11*d^3 - 3*a^
2*b^10*d^2*e + 3*a^3*b^9*d*e^2 - a^4*b^8*e^3)*x^4 + 10*(a^2*b^10*d^3 - 3*a^3*b^9*d^2*e + 3*a^4*b^8*d*e^2 - a^5
*b^7*e^3)*x^3 + 10*(a^3*b^9*d^3 - 3*a^4*b^8*d^2*e + 3*a^5*b^7*d*e^2 - a^6*b^6*e^3)*x^2 + 5*(a^4*b^8*d^3 - 3*a^
5*b^7*d^2*e + 3*a^6*b^6*d*e^2 - a^7*b^5*e^3)*x)]

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giac [B]  time = 0.23, size = 384, normalized size = 1.94 \begin {gather*} \frac {3 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{5}}{128 \, {\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {15 \, {\left (x e + d\right )}^{\frac {9}{2}} b^{4} e^{5} - 70 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{4} d e^{5} - 128 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{5} + 70 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{5} - 15 \, \sqrt {x e + d} b^{4} d^{4} e^{5} + 70 \, {\left (x e + d\right )}^{\frac {7}{2}} a b^{3} e^{6} + 256 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{3} d e^{6} - 210 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{6} + 60 \, \sqrt {x e + d} a b^{3} d^{3} e^{6} - 128 \, {\left (x e + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{7} + 210 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{7} - 90 \, \sqrt {x e + d} a^{2} b^{2} d^{2} e^{7} - 70 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{3} b e^{8} + 60 \, \sqrt {x e + d} a^{3} b d e^{8} - 15 \, \sqrt {x e + d} a^{4} e^{9}}{640 \, {\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

3/128*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^5/((b^5*d^2 - 2*a*b^4*d*e + a^2*b^3*e^2)*sqrt(-b^2*d + a*
b*e)) + 1/640*(15*(x*e + d)^(9/2)*b^4*e^5 - 70*(x*e + d)^(7/2)*b^4*d*e^5 - 128*(x*e + d)^(5/2)*b^4*d^2*e^5 + 7
0*(x*e + d)^(3/2)*b^4*d^3*e^5 - 15*sqrt(x*e + d)*b^4*d^4*e^5 + 70*(x*e + d)^(7/2)*a*b^3*e^6 + 256*(x*e + d)^(5
/2)*a*b^3*d*e^6 - 210*(x*e + d)^(3/2)*a*b^3*d^2*e^6 + 60*sqrt(x*e + d)*a*b^3*d^3*e^6 - 128*(x*e + d)^(5/2)*a^2
*b^2*e^7 + 210*(x*e + d)^(3/2)*a^2*b^2*d*e^7 - 90*sqrt(x*e + d)*a^2*b^2*d^2*e^7 - 70*(x*e + d)^(3/2)*a^3*b*e^8
+ 60*sqrt(x*e + d)*a^3*b*d*e^8 - 15*sqrt(x*e + d)*a^4*e^9)/((b^5*d^2 - 2*a*b^4*d*e + a^2*b^3*e^2)*((x*e + d)*
b - b*d + a*e)^5)

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maple [A]  time = 0.07, size = 305, normalized size = 1.54 \begin {gather*} -\frac {3 \sqrt {e x +d}\, a^{2} e^{7}}{128 \left (b e x +a e \right )^{5} b^{3}}+\frac {3 \sqrt {e x +d}\, a d \,e^{6}}{64 \left (b e x +a e \right )^{5} b^{2}}+\frac {3 \left (e x +d \right )^{\frac {9}{2}} b \,e^{5}}{128 \left (b e x +a e \right )^{5} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {3 \sqrt {e x +d}\, d^{2} e^{5}}{128 \left (b e x +a e \right )^{5} b}-\frac {7 \left (e x +d \right )^{\frac {3}{2}} a \,e^{6}}{64 \left (b e x +a e \right )^{5} b^{2}}+\frac {7 \left (e x +d \right )^{\frac {3}{2}} d \,e^{5}}{64 \left (b e x +a e \right )^{5} b}+\frac {7 \left (e x +d \right )^{\frac {7}{2}} e^{5}}{64 \left (b e x +a e \right )^{5} \left (a e -b d \right )}-\frac {\left (e x +d \right )^{\frac {5}{2}} e^{5}}{5 \left (b e x +a e \right )^{5} b}+\frac {3 e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{128 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

3/128*e^5/(b*e*x+a*e)^5*b/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(9/2)+7/64*e^5/(b*e*x+a*e)^5/(a*e-b*d)*(e*x+d)^(
7/2)-1/5*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(5/2)-7/64*e^6/(b*e*x+a*e)^5/b^2*(e*x+d)^(3/2)*a+7/64*e^5/(b*e*x+a*e)^5/b
*(e*x+d)^(3/2)*d-3/128*e^7/(b*e*x+a*e)^5/b^3*(e*x+d)^(1/2)*a^2+3/64*e^6/(b*e*x+a*e)^5/b^2*(e*x+d)^(1/2)*a*d-3/
128*e^5/(b*e*x+a*e)^5/b*(e*x+d)^(1/2)*d^2+3/128*e^5/b^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)/((a*e-b*d)*b)^(1/2)*arctan
((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)

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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((e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^3,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 = 0.17, size = 411, normalized size = 2.08 \begin {gather*} \frac {3\,e^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{128\,b^{7/2}\,{\left (a\,e-b\,d\right )}^{5/2}}-\frac {\frac {e^5\,{\left (d+e\,x\right )}^{5/2}}{5\,b}-\frac {7\,e^5\,{\left (d+e\,x\right )}^{7/2}}{64\,\left (a\,e-b\,d\right )}+\frac {3\,e^5\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{128\,b^3}+\frac {7\,e^5\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{64\,b^2}-\frac {3\,b\,e^5\,{\left (d+e\,x\right )}^{9/2}}{128\,{\left (a\,e-b\,d\right )}^2}}{\left (d+e\,x\right )\,\left (5\,a^4\,b\,e^4-20\,a^3\,b^2\,d\,e^3+30\,a^2\,b^3\,d^2\,e^2-20\,a\,b^4\,d^3\,e+5\,b^5\,d^4\right )-{\left (d+e\,x\right )}^2\,\left (-10\,a^3\,b^2\,e^3+30\,a^2\,b^3\,d\,e^2-30\,a\,b^4\,d^2\,e+10\,b^5\,d^3\right )+b^5\,{\left (d+e\,x\right )}^5-\left (5\,b^5\,d-5\,a\,b^4\,e\right )\,{\left (d+e\,x\right )}^4+a^5\,e^5-b^5\,d^5+{\left (d+e\,x\right )}^3\,\left (10\,a^2\,b^3\,e^2-20\,a\,b^4\,d\,e+10\,b^5\,d^2\right )-10\,a^2\,b^3\,d^3\,e^2+10\,a^3\,b^2\,d^2\,e^3+5\,a\,b^4\,d^4\,e-5\,a^4\,b\,d\,e^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)

[Out]

(3*e^5*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(128*b^(7/2)*(a*e - b*d)^(5/2)) - ((e^5*(d + e*x)^(5
/2))/(5*b) - (7*e^5*(d + e*x)^(7/2))/(64*(a*e - b*d)) + (3*e^5*(d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e)
)/(128*b^3) + (7*e^5*(a*e - b*d)*(d + e*x)^(3/2))/(64*b^2) - (3*b*e^5*(d + e*x)^(9/2))/(128*(a*e - b*d)^2))/((
d + e*x)*(5*b^5*d^4 + 5*a^4*b*e^4 - 20*a^3*b^2*d*e^3 + 30*a^2*b^3*d^2*e^2 - 20*a*b^4*d^3*e) - (d + e*x)^2*(10*
b^5*d^3 - 10*a^3*b^2*e^3 + 30*a^2*b^3*d*e^2 - 30*a*b^4*d^2*e) + b^5*(d + e*x)^5 - (5*b^5*d - 5*a*b^4*e)*(d + e
*x)^4 + a^5*e^5 - b^5*d^5 + (d + e*x)^3*(10*b^5*d^2 + 10*a^2*b^3*e^2 - 20*a*b^4*d*e) - 10*a^2*b^3*d^3*e^2 + 10
*a^3*b^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4)

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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((e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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