3.2086 \(\int \frac {(a+b x) (d+e x)^{5/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=162 \[ \frac {5 e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2}}-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (a+b x) (b d-a e)}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x)^2}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4} \]

[Out]

-5/24*e*(e*x+d)^(3/2)/b^2/(b*x+a)^3-1/4*(e*x+d)^(5/2)/b/(b*x+a)^4+5/64*e^4*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e
+b*d)^(1/2))/b^(7/2)/(-a*e+b*d)^(3/2)-5/32*e^2*(e*x+d)^(1/2)/b^3/(b*x+a)^2-5/64*e^3*(e*x+d)^(1/2)/b^3/(-a*e+b*
d)/(b*x+a)

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Rubi [A]  time = 0.09, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 47, 51, 63, 208} \[ -\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (a+b x) (b d-a e)}-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x)^2}+\frac {5 e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2}}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*e^2*Sqrt[d + e*x])/(32*b^3*(a + b*x)^2) - (5*e^3*Sqrt[d + e*x])/(64*b^3*(b*d - a*e)*(a + b*x)) - (5*e*(d +
e*x)^(3/2))/(24*b^2*(a + b*x)^3) - (d + e*x)^(5/2)/(4*b*(a + b*x)^4) + (5*e^4*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])
/Sqrt[b*d - a*e]])/(64*b^(7/2)*(b*d - a*e)^(3/2))

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 {(a+b x) (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)^5} \, dx\\ &=-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac {\left (5 e^2\right ) \int \frac {\sqrt {d+e x}}{(a+b x)^3} \, dx}{16 b^2}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x)^2}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac {\left (5 e^3\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{64 b^3}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x)^2}-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4}-\frac {\left (5 e^4\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{128 b^3 (b d-a e)}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x)^2}-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4}-\frac {\left (5 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 )}{64 b^3 (b d-a e)}\\ &=-\frac {5 e^2 \sqrt {d+e x}}{32 b^3 (a+b x)^2}-\frac {5 e^3 \sqrt {d+e x}}{64 b^3 (b d-a e) (a+b x)}-\frac {5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac {5 e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.32 \[ \frac {2 e^4 (d+e x)^{7/2} \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};-\frac {b (d+e x)}{a e-b d}\right )}{7 (a e-b d)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^4*(d + e*x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(7*(-(b*d) + a*e)^5)

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fricas [B]  time = 0.98, size = 894, normalized size = 5.52 \[ \left [-\frac {15 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (48 \, b^{5} d^{4} - 56 \, a b^{4} d^{3} e - 2 \, a^{2} b^{3} d^{2} e^{2} - 5 \, a^{3} b^{2} d e^{3} + 15 \, a^{4} b e^{4} + 15 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (118 \, b^{5} d^{2} e^{2} - 191 \, a b^{4} d e^{3} + 73 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (136 \, b^{5} d^{3} e - 172 \, a b^{4} d^{2} e^{2} - 19 \, a^{2} b^{3} d e^{3} + 55 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{384 \, {\left (a^{4} b^{6} d^{2} - 2 \, a^{5} b^{5} d e + a^{6} b^{4} e^{2} + {\left (b^{10} d^{2} - 2 \, a b^{9} d e + a^{2} b^{8} e^{2}\right )} x^{4} + 4 \, {\left (a b^{9} d^{2} - 2 \, a^{2} b^{8} d e + a^{3} b^{7} e^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{2} - 2 \, a^{3} b^{7} d e + a^{4} b^{6} e^{2}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{2} - 2 \, a^{4} b^{6} d e + a^{5} b^{5} e^{2}\right )} x\right )}}, -\frac {15 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (48 \, b^{5} d^{4} - 56 \, a b^{4} d^{3} e - 2 \, a^{2} b^{3} d^{2} e^{2} - 5 \, a^{3} b^{2} d e^{3} + 15 \, a^{4} b e^{4} + 15 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (118 \, b^{5} d^{2} e^{2} - 191 \, a b^{4} d e^{3} + 73 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (136 \, b^{5} d^{3} e - 172 \, a b^{4} d^{2} e^{2} - 19 \, a^{2} b^{3} d e^{3} + 55 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{192 \, {\left (a^{4} b^{6} d^{2} - 2 \, a^{5} b^{5} d e + a^{6} b^{4} e^{2} + {\left (b^{10} d^{2} - 2 \, a b^{9} d e + a^{2} b^{8} e^{2}\right )} x^{4} + 4 \, {\left (a b^{9} d^{2} - 2 \, a^{2} b^{8} d e + a^{3} b^{7} e^{2}\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d^{2} - 2 \, a^{3} b^{7} d e + a^{4} b^{6} e^{2}\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d^{2} - 2 \, a^{4} b^{6} d e + a^{5} b^{5} e^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*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*(48*b^5*d^4 - 56*a*b^4*d^3*e -
2*a^2*b^3*d^2*e^2 - 5*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (118*b^5*d^2*e^2 - 191*a
*b^4*d*e^3 + 73*a^2*b^3*e^4)*x^2 + (136*b^5*d^3*e - 172*a*b^4*d^2*e^2 - 19*a^2*b^3*d*e^3 + 55*a^3*b^2*e^4)*x)*
sqrt(e*x + d))/(a^4*b^6*d^2 - 2*a^5*b^5*d*e + a^6*b^4*e^2 + (b^10*d^2 - 2*a*b^9*d*e + a^2*b^8*e^2)*x^4 + 4*(a*
b^9*d^2 - 2*a^2*b^8*d*e + a^3*b^7*e^2)*x^3 + 6*(a^2*b^8*d^2 - 2*a^3*b^7*d*e + a^4*b^6*e^2)*x^2 + 4*(a^3*b^7*d^
2 - 2*a^4*b^6*d*e + a^5*b^5*e^2)*x), -1/192*(15*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e
^4*x + a^4*e^4)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (48*b^5*d^4 -
56*a*b^4*d^3*e - 2*a^2*b^3*d^2*e^2 - 5*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (118*b^
5*d^2*e^2 - 191*a*b^4*d*e^3 + 73*a^2*b^3*e^4)*x^2 + (136*b^5*d^3*e - 172*a*b^4*d^2*e^2 - 19*a^2*b^3*d*e^3 + 55
*a^3*b^2*e^4)*x)*sqrt(e*x + d))/(a^4*b^6*d^2 - 2*a^5*b^5*d*e + a^6*b^4*e^2 + (b^10*d^2 - 2*a*b^9*d*e + a^2*b^8
*e^2)*x^4 + 4*(a*b^9*d^2 - 2*a^2*b^8*d*e + a^3*b^7*e^2)*x^3 + 6*(a^2*b^8*d^2 - 2*a^3*b^7*d*e + a^4*b^6*e^2)*x^
2 + 4*(a^3*b^7*d^2 - 2*a^4*b^6*d*e + a^5*b^5*e^2)*x)]

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giac [A]  time = 0.21, size = 265, normalized size = 1.64 \[ -\frac {5 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{4}}{64 \, {\left (b^{4} d - a b^{3} e\right )} \sqrt {-b^{2} d + a b e}} - \frac {15 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{3} e^{4} + 73 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} d e^{4} - 55 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} + 15 \, \sqrt {x e + d} b^{3} d^{3} e^{4} - 73 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{2} e^{5} + 110 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} - 45 \, \sqrt {x e + d} a b^{2} d^{2} e^{5} - 55 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{6} + 45 \, \sqrt {x e + d} a^{2} b d e^{6} - 15 \, \sqrt {x e + d} a^{3} e^{7}}{192 \, {\left (b^{4} d - a b^{3} e\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^4*d - a*b^3*e)*sqrt(-b^2*d + a*b*e)) - 1/192*(15*(x
*e + d)^(7/2)*b^3*e^4 + 73*(x*e + d)^(5/2)*b^3*d*e^4 - 55*(x*e + d)^(3/2)*b^3*d^2*e^4 + 15*sqrt(x*e + d)*b^3*d
^3*e^4 - 73*(x*e + d)^(5/2)*a*b^2*e^5 + 110*(x*e + d)^(3/2)*a*b^2*d*e^5 - 45*sqrt(x*e + d)*a*b^2*d^2*e^5 - 55*
(x*e + d)^(3/2)*a^2*b*e^6 + 45*sqrt(x*e + d)*a^2*b*d*e^6 - 15*sqrt(x*e + d)*a^3*e^7)/((b^4*d - a*b^3*e)*((x*e
+ d)*b - b*d + a*e)^4)

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maple [A]  time = 0.07, size = 246, normalized size = 1.52 \[ -\frac {5 \sqrt {e x +d}\, a^{2} e^{6}}{64 \left (b e x +a e \right )^{4} b^{3}}+\frac {5 \sqrt {e x +d}\, a d \,e^{5}}{32 \left (b e x +a e \right )^{4} b^{2}}-\frac {5 \sqrt {e x +d}\, d^{2} e^{4}}{64 \left (b e x +a e \right )^{4} b}-\frac {55 \left (e x +d \right )^{\frac {3}{2}} a \,e^{5}}{192 \left (b e x +a e \right )^{4} b^{2}}+\frac {55 \left (e x +d \right )^{\frac {3}{2}} d \,e^{4}}{192 \left (b e x +a e \right )^{4} b}+\frac {5 \left (e x +d \right )^{\frac {7}{2}} e^{4}}{64 \left (b e x +a e \right )^{4} \left (a e -b d \right )}-\frac {73 \left (e x +d \right )^{\frac {5}{2}} e^{4}}{192 \left (b e x +a e \right )^{4} b}+\frac {5 e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{64 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*e^4/(b*e*x+a*e)^4/(a*e-b*d)*(e*x+d)^(7/2)-73/192*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(5/2)-55/192*e^5/(b*e*x+a*e)
^4/b^2*(e*x+d)^(3/2)*a+55/192*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(3/2)*d-5/64*e^6/(b*e*x+a*e)^4/b^3*(e*x+d)^(1/2)*a^2
+5/32*e^5/(b*e*x+a*e)^4/b^2*(e*x+d)^(1/2)*a*d-5/64*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(1/2)*d^2+5/64*e^4/(a*e-b*d)/b^
3/((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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(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 = 2.12, size = 309, normalized size = 1.91 \[ \frac {5\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{64\,b^{7/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {73\,e^4\,{\left (d+e\,x\right )}^{5/2}}{192\,b}-\frac {5\,e^4\,{\left (d+e\,x\right )}^{7/2}}{64\,\left (a\,e-b\,d\right )}+\frac {5\,e^4\,\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{64\,b^3}+\frac {55\,e^4\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{192\,b^2}}{b^4\,{\left (d+e\,x\right )}^4-\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,{\left (d+e\,x\right )}^3-\left (d+e\,x\right )\,\left (-4\,a^3\,b\,e^3+12\,a^2\,b^2\,d\,e^2-12\,a\,b^3\,d^2\,e+4\,b^4\,d^3\right )+a^4\,e^4+b^4\,d^4+{\left (d+e\,x\right )}^2\,\left (6\,a^2\,b^2\,e^2-12\,a\,b^3\,d\,e+6\,b^4\,d^2\right )+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e-4\,a^3\,b\,d\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*e^4*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(64*b^(7/2)*(a*e - b*d)^(3/2)) - ((73*e^4*(d + e*x)^
(5/2))/(192*b) - (5*e^4*(d + e*x)^(7/2))/(64*(a*e - b*d)) + (5*e^4*(d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*
d*e))/(64*b^3) + (55*e^4*(a*e - b*d)*(d + e*x)^(3/2))/(192*b^2))/(b^4*(d + e*x)^4 - (4*b^4*d - 4*a*b^3*e)*(d +
e*x)^3 - (d + e*x)*(4*b^4*d^3 - 4*a^3*b*e^3 + 12*a^2*b^2*d*e^2 - 12*a*b^3*d^2*e) + a^4*e^4 + b^4*d^4 + (d + e
*x)^2*(6*b^4*d^2 + 6*a^2*b^2*e^2 - 12*a*b^3*d*e) + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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