3.2082 \(\int \frac {a+b x}{(d+e x)^{7/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=196 \[ -\frac {63 b^{5/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2}}+\frac {63 b^2 e^2}{4 \sqrt {d+e x} (b d-a e)^5}+\frac {21 b e^2}{4 (d+e x)^{3/2} (b d-a e)^4}+\frac {63 e^2}{20 (d+e x)^{5/2} (b d-a e)^3}+\frac {9 e}{4 (a+b x) (d+e x)^{5/2} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)} \]

[Out]

63/20*e^2/(-a*e+b*d)^3/(e*x+d)^(5/2)-1/2/(-a*e+b*d)/(b*x+a)^2/(e*x+d)^(5/2)+9/4*e/(-a*e+b*d)^2/(b*x+a)/(e*x+d)
^(5/2)+21/4*b*e^2/(-a*e+b*d)^4/(e*x+d)^(3/2)-63/4*b^(5/2)*e^2*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/
(-a*e+b*d)^(11/2)+63/4*b^2*e^2/(-a*e+b*d)^5/(e*x+d)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \[ \frac {63 b^2 e^2}{4 \sqrt {d+e x} (b d-a e)^5}-\frac {63 b^{5/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2}}+\frac {21 b e^2}{4 (d+e x)^{3/2} (b d-a e)^4}+\frac {63 e^2}{20 (d+e x)^{5/2} (b d-a e)^3}+\frac {9 e}{4 (a+b x) (d+e x)^{5/2} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(63*e^2)/(20*(b*d - a*e)^3*(d + e*x)^(5/2)) - 1/(2*(b*d - a*e)*(a + b*x)^2*(d + e*x)^(5/2)) + (9*e)/(4*(b*d -
a*e)^2*(a + b*x)*(d + e*x)^(5/2)) + (21*b*e^2)/(4*(b*d - a*e)^4*(d + e*x)^(3/2)) + (63*b^2*e^2)/(4*(b*d - a*e)
^5*Sqrt[d + e*x]) - (63*b^(5/2)*e^2*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(11/2))

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

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.27 \[ -\frac {2 e^2 \, _2F_1\left (-\frac {5}{2},3;-\frac {3}{2};-\frac {b (d+e x)}{a e-b d}\right )}{5 (d+e x)^{5/2} (a e-b d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.32, size = 1858, normalized size = 9.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/40*(315*(b^4*e^5*x^5 + a^2*b^2*d^3*e^2 + (3*b^4*d*e^4 + 2*a*b^3*e^5)*x^4 + (3*b^4*d^2*e^3 + 6*a*b^3*d*e^4
+ a^2*b^2*e^5)*x^3 + (b^4*d^3*e^2 + 6*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4)*x^2 + (2*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*
e^3)*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*(315*b^4*e^4*x^4 - 10*b^4*d^4 + 85*a*b^3*d^3*e + 288*a^2*b^2*d^2*e^2 - 56*a^3*b*d*e^3 + 8*a^4*e^4 + 1
05*(7*b^4*d*e^3 + 5*a*b^3*e^4)*x^3 + 21*(23*b^4*d^2*e^2 + 59*a*b^3*d*e^3 + 8*a^2*b^2*e^4)*x^2 + 3*(15*b^4*d^3*
e + 277*a*b^3*d^2*e^2 + 136*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d^7*e + 10
*a^4*b^3*d^6*e^2 - 10*a^5*b^2*d^5*e^3 + 5*a^6*b*d^4*e^4 - a^7*d^3*e^5 + (b^7*d^5*e^3 - 5*a*b^6*d^4*e^4 + 10*a^
2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6 + 5*a^4*b^3*d*e^7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d^5*e^3 +
20*a^2*b^5*d^4*e^4 - 10*a^3*b^4*d^3*e^5 - 5*a^4*b^3*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2*a^6*b*e^8)*x^4 + (3*b^7*d^7*
e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*e^3 + 25*a^3*b^4*d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6 - a^6*b*d
*e^7 - a^7*e^8)*x^3 + (b^7*d^8 + a*b^6*d^7*e - 17*a^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e^3 - 25*a^4*b^3*d^4*e^4 -
a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6 - 3*a^7*d*e^7)*x^2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6*e^2 + 10
*a^4*b^3*d^5*e^3 - 20*a^5*b^2*d^4*e^4 + 13*a^6*b*d^3*e^5 - 3*a^7*d^2*e^6)*x), -1/20*(315*(b^4*e^5*x^5 + a^2*b^
2*d^3*e^2 + (3*b^4*d*e^4 + 2*a*b^3*e^5)*x^4 + (3*b^4*d^2*e^3 + 6*a*b^3*d*e^4 + a^2*b^2*e^5)*x^3 + (b^4*d^3*e^2
+ 6*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4)*x^2 + (2*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3)*x)*sqrt(-b/(b*d - a*e))*arct
an(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (315*b^4*e^4*x^4 - 10*b^4*d^4 + 85*a*b^3*d
^3*e + 288*a^2*b^2*d^2*e^2 - 56*a^3*b*d*e^3 + 8*a^4*e^4 + 105*(7*b^4*d*e^3 + 5*a*b^3*e^4)*x^3 + 21*(23*b^4*d^2
*e^2 + 59*a*b^3*d*e^3 + 8*a^2*b^2*e^4)*x^2 + 3*(15*b^4*d^3*e + 277*a*b^3*d^2*e^2 + 136*a^2*b^2*d*e^3 - 8*a^3*b
*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d^7*e + 10*a^4*b^3*d^6*e^2 - 10*a^5*b^2*d^5*e^3 + 5*a^6*b*d^4
*e^4 - a^7*d^3*e^5 + (b^7*d^5*e^3 - 5*a*b^6*d^4*e^4 + 10*a^2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6 + 5*a^4*b^3*d*e^
7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d^5*e^3 + 20*a^2*b^5*d^4*e^4 - 10*a^3*b^4*d^3*e^5 - 5*a^4*b^3
*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2*a^6*b*e^8)*x^4 + (3*b^7*d^7*e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*e^3 + 25*a^3*b^4*
d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6 - a^6*b*d*e^7 - a^7*e^8)*x^3 + (b^7*d^8 + a*b^6*d^7*e - 17*a
^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e^3 - 25*a^4*b^3*d^4*e^4 - a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6 - 3*a^7*d*e^7)*x^
2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6*e^2 + 10*a^4*b^3*d^5*e^3 - 20*a^5*b^2*d^4*e^4 + 13*a^6*b*d^
3*e^5 - 3*a^7*d^2*e^6)*x)]

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giac [B]  time = 0.23, size = 379, normalized size = 1.93 \[ \frac {63 \, b^{3} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt {-b^{2} d + a b e}} + \frac {15 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} e^{2} - 17 \, \sqrt {x e + d} b^{4} d e^{2} + 17 \, \sqrt {x e + d} a b^{3} e^{3}}{4 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} + \frac {2 \, {\left (30 \, {\left (x e + d\right )}^{2} b^{2} e^{2} + 5 \, {\left (x e + d\right )} b^{2} d e^{2} + b^{2} d^{2} e^{2} - 5 \, {\left (x e + d\right )} a b e^{3} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}}{5 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

63/4*b^3*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*
a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*sqrt(-b^2*d + a*b*e)) + 1/4*(15*(x*e + d)^(3/2)*b^4*e^2 - 17*sqrt(x
*e + d)*b^4*d*e^2 + 17*sqrt(x*e + d)*a*b^3*e^3)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^
2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*((x*e + d)*b - b*d + a*e)^2) + 2/5*(30*(x*e + d)^2*b^2*e^2 + 5*(x*e + d)*b^2*
d*e^2 + b^2*d^2*e^2 - 5*(x*e + d)*a*b*e^3 - 2*a*b*d*e^3 + a^2*e^4)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*
e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*(x*e + d)^(5/2))

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maple [A]  time = 0.07, size = 231, normalized size = 1.18 \[ -\frac {17 \sqrt {e x +d}\, a \,b^{3} e^{3}}{4 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{2}}+\frac {17 \sqrt {e x +d}\, b^{4} d \,e^{2}}{4 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{2}}-\frac {15 \left (e x +d \right )^{\frac {3}{2}} b^{4} e^{2}}{4 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{2}}-\frac {63 b^{3} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right )^{5} \sqrt {\left (a e -b d \right ) b}}-\frac {12 b^{2} e^{2}}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {2 b \,e^{2}}{\left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 e^{2}}{5 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*e^2/(a*e-b*d)^5*b^4/(b*e*x+a*e)^2*(e*x+d)^(3/2)-17/4*e^3/(a*e-b*d)^5*b^3/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a+1
7/4*e^2/(a*e-b*d)^5*b^4/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d-63/4*e^2/(a*e-b*d)^5*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^2/(a*e-b*d)^3/(e*x+d)^(5/2)-12*e^2/(a*e-b*d)^5*b^2/(e*x+d)^(1/2)+2*e^2/
(a*e-b*d)^4*b/(e*x+d)^(3/2)

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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)^(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 = 2.54, size = 284, normalized size = 1.45 \[ -\frac {\frac {2\,e^2}{5\,\left (a\,e-b\,d\right )}+\frac {42\,b^2\,e^2\,{\left (d+e\,x\right )}^2}{5\,{\left (a\,e-b\,d\right )}^3}+\frac {105\,b^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e-b\,d\right )}^4}+\frac {63\,b^4\,e^2\,{\left (d+e\,x\right )}^4}{4\,{\left (a\,e-b\,d\right )}^5}-\frac {6\,b\,e^2\,\left (d+e\,x\right )}{5\,{\left (a\,e-b\,d\right )}^2}}{b^2\,{\left (d+e\,x\right )}^{9/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{7/2}+{\left (d+e\,x\right )}^{5/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}-\frac {63\,b^{5/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^5\,e^5-5\,a^4\,b\,d\,e^4+10\,a^3\,b^2\,d^2\,e^3-10\,a^2\,b^3\,d^3\,e^2+5\,a\,b^4\,d^4\,e-b^5\,d^5\right )}{{\left (a\,e-b\,d\right )}^{11/2}}\right )}{4\,{\left (a\,e-b\,d\right )}^{11/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*e^2)/(5*(a*e - b*d)) + (42*b^2*e^2*(d + e*x)^2)/(5*(a*e - b*d)^3) + (105*b^3*e^2*(d + e*x)^3)/(4*(a*e -
b*d)^4) + (63*b^4*e^2*(d + e*x)^4)/(4*(a*e - b*d)^5) - (6*b*e^2*(d + e*x))/(5*(a*e - b*d)^2))/(b^2*(d + e*x)^(
9/2) - (2*b^2*d - 2*a*b*e)*(d + e*x)^(7/2) + (d + e*x)^(5/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e)) - (63*b^(5/2)*e^
2*atan((b^(1/2)*(d + e*x)^(1/2)*(a^5*e^5 - b^5*d^5 - 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))/(a*e - b*d)^(11/2)))/(4*(a*e - b*d)^(11/2))

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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)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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