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

Optimal. Leaf size=233 \[ -\frac {1155 b^{3/2} e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 (b d-a e)^{13/2}}+\frac {1155 b e^4}{64 \sqrt {d+e x} (b d-a e)^6}+\frac {385 e^4}{64 (d+e x)^{3/2} (b d-a e)^5}+\frac {231 e^3}{64 (a+b x) (d+e x)^{3/2} (b d-a e)^4}-\frac {33 e^2}{32 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^3}+\frac {11 e}{24 (a+b x)^3 (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{4 (a+b x)^4 (d+e x)^{3/2} (b d-a e)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \begin {gather*} -\frac {1155 b^{3/2} e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 (b d-a e)^{13/2}}+\frac {1155 b e^4}{64 \sqrt {d+e x} (b d-a e)^6}+\frac {385 e^4}{64 (d+e x)^{3/2} (b d-a e)^5}+\frac {231 e^3}{64 (a+b x) (d+e x)^{3/2} (b d-a e)^4}-\frac {33 e^2}{32 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^3}+\frac {11 e}{24 (a+b x)^3 (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{4 (a+b x)^4 (d+e x)^{3/2} (b d-a e)} \end {gather*}

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]

(385*e^4)/(64*(b*d - a*e)^5*(d + e*x)^(3/2)) - 1/(4*(b*d - a*e)*(a + b*x)^4*(d + e*x)^(3/2)) + (11*e)/(24*(b*d
- a*e)^2*(a + b*x)^3*(d + e*x)^(3/2)) - (33*e^2)/(32*(b*d - a*e)^3*(a + b*x)^2*(d + e*x)^(3/2)) + (231*e^3)/(
64*(b*d - a*e)^4*(a + b*x)*(d + e*x)^(3/2)) + (1155*b*e^4)/(64*(b*d - a*e)^6*Sqrt[d + e*x]) - (1155*b^(3/2)*e^
4*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*(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 {a+b x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac {1}{(a+b x)^5 (d+e x)^{5/2}} \, dx\\ &=-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}-\frac {(11 e) \int \frac {1}{(a+b x)^4 (d+e x)^{5/2}} \, dx}{8 (b d-a e)}\\ &=-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}+\frac {\left (33 e^2\right ) \int \frac {1}{(a+b x)^3 (d+e x)^{5/2}} \, dx}{16 (b d-a e)^2}\\ &=-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac {33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}-\frac {\left (231 e^3\right ) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{64 (b d-a e)^3}\\ &=-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac {33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac {231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac {\left (1155 e^4\right ) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{128 (b d-a e)^4}\\ &=\frac {385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac {33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac {231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac {\left (1155 b e^4\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{128 (b d-a e)^5}\\ &=\frac {385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac {33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac {231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac {1155 b e^4}{64 (b d-a e)^6 \sqrt {d+e x}}+\frac {\left (1155 b^2 e^4\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{128 (b d-a e)^6}\\ &=\frac {385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac {33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac {231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac {1155 b e^4}{64 (b d-a e)^6 \sqrt {d+e x}}+\frac {\left (1155 b^2 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 d-a e)^6}\\ &=\frac {385 e^4}{64 (b d-a e)^5 (d+e x)^{3/2}}-\frac {1}{4 (b d-a e) (a+b x)^4 (d+e x)^{3/2}}+\frac {11 e}{24 (b d-a e)^2 (a+b x)^3 (d+e x)^{3/2}}-\frac {33 e^2}{32 (b d-a e)^3 (a+b x)^2 (d+e x)^{3/2}}+\frac {231 e^3}{64 (b d-a e)^4 (a+b x) (d+e x)^{3/2}}+\frac {1155 b e^4}{64 (b d-a e)^6 \sqrt {d+e x}}-\frac {1155 b^{3/2} e^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 (b d-a e)^{13/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 52, normalized size = 0.22 \begin {gather*} -\frac {2 e^4 \, _2F_1\left (-\frac {3}{2},5;-\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{3 (d+e x)^{3/2} (a e-b d)^5} \end {gather*}

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*Hypergeometric2F1[-3/2, 5, -1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(3*(-(b*d) + a*e)^5*(d + e*x)^(3/2)
)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.58, size = 406, normalized size = 1.74 \begin {gather*} -\frac {e^4 \left (128 a^5 e^5-1408 a^4 b e^4 (d+e x)-640 a^4 b d e^4+1280 a^3 b^2 d^2 e^3-9207 a^3 b^2 e^3 (d+e x)^2+5632 a^3 b^2 d e^3 (d+e x)-1280 a^2 b^3 d^3 e^2-8448 a^2 b^3 d^2 e^2 (d+e x)-16863 a^2 b^3 e^2 (d+e x)^3+27621 a^2 b^3 d e^2 (d+e x)^2+640 a b^4 d^4 e+5632 a b^4 d^3 e (d+e x)-27621 a b^4 d^2 e (d+e x)^2-12705 a b^4 e (d+e x)^4+33726 a b^4 d e (d+e x)^3-128 b^5 d^5-1408 b^5 d^4 (d+e x)+9207 b^5 d^3 (d+e x)^2-16863 b^5 d^2 (d+e x)^3-3465 b^5 (d+e x)^5+12705 b^5 d (d+e x)^4\right )}{192 (d+e x)^{3/2} (b d-a e)^6 (-a e-b (d+e x)+b d)^4}-\frac {1155 b^{3/2} e^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{64 (a e-b d)^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(e^4*(-128*b^5*d^5 + 640*a*b^4*d^4*e - 1280*a^2*b^3*d^3*e^2 + 1280*a^3*b^2*d^2*e^3 - 640*a^4*b*d*e^4 +
128*a^5*e^5 - 1408*b^5*d^4*(d + e*x) + 5632*a*b^4*d^3*e*(d + e*x) - 8448*a^2*b^3*d^2*e^2*(d + e*x) + 5632*a^3*
b^2*d*e^3*(d + e*x) - 1408*a^4*b*e^4*(d + e*x) + 9207*b^5*d^3*(d + e*x)^2 - 27621*a*b^4*d^2*e*(d + e*x)^2 + 27
621*a^2*b^3*d*e^2*(d + e*x)^2 - 9207*a^3*b^2*e^3*(d + e*x)^2 - 16863*b^5*d^2*(d + e*x)^3 + 33726*a*b^4*d*e*(d
+ e*x)^3 - 16863*a^2*b^3*e^2*(d + e*x)^3 + 12705*b^5*d*(d + e*x)^4 - 12705*a*b^4*e*(d + e*x)^4 - 3465*b^5*(d +
e*x)^5))/((b*d - a*e)^6*(d + e*x)^(3/2)*(b*d - a*e - b*(d + e*x))^4) - (1155*b^(3/2)*e^4*ArcTan[(Sqrt[b]*Sqrt
[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(64*(-(b*d) + a*e)^(13/2))

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 2494, normalized size = 10.70

result too large to display

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*(3465*(b^5*e^6*x^6 + a^4*b*d^2*e^4 + 2*(b^5*d*e^5 + 2*a*b^4*e^6)*x^5 + (b^5*d^2*e^4 + 8*a*b^4*d*e^5 + 6
*a^2*b^3*e^6)*x^4 + 4*(a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 + a^3*b^2*e^6)*x^3 + (6*a^2*b^3*d^2*e^4 + 8*a^3*b^2*d*e
^5 + a^4*b*e^6)*x^2 + 2*(2*a^3*b^2*d^2*e^4 + a^4*b*d*e^5)*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 - 48*b^5*d^5 + 328*a*b^4*d^4*e
- 1030*a^2*b^3*d^3*e^2 + 2295*a^3*b^2*d^2*e^3 + 2048*a^4*b*d*e^4 - 128*a^5*e^5 + 1155*(4*b^5*d*e^4 + 11*a*b^4
*e^5)*x^4 + 231*(3*b^5*d^2*e^3 + 74*a*b^4*d*e^4 + 73*a^2*b^3*e^5)*x^3 - 99*(2*b^5*d^3*e^2 - 27*a*b^4*d^2*e^3 -
232*a^2*b^3*d*e^4 - 93*a^3*b^2*e^5)*x^2 + 11*(8*b^5*d^4*e - 68*a*b^4*d^3*e^2 + 345*a^2*b^3*d^2*e^3 + 1162*a^3
*b^2*d*e^4 + 128*a^4*b*e^5)*x)*sqrt(e*x + d))/(a^4*b^6*d^8 - 6*a^5*b^5*d^7*e + 15*a^6*b^4*d^6*e^2 - 20*a^7*b^3
*d^5*e^3 + 15*a^8*b^2*d^4*e^4 - 6*a^9*b*d^3*e^5 + a^10*d^2*e^6 + (b^10*d^6*e^2 - 6*a*b^9*d^5*e^3 + 15*a^2*b^8*
d^4*e^4 - 20*a^3*b^7*d^3*e^5 + 15*a^4*b^6*d^2*e^6 - 6*a^5*b^5*d*e^7 + a^6*b^4*e^8)*x^6 + 2*(b^10*d^7*e - 4*a*b
^9*d^6*e^2 + 3*a^2*b^8*d^5*e^3 + 10*a^3*b^7*d^4*e^4 - 25*a^4*b^6*d^3*e^5 + 24*a^5*b^5*d^2*e^6 - 11*a^6*b^4*d*e
^7 + 2*a^7*b^3*e^8)*x^5 + (b^10*d^8 + 2*a*b^9*d^7*e - 27*a^2*b^8*d^6*e^2 + 64*a^3*b^7*d^5*e^3 - 55*a^4*b^6*d^4
*e^4 - 6*a^5*b^5*d^3*e^5 + 43*a^6*b^4*d^2*e^6 - 28*a^7*b^3*d*e^7 + 6*a^8*b^2*e^8)*x^4 + 4*(a*b^9*d^8 - 3*a^2*b
^8*d^7*e - 2*a^3*b^7*d^6*e^2 + 19*a^4*b^6*d^5*e^3 - 30*a^5*b^5*d^4*e^4 + 19*a^6*b^4*d^3*e^5 - 2*a^7*b^3*d^2*e^
6 - 3*a^8*b^2*d*e^7 + a^9*b*e^8)*x^3 + (6*a^2*b^8*d^8 - 28*a^3*b^7*d^7*e + 43*a^4*b^6*d^6*e^2 - 6*a^5*b^5*d^5*
e^3 - 55*a^6*b^4*d^4*e^4 + 64*a^7*b^3*d^3*e^5 - 27*a^8*b^2*d^2*e^6 + 2*a^9*b*d*e^7 + a^10*e^8)*x^2 + 2*(2*a^3*
b^7*d^8 - 11*a^4*b^6*d^7*e + 24*a^5*b^5*d^6*e^2 - 25*a^6*b^4*d^5*e^3 + 10*a^7*b^3*d^4*e^4 + 3*a^8*b^2*d^3*e^5
- 4*a^9*b*d^2*e^6 + a^10*d*e^7)*x), -1/192*(3465*(b^5*e^6*x^6 + a^4*b*d^2*e^4 + 2*(b^5*d*e^5 + 2*a*b^4*e^6)*x^
5 + (b^5*d^2*e^4 + 8*a*b^4*d*e^5 + 6*a^2*b^3*e^6)*x^4 + 4*(a*b^4*d^2*e^4 + 3*a^2*b^3*d*e^5 + a^3*b^2*e^6)*x^3
+ (6*a^2*b^3*d^2*e^4 + 8*a^3*b^2*d*e^5 + a^4*b*e^6)*x^2 + 2*(2*a^3*b^2*d^2*e^4 + a^4*b*d*e^5)*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 - 48*b^5*d^5
+ 328*a*b^4*d^4*e - 1030*a^2*b^3*d^3*e^2 + 2295*a^3*b^2*d^2*e^3 + 2048*a^4*b*d*e^4 - 128*a^5*e^5 + 1155*(4*b^
5*d*e^4 + 11*a*b^4*e^5)*x^4 + 231*(3*b^5*d^2*e^3 + 74*a*b^4*d*e^4 + 73*a^2*b^3*e^5)*x^3 - 99*(2*b^5*d^3*e^2 -
27*a*b^4*d^2*e^3 - 232*a^2*b^3*d*e^4 - 93*a^3*b^2*e^5)*x^2 + 11*(8*b^5*d^4*e - 68*a*b^4*d^3*e^2 + 345*a^2*b^3*
d^2*e^3 + 1162*a^3*b^2*d*e^4 + 128*a^4*b*e^5)*x)*sqrt(e*x + d))/(a^4*b^6*d^8 - 6*a^5*b^5*d^7*e + 15*a^6*b^4*d^
6*e^2 - 20*a^7*b^3*d^5*e^3 + 15*a^8*b^2*d^4*e^4 - 6*a^9*b*d^3*e^5 + a^10*d^2*e^6 + (b^10*d^6*e^2 - 6*a*b^9*d^5
*e^3 + 15*a^2*b^8*d^4*e^4 - 20*a^3*b^7*d^3*e^5 + 15*a^4*b^6*d^2*e^6 - 6*a^5*b^5*d*e^7 + a^6*b^4*e^8)*x^6 + 2*(
b^10*d^7*e - 4*a*b^9*d^6*e^2 + 3*a^2*b^8*d^5*e^3 + 10*a^3*b^7*d^4*e^4 - 25*a^4*b^6*d^3*e^5 + 24*a^5*b^5*d^2*e^
6 - 11*a^6*b^4*d*e^7 + 2*a^7*b^3*e^8)*x^5 + (b^10*d^8 + 2*a*b^9*d^7*e - 27*a^2*b^8*d^6*e^2 + 64*a^3*b^7*d^5*e^
3 - 55*a^4*b^6*d^4*e^4 - 6*a^5*b^5*d^3*e^5 + 43*a^6*b^4*d^2*e^6 - 28*a^7*b^3*d*e^7 + 6*a^8*b^2*e^8)*x^4 + 4*(a
*b^9*d^8 - 3*a^2*b^8*d^7*e - 2*a^3*b^7*d^6*e^2 + 19*a^4*b^6*d^5*e^3 - 30*a^5*b^5*d^4*e^4 + 19*a^6*b^4*d^3*e^5
- 2*a^7*b^3*d^2*e^6 - 3*a^8*b^2*d*e^7 + a^9*b*e^8)*x^3 + (6*a^2*b^8*d^8 - 28*a^3*b^7*d^7*e + 43*a^4*b^6*d^6*e^
2 - 6*a^5*b^5*d^5*e^3 - 55*a^6*b^4*d^4*e^4 + 64*a^7*b^3*d^3*e^5 - 27*a^8*b^2*d^2*e^6 + 2*a^9*b*d*e^7 + a^10*e^
8)*x^2 + 2*(2*a^3*b^7*d^8 - 11*a^4*b^6*d^7*e + 24*a^5*b^5*d^6*e^2 - 25*a^6*b^4*d^5*e^3 + 10*a^7*b^3*d^4*e^4 +
3*a^8*b^2*d^3*e^5 - 4*a^9*b*d^2*e^6 + a^10*d*e^7)*x)]

________________________________________________________________________________________

giac [B]  time = 0.27, size = 500, normalized size = 2.15 \begin {gather*} \frac {1155 \, b^{2} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{4}}{64 \, {\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 (15 \, {\left (x e + d\right )} b e^{4} + b d e^{4} - a e^{5}\right )}}{3 \, {\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 {3}{2}}} + \frac {1545 \, {\left (x e + d\right )}^{\frac {7}{2}} b^{5} e^{4} - 5153 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{5} d e^{4} + 5855 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{5} d^{2} e^{4} - 2295 \, \sqrt {x e + d} b^{5} d^{3} e^{4} + 5153 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{4} e^{5} - 11710 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{4} d e^{5} + 6885 \, \sqrt {x e + d} a b^{4} d^{2} e^{5} + 5855 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{3} e^{6} - 6885 \, \sqrt {x e + d} a^{2} b^{3} d e^{6} + 2295 \, \sqrt {x e + d} a^{3} b^{2} e^{7}}{192 \, {\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 )}^{4}} \end {gather*}

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]

1155/64*b^2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((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)*sqrt(-b^2*d + a*b*e)) + 2/3*(15*(x*e + d)*b
*e^4 + b*d*e^4 - a*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)^(3/2)) + 1/192*(1545*(x*e + d)^(7/2)*b^5*e^4 - 5153*(x*e + d)^(5/2)*b^
5*d*e^4 + 5855*(x*e + d)^(3/2)*b^5*d^2*e^4 - 2295*sqrt(x*e + d)*b^5*d^3*e^4 + 5153*(x*e + d)^(5/2)*a*b^4*e^5 -
11710*(x*e + d)^(3/2)*a*b^4*d*e^5 + 6885*sqrt(x*e + d)*a*b^4*d^2*e^5 + 5855*(x*e + d)^(3/2)*a^2*b^3*e^6 - 688
5*sqrt(x*e + d)*a^2*b^3*d*e^6 + 2295*sqrt(x*e + d)*a^3*b^2*e^7)/((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)^4)

________________________________________________________________________________________

maple [B]  time = 0.08, size = 473, normalized size = 2.03 \begin {gather*} \frac {765 \sqrt {e x +d}\, a^{3} b^{2} e^{7}}{64 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}-\frac {2295 \sqrt {e x +d}\, a^{2} b^{3} d \,e^{6}}{64 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}+\frac {2295 \sqrt {e x +d}\, a \,b^{4} d^{2} e^{5}}{64 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}-\frac {765 \sqrt {e x +d}\, b^{5} d^{3} e^{4}}{64 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}+\frac {5855 \left (e x +d \right )^{\frac {3}{2}} a^{2} b^{3} e^{6}}{192 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}-\frac {5855 \left (e x +d \right )^{\frac {3}{2}} a \,b^{4} d \,e^{5}}{96 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}+\frac {5855 \left (e x +d \right )^{\frac {3}{2}} b^{5} d^{2} e^{4}}{192 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}+\frac {5153 \left (e x +d \right )^{\frac {5}{2}} a \,b^{4} e^{5}}{192 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}-\frac {5153 \left (e x +d \right )^{\frac {5}{2}} b^{5} d \,e^{4}}{192 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}+\frac {515 \left (e x +d \right )^{\frac {7}{2}} b^{5} e^{4}}{64 \left (a e -b d \right )^{6} \left (b e x +a e \right )^{4}}+\frac {1155 b^{2} 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 )^{6} \sqrt {\left (a e -b d \right ) b}}+\frac {10 b \,e^{4}}{\left (a e -b d \right )^{6} \sqrt {e x +d}}-\frac {2 e^{4}}{3 \left (a e -b d \right )^{5} \left (e x +d \right )^{\frac {3}{2}}} \end {gather*}

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]

515/64*e^4/(a*e-b*d)^6*b^5/(b*e*x+a*e)^4*(e*x+d)^(7/2)+5153/192*e^5/(a*e-b*d)^6*b^4/(b*e*x+a*e)^4*(e*x+d)^(5/2
)*a-5153/192*e^4/(a*e-b*d)^6*b^5/(b*e*x+a*e)^4*(e*x+d)^(5/2)*d+5855/192*e^6/(a*e-b*d)^6*b^3/(b*e*x+a*e)^4*(e*x
+d)^(3/2)*a^2-5855/96*e^5/(a*e-b*d)^6*b^4/(b*e*x+a*e)^4*(e*x+d)^(3/2)*a*d+5855/192*e^4/(a*e-b*d)^6*b^5/(b*e*x+
a*e)^4*(e*x+d)^(3/2)*d^2+765/64*e^7/(a*e-b*d)^6*b^2/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^3-2295/64*e^6/(a*e-b*d)^6*b^
3/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a^2*d+2295/64*e^5/(a*e-b*d)^6*b^4/(b*e*x+a*e)^4*(e*x+d)^(1/2)*a*d^2-765/64*e^4/(
a*e-b*d)^6*b^5/(b*e*x+a*e)^4*(e*x+d)^(1/2)*d^3+1155/64*e^4/(a*e-b*d)^6*b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^
(1/2)/((a*e-b*d)*b)^(1/2)*b)-2/3*e^4/(a*e-b*d)^5/(e*x+d)^(3/2)+10*e^4/(a*e-b*d)^6*b/(e*x+d)^(1/2)

________________________________________________________________________________________

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((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?

________________________________________________________________________________________

mupad [B]  time = 2.64, size = 436, normalized size = 1.87 \begin {gather*} \frac {\frac {3069\,b^2\,e^4\,{\left (d+e\,x\right )}^2}{64\,{\left (a\,e-b\,d\right )}^3}-\frac {2\,e^4}{3\,\left (a\,e-b\,d\right )}+\frac {5621\,b^3\,e^4\,{\left (d+e\,x\right )}^3}{64\,{\left (a\,e-b\,d\right )}^4}+\frac {4235\,b^4\,e^4\,{\left (d+e\,x\right )}^4}{64\,{\left (a\,e-b\,d\right )}^5}+\frac {1155\,b^5\,e^4\,{\left (d+e\,x\right )}^5}{64\,{\left (a\,e-b\,d\right )}^6}+\frac {22\,b\,e^4\,\left (d+e\,x\right )}{3\,{\left (a\,e-b\,d\right )}^2}}{b^4\,{\left (d+e\,x\right )}^{11/2}-\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,{\left (d+e\,x\right )}^{9/2}+{\left (d+e\,x\right )}^{3/2}\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )+{\left (d+e\,x\right )}^{7/2}\,\left (6\,a^2\,b^2\,e^2-12\,a\,b^3\,d\,e+6\,b^4\,d^2\right )-{\left (d+e\,x\right )}^{5/2}\,\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 )}+\frac {1155\,b^{3/2}\,e^4\,\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 )}{64\,{\left (a\,e-b\,d\right )}^{13/2}} \end {gather*}

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]

((3069*b^2*e^4*(d + e*x)^2)/(64*(a*e - b*d)^3) - (2*e^4)/(3*(a*e - b*d)) + (5621*b^3*e^4*(d + e*x)^3)/(64*(a*e
- b*d)^4) + (4235*b^4*e^4*(d + e*x)^4)/(64*(a*e - b*d)^5) + (1155*b^5*e^4*(d + e*x)^5)/(64*(a*e - b*d)^6) + (
22*b*e^4*(d + e*x))/(3*(a*e - b*d)^2))/(b^4*(d + e*x)^(11/2) - (4*b^4*d - 4*a*b^3*e)*(d + e*x)^(9/2) + (d + e*
x)^(3/2)*(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3) + (d + e*x)^(7/2)*(6*b^4*d^2
+ 6*a^2*b^2*e^2 - 12*a*b^3*d*e) - (d + e*x)^(5/2)*(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
)) + (1155*b^(3/2)*e^4*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)))/(64*(a*e - b*d)^(13/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________