3.17.1 \(\int \frac {A+B x}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=209 \[ -\frac {e^2 (-a B e-5 A b e+6 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}}+\frac {e \sqrt {d+e x} (-a B e-5 A b e+6 b B d)}{8 b (a+b x) (b d-a e)^3}-\frac {\sqrt {d+e x} (-a B e-5 A b e+6 b B d)}{12 b (a+b x)^2 (b d-a e)^2}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 78, 51, 63, 208} \begin {gather*} -\frac {e^2 (-a B e-5 A b e+6 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}}+\frac {e \sqrt {d+e x} (-a B e-5 A b e+6 b B d)}{8 b (a+b x) (b d-a e)^3}-\frac {\sqrt {d+e x} (-a B e-5 A b e+6 b B d)}{12 b (a+b x)^2 (b d-a e)^2}-\frac {\sqrt {d+e x} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b - a*B)*Sqrt[d + e*x])/(3*b*(b*d - a*e)*(a + b*x)^3) - ((6*b*B*d - 5*A*b*e - a*B*e)*Sqrt[d + e*x])/(12*b
*(b*d - a*e)^2*(a + b*x)^2) + (e*(6*b*B*d - 5*A*b*e - a*B*e)*Sqrt[d + e*x])/(8*b*(b*d - a*e)^3*(a + b*x)) - (e
^2*(6*b*B*d - 5*A*b*e - a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(3/2)*(b*d - a*e)^(7/2))

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {A+B x}{(a+b x)^4 \sqrt {d+e x}} \, dx\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{3 b (b d-a e) (a+b x)^3}+\frac {(6 b B d-5 A b e-a B e) \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{6 b (b d-a e)}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac {(6 b B d-5 A b e-a B e) \sqrt {d+e x}}{12 b (b d-a e)^2 (a+b x)^2}-\frac {(e (6 b B d-5 A b e-a B e)) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{8 b (b d-a e)^2}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac {(6 b B d-5 A b e-a B e) \sqrt {d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac {e (6 b B d-5 A b e-a B e) \sqrt {d+e x}}{8 b (b d-a e)^3 (a+b x)}+\frac {\left (e^2 (6 b B d-5 A b e-a B e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b (b d-a e)^3}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac {(6 b B d-5 A b e-a B e) \sqrt {d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac {e (6 b B d-5 A b e-a B e) \sqrt {d+e x}}{8 b (b d-a e)^3 (a+b x)}+\frac {(e (6 b B d-5 A b e-a B e)) \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 (b d-a e)^3}\\ &=-\frac {(A b-a B) \sqrt {d+e x}}{3 b (b d-a e) (a+b x)^3}-\frac {(6 b B d-5 A b e-a B e) \sqrt {d+e x}}{12 b (b d-a e)^2 (a+b x)^2}+\frac {e (6 b B d-5 A b e-a B e) \sqrt {d+e x}}{8 b (b d-a e)^3 (a+b x)}-\frac {e^2 (6 b B d-5 A b e-a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 97, normalized size = 0.46 \begin {gather*} \frac {\sqrt {d+e x} \left (\frac {e^2 (a B e+5 A b e-6 b B d) \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^3}+\frac {a B-A b}{(a+b x)^3}\right )}{3 b (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*((-(A*b) + a*B)/(a + b*x)^3 + (e^2*(-6*b*B*d + 5*A*b*e + a*B*e)*Hypergeometric2F1[1/2, 3, 3/2,
(b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^3))/(3*b*(b*d - a*e))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.89, size = 325, normalized size = 1.56 \begin {gather*} \frac {e^2 \sqrt {d+e x} \left (-3 a^3 B e^3+33 a^2 A b e^3+8 a^2 b B e^2 (d+e x)-24 a^2 b B d e^2+40 a A b^2 e^2 (d+e x)-66 a A b^2 d e^2+57 a b^2 B d^2 e+3 a b^2 B e (d+e x)^2-56 a b^2 B d e (d+e x)+33 A b^3 d^2 e+15 A b^3 e (d+e x)^2-40 A b^3 d e (d+e x)-30 b^3 B d^3+48 b^3 B d^2 (d+e x)-18 b^3 B d (d+e x)^2\right )}{24 b (b d-a e)^3 (-a e-b (d+e x)+b d)^3}+\frac {\left (a B e^3+5 A b e^3-6 b B d e^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{3/2} (b d-a e)^3 \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

(e^2*Sqrt[d + e*x]*(-30*b^3*B*d^3 + 33*A*b^3*d^2*e + 57*a*b^2*B*d^2*e - 66*a*A*b^2*d*e^2 - 24*a^2*b*B*d*e^2 +
33*a^2*A*b*e^3 - 3*a^3*B*e^3 + 48*b^3*B*d^2*(d + e*x) - 40*A*b^3*d*e*(d + e*x) - 56*a*b^2*B*d*e*(d + e*x) + 40
*a*A*b^2*e^2*(d + e*x) + 8*a^2*b*B*e^2*(d + e*x) - 18*b^3*B*d*(d + e*x)^2 + 15*A*b^3*e*(d + e*x)^2 + 3*a*b^2*B
*e*(d + e*x)^2))/(24*b*(b*d - a*e)^3*(b*d - a*e - b*(d + e*x))^3) + ((-6*b*B*d*e^2 + 5*A*b*e^3 + a*B*e^3)*ArcT
an[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(8*b^(3/2)*(b*d - a*e)^3*Sqrt[-(b*d) + a*e])

________________________________________________________________________________________

fricas [B]  time = 0.48, size = 1337, normalized size = 6.40

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*(6*B*a^3*b*d*e^2 - (B*a^4 + 5*A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (B*a*b^3 + 5*A*b^4)*e^3)*x^3 + 3*(6*B*a
*b^3*d*e^2 - (B*a^2*b^2 + 5*A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (B*a^3*b + 5*A*a^2*b^2)*e^3)*x)*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*(4*(B*a*b^4 + 2*A*
b^5)*d^3 - 2*(10*B*a^2*b^3 + 17*A*a*b^4)*d^2*e + (13*B*a^3*b^2 + 59*A*a^2*b^3)*d*e^2 + 3*(B*a^4*b - 11*A*a^3*b
^2)*e^3 - 3*(6*B*b^5*d^2*e - (7*B*a*b^4 + 5*A*b^5)*d*e^2 + (B*a^2*b^3 + 5*A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 -
(31*B*a*b^4 + 5*A*b^5)*d^2*e + (29*B*a^2*b^3 + 25*A*a*b^4)*d*e^2 - 4*(B*a^3*b^2 + 5*A*a^2*b^3)*e^3)*x)*sqrt(e
*x + d))/(a^3*b^6*d^4 - 4*a^4*b^5*d^3*e + 6*a^5*b^4*d^2*e^2 - 4*a^6*b^3*d*e^3 + a^7*b^2*e^4 + (b^9*d^4 - 4*a*b
^8*d^3*e + 6*a^2*b^7*d^2*e^2 - 4*a^3*b^6*d*e^3 + a^4*b^5*e^4)*x^3 + 3*(a*b^8*d^4 - 4*a^2*b^7*d^3*e + 6*a^3*b^6
*d^2*e^2 - 4*a^4*b^5*d*e^3 + a^5*b^4*e^4)*x^2 + 3*(a^2*b^7*d^4 - 4*a^3*b^6*d^3*e + 6*a^4*b^5*d^2*e^2 - 4*a^5*b
^4*d*e^3 + a^6*b^3*e^4)*x), 1/24*(3*(6*B*a^3*b*d*e^2 - (B*a^4 + 5*A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (B*a*b^3 + 5
*A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (B*a^2*b^2 + 5*A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (B*a^3*b + 5
*A*a^2*b^2)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (4*(B*a*b^
4 + 2*A*b^5)*d^3 - 2*(10*B*a^2*b^3 + 17*A*a*b^4)*d^2*e + (13*B*a^3*b^2 + 59*A*a^2*b^3)*d*e^2 + 3*(B*a^4*b - 11
*A*a^3*b^2)*e^3 - 3*(6*B*b^5*d^2*e - (7*B*a*b^4 + 5*A*b^5)*d*e^2 + (B*a^2*b^3 + 5*A*a*b^4)*e^3)*x^2 + 2*(6*B*b
^5*d^3 - (31*B*a*b^4 + 5*A*b^5)*d^2*e + (29*B*a^2*b^3 + 25*A*a*b^4)*d*e^2 - 4*(B*a^3*b^2 + 5*A*a^2*b^3)*e^3)*x
)*sqrt(e*x + d))/(a^3*b^6*d^4 - 4*a^4*b^5*d^3*e + 6*a^5*b^4*d^2*e^2 - 4*a^6*b^3*d*e^3 + a^7*b^2*e^4 + (b^9*d^4
- 4*a*b^8*d^3*e + 6*a^2*b^7*d^2*e^2 - 4*a^3*b^6*d*e^3 + a^4*b^5*e^4)*x^3 + 3*(a*b^8*d^4 - 4*a^2*b^7*d^3*e + 6
*a^3*b^6*d^2*e^2 - 4*a^4*b^5*d*e^3 + a^5*b^4*e^4)*x^2 + 3*(a^2*b^7*d^4 - 4*a^3*b^6*d^3*e + 6*a^4*b^5*d^2*e^2 -
4*a^5*b^4*d*e^3 + a^6*b^3*e^4)*x)]

________________________________________________________________________________________

giac [B]  time = 0.21, size = 429, normalized size = 2.05 \begin {gather*} \frac {{\left (6 \, B b d e^{2} - B a e^{3} - 5 \, A b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {18 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{3} d e^{2} - 48 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e^{2} + 30 \, \sqrt {x e + d} B b^{3} d^{3} e^{2} - 3 \, {\left (x e + d\right )}^{\frac {5}{2}} B a b^{2} e^{3} - 15 \, {\left (x e + d\right )}^{\frac {5}{2}} A b^{3} e^{3} + 56 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} d e^{3} + 40 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} d e^{3} - 57 \, \sqrt {x e + d} B a b^{2} d^{2} e^{3} - 33 \, \sqrt {x e + d} A b^{3} d^{2} e^{3} - 8 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b e^{4} - 40 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{2} e^{4} + 24 \, \sqrt {x e + d} B a^{2} b d e^{4} + 66 \, \sqrt {x e + d} A a b^{2} d e^{4} + 3 \, \sqrt {x e + d} B a^{3} e^{5} - 33 \, \sqrt {x e + d} A a^{2} b e^{5}}{24 \, {\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\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((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

1/8*(6*B*b*d*e^2 - B*a*e^3 - 5*A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^3 - 3*a*b^3*d^2*e
+ 3*a^2*b^2*d*e^2 - a^3*b*e^3)*sqrt(-b^2*d + a*b*e)) + 1/24*(18*(x*e + d)^(5/2)*B*b^3*d*e^2 - 48*(x*e + d)^(3
/2)*B*b^3*d^2*e^2 + 30*sqrt(x*e + d)*B*b^3*d^3*e^2 - 3*(x*e + d)^(5/2)*B*a*b^2*e^3 - 15*(x*e + d)^(5/2)*A*b^3*
e^3 + 56*(x*e + d)^(3/2)*B*a*b^2*d*e^3 + 40*(x*e + d)^(3/2)*A*b^3*d*e^3 - 57*sqrt(x*e + d)*B*a*b^2*d^2*e^3 - 3
3*sqrt(x*e + d)*A*b^3*d^2*e^3 - 8*(x*e + d)^(3/2)*B*a^2*b*e^4 - 40*(x*e + d)^(3/2)*A*a*b^2*e^4 + 24*sqrt(x*e +
d)*B*a^2*b*d*e^4 + 66*sqrt(x*e + d)*A*a*b^2*d*e^4 + 3*sqrt(x*e + d)*B*a^3*e^5 - 33*sqrt(x*e + d)*A*a^2*b*e^5)
/((b^4*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*((x*e + d)*b - b*d + a*e)^3)

________________________________________________________________________________________

maple [B]  time = 0.08, size = 679, normalized size = 3.25 \begin {gather*} \frac {5 \left (e x +d \right )^{\frac {5}{2}} A \,b^{2} e^{3}}{8 \left (b e x +a e \right )^{3} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {\left (e x +d \right )^{\frac {5}{2}} B a b \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}-\frac {3 \left (e x +d \right )^{\frac {5}{2}} B \,b^{2} d \,e^{2}}{4 \left (b e x +a e \right )^{3} \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {5 \left (e x +d \right )^{\frac {3}{2}} A b \,e^{3}}{3 \left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {5 A \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (a e -b d \right ) b}}+\frac {B a \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} B a \,e^{3}}{3 \left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} B b d \,e^{2}}{\left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {3 B d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (a e -b d \right ) b}}+\frac {11 \sqrt {e x +d}\, A \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a e -b d \right )}-\frac {\sqrt {e x +d}\, B a \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a e -b d \right ) b}-\frac {5 \sqrt {e x +d}\, B d \,e^{2}}{4 \left (b e x +a e \right )^{3} \left (a e -b d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*e^3/(b*e*x+a*e)^3*b^2/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(5/2)*A+1/8*e^3/(b*e*x+a*e)^3*
b/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(5/2)*a*B-3/4*e^2/(b*e*x+a*e)^3*b^2/(a^3*e^3-3*a^2*b*d
*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(5/2)*B*d+5/3*e^3/(b*e*x+a*e)^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)*
A*b+1/3*e^3/(b*e*x+a*e)^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(3/2)*a*B-2*e^2/(b*e*x+a*e)^3/(a^2*e^2-2*a*b*d*e
+b^2*d^2)*(e*x+d)^(3/2)*B*b*d+11/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(1/2)*A-1/8*e^3/(b*e*x+a*e)^3/b/(a*e-b*
d)*(e*x+d)^(1/2)*a*B-5/4*e^2/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(1/2)*B*d+5/8*e^3/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*
d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A+1/8*e^3/b/(a^3*e^3-3*a^2*b*d*
e^2+3*a*b^2*d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*a*B-3/4*e^2/(a^3*e^
3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d

________________________________________________________________________________________

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

________________________________________________________________________________________

mupad [B]  time = 2.13, size = 331, normalized size = 1.58 \begin {gather*} \frac {\frac {{\left (d+e\,x\right )}^{3/2}\,\left (5\,A\,b\,e^3+B\,a\,e^3-6\,B\,b\,d\,e^2\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {b\,{\left (d+e\,x\right )}^{5/2}\,\left (5\,A\,b\,e^3+B\,a\,e^3-6\,B\,b\,d\,e^2\right )}{8\,{\left (a\,e-b\,d\right )}^3}-\frac {\sqrt {d+e\,x}\,\left (B\,a\,e^3-11\,A\,b\,e^3+10\,B\,b\,d\,e^2\right )}{8\,b\,\left (a\,e-b\,d\right )}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2}+\frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {d+e\,x}\,\left (5\,A\,b\,e+B\,a\,e-6\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (5\,A\,b\,e^3+B\,a\,e^3-6\,B\,b\,d\,e^2\right )}\right )\,\left (5\,A\,b\,e+B\,a\,e-6\,B\,b\,d\right )}{8\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((d + e*x)^(3/2)*(5*A*b*e^3 + B*a*e^3 - 6*B*b*d*e^2))/(3*(a*e - b*d)^2) + (b*(d + e*x)^(5/2)*(5*A*b*e^3 + B*a
*e^3 - 6*B*b*d*e^2))/(8*(a*e - b*d)^3) - ((d + e*x)^(1/2)*(B*a*e^3 - 11*A*b*e^3 + 10*B*b*d*e^2))/(8*b*(a*e - b
*d)))/((d + e*x)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e) + b^3*(d + e*x)^3 - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^2
+ a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2) + (e^2*atan((b^(1/2)*e^2*(d + e*x)^(1/2)*(5*A*b*e + B*a*
e - 6*B*b*d))/((a*e - b*d)^(1/2)*(5*A*b*e^3 + B*a*e^3 - 6*B*b*d*e^2)))*(5*A*b*e + B*a*e - 6*B*b*d))/(8*b^(3/2)
*(a*e - b*d)^(7/2))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________