3.1820 \(\int \frac {(A+B x) \sqrt {d+e x}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=209 \[ \frac {e^2 (-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{5/2}}-\frac {e \sqrt {d+e x} (-a B e-A b e+2 b B d)}{8 b^2 (a+b x) (b d-a e)^2}-\frac {\sqrt {d+e x} (-a B e-A b e+2 b B d)}{4 b^2 (a+b x)^2 (b d-a e)}-\frac {(d+e x)^{3/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {27, 78, 47, 51, 63, 208} \[ \frac {e^2 (-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{5/2}}-\frac {e \sqrt {d+e x} (-a B e-A b e+2 b B d)}{8 b^2 (a+b x) (b d-a e)^2}-\frac {\sqrt {d+e x} (-a B e-A b e+2 b B d)}{4 b^2 (a+b x)^2 (b d-a e)}-\frac {(d+e x)^{3/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

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]

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

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Mathematica [C]  time = 0.06, size = 98, normalized size = 0.47 \[ \frac {(d+e x)^{3/2} \left (\frac {3 e^2 (a B e+A b e-2 b B d) \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^3}+\frac {3 a B-3 A b}{(a+b x)^3}\right )}{9 b (b d-a e)} \]

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]

((d + e*x)^(3/2)*((-3*A*b + 3*a*B)/(a + b*x)^3 + (3*e^2*(-2*b*B*d + A*b*e + a*B*e)*Hypergeometric2F1[3/2, 3, 5
/2, (b*(d + e*x))/(b*d - a*e)])/(b*d - a*e)^3))/(9*b*(b*d - a*e))

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fricas [B]  time = 0.74, size = 1218, normalized size = 5.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.25, size = 386, normalized size = 1.85 \[ -\frac {{\left (2 \, B b d e^{2} - B a e^{3} - A b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} - \frac {6 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{3} d e^{2} - 6 \, \sqrt {x e + d} B b^{3} d^{3} e^{2} - 3 \, {\left (x e + d\right )}^{\frac {5}{2}} B a b^{2} e^{3} - 3 \, {\left (x e + d\right )}^{\frac {5}{2}} A b^{3} e^{3} - 8 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} d e^{3} + 8 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} d e^{3} + 15 \, \sqrt {x e + d} B a b^{2} d^{2} e^{3} + 3 \, \sqrt {x e + d} A b^{3} d^{2} e^{3} + 8 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b e^{4} - 8 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{2} e^{4} - 12 \, \sqrt {x e + d} B a^{2} b d e^{4} - 6 \, \sqrt {x e + d} A a b^{2} d e^{4} + 3 \, \sqrt {x e + d} B a^{3} e^{5} + 3 \, \sqrt {x e + d} A a^{2} b e^{5}}{24 \, {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*B*b*d*e^2 - B*a*e^3 - A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^2 - 2*a*b^3*d*e +
a^2*b^2*e^2)*sqrt(-b^2*d + a*b*e)) - 1/24*(6*(x*e + d)^(5/2)*B*b^3*d*e^2 - 6*sqrt(x*e + d)*B*b^3*d^3*e^2 - 3*(
x*e + d)^(5/2)*B*a*b^2*e^3 - 3*(x*e + d)^(5/2)*A*b^3*e^3 - 8*(x*e + d)^(3/2)*B*a*b^2*d*e^3 + 8*(x*e + d)^(3/2)
*A*b^3*d*e^3 + 15*sqrt(x*e + d)*B*a*b^2*d^2*e^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 - 8*(x*e + d)^(3/2)*A*a*b^2*e^4 - 12*sqrt(x*e + d)*B*a^2*b*d*e^4 - 6*sqrt(x*e + d)*A*a*b^2*d*e^4 + 3*sqrt(x*
e + d)*B*a^3*e^5 + 3*sqrt(x*e + d)*A*a^2*b*e^5)/((b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*((x*e + d)*b - b*d + a*
e)^3)

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maple [B]  time = 0.08, size = 494, normalized size = 2.36 \[ \frac {\left (e x +d \right )^{\frac {5}{2}} A b \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\left (e x +d \right )^{\frac {5}{2}} B a \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {\left (e x +d \right )^{\frac {5}{2}} B b d \,e^{2}}{4 \left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {A \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} A \,e^{3}}{3 \left (b e x +a e \right )^{3} \left (a e -b d \right )}-\frac {\left (e x +d \right )^{\frac {3}{2}} B a \,e^{3}}{3 \left (b e x +a e \right )^{3} \left (a e -b d \right ) b}+\frac {B a \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {B d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b}-\frac {\sqrt {e x +d}\, A \,e^{3}}{8 \left (b e x +a e \right )^{3} b}-\frac {\sqrt {e x +d}\, B a \,e^{3}}{8 \left (b e x +a e \right )^{3} b^{2}}+\frac {\sqrt {e x +d}\, B d \,e^{2}}{4 \left (b e x +a e \right )^{3} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(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 = 2.13, size = 310, normalized size = 1.48 \[ \frac {\frac {{\left (d+e\,x\right )}^{5/2}\,\left (A\,b\,e^3+B\,a\,e^3-2\,B\,b\,d\,e^2\right )}{8\,{\left (a\,e-b\,d\right )}^2}-\frac {\sqrt {d+e\,x}\,\left (A\,b\,e^3+B\,a\,e^3-2\,B\,b\,d\,e^2\right )}{8\,b^2}+\frac {\left (A\,b\,e^3-B\,a\,e^3\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b\,\left (a\,e-b\,d\right )}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2}+\frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {d+e\,x}\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^3+B\,a\,e^3-2\,B\,b\,d\,e^2\right )}\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{8\,b^{5/2}\,{\left (a\,e-b\,d\right )}^{5/2}} \]

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

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

[Out]

Timed out

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