3.1870 \(\int \frac {A+B x}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=215 \[ -\frac {\sqrt {d+e x} (A b-a B)}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {\sqrt {d+e x} (-a B e-3 A b e+4 b B d)}{4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}+\frac {e (a+b x) (-a B e-3 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {770, 78, 51, 63, 208} \[ -\frac {\sqrt {d+e x} (A b-a B)}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {\sqrt {d+e x} (-a B e-3 A b e+4 b B d)}{4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}+\frac {e (a+b x) (-a B e-3 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*b*B*d - 3*A*b*e - a*B*e)*Sqrt[d + e*x])/(4*b*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((A*b - a*B)*
Sqrt[d + e*x])/(2*b*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e*(4*b*B*d - 3*A*b*e - a*B*e)*(a +
b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*b^(3/2)*(b*d - a*e)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2])

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]

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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

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Mathematica [A]  time = 0.26, size = 170, normalized size = 0.79 \[ \frac {(a+b x) \sqrt {d+e x} \left (\frac {(a+b x) (-a B e-3 A b e+4 b B d) \left (\sqrt {b} \sqrt {d+e x} (a e-b d)+e (a+b x) \sqrt {a e-b d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )\right )}{2 \sqrt {b} \sqrt {d+e x} (b d-a e)^2}+a B-A b\right )}{2 b \left ((a+b x)^2\right )^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*Sqrt[d + e*x]*(-(A*b) + a*B + ((4*b*B*d - 3*A*b*e - a*B*e)*(a + b*x)*(Sqrt[b]*(-(b*d) + a*e)*Sqrt[d
+ e*x] + e*Sqrt[-(b*d) + a*e]*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(2*Sqrt[b]*(b*d
- a*e)^2*Sqrt[d + e*x])))/(2*b*(b*d - a*e)*((a + b*x)^2)^(3/2))

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fricas [B]  time = 0.96, size = 808, normalized size = 3.76 \[ \left [-\frac {{\left (4 \, B a^{2} b d e - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (B a b^{2} + 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (B a^{2} b + 3 \, A a b^{2}\right )} e^{2}\right )} x\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 (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (B a^{2} b^{2} + 7 \, A a b^{3}\right )} d e - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (5 \, B a b^{3} + 3 \, A b^{4}\right )} d e + {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{5} d^{3} - 3 \, a^{3} b^{4} d^{2} e + 3 \, a^{4} b^{3} d e^{2} - a^{5} b^{2} e^{3} + {\left (b^{7} d^{3} - 3 \, a b^{6} d^{2} e + 3 \, a^{2} b^{5} d e^{2} - a^{3} b^{4} e^{3}\right )} x^{2} + 2 \, {\left (a b^{6} d^{3} - 3 \, a^{2} b^{5} d^{2} e + 3 \, a^{3} b^{4} d e^{2} - a^{4} b^{3} e^{3}\right )} x\right )}}, -\frac {{\left (4 \, B a^{2} b d e - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (B a b^{2} + 3 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (B a^{2} b + 3 \, A a b^{2}\right )} e^{2}\right )} x\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 (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (B a^{2} b^{2} + 7 \, A a b^{3}\right )} d e - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (5 \, B a b^{3} + 3 \, A b^{4}\right )} d e + {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{5} d^{3} - 3 \, a^{3} b^{4} d^{2} e + 3 \, a^{4} b^{3} d e^{2} - a^{5} b^{2} e^{3} + {\left (b^{7} d^{3} - 3 \, a b^{6} d^{2} e + 3 \, a^{2} b^{5} d e^{2} - a^{3} b^{4} e^{3}\right )} x^{2} + 2 \, {\left (a b^{6} d^{3} - 3 \, a^{2} b^{5} d^{2} e + 3 \, a^{3} b^{4} d e^{2} - a^{4} b^{3} e^{3}\right )} x\right )}}\right ] \]

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)^(3/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 402, normalized size = 1.87 \[ -\frac {{\left (4 \, B b d e^{2} - B a e^{3} - 3 \, A b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{3} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b^{2} d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} b e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e^{2} - 4 \, \sqrt {x e + d} B b^{2} d^{2} e^{2} - {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{3} - 3 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{3} + 3 \, \sqrt {x e + d} B a b d e^{3} + 5 \, \sqrt {x e + d} A b^{2} d e^{3} + \sqrt {x e + d} B a^{2} e^{4} - 5 \, \sqrt {x e + d} A a b e^{4}}{4 \, {\left (b^{3} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b^{2} d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} b e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \]

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)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.07, size = 556, normalized size = 2.59 \[ \frac {\left (b x +a \right ) \left (3 A \,b^{3} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+B a \,b^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-4 B \,b^{3} d \,e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+6 A a \,b^{2} e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+2 B \,a^{2} b \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-8 B a \,b^{2} d \,e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 A \,a^{2} b \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+B \,a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-4 B \,a^{2} b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+5 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A a b \,e^{2}-5 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A \,b^{2} d e -\sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,a^{2} e^{2}-3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B a b d e +4 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,b^{2} d^{2}+3 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} A \,b^{2} e +\sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} B a b e -4 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} B \,b^{2} d \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b e} \]

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)^(3/2)/(e*x+d)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \]

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)^(3/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*sqrt(e*x + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{\sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2)),x)

[Out]

int((A + B*x)/((d + e*x)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)

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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)/(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(1/2),x)

[Out]

Timed out

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