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

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

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Rubi [A]  time = 0.19, 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, 47, 63, 208} \begin {gather*} -\frac {(d+e x)^{3/2} (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} (-3 a B e-A b e+4 b B d)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {e (a+b x) (-3 a B e-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^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}} \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)^(3/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.33, size = 167, normalized size = 0.78 \begin {gather*} \frac {(a+b x) \left (\frac {(a+b x) (-3 a B e-A b e+4 b B d) \left (\sqrt {b} e (a+b x) \sqrt {d+e x} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )-b (d+e x) \sqrt {a e-b d}\right )}{\sqrt {a e-b d}}-2 b^2 (d+e x)^2 (A b-a B)\right )}{4 b^3 \left ((a+b x)^2\right )^{3/2} \sqrt {d+e x} (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)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 42.64, size = 243, normalized size = 1.13 \begin {gather*} \frac {(-a e-b e x) \left (\frac {e \sqrt {d+e x} \left (-3 a^2 B e^2-a A b e^2-5 a b B e (d+e x)+7 a b B d e+A b^2 e (d+e x)+A b^2 d e-4 b^2 B d^2+4 b^2 B d (d+e x)\right )}{4 b^2 (b d-a e) (-a e-b (d+e x)+b d)^2}+\frac {\left (-3 a B e^2-A b e^2+4 b B d e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{5/2} (b d-a e) \sqrt {a e-b d}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \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)^(3/2),x]

[Out]

((-(a*e) - b*e*x)*((e*Sqrt[d + e*x]*(-4*b^2*B*d^2 + A*b^2*d*e + 7*a*b*B*d*e - a*A*b*e^2 - 3*a^2*B*e^2 + 4*b^2*
B*d*(d + e*x) + A*b^2*e*(d + e*x) - 5*a*b*B*e*(d + e*x)))/(4*b^2*(b*d - a*e)*(b*d - a*e - b*(d + e*x))^2) + ((
4*b*B*d*e - A*b*e^2 - 3*a*B*e^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(4*b^(5/2)*(b
*d - a*e)*Sqrt[-(b*d) + a*e])))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [B]  time = 0.46, size = 721, normalized size = 3.35 \begin {gather*} \left [\frac {{\left (4 \, B a^{2} b d e - {\left (3 \, B a^{3} + A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (3 \, B a^{2} b + 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 (5 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} d e + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (9 \, B a b^{3} - A b^{4}\right )} d e + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{5} d^{2} - 2 \, a^{3} b^{4} d e + a^{4} b^{3} e^{2} + {\left (b^{7} d^{2} - 2 \, a b^{6} d e + a^{2} b^{5} e^{2}\right )} x^{2} + 2 \, {\left (a b^{6} d^{2} - 2 \, a^{2} b^{5} d e + a^{3} b^{4} e^{2}\right )} x\right )}}, \frac {{\left (4 \, B a^{2} b d e - {\left (3 \, B a^{3} + A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (3 \, B a^{2} b + 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 (5 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} d e + {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} e^{2} + {\left (4 \, B b^{4} d^{2} - {\left (9 \, B a b^{3} - A b^{4}\right )} d e + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{5} d^{2} - 2 \, a^{3} b^{4} d e + a^{4} b^{3} e^{2} + {\left (b^{7} d^{2} - 2 \, a b^{6} d e + a^{2} b^{5} e^{2}\right )} x^{2} + 2 \, {\left (a b^{6} d^{2} - 2 \, a^{2} b^{5} d e + a^{3} b^{4} e^{2}\right )} x\right )}}\right ] \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.36, size = 337, normalized size = 1.57 \begin {gather*} \frac {{\left (4 \, B b d e^{2} - 3 \, 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 )}{4 \, {\left (b^{3} d e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{2} e^{2} \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} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{3} + {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{3} + 7 \, \sqrt {x e + d} B a b d e^{3} + \sqrt {x e + d} A b^{2} d e^{3} - 3 \, \sqrt {x e + d} B a^{2} e^{4} - \sqrt {x e + d} A a b e^{4}}{4 \, {\left (b^{3} d e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b^{2} e^{2} \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}} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

1/4*(4*B*b*d*e^2 - 3*B*a*e^3 - A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d*e*sgn((x*e + d)*b
*e - b*d*e + a*e^2) - a*b^2*e^2*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 - 5*(x*e + d)^(3/2)*B*a*b*e^3 + (x*e + d)^(3/2)*A*b^2*e^3 + 7
*sqrt(x*e + d)*B*a*b*d*e^3 + sqrt(x*e + d)*A*b^2*d*e^3 - 3*sqrt(x*e + d)*B*a^2*e^4 - sqrt(x*e + d)*A*a*b*e^4)/
((b^3*d*e*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b^2*e^2*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 = 555, normalized size = 2.58 \begin {gather*} \frac {\left (A \,b^{3} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 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 )+2 A a \,b^{2} e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+6 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 )+A \,a^{2} b \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+3 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 )-\sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A a b \,e^{2}+\sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A \,b^{2} d e -3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,a^{2} e^{2}+7 \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}+\sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} A \,b^{2} e -5 \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 ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{2} e} \end {gather*}

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)^(3/2),x)

[Out]

1/4*(A*b^3*e^3*x^2*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+3*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)+2*A*a*b^2*e^3*x*arctan((e*x+d)^(1
/2)/((a*e-b*d)*b)^(1/2)*b)+6*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*arcta
n((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)*A*b^2*e+A*a^2*b*e^3*arctan((e*x+d)^(1
/2)/((a*e-b*d)*b)^(1/2)*b)-5*((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+3*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)^(1/2)*A*a*b*e^2+((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*A*b^2*d*e-3*((a*e-b*d
)*b)^(1/2)*(e*x+d)^(1/2)*B*a^2*e^2+7*((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)/e*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^2/(a*e-b*d)/((b*x+a)^2)^(3/2)

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {d+e\,x}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \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)^(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]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \sqrt {d + e x}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)**(3/2),x)

[Out]

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

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