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

Optimal. Leaf size=140 \[ -\frac {(-3 a B e+A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2} \sqrt {b d-a e}}+\frac {\sqrt {d+e x} (-3 a B e+A b e+2 b B d)}{b^2 (b d-a e)}-\frac {(d+e x)^{3/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

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Rubi [A]  time = 0.12, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 78, 50, 63, 208} \begin {gather*} \frac {\sqrt {d+e x} (-3 a B e+A b e+2 b B d)}{b^2 (b d-a e)}-\frac {(-3 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 )}{b^{5/2} \sqrt {b d-a e}}-\frac {(d+e x)^{3/2} (A b-a B)}{b (a+b x) (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),x]

[Out]

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 119, normalized size = 0.85 \begin {gather*} \frac {\frac {(-3 a B e+A b e+2 b B d) \left (\sqrt {b} \sqrt {d+e x}-\sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{b^{3/2}}+\frac {(d+e x)^{3/2} (a B-A b)}{a+b x}}{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),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 128, normalized size = 0.91 \begin {gather*} \frac {(3 a B e-A b e-2 b B d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{5/2} \sqrt {a e-b d}}+\frac {\sqrt {d+e x} (3 a B e-A b e+2 b B (d+e x)-2 b B d)}{b^2 (a e+b (d+e x)-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),x]

[Out]

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

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fricas [A]  time = 0.45, size = 393, normalized size = 2.81 \begin {gather*} \left [\frac {{\left (2 \, B a b d - {\left (3 \, B a^{2} - A a b\right )} e + {\left (2 \, B b^{2} d - {\left (3 \, B a b - A b^{2}\right )} e\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 ({\left (3 \, B a b^{2} - A b^{3}\right )} d - {\left (3 \, B a^{2} b - A a b^{2}\right )} e + 2 \, {\left (B b^{3} d - B a b^{2} e\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (a b^{4} d - a^{2} b^{3} e + {\left (b^{5} d - a b^{4} e\right )} x\right )}}, \frac {{\left (2 \, B a b d - {\left (3 \, B a^{2} - A a b\right )} e + {\left (2 \, B b^{2} d - {\left (3 \, B a b - A b^{2}\right )} e\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 ({\left (3 \, B a b^{2} - A b^{3}\right )} d - {\left (3 \, B a^{2} b - A a b^{2}\right )} e + 2 \, {\left (B b^{3} d - B a b^{2} e\right )} x\right )} \sqrt {e x + d}}{a b^{4} d - a^{2} b^{3} e + {\left (b^{5} d - a b^{4} e\right )} x}\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),x, algorithm="fricas")

[Out]

[1/2*((2*B*a*b*d - (3*B*a^2 - A*a*b)*e + (2*B*b^2*d - (3*B*a*b - A*b^2)*e)*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*((3*B*a*b^2 - A*b^3)*d - (3*B*a^2*b - A*a*b
^2)*e + 2*(B*b^3*d - B*a*b^2*e)*x)*sqrt(e*x + d))/(a*b^4*d - a^2*b^3*e + (b^5*d - a*b^4*e)*x), ((2*B*a*b*d - (
3*B*a^2 - A*a*b)*e + (2*B*b^2*d - (3*B*a*b - A*b^2)*e)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqr
t(e*x + d)/(b*e*x + b*d)) + ((3*B*a*b^2 - A*b^3)*d - (3*B*a^2*b - A*a*b^2)*e + 2*(B*b^3*d - B*a*b^2*e)*x)*sqrt
(e*x + d))/(a*b^4*d - a^2*b^3*e + (b^5*d - a*b^4*e)*x)]

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giac [A]  time = 0.22, size = 126, normalized size = 0.90 \begin {gather*} \frac {2 \, \sqrt {x e + d} B}{b^{2}} + \frac {{\left (2 \, B b d - 3 \, B a e + A b e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {\sqrt {x e + d} B a e - \sqrt {x e + d} A b e}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 186, normalized size = 1.33 \begin {gather*} \frac {A e \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}\, b}-\frac {3 B a e \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}\, b^{2}}+\frac {2 B d \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}\, b}-\frac {\sqrt {e x +d}\, A e}{\left (b e x +a e \right ) b}+\frac {\sqrt {e x +d}\, B a e}{\left (b e x +a e \right ) b^{2}}+\frac {2 \sqrt {e x +d}\, B}{b^{2}} \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),x)

[Out]

2*B/b^2*(e*x+d)^(1/2)-1/b*(e*x+d)^(1/2)/(b*e*x+a*e)*A*e+1/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*a*B*e+1/b/((a*e-b*d)*b
)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*e-3/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b
*d)*b)^(1/2)*b)*a*B*e+2/b/((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 \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)^(1/2)/(b^2*x^2+2*a*b*x+a^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 = 2.01, size = 108, normalized size = 0.77 \begin {gather*} \frac {2\,B\,\sqrt {d+e\,x}}{b^2}-\frac {\left (A\,b\,e-B\,a\,e\right )\,\sqrt {d+e\,x}}{b^3\,\left (d+e\,x\right )-b^3\,d+a\,b^2\,e}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )\,\left (A\,b\,e-3\,B\,a\,e+2\,B\,b\,d\right )}{b^{5/2}\,\sqrt {a\,e-b\,d}} \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),x)

[Out]

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

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sympy [B]  time = 107.95, size = 1251, normalized size = 8.94

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),x)

[Out]

-2*A*a*e**2*sqrt(d + e*x)/(2*a**2*b*e**2 - 2*a*b**2*d*e + 2*a*b**2*e**2*x - 2*b**3*d*e*x) + A*a*e**2*sqrt(-1/(
b*(a*e - b*d)**3))*log(-a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) - b**2*d
**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/(2*b) - A*a*e**2*sqrt(-1/(b*(a*e - b*d)**3))*log(a**2*e**2*sq
rt(-1/(b*(a*e - b*d)**3)) - 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sq
rt(d + e*x))/(2*b) - A*d*e*sqrt(-1/(b*(a*e - b*d)**3))*log(-a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*
sqrt(-1/(b*(a*e - b*d)**3)) - b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/2 + A*d*e*sqrt(-1/(b*(a*e
- b*d)**3))*log(a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) - 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqr
t(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/2 + 2*A*d*e*sqrt(d + e*x)/(2*a**2*e**2 - 2*a*b*d*e + 2*a*b*e**2*x -
2*b**2*d*e*x) + 2*A*e*atan(sqrt(d + e*x)/sqrt(a*e/b - d))/(b**2*sqrt(a*e/b - d)) + 2*B*a**2*e**2*sqrt(d + e*x)
/(2*a**2*b**2*e**2 - 2*a*b**3*d*e + 2*a*b**3*e**2*x - 2*b**4*d*e*x) - B*a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3))*
log(-a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) - b**2*d**2*sqrt(-1/(b*(a*e
- b*d)**3)) + sqrt(d + e*x))/(2*b**2) + B*a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3))*log(a**2*e**2*sqrt(-1/(b*(a*e
- b*d)**3)) - 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/
(2*b**2) - 2*B*a*d*e*sqrt(d + e*x)/(2*a**2*b*e**2 - 2*a*b**2*d*e + 2*a*b**2*e**2*x - 2*b**3*d*e*x) + B*a*d*e*s
qrt(-1/(b*(a*e - b*d)**3))*log(-a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3))
- b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/(2*b) - B*a*d*e*sqrt(-1/(b*(a*e - b*d)**3))*log(a**2*
e**2*sqrt(-1/(b*(a*e - b*d)**3)) - 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3
)) + sqrt(d + e*x))/(2*b) - 4*B*a*e*atan(sqrt(d + e*x)/sqrt(a*e/b - d))/(b**3*sqrt(a*e/b - d)) + 2*B*d*atan(sq
rt(d + e*x)/sqrt(a*e/b - d))/(b**2*sqrt(a*e/b - d)) + 2*B*sqrt(d + e*x)/b**2

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