3.1754 \(\int \frac {A+B x}{(a+b x)^2 \sqrt {d+e x}} \, dx\)

Optimal. Leaf size=103 \[ -\frac {(-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x} (A b-a B)}{b (a+b x) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {78, 63, 208} \[ -\frac {(-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x} (A b-a B)}{b (a+b x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.10, size = 102, normalized size = 0.99 \[ \frac {\sqrt {d+e x} (a B-A b)}{b (a+b x) (b d-a e)}-\frac {(-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} (b d-a e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.69, size = 397, normalized size = 3.85 \[ \left [\frac {{\left (2 \, B a b d - {\left (B a^{2} + A a b\right )} e + {\left (2 \, B b^{2} d - {\left (B a b + A b^{2}\right )} e\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d - {\left (B a^{2} b - A a b^{2}\right )} e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{4} d^{2} - 2 \, a^{2} b^{3} d e + a^{3} b^{2} e^{2} + {\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} x\right )}}, \frac {{\left (2 \, B a b d - {\left (B a^{2} + A a b\right )} e + {\left (2 \, B b^{2} d - {\left (B a b + A b^{2}\right )} e\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left ({\left (B a b^{2} - A b^{3}\right )} d - {\left (B a^{2} b - A a b^{2}\right )} e\right )} \sqrt {e x + d}}{a b^{4} d^{2} - 2 \, a^{2} b^{3} d e + a^{3} b^{2} e^{2} + {\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.26, size = 135, normalized size = 1.31 \[ \frac {{\left (2 \, B b d - B a e - A b e\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d - a b e\right )} \sqrt {-b^{2} d + a b e}} + \frac {\sqrt {x e + d} B a e - \sqrt {x e + d} A b e}{{\left (b^{2} d - a b e\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 195, normalized size = 1.89 \[ \frac {A e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}+\frac {B a e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b}-\frac {2 B d \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}+\frac {\left (A b -B a \right ) \sqrt {e x +d}\, e}{\left (a e -b d \right ) \left (a e -b d +\left (e x +d \right ) b \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*(A*b-B*a)/(a*e-b*d)/b*(e*x+d)^(1/2)/(b*(e*x+d)+a*e-b*d)+1/(a*e-b*d)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)
/((a*e-b*d)*b)^(1/2)*b)*A*e+1/(a*e-b*d)/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a*
e-2/(a*e-b*d)/((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)/(b*x+a)^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?

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mupad [B]  time = 1.29, size = 99, normalized size = 0.96 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{b^{3/2}\,{\left (a\,e-b\,d\right )}^{3/2}}+\frac {\left (A\,b\,e-B\,a\,e\right )\,\sqrt {d+e\,x}}{b\,\left (a\,e-b\,d\right )\,\left (a\,e-b\,d+b\,\left (d+e\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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