3.1863 \(\int \frac {A+B x}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac {2 (a+b x) (B d-A e)}{e \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}-\frac {2 (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {770, 78, 63, 208} \[ -\frac {2 (a+b x) (B d-A e)}{e \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}-\frac {2 (a+b x) (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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}{(d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {A+B x}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{e (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((A b-a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{e (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 (A b-a B) \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 )}{e (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {2 (B d-A e) (a+b x)}{e (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (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 = 0.08, size = 128, normalized size = 0.93 \[ \frac {2 (a+b x) \left (e \sqrt {d+e x} (a B-A b) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )-\sqrt {b} (b d-a e) (B d-A e)\right )}{\sqrt {b} e \sqrt {(a+b x)^2} \sqrt {d+e x} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 363, normalized size = 2.63 \[ \left [\frac {{\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (B b^{2} d^{2} + A a b e^{2} - {\left (B a b + A b^{2}\right )} d e\right )} \sqrt {e x + d}}{b^{3} d^{3} e - 2 \, a b^{2} d^{2} e^{2} + a^{2} b d e^{3} + {\left (b^{3} d^{2} e^{2} - 2 \, a b^{2} d e^{3} + a^{2} b e^{4}\right )} x}, -\frac {2 \, {\left ({\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (B b^{2} d^{2} + A a b e^{2} - {\left (B a b + A b^{2}\right )} d e\right )} \sqrt {e x + d}\right )}}{b^{3} d^{3} e - 2 \, a b^{2} d^{2} e^{2} + a^{2} b d e^{3} + {\left (b^{3} d^{2} e^{2} - 2 \, a b^{2} d e^{3} + a^{2} b e^{4}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 117, normalized size = 0.85 \[ -\frac {2 \, {\left (B a \mathrm {sgn}\left (b x + a\right ) - A b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} - \frac {2 \, {\left (B d \mathrm {sgn}\left (b x + a\right ) - A e \mathrm {sgn}\left (b x + a\right )\right )}}{{\left (b d e - a e^{2}\right )} \sqrt {x e + d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 148, normalized size = 1.07 \[ -\frac {2 \left (b x +a \right ) \left (\sqrt {e x +d}\, A b e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-\sqrt {e x +d}\, B a e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+\sqrt {\left (a e -b d \right ) b}\, A e -\sqrt {\left (a e -b d \right ) b}\, B d \right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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