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

Optimal. Leaf size=114 \[ \frac {e (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

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Rubi [A]  time = 0.09, antiderivative size = 114, 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, 21, 51, 63, 208} \begin {gather*} \frac {e (a+b x) \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}}-\frac {\sqrt {d+e x}}{\sqrt {a^2+2 a b x+b^2 x^2} (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)^(3/2)),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 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 {\left (a b+b^2 x\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {\sqrt {d+e x}}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (e \left (a b+b^2 x\right )\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {\sqrt {d+e x}}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (a b+b^2 x\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {\sqrt {d+e x}}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (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 = 85, normalized size = 0.75 \begin {gather*} \frac {\frac {\sqrt {d+e x}}{a e-b d}+\frac {e (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )}{\sqrt {b} (a e-b d)^{3/2}}}{\sqrt {(a+b x)^2}} \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]

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

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IntegrateAlgebraic [A]  time = 13.88, size = 130, normalized size = 1.14 \begin {gather*} \frac {(-a e-b e x) \left (\frac {e \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{\sqrt {b} (a e-b d)^{3/2}}-\frac {e \sqrt {d+e x}}{(b d-a e) (-a e-b (d+e x)+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])/((b*d - a*e)*(b*d - a*e - b*(d + e*x)))) + (e*ArcTan[(Sqrt[b]*Sqrt[-(b*
d) + a*e]*Sqrt[d + e*x])/(b*d - a*e)])/(Sqrt[b]*(-(b*d) + a*e)^(3/2))))/(e*Sqrt[(a*e + b*e*x)^2/e^2])

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fricas [A]  time = 0.45, size = 280, normalized size = 2.46 \begin {gather*} \left [-\frac {\sqrt {b^{2} d - a b e} {\left (b e x + a e\right )} \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^{2} d - a b e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x\right )}}, -\frac {\sqrt {-b^{2} d + a b e} {\left (b e x + a e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x}\right ] \end {gather*}

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]

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

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giac [B]  time = 0.24, size = 189, normalized size = 1.66 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{{\left (b d e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a 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 {\sqrt {x e + d} e^{2}}{{\left (b d e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a 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 )}} \end {gather*}

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]

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

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maple [A]  time = 0.06, size = 112, normalized size = 0.98 \begin {gather*} \frac {\left (b e x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+a e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+\sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )^{2}}{\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}}} \end {gather*}

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]

(b*e*x*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+a*e*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)+(e*x+d)^(1/
2)*((a*e-b*d)*b)^(1/2))*(b*x+a)^2/((a*e-b*d)*b)^(1/2)/(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 {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} \sqrt {e x + d}}\,{d x} \end {gather*}

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 = -0.01 \begin {gather*} \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 \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 {a + b x}{\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)/(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(1/2),x)

[Out]

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

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