3.2080 \(\int \frac{a+b x}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=140 \[ \frac{15 e^2}{4 \sqrt{d+e x} (b d-a e)^3}-\frac{15 \sqrt{b} e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{7/2}}+\frac{5 e}{4 (a+b x) \sqrt{d+e x} (b d-a e)^2}-\frac{1}{2 (a+b x)^2 \sqrt{d+e x} (b d-a e)} \]

[Out]

(15*e^2)/(4*(b*d - a*e)^3*Sqrt[d + e*x]) - 1/(2*(b*d - a*e)*(a + b*x)^2*Sqrt[d + e*x]) + (5*e)/(4*(b*d - a*e)^
2*(a + b*x)*Sqrt[d + e*x]) - (15*Sqrt[b]*e^2*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^
(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0631258, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \[ \frac{15 e^2}{4 \sqrt{d+e x} (b d-a e)^3}-\frac{15 \sqrt{b} e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{7/2}}+\frac{5 e}{4 (a+b x) \sqrt{d+e x} (b d-a e)^2}-\frac{1}{2 (a+b x)^2 \sqrt{d+e x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*e^2)/(4*(b*d - a*e)^3*Sqrt[d + e*x]) - 1/(2*(b*d - a*e)*(a + b*x)^2*Sqrt[d + e*x]) + (5*e)/(4*(b*d - a*e)^
2*(a + b*x)*Sqrt[d + e*x]) - (15*Sqrt[b]*e^2*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^
(7/2))

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 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]

Rubi steps

\begin{align*} \int \frac{a+b x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{1}{(a+b x)^3 (d+e x)^{3/2}} \, dx\\ &=-\frac{1}{2 (b d-a e) (a+b x)^2 \sqrt{d+e x}}-\frac{(5 e) \int \frac{1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e)}\\ &=-\frac{1}{2 (b d-a e) (a+b x)^2 \sqrt{d+e x}}+\frac{5 e}{4 (b d-a e)^2 (a+b x) \sqrt{d+e x}}+\frac{\left (15 e^2\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^2}\\ &=\frac{15 e^2}{4 (b d-a e)^3 \sqrt{d+e x}}-\frac{1}{2 (b d-a e) (a+b x)^2 \sqrt{d+e x}}+\frac{5 e}{4 (b d-a e)^2 (a+b x) \sqrt{d+e x}}+\frac{\left (15 b e^2\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{8 (b d-a e)^3}\\ &=\frac{15 e^2}{4 (b d-a e)^3 \sqrt{d+e x}}-\frac{1}{2 (b d-a e) (a+b x)^2 \sqrt{d+e x}}+\frac{5 e}{4 (b d-a e)^2 (a+b x) \sqrt{d+e x}}+\frac{(15 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 )}{4 (b d-a e)^3}\\ &=\frac{15 e^2}{4 (b d-a e)^3 \sqrt{d+e x}}-\frac{1}{2 (b d-a e) (a+b x)^2 \sqrt{d+e x}}+\frac{5 e}{4 (b d-a e)^2 (a+b x) \sqrt{d+e x}}-\frac{15 \sqrt{b} e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0139499, size = 50, normalized size = 0.36 \[ -\frac{2 e^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{b (d+e x)}{a e-b d}\right )}{\sqrt{d+e x} (a e-b d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^2*Hypergeometric2F1[-1/2, 3, 1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e)^3*Sqrt[d + e*x])

________________________________________________________________________________________

Maple [A]  time = 0.016, size = 179, normalized size = 1.3 \begin{align*} -2\,{\frac{{e}^{2}}{ \left ( ae-bd \right ) ^{3}\sqrt{ex+d}}}-{\frac{7\,{b}^{2}{e}^{2}}{4\, \left ( ae-bd \right ) ^{3} \left ( bex+ae \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{9\,ab{e}^{3}}{4\, \left ( ae-bd \right ) ^{3} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}+{\frac{9\,{b}^{2}d{e}^{2}}{4\, \left ( ae-bd \right ) ^{3} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}-{\frac{15\,b{e}^{2}}{4\, \left ( ae-bd \right ) ^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^2/(a*e-b*d)^3/(e*x+d)^(1/2)-7/4*e^2/(a*e-b*d)^3*b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)-9/4*e^3/(a*e-b*d)^3*b/(b*
e*x+a*e)^2*(e*x+d)^(1/2)*a+9/4*e^2/(a*e-b*d)^3*b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d-15/4*e^2/(a*e-b*d)^3*b/((a*e-
b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.07289, size = 1582, normalized size = 11.3 \begin{align*} \left [-\frac{15 \,{\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} +{\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} +{\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} x\right )} \sqrt{\frac{b}{b d - a e}} \log \left (\frac{b e x + 2 \, b d - a e + 2 \,{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{\frac{b}{b d - a e}}}{b x + a}\right ) - 2 \,{\left (15 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \,{\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} +{\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} +{\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} +{\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}, -\frac{15 \,{\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} +{\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} +{\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} x\right )} \sqrt{-\frac{b}{b d - a e}} \arctan \left (-\frac{{\left (b d - a e\right )} \sqrt{e x + d} \sqrt{-\frac{b}{b d - a e}}}{b e x + b d}\right ) -{\left (15 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \,{\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} +{\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} +{\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} +{\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(15*(b^2*e^3*x^3 + a^2*d*e^2 + (b^2*d*e^2 + 2*a*b*e^3)*x^2 + (2*a*b*d*e^2 + a^2*e^3)*x)*sqrt(b/(b*d - a*
e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(15*b^2*e^2*x^2
- 2*b^2*d^2 + 9*a*b*d*e + 8*a^2*e^2 + 5*(b^2*d*e + 5*a*b*e^2)*x)*sqrt(e*x + d))/(a^2*b^3*d^4 - 3*a^3*b^2*d^3*
e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^3*b^2*e^4)*x^3 + (b^5*d^4
- a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^4 - 5*a^2*b^3*d^3*e + 3*a
^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x), -1/4*(15*(b^2*e^3*x^3 + a^2*d*e^2 + (b^2*d*e^2 + 2*a*b*e^3)*x^2 +
(2*a*b*d*e^2 + a^2*e^3)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x
+ b*d)) - (15*b^2*e^2*x^2 - 2*b^2*d^2 + 9*a*b*d*e + 8*a^2*e^2 + 5*(b^2*d*e + 5*a*b*e^2)*x)*sqrt(e*x + d))/(a^2
*b^3*d^4 - 3*a^3*b^2*d^3*e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^
3*b^2*e^4)*x^3 + (b^5*d^4 - a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^
4 - 5*a^2*b^3*d^3*e + 3*a^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.13659, size = 317, normalized size = 2.26 \begin{align*} \frac{15 \, b \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{-b^{2} d + a b e}} + \frac{2 \, e^{2}}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{x e + d}} + \frac{7 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} e^{2} - 9 \, \sqrt{x e + d} b^{2} d e^{2} + 9 \, \sqrt{x e + d} a b e^{3}}{4 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*s
qrt(-b^2*d + a*b*e)) + 2*e^2/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(x*e + d)) + 1/4*(7*(x*e
+ d)^(3/2)*b^2*e^2 - 9*sqrt(x*e + d)*b^2*d*e^2 + 9*sqrt(x*e + d)*a*b*e^3)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b
*d*e^2 - a^3*e^3)*((x*e + d)*b - b*d + a*e)^2)